text
stringlengths 83
79.5k
|
|---|
H: Wrong result in spice simulation of DC Offset at AC coupled diode
I'm playing around with a diode as AM demodulator and have stumbled upon a phenomenon that seems logical but I can't reproduce it in LTSpice. My setup is the following:
simulate this circuit – Schematic created using CircuitLab
U_in is a sine wave with amplitude of about 0.5V and a DC-Offset of 2.5V.
$$ U_{in} = 2.5V + 0.5V * sin(wt) $$
Now when I measure the voltage after the decoupling capacitor, which is supposed to strip the DC bias, I get a sine wave with \$ U_{max} = 200 mV \$ and \$ U_{min} = -800 mV \$.
So far, so good. My interpretation of this is that the positive half wave is passed through the diode above the forward voltage of about 200 mV while the negative half wave is blocked by the diode and generates a negative DC offset on the right side of the capacitor.
I know that I can align the signal around 0V by adding a DC return path after the decoupling cap.
The question is why I cannot reproduce this behaviour in LTSpice. Even when using a detailed capacitor model with ESR and parallel resistance, adding the scope input (1Meg, 15p, 10:1) to the simulation and trying different diodes, the result will always be a sine wave that is offset by at most 10 mV. If it wasn't for the measuring utility, I'd say there wasn't any DC bias at all.
Am I missing something? Is there some parasitic parameter I'm not taking into account?
Here is a screenshot of my simulation, the green trace is the (wrong) simulation, the blue trace is what I see on the real world scope:
Any help would be greatly appreciated!
Kind Regards
AI: You should know that, because of the reactive elements, there can be some transients that can confuse if left only for a short time. In this case, you can run the simulation with .tran 500u uic (to be sure it starts everything from zero), and make that end resistance 100, as in the original picture, otherwise you'll have to wait even longer. I didn't have your diode, so I just used the ol' BAT54, but that shouldn't make too much of a difference in this case.
|
H: What percentage error?
A transducer with an internal resistance of 10 k Ohm generates an output voltage
which is sensed by a voltmeter having an internal resistance of 30 k Ohm. What
percentage error is to be expected? How can Engineer fix this error?
AI: normally one chooses a better DMM than using an analog meter with a certain Ohm/volt inverse current transfer function or in this case a fixed R.
An Engineer would simply correct the Load regulation error to compute the correct value using inverse impedance ratio..
So if true sensor out was 1V with 10k then a 30k load would divide the and you expect sense voltage Vs= (30/30+10) * 1= 750mV = 3/4 * Vs thus corrected reading is 4/3 is the correction factor.
or use a 1MOhm DMM then your correction factor is 1% higher which is slightly larger or significant than the accuracy of cheap devices nut perhaps less significant on the resistor tolerance, so often we ignore it unless accuracy is critical.
Does that make sense?
|
H: Combining dirty 12V inputs with delicate micro controller
I'm building a (LED) blinker circuit for my motorcycle. It will have the basic on/off blinking for left and right (J1 & J2) and an "emergency" mode (J3) for more "aggressive" LED blinking, which will increase the visibility when driving between lanes during traffic jams.
J1, J2 and J3 are the inputs for the switches (L, R, both). J5 and J6 are the outputs to the LED blinkers. J4 and J7 are GND and +12V from the motorcycle.
There's a ATTiny45 which reads the inputs L or R. If both are high, it will go into emergency blinking mode. It uses two output pins with pulses that will blink the LEDs. (if anyone is interested in the code, I'd be happy to share it).
Now, the circuit works on my breadboard (and I hope I correctly transferred it into Fritzing). But since this is my first project in electronics I have a ton of questions and doubts. Now, I'm not asking you guys to review my project (however, would be appreciated), but want to focus on my biggest question for this circuit, the isolation between the "dirty" 11-14V inputs and the "delicate" ATTiny micro controller.
On the input side I've used an opto coupler. But I'm actually wondering if this de-couples anything, since there's a common ground.
On the output side I've used transistors + MOSFETS. I couldn't drive the MOSFETS directly, although I forgot why. I'm wondering if it's not safer to use an opto coupler here in stead of the transistors.
All parts used are in the diagram.
Not sure if it matters much, but the environment on a motorcycle is relatively
hostile, vibrations, voltage spikes, etc.
I hope I didn't make too many stupid mistakes!
Many thanks for your input!
Thanks,
Sjoerd (NL)
--- EDIT, REPLY TO ASMYLDOF ---
Hi Asmyldof,
Thanks for your time to write this amazing reply. This is really helping me learn creating good circuits. I really appreciate it! I've spent quite a number of hours reading and researching the things you wrote down. In return:
The KA78M05TU is rated at max. 30V. Doing some research I found in a TI white paper that 12V systems may have pulses up to 87V, and 24V systems up to 174V (and I was planning a unit for my 24V Defender). I could not find any through hole LDO's at Conrad or Farnell with such a high max. input voltage. But perhaps the max. input voltage is not the spec I should be looking for here (?).
Could perhaps the transient and negative voltage problem be adequately resolved with a TVS-diode? Vishay has the P4KA series TVS, especially designed for automotive conditions. http://www.vishay.com/docs/88364/p4ka.pdf
Good point, added the N4148's for added protection.
The idea for the existing N4148's in the diagram was to stop current flowing into both inputs, as they're connected because of the "hazard" switch, that signals to both the inputs.
Reversed. (they were correct on my breadboard, hence the circuit worked in real life).
Added. (they were present on my breadboard, hence the
circuit worked in real life).
Interesting point. I'd have to read more about it to fully understand it. For now, I think I will leave the opto-coupler out, to keep it simple, especially if chances for problems are quite low.
I've upped them factor 10 to 4.7k. Although I'm not fully understanding the math behind the base resistors. From what I've read, transistors are current controlled (vs. MOSFETS, which are voltage controlled). Looking in the 2N3904 I cannot find the value that determines at what current the transistor reliably switches.
I added the 22ohm resistor on the ATTiny VCC to make sure it didn't get more than the specified 200mA. Is this unneccessary, or is your point to add a capacitor across the VCC and GND of the microcontroller? How would I calculate the capacitor value?
Ah. Good point, in my case not relevant, as I'm driving quite a bare bone motorcycle; the Yamaha XT600 (kickstart version). I was thinking of making a few of these for some friends of mine, so I will look into this.
At some point I have been thinking to source the input voltage from the switches, as I noticed the ATTiny had practically no boot up time. But I abandoned it since I would have to beef up the switch wires somewhat. The upside however is that I would need one less connector, have no sleep current and would mitigate your point raised.
About the input clamp: See point #1.
About the opto-coupler protection: I will dive into this deeper. I'm leaving it out for now.
Power supply filtering: I'm not neccesarily sticking to the bare minimum, space is a concern on a motorcycle, when I'm leaving certain protection/filtering/etc. out, it's more because of ignorence and lack of experience on my side :). I'd say I'm aiming for a somewhat overengineered circuit, without going overboard :). Having said that, again, see point #1, could this be adequately solved by TVS diode?
I took the values of the two capacitors on the regulator from the datasheet, on my breadboard I'm actually using higher values. You explain that C2 must be quite a bit smaller than C1. But what would be a good way to calculate the values. Is it possible to use values that are too high?
I'm sorry if I'm stealing your time. You've been a great help so far. It's OK to be brief in your answer and give me the pointers to continue the research on my own. Again, many thanks, Sjoerd.
AI: Let's start with the Technical (built my own 2-pin electronic blinker for 3W ~ 300W for my Div900 years ago - if that helps):
Your setup isn't safe enough if you want to be assured of use of your indicators at all times. Which you do.
Why not? Because of the following reasons:
The 78x05 type regulators are hardly ever specified for high-transients (supra-40V), nor are they guaranteed to perform correctly for negative input, while not as likely as it was in 1960's cars, it's not impossible and broken is broken.
Similarly you want your opto-couplers protected from inversion by an extra, at least 100V capable diode. Nor do you want one switch to turn on another in vehicles, if you do it properly, so adding more 4148's to the other inputs seems best. (Your forward current at 30V seems to stay below limits, so that seems ok).
Your MOSFETs are Connected the wrong way around. They show P-Type, but their source is connected to the low-side load. This means your lights are always on.
How do your MOSFETs turn off (once you have flipped them)? There should be a pull-up to their gates, or your lamps will turn on and then maybe, once, in the distant future through leakage turn off.
Be aware (when you start debugging) that your MOSFETs are connected directly to the dirty voltage, and though it's much less likely to be problematic, there's a chance when pull-up's are there some noise may filter back in. I've never had trouble with that myself with AVRs though, as their output stages are rock solid. If you do use an Opto-Coupler there, beware to calculate the drive currents and turn/on turn off times, you may still need an extra transistor there.
Your base resistors for your NPNs can be a factor 10 larger, this will still switch exceptionally well, but it's not mandatory (though I'd do it - for the current consumption).
Without any extra capacitance that 22 Ohm on your Vcc will only make your MCU more susceptible to supply noise that's coupling in. Any switching current to its core will need to go through 22 Ohm + Board inductance without any nearby help.
Not so much a problem, but a tip: Make sure you hook the power wire up to something that doesn't turn on in any setting you don't want it on in. Since you estimate a pretty appreciable steady-state current use by your relay. Some bikes (mine included) power the indicators by default in more settings than just the one where you're driving / running the engine.
So, think about the points above and then I suggest you also put in an input-clamp, that clamps a while before any component hits their limit.
You can consider splitting the opto-coupler resistance in two parts, if you want full protection, and protect the upper half + diode with a 15V zener. If you have that zener and a 350-ish Ohm resistor in the protected chain, you're certain the LED will not see more than a 40mA share. If you do, though, do use a decent zener, not one of those 0402 grains of salt, in case spikes come often and close together those will evaporate just as happily as your couplers. I don't expect it to be needed, but it'd look like this:
simulate this circuit – Schematic created using CircuitLab
You can add some more filtering on supply leading only to your regulator as well, since you seem to be sticking to the bare minimum. The blinker cutting out once every few seconds isn't that harmful if you blink at standard 1.5Hz-ish rate, but if it happens every 10ms you may not be blinking all that much, ever.
Also, be aware that with no or low inductance, 300nF does nearly nothing to compete with very heavy switching loads and sources in your system. They laugh at anything expressed in nanos. Unless you choose a 0.05mm2 wire to power the set-up... In which case even LED lights turning on will make your relay reset.
What I did myself was a well calculated version of this:
simulate this circuit
This isn't Ideal, as your clamp is in a high-current path, and who knows. It'd have been better encapsulated in a somewhat safer frame with a limited current. But I wanted to also offer some protection to the down-stream LED lights, as they were ready-made after-market back then, and those lights are teh poop with safety. Hence also the need to not dampen too much, for fear of having to swap in a 21W regular along the side of the road. These days almost everything on the bike is home made or maintained, so some tweaks would be possible, had I not encased it in military grade resin.
C2 is (much) smaller than C1, because you want the filter effect to be verifiably stronger than the boost effect. If the main body of capacitance is C2, you are potentially building a boost converter powered by your alternator frequency.
R is dampening for unwanted effects, such as the boost work or oscillations at unexpected frequencies, and it doesn't hurt in front of a linear regulator anyway.
My regulator was low-leakage as my MCU spends most of its life (even while blinking) in a sub 1uA sleep, having 3mA drain by the regulator would have been a point of great :'(.
L and C1 are selected for a good filtering responses at the frequencies expected in your device at realistic drains from the regulator. For frequency my suggestion is to aim at 35Hz to 2.5kHz for most Touring/Sport-Tour Bikes and up to 5kHz for Sport and SuperSport (peak-RPM/60 * 3 * 7-ish). Where obviously the higher frequencies should only come in when you see if you're not creating any unwanted oscillations anywhere.
If you can get it so that C2 is still at least 100nF and C1 is at least 10 times that, or 25 times... That's where at my lab the scope and the bike meet, to see how well my guesstimation worked.
Of course, all parts between a positive voltage and ground in my design were chosen for low leakage as well, as in all my automotive/bike-o-motive designs.
(Mind you, none of the above contains any maths other than quick top-of-head)
And one more note, before the comments come in: Of course I should remove the point about the virtual magic ground, as with a true virtual ground the TVS wouldn't work, etc etc, but I wanted to show off a little for the hours of typing and putting up with the circuit editor. In reality the TVS is even more cleverly placed, but that'd give away some parts I signed away inside the dotted box.
Then, I would like to add a little off-topic, since you brought it up yourself;
Be aware that several, if not many, calamity professionals in west Europe, NL especially (my home too), strongly suggest you not use emergency lights in between traffic. It causes confusion (to those only seeing one of your sides) as well as aggravation. It hurts more than it helps. While not on-topic as a comment, I felt I should let you know, since confusion and aggravation are leading causes for car drivers to collide with motorcyclists. Just use your headlight, patience and navigational skill.
To date I have spent many hundreds of hours in traffic jams and my "calamity indicator" has been on for a total I estimate below 120minutes, the bulk of which when a screw in my tyre forced me to park "unsafely".
Brief responses to your added questions:
For longer work, you'll have to aim more at new questions, this shouldn't become a complex state-full thread themed "Help Sjoerd finish his design from start to finish"
The main reason I wouldn't use the 78-types is the high current it draws itself, it can be protected sufficiently by a TVS and a filter. It may be wise for a 24V design to find something a bit more in the direction of 40Vmax, to allow you vehicle voltage to fluctuate upto at least 34V before your protection starts clamping.
1b. Those TVS should work, but make sure you get their clamping voltage a ways beyond your maximum expected normal use voltage, but at least a few volts below the lowest Vmax of your electronics. Try to think about elements that limit the peak pulse to, or preferably way below, 40A. You can use a polyfuse, or the added filter-damping fuse for that.
It should be fine without opto's, if you have a nice noise source you can stress test it in-lab. A drill driving a medium-size DC motor with a low-voltage source in series might give you a nice worst-case test signal. Or it might not.
The transistors switch in a sense when you let them saturate. With 4.7k they should be able to pull much more than 70mA and still saturate, when the resistors are connected to a 5V signal. If you want more details there's a lot about that on this site.
You can add a small resistor as a sort of extra filter, with or without a ferrite bead for the high frequency stuff. But that will require extra capacitance. It is smart to add a 22nF to 100nF bypass capacitor to any digital chip, to supply switching currents locally. There's also a lot about this on this website. Adding a current-limiting resistor to an MCU is a very strange practise. Especially for 200mA on a Tiny45. If your Tiny45 is pulling an average current of 200mA it's broken and your relay can be discarded anyway. Digital chips, if not filtered, get connected directly, because "we" trust them to behave. Unless we know otherwise.
Input filtering and all that:
The capacitors and the inductor in that image get calculated to filter at the appropriate points without causing oscillations. This post is already too long to detail that, but again, there's a lot on filter design on this website already anyway. The reason I mention the C2 to C1 relationship is that if you don't stick to that, there's a risk you are building a boost converter using your alternator ripple, which might boost the voltage to 1.5x, 2x, or even more. This is not the likeliest thing to happen if you're experienced, but when starting out, it's best to stick to the rule for C1 and C2.
The diode is protection from ms-scale cut-outs and input voltage reversal. To note, as well, that if input reversal is a likely possibility, it's best to have all caps before the diode be ceramic or other bipolar types. For longevity.
For the caps you can over-do it, costing money and space and possibly creating high inrush currents, or possibly the opposite: High ESR in the caps, which would dampen the filter even more than the dampening resistor. If you move the dampening resistor before the TVS, when the TVS only protects your electronics, the filter will work better, you can then also look into pi-filtering, but you can also over-do filters quite easily. A simple L-C filter in the ballpark, a small safety resistor and a TVS should get you so close to Automotive spec that it'll probably do well without actual full testing for all but commercial production runs.
|
H: How to convert an existing design from USB Mini/Micro to use USB Type-C connector?
What does it convert a design from USB Mini or Micro connector to a USB type-C connector?
To have a more specific question, take for example FTDI Basic (schematic below) which is USB/Serial converter based on FTDI232RL and has a USB Mini connector. Converting this design a USB Micro connector is trivial, just rewiring the 4 connections to the USB Micro connector. What about a USB Type C connector, is it just a clever rewiring? Does it requires additional ICs and functionality to be type C compatible?
All I care about is modifying my existing custom PCB designs to have a functional USB type C connector. Having additional Type C features such as higher speed are non goal. Also, I am not looking for an off the shelf Type C USB/Serial converter but am more interested in converting existing designs to Type C. This is for a hobby/one-of/open-source designs if it matters.
AI: The FT232 is full-speed (12mbps) USB device, so you can't possibly take any advantage of Type-C high-speed connector.
To make your FT232 design with Type-C receptacle, all you need is to wire your D+/D- to PAIR of D+/D- pins on Type-C (to both pairs in parallel) receptacle, and pull down both CC1 and CC2 pins to ground with 5.1k resistors.
|
H: Why button battery LR1154 is called LR44?
It seems I cannot find complete explanation for this question everywhere on the internet.
So far, I know the followings:
L - Alkaline as +/Zinc as -
R - round
11 - battery's diameter in mm (in fact, the standard states, that the actual diameter is 11.6 mm[1])
54 - battery's height in 0.1 mm
What's the meaning of 44?
AI: LR44 is some companies proprietary name for the IEC LR1144 battery.
https://en.wikipedia.org/wiki/List_of_battery_sizes
Manufacturers may assign proprietary names and numbers to their
batteries, disregarding common, colloquial, IEC, and ANSI naming
conventions (see LR44 battery as an example). Often this is done to
steer customers towards a specific brand, and away from competing or
generic brands, by obfuscating the common name.
The table in the middle of https://en.wikipedia.org/wiki/List_of_battery_sizes#Silver_oxide_and_alkaline_cells shows that the LR1144 IEC designation corresponds to the "most common" name of SR44, otherwise called the LR44
|
H: Simulation went wrong
Here is a simple dickson inverter:
The simulation shows this:
Is it a problem with the simulator? I don't see why the capactior was charged to the voltage of the square wave signal? Why is the push-pull pair not working?
UPDATE: I have replaced the push-pull pair with a single NPN transistor, as suggested.
It works as expected, but the current is limited.
Is there a way to do this without a series resistor and single supply? (E.g. eliminate power disciplation on the resistor).
UPDATE: finally! Frequency increased to 100K, voltage source drives a voltage amplifier, and then the push-pull pair.
Simulation result:
Frequency had to be increased because the load resistor was changed to 100 ohms. Now it can deliver about 150mA. Great!
AI: V1 in your push pull circuit should be ranging from 0V to +15 volt (15 Vp-p) to properly drive the push-pull pair. These are configured as emitter followers and you are probably only driving them at the moment with a 5 Vp-p square wave hence you only see about -3 volts coming from the charge pump.
With just Q1 used, you have wired it as a common emitter and hence the collector voltage will drive nearly from 0 volts to +15 volts.
|
H: Is there any particular criterion that makes a CPU be called "16 bits" CPU for example
What makes a CPU be called for example: "this CPU is n bits"?
AI: Short answer: There is no universally accepted definition.
Less short answer: If the CPU supports all of the basic primitives on a 16-bit datatype, then it would probably be considered sixteen bits by the majority of users.
This has been a "holy war" since 1976 or so, and there is no "right" or "wrong" answer. Was the 8088 sixteen bits? Probably. So, was the Z80? It had some sixteen bit math, and an 8-bit databus. (Probably not -- the Z80 had no native 16-bit logical instructions, only add and subtract).
The question surged again when the 68000 with its 32-bit registers, and rich set of 32-bit operations appeared, but an internal 16-bit ALU and an external 16-bit databus (and then just to throw MORE confusion, the 68008 variant, with an 8-bit databus).
|
H: How to set initial state of 4 bit "exclusive output" Latch?
I have extended SR flip-flop to have four inputs and four outputs.
It works well once an input has been pressed but I'm not sure how to set the initial state.
I saw a schematic online where a capacitor was placed in parallel with the pull down resistor that connects to my preferred initial states NAND gate. On my breadboard this appears to work. On the simulator however, if I leave the initial state and return to it I get a "convergence failed" error.
My question is, does the "convergence failed" error highlight a real problem with my circuit, and if so is there a better way to set the initial state of the circuit?
Circuit diagram:
My netlist code is below:
$ 1 0.000005 10.20027730826997 50 5 50
151 688 80 800 80 0 2 0 5
151 688 208 800 208 0 2 5.000000000000002 5
151 688 336 800 336 0 2 5 5
151 688 464 800 464 0 2 5 5
w 800 80 800 160 0
w 800 208 800 288 0
w 800 336 800 416 0
w 800 464 800 544 0
162 800 80 848 80 1 2.1024259 1 0 0 0.01
162 800 208 848 208 1 2.1024259 1 0 0 0.01
162 800 336 848 336 1 2.1024259 1 0 0 0.01
162 800 464 848 464 1 2.1024259 1 0 0 0.01
r 848 80 912 80 0 1000
r 848 208 912 208 0 1000
r 848 336 912 336 0 1000
r 848 464 912 464 0 1000
g 912 80 912 96 0
g 912 208 912 224 0
g 912 336 912 352 0
g 912 464 912 480 0
R 352 96 352 64 0 0 40 5 0 0 0.5
w 688 64 416 64 0
w 688 192 416 192 0
w 416 64 416 96 0
r 352 96 416 96 0 1000
s 416 96 416 144 0 1 true
g 416 144 416 160 0
w 416 192 416 224 0
w 688 320 416 320 0
w 416 320 416 352 0
w 688 448 416 448 0
w 416 448 416 480 0
s 416 224 416 272 0 1 true
s 416 352 416 400 0 1 true
s 416 480 416 528 0 1 true
g 416 272 416 288 0
g 416 400 416 416 0
g 416 528 416 544 0
w 352 96 352 192 0
w 352 192 352 320 0
w 352 320 352 448 0
r 352 192 416 192 0 1000
r 352 320 416 320 0 1000
r 352 448 416 448 0 1000
r 688 128 752 128 0 1000
g 752 128 768 128 0
r 688 256 752 256 0 1000
r 688 384 752 384 0 1000
r 688 512 752 512 0 1000
g 752 256 768 256 0
g 752 384 768 384 0
g 752 512 768 512 0
d 640 80 688 80 1 0.805904783
d 640 96 688 96 1 0.805904783
d 640 112 688 112 1 0.805904783
w 688 128 688 112 0
w 688 112 688 96 0
w 688 80 688 96 0
d 640 208 688 208 1 0.805904783
d 640 224 688 224 1 0.805904783
d 640 240 688 240 1 0.805904783
d 640 336 688 336 1 0.805904783
d 640 368 688 368 1 0.805904783
d 640 352 688 352 1 0.805904783
d 640 464 688 464 1 0.805904783
d 640 480 688 480 1 0.805904783
d 640 496 688 496 1 0.805904783
w 688 512 688 496 0
w 688 496 688 480 0
w 688 480 688 464 0
w 688 384 688 368 0
w 688 368 688 352 0
w 688 352 688 336 0
w 688 256 688 240 0
w 688 240 688 224 0
w 688 224 688 208 0
w 800 160 608 160 0
w 608 160 608 208 0
w 608 208 640 208 0
w 608 208 608 336 0
w 608 336 640 336 0
w 608 336 608 464 0
w 608 464 640 464 0
w 800 288 576 288 0
w 576 288 576 224 0
w 576 224 576 96 0
w 576 96 640 96 0
w 576 288 576 352 0
w 576 352 640 352 0
w 576 352 576 480 0
w 576 480 640 480 0
w 800 416 544 416 0
w 544 416 544 240 0
w 544 240 640 240 0
w 544 240 544 112 0
w 544 112 640 112 0
w 544 416 544 496 0
w 544 496 640 496 0
w 800 544 512 544 0
w 512 544 512 368 0
w 512 368 640 368 0
w 512 368 512 224 0
w 512 224 640 224 0
w 512 224 512 80 0
w 512 80 640 80 0
c 688 144 752 144 0 0.00001 4.275735888810727
w 688 128 688 144 0
w 752 128 752 144 0
AI: This is how everyone's specs ( or question) should be written as follows:
A Hierarchical Input Process Output or HIPO descrption. Same for documenting software routines. INs OUTs and FUNCTIONS
Inputs: 4 momentary SET switches (normally=0)
Process: Each switch acts as a SET to it's own Register AND a RESET to all others, thus creating an exclusive N bit SET function to Register
This is an async. latch function, not a edge-triggered synchronous type.
Initial Condition options; 1000, 0100, 0010, 0001, or 1111. illegal=0000.
IN=0000 does not converge to steady state
OUTPUT Latch follows INPUT exclusive switch with last switch pressed
If multiple inputs are pressed ( multiple outputs occur in same sequence but REMEMBERS Last SWITCH INPUT released with Exclusive Output.
Outputs: 4 Bit Latch with exclusive state ( only 1 output SET after nano delay )
Proof of Concept in slow motion ( 0.1ns per second )
https://goo.gl/F3H7PT ( Needs Java Approval in browser )
|
H: Difference between RF ferrite and normal ferrite
Why oscillation stops in a RF oscillator(over 100MHz) if I replace the air core Inductor with a normal ferrite core inductor ?
But starts oscillating again if I rewind the inductor over a RF ferrite core. (salvaged from TV balun)
So what's the basic difference between RF ferrite and normal ferrite ?
AI: Ferrites come in a variety of flavors. The main groups are ferrites for
Filters (for signal filtering, e.g. in telecommunications)
Power Conversion
Interference suppression (EMI filters, not to be confused with signal filters)
Ferrites for signal filters need to have low losses and must have a defined temperature factor in order to make the LC combination work well over temperature. Also, they should be stable over time.
Ferrites for power conversion must have low losses and a high saturation value.
A ferrite core used for EMI filters is usually quite lossy but has high permeability, allowing for a high impedance in the desired frequency range.
Your oscillator would go into the first group (signal filtering), and if you substitute your ferrite with one salvaged from an EMI component, you might end up having fairly high losses, preventing your circuit to have a good-enough Q-factor.
Then, there's the pitfall of using an iron powder core by mistake. If varnished with epoxy, they look quite similar to ferrites, but will not work for RF signal applications (at all, in any reasonable way) due to the comparatively huge eddy current losses they have because of the fairly large iron particles they're made of.
Good sources for further reading are the introductory chapters of these two large data books:
Ferroxcube/Yageo Data Book (Soft Ferrites and Accessories)
TDK/Epcos Data Book (Ferrites and Accessories)
And, above all, the best book about ferrites I could find so far:
E.C. Snelling, Soft Ferrites, Properties and Applications, 1969
I came across this book because it's in the literature list on this great page about producing your wound components, worth spending some hours reading through. Quoting the site's author: "If you are using a ferrite core and this book doesn't have the answer then you are in trouble. Wide ranging both in theory and practice. The maths is intelligible and (bliss o' joy) uses SI units."
|
H: tube amplifier build power supply design
Hello Im just starting to work with valves but have a question on the power supply side. The question is the amplifier I am building needs 300 VDC. As I understand voltage after bridge rectifiers across the smoothing capacitor increases.
The transformer I have is 660 volts AC center tapped 120 milliamps.
The amp requires a single ended power supply. The multi meter shows 330 volts AC no load. but where I bought the transformer it says 290 volts ac and it could power the tubes Iam using. But the real question is will that voltage come down under load on the AC side and the DC side.
Im not sure if 330 ac is to much for the tubes. So what I was wondering if I needed something as low as 250 Volts AC. Then that would go up after rectifier and capacitor.
The bottom line is what ac voltage do I need to convert to dc 300 v. Is it the same or is the ac part lower ?
Also what would be the best way to wire the center taped output to get the most current out of the transformer because I only need one x 300 volts supply.
The diodes I am using for rectification are 1n4007.
Im confused with this RMS and peak power. I know peak is higher and do the caps need to be at the peak voltage. The caps are rated at 450 volts The tubes Iam using are 6v6s and there will be 2 of those to make a stereo amp. There is also 2 x 6j5s which are driver tubes. A total of 4 valves will be used. The SE output transformer is 5000 ohms so Iam not sure if there is enough current. Would a larger capacitor and choke compensate the current demand at peak power?
sorry if this is long but any help would be much apreciated.
Thanks
AI: So you have a transformer with CT output then of course you can use only one side for rectification and leave another side open. Since it will show 290Vrms under full load and you don't have that much load, we can assume it will show 300Vac. As stated by RoyC, peak output voltage after rectification will be 1.4 times secondary RMS but we leave some margin (due to ripple) and let's say the output will be 300 x 1.35 = 405VDC under full load. But for 300VDC, you'll need 300/1.35=220Vac output. But you have 300Vac, so let me check if you can still use it.
For driver stage, since it has a triode and Class-A config, supply voltage directly affects on operating point (thus the harmonic content and distortion). So, if your circuit is designed for 300VDC then you'll need 300VDC. Note that ±20V is a fair margin for a Class A driver's supply. For power stage, it's not that important but 405VDC is dangerous for that stage.
Im not sure if 330 ac is to much for the tubes.
Let's check: According to 6J5 datasheet and the circuit given, quiescent current is 1.5mA (blue point) for 300VDC:
You can see that quiescent anode (plate) voltage is around 160VDC.
If you directly apply 405VDC to the driver stage without any modification, operating point will be like in the following (blue point):
As you can see, quiescent anode (plate) voltage will be around 210VDC. According to datasheet, maximum allowable anode (plate) voltage is 300VDC. Besides, voltage gain will be a bit different. And also, harmonic content for high (10Vpp) input signals will be quite different, but I think input won't be higher than 2Vpp in your application even if it's a guitar amp. So, you're still good.
But, I personally recommend this:
After rectification, put a RC filter with R = 22k/1W and C = 10uF/450V. This will provide a voltage drop for preamp stage and nice filtering for 100Hz ripples. You can also play with R value until you get enough performance.
|
H: Is it possible to explain a transistor (or vacuum tube or relay) using a marble-based switch for students?
In Japan I saw a simulation of internet based communication in the The National Museum of Emerging Science and Innovation (Miraikan). It attempted to replicate ascii characters moving along a 8 part wire using combinations of marbles rolling along.
I've just finished reading Charles Petzold's book, Code: The Hidden Language of Computer Hardware and Software. In it Charles explains building relays into gates, gates into logic components, and logic components into computing machines.
I wondered if a mechanism existed that simulated a switch using marbles.
My question is: Is it possible to explain a transistor (or vacuum tube or relay) using a marble-based switch for students?
AI: I'd be thinking of something like this:
The top hopper contains blue marbles (so you can see where things come from). A spring loaded flap prevents anything falling.
Drop a red marble into 'base' tube ( = small base current) - it opens the flap momentarily, marbles drop from the 'collector' hopper and the red ('base' marble) and blue ('collector') marbles accumulate in "emitter' container below.
Flap then closes due to spring action (that's the tricky mechanical part).
The gain of the 'transistor' is then determined by how many blue marbles fell for a single red marble. Analogy gain = collector current/base current.
We also can see Kirchoff's current law in action -
the number of 'emitter' marbles = number of collector marbles + the base marble
or the analogy emitter current = base current + collector current
Hope this gives you some ideas.
|
H: Transistor current amplifier for 20W speaker
I have a 5 Vpp max audio signal in my circuit which I obtain from earlier stages of processing. I would like to deliver the signal to a 20W 4 ohms speaker. But since an OPAMP circuit generates the signal, it won't be able to supply enough current. How can I use transistors to amplify the current delivered? I don't think I need a common emitter voltage amplifier since the voltage is already large. What amplifier topology should I use? What kind of transistors do I need for the purpose?
I'm trying to refrain from using specific ICs and use transistors instead, and I don't need a very complicated adjustable amplifier, my only goal is be able to listen to the contents of the signal. The 20 W speaker is necessary for my application, a weaker speaker will not do. I already have a 12 V and 30 V power supply used in the previous stages.
AI: 20 watts into 4 ohms implies an RMS sine wave voltage of \$\sqrt{80}\$ volts = 8.94 V RMS. The sine wave will peak at 12.65 volts and have a peak-to-peak value of 25.3 Vp-p.
For fairly straightforward amplifiers, this means having a total power supply voltage that exceeds 25.3 volts so maybe a +/- 15 volt power rail is worth considering. This leaves a little bit of headroom because most op-amps on +/-15 volt rails will probably only be able to swing to +/- 13 volts (26 Vp-p) and, with a push-pull emitter follower stage you should be able to achieve about 1 volt lower (under load) with a little bit of care.
So, a +/- 15 V supply will just about suit reproducing a sine wave of 20 watts power into a 4 ohm speaker.
However, if you are talking about music reproduction then it's a different story. Music tends to have a "crest factor" that makes delivering 20 watts a bigger problem. Compare a sine wave and music: -
For the same peak power delivered (2 watts) a sine wave would be delivering 1 watt (average) whilst music would deliver significantly less at below 0.1 watts.
So, you have to decide what you are talking about when you say 20 watts.
Anyway, the circuit is the same but the power rails are different so, toplogically, if you don't want to use a ready-made chip solution then consider adding an NPN and PNP transistor to an op-amp like this: -
The diagram above was taken from the DS of the OPA454.
It shows power supplies of +/- 50 volts but these could be more like +/- 16 volts for just delivering a sine wave of 20 watts to 4 ohms. I don't think it's useful to go into further details until you decide what you want (20 watt sine or 20 watt music).
|
H: Is JTAG the standard way to program ARM processors?
Some electrical engineer once told me, that each ARM M3 processor can be flashed the same way, no matter which manufacturer it came from.
I suppose he meant using JTAG, or is there another way?
Btw, I'm meaning to flash an empty ARM processor without a bootloader.
AI: I suppose he meant using JTAG, or is there another way?
Probably he didn't. ARM has their own debugging bus standard – SWD (single wire debug), that is very well-specified. JTAG, on the other hand, is merely a electrical and shift-register-level standard, and it's up to device manufacturers to give JTAG endpoints and actions a meaning.
SWD programmers can be had for <5€. Look for products called "STLink v2 compatible" or so. The ST in the name stems from the fact that they are based on the protocol that STmicro USB-to-SWD adapters speak between host computer and adapter, but since they only "transport" SWD, they work with every SWD-compatible microcontroller.
On most development systems, you'd want to use OpenOCD as "driver" for these devices, so that you can easily flash images (either directly through OpenOCD or using e.g. GDB's load). If you're stuck with an absurd OS that needs specific drivers even for generic devices, you'd probably have to install ST's tools.
Btw, I'm meaning to flash an empty ARM processor without a bootloader.
Yep, sounds like a classical case for SWD – ARM offers hardware for that in their cores, and most manufacturers choose to use that and assign pins.
Also note that most manufacturers (including ST) ship their "empty" ICs with some kind of bootloader, over which you can load firmware through a serial port or even USB into the device – suuuuper handy for manufacturing.
For a bit of discussion on SWD, I actually recommend (for the pure fun it is to read) PoC||GTFO 0x10, pp. 26, which begins with an introduction of ARM's debugging infrastructure and SWD as a protocol, and then goes on to explain how to use SWD-connected ARMs as sophisticated IO expanders rather than independent MCUs.
|
H: How does the CD4066 Work?
I am learning electronics so this is a beginners question. Its about the CD4066 IC as mentioned here
https://en.wikibooks.org/wiki/Practical_Electronics/IC/4066
Is the 4066 equivalent to four independently controllable relays, where one relay switches on the pair X1,Y1?-Is this the purpose of this IC? So four relays are required to switch (the 4 pairs of audio signals)
X1,Y1
X2,Y2
X3,Y3
X4,Y4
Because in the IC there are four switches that can control the four pairs above. So with this IC 4 speakers are controllable by a digital CPU.
AI: You can regard the 4066 as 4 relay contacts but there are a bunch of constraints that limit this analogy. The main ones are: -
On resistance is massively higher than a relay contact (typically 125 ohm)
Ohmic linearity with input signal voltage means signal distortion may occur (unlike normal relay contacts).
Any input / output signal must be constrained to the power rails of the device. In other words, there is *some" isolation capability that is useful but don't expect truly AC signals to do anything other than either badly distort or burn the chip.
Maximum current is limited to a few mA
Open circuit capacitance is quite high thus limiting the ability to "open-circuit" high speed signals.
|
H: Is it sufficient to disconnect the negative terminal of a voltage source?
I have a scenario where I would like to use N-channel MOSFETs, because they have lower on-resistance and are cheaper.
simulate this circuit – Schematic created using CircuitLab
Would there be any problems with this approach?
I've read this answer but I understand the main concern is somehow accidentally forming a closed loop. But if the MOSFET is safely at the negative terminal of each voltage source, I don't see how there would be any risk?
Am I correct that the positive terminal of either voltage source cannot conduct until the negative terminal is also connected? In effect, as if it wasn't connected at all?
AI: Disconnecting the negative side is sufficient provided that the MOSFET is the only connection to the negative side of the supply. I have designed electronic circuit breakers for batteries that work that way. They worked fine and were even certified for use in an explosive atmosphere.
|
H: Calculating self-induction current at specific time
I've been studying Circuit Analysis, and there was this problem I don't know how to solve. I don't know if this kind of question does not fit to this site's format. I'm sorry if it doesn't but I need answer as soon as possible because there is exam and the first place I could ask was here. I need to see how it is solved.
The switch opens at \$ t = 0 \$ . Find \$ i_{1}(t) \$ for \$ t > 0 \$.
AI: I1(t) = I10 * exp(-t/tau) t > 0
I10 = V/(R1//R2) * R1/(R1+R2)
tau = L/(R1+R2)
R1 = 6 Ohm
R2 = 12 Ohm
This is the solution, it is very mechanical, but try to think it through. The circuit is stationary for t < 0 so the inductor is equivalent to a short circuit. At t = 0 the switch opens, the current in the inductor cannot change suddently (that would imply a finite change in energy has occurred in an infinitesimal time), then you have a current circulating in a resistive circuit, it will eventually dissipate all the available energy (which was stored in the inductor). How fast? that depends on the time constant of the circuit.
Since all the remaining sources are stationary (there are no other current/voltage sources) the final solution is stationary and you can calculate that by simply considering the inductor a short circuit in the circuit at time > 0. But in this case it is pretty obvious that the current is going to go to zero.
|
H: Why do real LC oscillating circuits not oscillate at the resonance frequency?
Consider these two simple LC oscillators with ideal op-amps:
simulate this circuit – Schematic created using CircuitLab
The resonance frequency is for both circuits about 159 kHz, however neither of the two will oscillate at that frequency in both the simulations and the real application. Why is that?
For instance, the first circuit, when tested, oscillates with a frequency of about 132 kHz and 156 kHz in the LTspice simulation. Ok, components are not perfect, but here we are significantly off the expected value. Furthermore, taking into account the internal resistance of the capacitor and inductor, if you consider a dumped RLC circuit with dumped oscillations, the frequency should still be the resonance frequency.
In many other (more complex circuits) I often read that the LC couple will oscillate at the resonance frequency $$\frac{1}{2\pi\sqrt{LC}}$$. However, this is almost never the case, and the frequency is far off that value (again, why?).
AI: According to the oscillation condition (Barkhausen) a circuit with frequency-dependent feedback can oscillate only at a frequency where the loop gain is unity (or slightly larger). In particular, this means that the PHASE of the loop gain function must be ZERO.
For large frequencies (and the example frequency above 100 kHz can be considered as large) the phase shift of the used opamps must not be neglected. As a result, the frequency of oscillation (if the circuit does oscillate!) is the frequency where the passive network has a positive phase shift which exactly can cancel the (unwanted, but unavoidable) negative phase shift of the opamp. For this reason, the resulting frequency is less than the desired frequency.
(The passive LC circuit has a positive phase shift for frequencies BELOW the passive resonance point).
First Remark: The loop gain of a circuit with feedback is simply the product of the feedback network and the gain of the amplifier stage.
Second remark: The first circuit is not very common because it has no stabilizing negative DC feedback. The opamp is driven deep into saturation.
Both resistors in the positive feedback path of the second circuit are too large (to much damping, bad selectivity). Good results for R2=R3=10 Ohms.
|
H: RHP zero in boost converter
Below is the control to output transfer function of a boost converter.
(source: Switching Power Supplies A - Z by Sanjaya Maniktala here at page 286)
And this is one part talking about the RHP zero the book. I don't understand the last part from "Eventually...".
When output voltage dips, the controller adjusts to increase duty cycle. However, by increasing duty cycle the time to transfer energy to load decreases. If everything goes on like this, finally the duty cycle will go to 1 and no time for transferring energy to load. But the note below says that eventually the inductor current ramps up to the right level and the strange behavior gets corrected. Why duty cycle not go to 1?
Note: Intuitively, the RHP zero is often explained as follows — if we suddenly increase the load, the output dips slightly. This causes
the converter to increase its duty cycle in an effort to restore the
output. Unfortunately, for both the boost and the buck-boost, energy
is delivered to the load only during the switch off-time. So, an
increase in the duty cycle decreases the off-time, and there is now,
unfortunately, a smaller interval available for the stored inductor
energy to get transferred to the output. Therefore, the output
voltage, instead of increasing as we were hoping, dips even further
for a few cycles. This is the RHP zero in action. Eventually, the
current in the inductor does manage to ramp up over several successive
switching cycles to the new level consistent with the increased energy
demand, and so this strange situation gets corrected — provided full
instability has not already occurred!
AI: The continuous mode boost and buck-boost converters exhibit control-to-output transfer functions \$G_{vd}(s)=\hat{v}(s)/\hat{d}(s)\$ containing two poles and one RHS (right half-plane) zero, called zero of nonminimum phase.
Your original transfer funtion is:
$$G_{vd}(s)=\frac{1}{V_{RAMP}}\times \frac{V_{IN}}{\left ( 1-D \right )^2}\times \frac{1/\underline{L}C\times\left ( 1-s\left ( \underline{L}/R \right ) \right )}{s^2+s\left ( 1/RC \right )+1/\underline{L}C}$$
Starting from a simpler function (without the RHP zero), named \$ G_{vd}'\$:
$$G_{vd}'=\frac{1}{V_{RAMP}}\times \frac{V_{IN}}{\left ( 1-D \right )^2}\times \frac{1/\underline{L}C}{s^2+s\left ( 1/RC \right )+1/\underline{L}C}$$
Or, placed as a standard second order t.f:
$$G_{vd}'=K_{DC}\times \frac{1/\underline{L}C}{s^2+s\left ( 1/RC \right )+1/\underline{L}C}$$
where the DC gain is \$K_{DC} =\frac{1}{V_{RAMP}}\times \frac{V_{IN}}{\left ( 1-D \right )^2}\$.
The equation can be rewritten as:
$$G_{vd}'=K_{DC}\times \frac{1}{\left ( \frac{s}{\omega_0} \right )^2 + \frac{s}{\omega_0Q}+1 }$$
With \$\omega_0=1/\sqrt{\underline{L} C}\$ and \$\omega_0Q=R/\underline{L}\$
Similarly, the original transfer function can be expressed as (RHP zero included):
$$G_{vd}=K_{DC}\times \frac{\left (1-\frac{s}{\omega_{RHP}} \right )}{\left ( \frac{s}{\omega_0} \right )^2 + \frac{s}{\omega_0Q}+1 }$$
The response will be:
$$\hat{v}(s)= \frac{K_{DC}}{\left ( \frac{s}{\omega_0} \right )^2 + \frac{s}{\omega_0Q}+1}\times\hat{d}(s) - \frac{K_{DC}/\omega_{RHP}}{\left ( \frac{s}{\omega_0} \right )^2 + \frac{s}{\omega_0Q}+1 }\times\hat{d}(s)\times s$$
The response of the original system is the sum of two components: The first is equivalent to the modified system response (without the zero) and the second is the derivative (scaled) of that one. For the case of a stable system with a step input in \$t = 0\$, this last component will have substantial influence at the beginning and then will vanishes when \$t\rightarrow\infty \$. Note the negative sign leads to a momentary opposite effect in output (nonminimum phase).
UPDATE:
The presence of zero RHP in the model is explained as follows: For the output voltage to increase, the duty cycle must be increased in such a way that the inductor will be disconnected from the load for a long time, causing the output voltage to drop (i.e. in the opposite direction to the one desired). The controller must be designed to meet the project requirements and avoid oscillations while maintaining duty cycle below an undesirable 100% - being limited by the PWM integrated circuit itself.
|
H: Show that the power dissipated in a resistor over infinite time is equal to the energy stored in a discharging capacitor
I am learning about RC transient responses and if a charged capacitor is connected to a single resistor, its voltage will decrease exponentially. I understand this and how its equation is derived. I also understand how the energy stored in a capacitor is obtained.
So, given an infinite amount of time, all of the energy will be dissipated in the resistor. I would like to prove that the integral of p(t)dt between 0 and infinity is equal to 0.5CV₀². I've tried substituting in i²(t)R for p(t) and then equating i to -Cdv(t)/dt but I end up with something that isn't even an integral anymore, namely of the form RC²∫dv(t)/dt between 0 and infinity.
What sort of substitution should I make to allow this to be integrated? Thank you for reading.
AI: The voltage across a capacitor discharging (exponentially) through a resistor is: -
\$V = V_0\times e^{\dfrac{-t}{CR}}\$ where Vo is the voltage at t = 0
The current is the above voltage divided by R
So, power is \$\dfrac{V_0^2}{R}\times e^{\dfrac{-2t}{CR}}\$
Note that the exponential term now has a 2 in it because it became squared.
Now if you integrate power you accumulate the energy: -
\$\dfrac{V_0^2}{R}\times \dfrac{CR}{-2}\left[e^{\dfrac{-2t}{CR}}\right]_0^\infty\$
If you resolved the integral between 0 and infinity it comes to -1 hence energy taken by resistor is simply: -
\$\dfrac{CV_0^2}{2}\$
|
H: Understanding this AVR ASM routine
This is an abbreviated version of WS2812 LED control routine. I have removed all the no-ops for the sake of readability here. Can someone explain the role of the branch instruction (brne) towards the end of the routine?
asm volatile(
" ldi %0,8 \n\t"
"loop%=: \n\t"
" out %2,%3 \n\t" // '1' [01] '0' [01] - re " sbrs %1,7 \n\t" // '1' [03] '0' [02]
" out %2,%4 \n\t" // '1' [--] '0' [03] - fe-low
" lsl %1 \n\t" // '1' [04] '0' [04] " out %2,%4 \n\t" // '1' [+1] '0' [+1] - fe-high
" dec %0 \n\t" // '1' [+2] '0' [+2]
" brne loop%=\n\t" // '1' [+3] '0' [+4]
: "=&d" (ctr)
: "r" (curbyte), "I" (_SFR_IO_ADDR(ws2812_PORTREG)), "r" (maskhi), "r" (masklo)
);
AI: Variable %0 (ctr in the surrounding function) is counting the number of bits in the byte being transmitted. The branch causes the routine to loop until all 8 bits of %1 (curbyte) have been processed.
It looks like you may have removed more than just no-ops. There should be at least one branch based on the value of a bit in %1 that selects one of two paths through the body of the loop. The out instructions toggle an output pin high and low to create the PWM waveform that the WS2812 requires, and the no-ops control the relative timing of the high and low periods.
Ah, yes. Looking at the original code, it has:
asm volatile(
" ldi %0,8 \n\t"
"loop%=: \n\t"
" out %2,%3 \n\t" // '1' [01] '0' [01] - re
[ variable number of NOPs based on w1_nops ]
" sbrs %1,7 \n\t" // '1' [03] '0' [02]
" out %2,%4 \n\t" // '1' [--] '0' [03] - fe-low
" lsl %1 \n\t" // '1' [04] '0' [04]
[ variable number of NOPs based on w2_nops ]
" out %2,%4 \n\t" // '1' [+1] '0' [+1] - fe-high
[ variable number of NOPs based on w3_nops ]
" dec %0 \n\t" // '1' [+2] '0' [+2]
" brne loop%=\n\t" // '1' [+3] '0' [+4]
: "=&d" (ctr)
: "r" (curbyte), "I" (_SFR_IO_ADDR(ws2812_PORTREG)), "r" (maskhi), "r" (masklo)
);
The pin gets set high at the beginning of the loop. The sbrs instruction (Skip if Bit in Register is Set?) determines whether the pin gets set low after the w1_nops period, or not until after the w2_nops period as well. The total of w1_nops + w2_nops + w3_nops (plus the instructions shown here) determines the total period of each bit.
|
H: Maximum slope of a sine wave
For a signal with amplitude,
$$V = A\sin (\omega t)$$
any idea at what point the maximum slope (gradient, dv/dt) is? I have gone through the method of differentiating which yields
$$A\omega \cos(\omega t) \tag{i}$$
and then doing a second differential, which yields
$$-A \omega^2 \sin (\omega t) \tag{ii}$$
Equating (i) to 0 and then substituting back into (ii) to find which point is where I got stuck. I can do this with a normal equation but trigonometry sometimes confuses me. Any help will be appreciated.
AI: The maximum slope for a sine wave that has no offset and an ampliutde \$A_0\$ occurs exactly during the zero crossings. Its value is simply \$A_0 \omega\$
The derivation is
$$
\frac{d}{dt} A_0 \sin (\omega t) = A_0 \omega \cos(\omega t)
$$
which gives the slope of the sine wave.
The maximum of the cosine is 1. Therefore the maximum is \$A_0 \omega\$.
$$
\max \left\{ A_0 \omega \cos(\omega t) \right\}= A_0 \omega
$$
The results makes sense, since intuitively the slope has to increase with the amplitude as well as the frequency.
|
H: Altium - Deselect after compile error
When I compile my schematic, I get some errors in the compile window. When I double click them it zooms in on them and highlights them with a mask. How do I get rid of that mask?
AI: Click the "Clear" button at the bottom right of the editor window.
|
H: Power supply enclosure
I have bought enclosure for my power supply (datasheet), which looks like the follow:
How can i connect the plugs to my circiut? Soldering is not an option because the box is made from ABS.
AI: It looks like the pins are 4mm in diameter, so if the manufacturer does not offer a matching contact, you could conceivably pull the center contact out of an RCA jack (3.175 mm), spread it out a bit, and solder it to your PCB in the appropriate orientation.
|
H: Looking for a replacement relay
I have an relay in an old device (a heated display case), that seems to be failing.
The relay is a Midtex 157-32U200, and the few distributors that I could find that carry this relay refuse to sell it to me as it's supposed to be a "restricted item" (I guess only machine resellers are allowed to buy it).
I've found something similar at my local electronics place - it's a generic relay named JQX-53F 32. The ratings seem to be similar, and while this relay has an additional bank of connectors in the middle row, the original one only used the outer rows, so I don't know if that would be of any relevance.
Obviously, I'm not an electrical engineer, so I don't know the first thing about relays, but perhaps somebody can help me find a suitable replacement for my dying old one.
I've attached the wiring diagram of the machine, and pictures of the old and new relay, so perhaps that'll help more than my amateurish explanations...
AI: That ought to work fine, as long as your relay socket will accept the middle row of contacts. If not, see if you can find a JQX-53F 2Z, which ought to be a perfect match.
The only physical difference will be the lack of the mecbanical tab at the rear which is intended to lock the relay into the socket. However, since you have a stationary application with no shock or vibration this should not be an issue.
|
H: Zener Diode at Base of Transistor
simulate this circuit – Schematic created using CircuitLab
I am building a so called protection section for my mini project which involves a 3.9V zener diode parallel to a battery as a shunt voltage regulator.
Assuming the voltage at the 1N4002 diode is 4.1 V and that the battery has reach 3.9 V to cause the zener diode to conduct, will the voltage at base of the transistor be 0.2 V?
The transistor I intend to use is a BC549C low noise transistor instead of the general 2N2222.
Assume that the circuit is current limiting with a 20 mA into the battery, will this be enough to turn on the base of the transistor (without a sufficient 0.7 V across Vbe to turn it on)?
AI: It appears that you want the LED to be on when the battery voltage is above the limit set by the zener.
Unfortunately, the way you have it will not work as you expect.
You are using the transistor as a high side switch. To make it conduct, you must make the voltage on the base higher than the sum of the Vbe and the LED forward voltage. Then there's the 3.9V for the zener. So, 1.5V for the LED plus 0.7V for Vbe and 3.9V for the zener means the LED will only light when the battery voltage is above 6.1V.
The usual way would be to use the transistor as a low side switch (between the LED and ground.). Then the battery only has to reach 0.7V plus the zener voltage to make the LED light.
|
H: Smoothing modified sine wave inverter
I need a backup power source for my aquarium tank filtration system. A pure sine wave UPS seems to be an expensive overkill, so I'm looking for alternatives.
The filtration system pump runs of 230V, 50Hz AC motor with magnetic coupling (I assume synchronous?) and consumes somewhat less than 10W of power.
My idea was to buy a cheap modified sine wave power inverter from eBay, rated for 100W, and use a spare 12V sealed lead-acid motorcycle battery I already have.
The inverter came yesterday, I've tried it with the battery and my pump. It seems to work, at least in sense the motor pumps water properly, but it does not seem to like the modified sine wave much - it's very noisy and produces more heat than normal. I stopped it after a minute worrying about damaging my equipment. I suspect this problem is caused by sharp edges and high harmonics produced by the modified sine wave inverter.
Wouldn't it be possible to simply smooth the output of a modified sine wave inverter somewhat to make it more acceptable for the motor? How about using a some capacitor on the 230V output?
I've done some research and found some people arguing it was inefficient but possible, in principle, by using a capacitor on the output, parallel to the load. I don't need it to be very efficient, it's just 10W, nor I need the smoothing to be perfect, but I definitely need to make the output more sine-like. Is this idea viable? If so, how I compute the optimal capacity of the filtration capacitor?
AI: I would suggest that an inductor in series with the load may help.
Most times, a Modified Sine Wave inverter produces a simple square wave output with significant OFF-time. The peak voltage matches that of a sine-wave and the off-time is increased such that the average voltage is equal to that of a sine wave.
But it's still a square (rectangular) wave.
Placing a capacitor across the load is a bad idea because the peak current will be very high. Look at the leading edge of the waveform - it's almost vertical.
However, an inductor in series with the load may slow the edges sufficiently that the motor is "more happy" than with the plain square wave from the inverter.
Size, current rating, inductance value is "left as an exercise for the student".
|
H: Will this circuit suffice for low battery detection for any microcontroller?
I'm making a major digital circuit powered by two 3.6 volt cordless phone NiCD rechargeable batteries, and one thing I want to include is a low battery detector. To protect the rest of the circuit from overheating, I use a typical 7805 voltage regulator circuit to give all my circuit 5 volts.
What I want to do is detect if the source battery is too weak to power the rest of the circuit and allow a microcontroller to detect it via a GPIO pin's logic state.
Because digital circuits ideally run on 5VDC, I figured using a zener diode rated at 5.1V with a 10K resistor for R1 would be sufficient to detect the low voltage itself but I'm not sure if it would explode the GPIO pin even though its pulled up by 5V through R2 which I may also make as 10K.
Questions are:
Would this circuit have potential to work or would connecting fresh batteries equaling the same voltage explode a GPIO pin?
Also, is 10K an ok value for the resistors or should I choose something different?
The microcontroller in question is at89C2051.
AI: This circuit won't work - the transistor will be saturated, and the GPIO pin held low, until the battery voltage drops to ~ 0.7 volts.
Also, your thought of using an LM7805 to regulate your 7.2 volt battery down to 5 volts will only work with fully-charged batteries, as the 7805 need a minimum of 2 volts across it to maintain regulation.
|
H: Identifying Source of Periodic Artifact at Op-Amp Output
My MAX44251 dual op-amp has a very small unwanted 131KHz periodic artifact at the output, seemingly regardless of how it's configured.
My assumption was EMI, but I can't see this 131KHz signal on any other part of the circuit. I've also tested this in multiple buildings, with multiple probes, with all other electronics turned off, and surrounded by foil shielding.
What should I try to remove it? I would like to at least achieve a voltage follower with noise under 1mV.
The chip was originally used in a more complex circuit when I first noticed the problem. BUT, to isolate this issue I made a whole new test PCB with fresh components. I left extra pads to reconfigure the chip in different ways while testing.
Right now it is configured very simply:
simulate this circuit – Schematic created using CircuitLab
The bypass caps are on the bottom ground plane layer. Vias are hand soldered.
I have observed the effect through both the Agilent 10X passive probe (It's hard to see), and through a probe like the following, with which I can zoom all the way to 2mv/div. Originally, it was observed because the output is fed to a comparator, and the comparator output indicated the input signal amplitude was > the desired 2mV.
The waveform is periodic but kind of strange. Here's a few pics from different angles:
200 ns Stopped
50 ns Free Running
20 ns Free running
10 ns Stopped
AI: I can't really tell if this is a actually a symptom of what is described in the datasheet:
Notice how there's a spike that exceeds \$30 \frac{\text{nV}}{\sqrt{\text{Hz}}}\$ at 65kHz – pretty much half of the frequency you're observing your noise at; they didn't characterize up to 131.5kHz, however.
What should I try to remove it? I would like to at least achieve a voltage follower with noise under 1mV.
If you just need a low-bandwidth voltage follower: Use a low-pass filter.
If you need signal up to 65 kHz and above: An RLC notch (band-stop) would probably work best; a quick & lazy design run on my favourite passive filter design tool yielded R=0.16Ω, L=1µH, C=1.5µF as possible configuration.
Note that you could try to use the inverse circuit (RLC bandpass; swap the (L--C) with the R) in the feedback branch of your voltage follower.
|
H: Minimum current of a sensitive gate SCR
i have a few 2N5064 SCR that are puzzling me. No matter how many hundreds of thousands of ohms i put between their gates and a 3.2V battery source, it always starts to conduct. I even put a 10 V zener between them, and it still turns on. Looking at the datasheet, i am puzzled. Where minimum gate turn on current is, there is just a dash. Same for minimum voltage. What does this mean? Will they activate with pretty much any current?
Thanks
AI: 2N5064 is a sensitive-gate SCR. It triggers with a maximum of 200uA at room temperature (Tj = 25°C).
No minimum is specified in the official data, as you can see.
Edit: In order to keep it "off" guaranteed, you should keep the gate voltage to less than 100mV, which is guaranteed to not trigger up to Tj = 125°C or so, depending on the manufacturer. In practice that may mean that you'll have to put a resistor from cathode to gate to decrease the sensitivity.
If you really want something sensitive, play with the gate on a MOSFET such as a 2N7000. pA are more than enough typically.
|
H: Generating a current with ionized water droplets (in a cloud)
Playing mind games with myself tonight; I know some of the windiest spots in the world are at the tops of mountains and often socked in the clouds. To measure meteorological phenomena at these locations is a difficult task, with rime ice eliminating mechanical energy generation and solar very limited due to clouds.
With occasional ionized cloud droplets, would it be possible to generate electricity "indirectly" by inducing charge on a conductor without any moving parts? Would this be possible even if there were equal numbers of positive and negative ions?
AI: When ions travel in a magnetic field they are deflected due to the Lorentz force. Positive and negative ions experience opposite deflection.
This effect can be used to make an Electro Hydro Dynamic generator.
If ionized wind were blowing through an enclosed channel towards two metal plates that were separated by a gap. And a permanent magnet was placed up-wind of the plates then positive ions would deflect to one plate and negative to the other. A potential difference would develop between the plates due to the accumulation of charge. That difference could be used to generate electric current.
The amount of current would probably be quite small, and would depend on the charge density in the wind, magnetic field strength, wind velocity, and channel shape.
But in my opinion if its really the windiest spots in the world just use a small windmill to generate the electricity. You could probably cut the moving parts down to just one piece the blade (with a permanent magnet on it).
|
H: Open-Drain Through Schmitt Trigger
I am attempting to use a board that performs level shifting from a 5V GPIO pin to a 3.3V RSTn pin on a Decawave DWM1000 Module via a SN74LVC2G17DBVR.
The problem is, as discussed in Section 5.6.3 of the datasheet, the external source of the RSTn pin must be open-drained as soon as the chip has been reset. If it were directly connected to the GPIO pin, then it would simply require that the pin be configured as a high-impedance input after driving low. However, I suspect that the trigger is complicating things.
In order to initiate a reset, the GPIO pin is configured as an output with a low value. It then holds this value for 2 milliseconds and then is configured to a high-impedance input. I think that the trigger cannot allow an output to float, and so the reset can only be driven high or low.
So I guess my question is this: Is there a modification that I can make to the existing board that can address this issue? If not, what component should I use instead?
AI: I didn't see answers to any of my questions yet, but I think I might have worked-out (at least one of) the problem(s) you were asking about:
Using a 5V GPIO correctly as an open-drain output, won't be a problem to an input limited to 3.3V, which is expecting to be driven by an open-drain output. That's because the open-drain output either provides a low-resistance path to 0V, or else is effectively not there (apart from tiny leakage current). As an open-drain output, it doesn't output 5V (unlike a push-pull output), which seems to be a part of your concern.
It is clear from the Decawave DW1000 transceiver datasheet that its RSTn signal expects to be driven by an open-drain output:
An external source should open-drain the RSTn pin once the DW1000 has been reset.
Or it could be driven by a 3.3V (or 5V) GPIO output in push-pull mode, but only for driving the RSTn signal low, and then that MCU pin must be changed to become an input (in other words, it must not drive the signal high) as explained here:
If RSTn is controlled by a GPIO of an external micro-controller care should be taken to ensure that the GPIO is configured as high-impedance as soon as it is released from the LOW state.
However, the IC you mentioned you were using as a level converter (TI SN74LVC2G17 Dual Schmitt-Trigger Buffer) is not an open-drain output device!
Since driving RSTn low resets the Decawave IC, when you stop driving that signal low to the buffer (to end the reset), you are then driving that signal high (since that TI Schmitt-trigger is not an open drain output device). However the Decawave IC datasheet says:
RSTn should never be driven high by an external source.
Therefore you should not be driving the Decawave IC RSTn signal using that TI Schmitt-trigger buffer IC.
Options include:
drive the Decawave 3.3V RSTn signal input directly from the 5V GPIO which you mentioned, in open-drain mode, if you can be sure you will never drive the output high in push-pull mode, e.g. due to programming error or other causes outside your control;
or
use a suitable open-drain buffer IC between the 5V GPIO (this time configured in push-pull mode) and the 3.3V RSTn signal input.
Update: It's difficult to advice on a suitable open-drain buffer IC, without knowing the design constraints, and the reasons why the current 74LVC2G17 was chosen originally.
That 74LVC2G17 has two (non-open-drain) buffer gates per package. I found that NXP produce a 74LVC2G06 which has two open-drain, Schmitt-trigger, inverters. They seem to have the same pinout as your current 74LVC2G17 (but do double-check).
As the 74LVC2G06 gates are inverters, you would need to invert the logic of the GPIO pins driving them, in order to have equivalent behaviour to the 74LVC2G17 (and, of course, the 74LVC2G06 gates don't drive their outputs high), as shown here from the NXP datasheet.
Source NXP datasheet
At least that is a starting point in your search for a more-suitable alternative device.
If you need more open-drain gates, then the (relatively common) 74LVC06 devices may be worth investigating.
|
H: What are the uses of 400V capacitors?
I salvaged some 470uF 400V electrolytic capacitors from a junk yard. They look like
Why would a PCB contain 2 such capacitors on a board? A motor starter?
Are there any other uses of these capcitors? Say, making a filtered power for audio?
AI: In a SMPS, at the first filter stage after initial rectification. The high voltage is then chopped at a high frequency in order to allow use of a much smaller transformer than if the 220VAC input was reduced directly.
|
H: Making a Triangle wave on LTSpice using LM741 opamp
I'm trying to get a 1kHz, 15V triangle wave on LTSpice. For that, I'm using the following design:
So far, my Square Wave has the desired frequency and amplitude, but my triangle wave has a lower amplitude. How can I correct this so both waves match at +15v and -15v?
I'm using the following specs, sorted out by some formulas but also trying. There are some values that I don't see how affect what I get.
Vcc=15v
R1=150k
C1=3.03361n F
R2=R3=100k
R4=R5=1k
C2=1000n F
R6=3k
Thanks a lot for your help.
EDIT: [UPDATED] here's a caption of the actual simulation i'm running on LTSpice:
AI: There a few problems with your circuit:
1) The second stage will generate an exponential rather than a triangle.
2) The resistor values for feedback on the second stage are much too low - the amplitude may be limited by the output current capability of the 741. You can use something in the range 10k-100k.
3) The actual amplitude of the output will be limited by the opamp. The 741 os a very ancient opamp and is not as good as more modern ones - as Peter mentioned the output cannot get closer than ~2.5v to the rail.
This configuration should give better results:
It still uses two opamps and only needs one capacitor to set the timing.
The first stage is arranged as a schmitt trigger to detect when the output of the second stage reaches voltages determined by resistors R1 and R2. The second stage is configured as an integrator and will produce a triangular wave with linear slopes.
The square wave output is still determined by the saturation of the opamp as before, an opamp with rail to rail output will get very close to the supply rails.
|
H: Output resistance of non-inverting amplifier
I'm reading a book*, and it derives the output resistance of a non-inverting op-amp amplifier by grounding the input voltage, applying a test current source at the output, and measuring the resulting output voltage:
Here L is the loop gain (i.e. open-loop gain a x feedback gain b).
Here's my take on finding the output resistance, which gives a different answer. We can think of the closed-loop amplifier as a black box, as shown below.
We can measure the output resistance R_o by shorting the output and measuring the current that flows through. The resistance is then:
$$
R_o = \frac{Av_i}{i_{o,\text{short}}} \approx \frac{v_i}{bi_{o,\text{short}}}
$$
Where I have used A = 1/b, with b being the feedback gain.
If we short the output of the actual circuit (with v_I included), we have:
$$
i_{o,\text{short}}=\frac{av_D}{r_o} = \frac{av_I}{(1+L)r_o}
$$
(Note that \$R_1\$ and \$R_2\$ appear in series with each other and in parrallel with the short, and therefore do not affect the output current).
Then:
$$
R_o = \frac{(1+L)r_0}{ab} = \frac{1+L}{L}r_o \approx r_o
$$
*"Design with Operational Amplifiers and Analog Integrated Circuits", by Sergio Franco.
AI: Your analysis will work when the output is forced to 1V instead of 0V. I was able to get the correct answer (output impedance is indeed lowered by loop gain!) by analyzing a 1V voltage source on the output.
The way that I prevent this sort of mistake is by being strict about how I write my equations. Among other rules, I don't introduce new terms like L and b, I don't use any numbers until the end, and I don't make approximations in the algebra. When the algebra becomes tedious, I use the Maxima computer algebra program. I think there is also a Mathematica package somewhere that does this type of analysis correctly.
In this problem, the rule that was violated is that the Vout term was replaced by a number before the equations were solved. In this case the number choice was unfortunate and it ruined the equation.
The more intuitive explanation is that Zout=Vout/Iout, so Vout=0 is a problematic choice.
|
H: what is the difference between coin battery and alkaline battery
i'm quite curious with coin battery and alkaline battery. I have submerge 2 coin battery into the water and the battery short circuit and damaged after a while. While i put 2 alkaline battery into the water, but they still can perform well. i wonder why both type of battery is short circuit, but only coin battery will damage. Does anyone know why?
AI: "Coin" is a physical "form factor" or shape of a battery cell. The more common form factor is a cylinder (like a AA cell, etc.)
Alkaline is a type of CHEMISTRY of the battery cell. Different chemistries of batteries can be found in various form-factors. There is no particular correlation between the form-factor and the chemistry of a battery cell.
Whether a particular battery cell can survive under water has nothing to do with its form-factor or its chemistry. Most batteries are not designed to operate submerged under water. No matter what size, shape, or chemistry they are.
|
H: STM32: Busy flag is set after I2C initialization
For the reference: the same problem is described there, but the author's solution doesn't work for me - I2C busy flag strange behaviour
I used STM32CubeMX to generate project template with I2C peripherals initialization. Unfortunately it works somehow strange: after HAL_I2C_MspInit(I2C1) is being invoked, bus is considered permanently busy.
If I try to apply
__HAL_RCC_I2C1_FORCE_RESET();
HAL_Delay(1000);
__HAL_RCC_I2C1_RELEASE_RESET();
That resolves problem with BUSY flag, but causes problem - SB bit not being set after START is generated. According to debugger, I2C registers are cleared completely after the reset - I suspect this is the problem with that method.
I also confimed short voltage drop at SDA line during startup, that is probably the cause of the issue. I took a closer look at SDA/SCL pins initialization code generated by CubeMX:
void HAL_I2C_MspInit(I2C_HandleTypeDef* hi2c)
{
GPIO_InitTypeDef GPIO_InitStruct;
if(hi2c->Instance==I2C1)
{
/* USER CODE BEGIN I2C1_MspInit 0 */
/* USER CODE END I2C1_MspInit 0 */
/**I2C1 GPIO Configuration
PB6 ------> I2C1_SCL
PB7 ------> I2C1_SDA
*/
GPIO_InitStruct.Pin = GPIO_PIN_6|GPIO_PIN_7;
GPIO_InitStruct.Mode = GPIO_MODE_AF_OD;
GPIO_InitStruct.Speed = GPIO_SPEED_FREQ_LOW;
HAL_GPIO_Init(GPIOB, &GPIO_InitStruct);
/* Peripheral clock enable */
__HAL_RCC_I2C1_CLK_ENABLE();
/* USER CODE BEGIN I2C1_MspInit 1 */
/* USER CODE END I2C1_MspInit 1 */
}
}
I changed it to put clock enable before HAL_GPIO_Init() invocation and now my I2C communications works (at least I didn't noticed anything weird yet).
Finally, my question is - is there any better solution for this? CubeMX places clock enable code after the GPIO init method invocation. I can stay with two invocations of __HAL_RCC_I2C1_CLK_ENABLE(), but that's quite ugly in my opinion, so I am looking for any better solution, either software or hardware.
Device is STM32F100RB on STM32VLDiscovery board (with STLink v1), in case that matters.
AI: In my opinion STM32CubeMX code should not be considered as a ready to use code, but some as an example you can start with. With most of the microcontrollers it works, but there are some rare cases when it is not.
If you know it is not working and you have found the solution as well, you do not have to stick to the original code. In your case you can omit the __HAL_RCC_I2C1_CLK_ENABLE() call after the GPIO initialization, and leave the one before it. If it works, and you have said it works, then use the working way. Even ST's software can have bugs.
You are using an official board so the hardware should be OK, but you can check if the pull-up resistor values are correct. Or if a slave device does something during the initialization.
The best would be to run your code with everything disconnected from the Discovery (apart from the pull-ups), and check if it is still stuck in busy. If yes, it is fine if you replace that line in the generated code. It is not that big modification.
Unfortunately there is not any I2C exapmle in the STM32CubeF1 example package (this is not the code generator), under the STM32Cube_FW_F1_V1.4.0\Projects\STM32VL-Discovery\Examples. But if you check the MspInit functions of the UART or SPI. The clocks are enabled in both of them before the GPIO init.
void HAL_SPI_MspInit(SPI_HandleTypeDef *hspi)
{
GPIO_InitTypeDef GPIO_InitStruct;
if (hspi->Instance == SPIx)
{
/*##-1- Enable peripherals and GPIO Clocks #################################*/
/* Enable GPIO TX/RX clock */
SPIx_SCK_GPIO_CLK_ENABLE();
SPIx_MISO_GPIO_CLK_ENABLE();
SPIx_MOSI_GPIO_CLK_ENABLE();
/* Enable SPI clock */
SPIx_CLK_ENABLE();
/*##-2- Configure peripheral GPIO ##########################################*/
/* SPI SCK GPIO pin configuration */
GPIO_InitStruct.Pin = SPIx_SCK_PIN;
GPIO_InitStruct.Mode = GPIO_MODE_AF_PP;
GPIO_InitStruct.Pull = GPIO_PULLDOWN;
GPIO_InitStruct.Speed = GPIO_SPEED_FREQ_LOW;
HAL_GPIO_Init(SPIx_SCK_GPIO_PORT, &GPIO_InitStruct);
void HAL_UART_MspInit(UART_HandleTypeDef *huart)
{
GPIO_InitTypeDef GPIO_InitStruct;
/*##-1- Enable peripherals and GPIO Clocks #################################*/
/* Enable GPIO TX/RX clock */
USARTx_TX_GPIO_CLK_ENABLE();
USARTx_RX_GPIO_CLK_ENABLE();
/* Enable USARTx clock */
USARTx_CLK_ENABLE();
/*##-2- Configure peripheral GPIO ##########################################*/
/* UART TX GPIO pin configuration */
GPIO_InitStruct.Pin = USARTx_TX_PIN;
GPIO_InitStruct.Mode = GPIO_MODE_AF_PP;
GPIO_InitStruct.Pull = GPIO_PULLUP;
GPIO_InitStruct.Speed = GPIO_SPEED_FREQ_HIGH;
HAL_GPIO_Init(USARTx_TX_GPIO_PORT, &GPIO_InitStruct);
So I think your solution is perfectly fine.
|
H: Weird problem with ADXL345, Buzzer and Arduino
I've set up a Serial port via a Bluetooth module (HC05). I'm using a JavaFX app that I wrote to read the serial port, plot the acceleration values (via ADXL345) and write back to the serial port an alarm bit (1 for ON and 0 for OFF) whenever a certain function of accelX reaches over a value.
The weird part is, whenever Arduino detects ON bit and turns on the buzzer, my acceleration values become quite noisy! I've checked my codes both in Arduino side and Java side and couldn't find a problem.
Then I put a LED instead of buzzer and to my surprise it turned on without interfering with accel. readings !!
I really don't know what is the problem with buzzer. Can anyone help?
Here is my suspicion:
I'm using a break out board for ADXL345. I'm not sure whether it can tolerate 5 volts. I connected Uno's 3.3 volt to it's Vcc and I'm using I2C for communications. (with two 4.7k pull-up resistors connected between it's SDA and SCL pins and Uno's 3.3v).
Could it be that the buzzer when turned on somehow draws so much current that interferes with those pull up resistors?
Here is the photo of my setup (the lower red line is 3.3v and upper red line is 5 vols both coming from Uno. All grounds are connected together):
And here is the readings of accels. in IDLE:
And I get these reading out of Serial Port:
But whenever I tilt the breadboard (limit for alert is 20 degrees), it correctly beeps, but due to noisy accelerations, buzzer starts, stops, starts in a random pattern:
AI: To expand from my comment:
This could easily be due to the buzzer itself: either physical vibration or EM noise. The fact that replacing the buzzer with an LED supports this as the LED gives no physical movement, and draws a nice steady amount of current. As the buzzer is on the same bread board as the LED, it is mechanically rigid to the sensor, while also being electrically close, and so it is tricky to say which out of physical or EM noise is more likely (my hunch would be physical movement which is easier to cause and easier to cure).
To confirm that it is the buzzer that is causing the issue I would recommend connecting the buzzer via some flying leads. This will mechanically isolate the buzzer from the sensor. When the test is repeated, if the sensor does not give the "funny" reading you are seeing now, the mechanical movement of the buzzer is the cause. The cure for this is then simple: mechanically isolate the buzzer from the sensor (put one or other or both on flying leads). If that is not something you can do for your application, you can either dampen the buzzer signal (using rubber mountings) or you could change your software to filter out the noise (which is often done using over-sampling and taking the mean of the data).
You have replied to my comment saying that the flying leads fix the issue, so you now have various options:
Mechanical: isolate the buzzer from the sensor in any way you like (flying leads, rubber mountings)
Software: sample the data and filter out fast changing values (averaging over a set period of time is the easiest)
Hardware: use a different sensor that samples at a slower rate (so can't pick up the motion from the buzzer). If you are using a different sensor with an analogue output, you could filter the output, with a simple capacitor.
As this appears to be mainly a software project, I would go for a basic software filter. The specifications of it will depend on your application; speed of response to change, fasted change you want etc. You can go a long way with filtering, depending on your application, using; averaging, rate of change, frequency matching, debouncing.
|
H: Is it possible to connect LiPo level indicators in parallel?
I've never studied the terminology of electrics so I had a hard time searching for whatever I'm asking. If this is a duplicate please mark it as such :)
I've got four battery packs in series to produce a higher voltage output, it all works great. I also have four individual battery level indicators, however these are connected separately over each battery like so:
However, from what little I know it should be possible to connected them more according to this picture:
I'm thinking it might give the wrong readings if one cell is higher than lowest one?
The indicators are one of these classic alarm ones:
Again, I'm sorry if the descriptions are off or the terminology isn't there.. This isn't my main field and I'm a home brew enthusiast :)
AI: Easy: That is a definite NO
Your LiPo cells are in series. The orange and purple connections in your second drawing are the same as the connections directly above them where you switch the LiPos in series.
So there is a straight path: from purple wire down, purple wire up to the right, up to the battery contact hopping over to the battery where we started from. That's a DEAD SHORT.
You might have drawn the connections above and below the battery as separate connections but they're not.
I have drawn two of these shorting paths.
It will ruin your day ! So don't !
|
H: What is a more effective shield for magnetic fields between 300 and 500kHz Solid copper or copper mesh?
I am working on a PCB that is very crowded, and has high gain amplifiers working between 300kHz and 500kHz
Typically I would use Mu metal or similar for shielding at this frequency, but obviously nobody makes Mu metal PCBs. So I have a choice of solid or hatched pours. External shields are not an option.
I don't have any controlled impedance tracks.
My only worry is the high frequency AC magnetic fields. We use copper mesh shielding in our RF cages, which works rather better than I expected. I suspect this is due to the shorted turns.
I asked a couple of shielding companies, but they don't characterize their meshes for this sort of application.
Can someone point me to data that would indicate whether solid or meshed copper pours would perform better in this situation?
AI: Solid would perform better, all other things being equal, but perhaps not significantly better.
Since the 'holes' in your mesh will be a tiny fraction of a wavelength, the mesh should behave similarly to a thinner (higher resistivity) solid copper layer when measured from a relatively large distance away compared to the 'holes'.
The 'shorted turns' you mention are just eddy currents which will occur in either case.
|
H: Is there a standard/foolproof way to tell what input or output impedance is?
Sorry if this is a stupid question, but I'm in multiple electronics courses and I can't seem to understand input vs output impedance no matter who I ask.
I can always find the impedance between two points, if I am given the two points and I know all resistor/capacitor/inductor etc. values between those points along all possible paths between them. But when I'm asked to find input or output impedance of an entire circuit, I have no idea what those two points are. My professor seems to automatically understand which two points happen to correspond to input or output impedance in a given example, but to me, it seems completely arbitrary.
I understand that power supplies and loads have effective impedances, but often when I look at a circuit, I have no idea where the power supply ends and the load begins, or whether it's even relevant to think in terms of power supplies and loads... to me it's just a bunch of components jumbled together.
Here's some examples of worked homework problems (solutions in red):
Why is it that the input impedance of the circuit in part "a" is 10k? Obviously there's a 10k resistor on the op amp input, which seems extremely simple, but it's also extremely vague. Why don't I need to worry about the 500k resistor? Why should I care that the input has a 10k resistor when the signal also runs into a 500k resistor and an op amp with an enormous resistance?
Then in part "c" we add another op amp to greatly increase the input resistance. Now all of a sudden we do care about the large op amp resistance, just because we put one op amp in front of another.
It really seems to be that simple, but I feel like I'm just saying "I'll just throw a 10k resistor in front of this circuit and voila, there's you 10k input resistance, have a nice day." If it's really that simple, I at least want to know why it's a helpful idea.
Could someone please break down this concept and really explain it to me like I have never seen electricity in my life. For some reason it's so obvious to others, that they do a terrible job of understanding just how confused I am, and their explanation won't help.
AI: pjc50 summed up the solution in comments.
Input impedance is defined as the impedance you would see no matter what the input voltage/current. With the fact that ideal op-amp analysis pins both inputs of the op-amp to the same voltage, and one of the inputs pinned to ground, the other terminal can be treated as a virtual ground. Now you have an infinite current source/sink at that virtual ground. That means if you create a test input voltage on the input side of the circuit, all it will see is 10k to "ground". 1V/10k=100uAmps. Input impedance is the test voltage divided by the resulting current so 1V/100uAmps = 10k. So that's how they're coming up with 10k as an input impedance.
If you wanted to determine the output impedance, you could look at the tail end and try and inject a 1V supply at the output side of the op-amp. The output of an op-amp can source or sink infinite current (pseudo-ground) so you basically have a connection from the output wire directly to ground. A wire directly to ground is 0 ohms so the output impedance in this case is 0 (or very low).
From what I can tell, your confusion comes then from not understanding how an ideal op-amp works. The input side of an op-amp tries to pin its two inputs to the same voltage and the output can source/sink infinite current to create the desired voltage.
|
H: How do I use Type K Thermocouple?
I see a lot of applications using this type of couple but there is no document showing how to proper use it!
In my case I have a 10mm thick steel plate that get heated on one side, should a M6 hole 7mm long be enough for a M6 thermocouple like the one in the image to give proper temperature readings?
Is the end part of the thermocouple the sensor?
AI: This kind of thermocouple won't really be ideal. You would like to see an immersion depth of 10x the hole diameter or 60mm for best accuracy.
The sensitive bit is the knob at the end:
You would like that to press reasonably hard against the bottom of a blind hole so the hole depth should be a bit short of having it go all the way in.
I suggest drilling the hole in from the edge if possible (rather than on the top or bottom) to get the maximum thermal coupling. Running the wires along the plate a bit will also improve the accuracy.
|
H: Reverse polarity protecion on an electronic load
I want to add reverse polarity protection on my electronic load which will be capable of sinking 20A and with a max. voltage of 50V (max. power is limited to 250W). Because I want to go with low input voltages for e.g. testing batteries, a typical solution like a PFET at the positive load input terminal is not a solution because of the V_GS_th of the PFET.
Diodes are also not suitable because of the high current and the related power dissipation.
I came across the solution of Spehro Pefhany which is shown in the picture.
I got one question about it:
My electronic load input terminals (Vx) are going to be floating (no connection to mains earth because of a transformer). The metal case is going to be connected to mains earth.
Is there a possibility to screw anything up, if some strange connections are made?
E.g. connect the negative load input terminal to the case and attach there the positive terminal of my DUT. Now I connect the negative terminal of my DUT (which is mains earth referenced) to the positive load input terminal. That would short out Q2 and the current would flow through the body diode of Q1.
Is this the only case which could cause problems?
What would happen if I connect a mains earth referenced power supply to my load terminals in a normal way (positive to positive and negative to negative)? That would connect the negative floating input terminal to mains earth and would bridge the isolation. Could this be a problem?
AI: E.g. connect the negative load input terminal to the case...
Don't do that - problem solved!
connect a mains earth referenced power supply to my load terminals in
a normal way... would bridge the isolation. Could this be a problem?
If you connect a ground-referenced device it will ground one of your terminals and your circuit will no longer be floating, but it's not 'bridging the isolation'. Your electronic load's internal ground can still be at a different potential, which is the only reason you need to 'isolate' it.
|
H: Does current fundamentally determine the speed or the quantity of electrons in a curcit?
I am very much a beginner so I apologize for my ignorance. But this has been bothering me for a long time and I found little to no answers.
I know that the definition of Current is charge passing through a point per second measured in A.
My confusion is how the increase in charge passing through a point achieved when we increase the current.
My belief is by increasing the current we increase the speed of charge, thus allowing us to register charges passing through a point more frequently since they are moving in a circle. Would that be true?
I've done lots of googling but this is the only related thread that in a way confirms what i said:
https://physics.stackexchange.com/questions/183482/does-a-resistor-slow-down-the-flow-of-electrons-or-just-let-less-electrons-throu ( i apologize the link I provided the first time was wrong)
The answer does not have upvotes and I found no other proof so I am very not confident that this is true.
Could it be that increasing the current releases more particles from the terminal, thus increasing the quantity, instead of speed?
AI: Assume you have a conductor of volume \$A l\$ (cross-sectional area = \$A\$ and length = \$l\$)
Let \$n\$ be the number of free electrons per unit volume of the conductor, then the total number of free electrons in the conductor is given by \$Aln\$.
If \$e\$ is the charge, then net charge \$Q\$ is, \$Q = A l n e\$
Let \$v\$ be the velocity of electrons due to this potential difference. This is called as drift velocity of the electrons.
Let \$t\$ be the time taken by an electron to cross the length of the conductor.
Then, \$t = l/v\$.
As current, \$i = q/t\$, we get \$i = A l n e/t\$. On substituting \$t\$ from above equation, we get \$i = A n e v\$.
So current is directly proportional to drift velocity of the electrons
But this drift velocity is in terms of few millimeters per second. It is because of the large value of \$n\$ (no. of electrons per unit volume) that effect of velocity gets magnified.
|
H: Why does the LTspice op amp stability tutorial recommend open loop analysis?
I have performed a hand analysis on an amplifier circuit that I am suspicious of, and I am still suspicious, so I want to validate my hand analysis with LTspice.
I found this tutorial online:
http://www.linear.com/solutions/4449
It recommends to find the "open loop gain and phase response" to analyze the phase margin. It discredits the "closed loop response" from having the necessary information to analyze amplifier stability.
I can't post my exact circuit, but that shouldn't matter. Here's an example circuit, modified as instructed by the tutorial:
My prior understanding was that the open loop response of an op amp was just the op amp by itself with no feedback circuit, and that the closed loop response contained all the information needed to analyze an amplifier with its feedback circuit.
Can someone please explain what the LTspice tutorial is talking about?
Thanks
AI: Yes - sometimes the terms are somewhat misleading. In general, we have three different gain conventions for feedback circuits:
Open-loop gain of the amplifier (alone): Aol,
Closed-loop gain of the amplifier with feedback: Acl,
Loop gain Aloop, which is the gain of the complete loop (to be measured or simulated after breaking the loop at a suitable point). Hence, the loop gain is the product $$A_{loop}=(-A_{ol}*\beta)$$
with beta = feedback factor
Note that this definition includes the inverting sign at the opamp input.
Based on these conventions the closed-loop gain is $$A_{cl}=(A_{ol}*\alpha)/(1-A_{loop})$$with alpha = forward factor, if existent, otherwise unity
Note that stability margins (phase margin, gain margin) are defined for the loop gain Aloop only. For the purpose of determining the margins, the loop must be opened at a suitable point (opamp output or inv. input) for injecting a test signal. I hope this clarifies something.
It is the purpose of the following example to demonstrate the meaning (and the correct sign) of the term"alpha":
Example:
(a) Non-inverting amplifier (R2=feedback resistor): alpha=1, beta=R1/(R1+R2);
(b) Inverting amplifier: alpha=-R2/(R1+R2), beta=R1/(R1+R2).
|
H: boost 100ns pulse from 5 v to 10 v
I have to amplify or "step up" the voltage of a pulse generator.
I have a pulse generator, which generates 5 volt pulses with on times ranging from 100nano seconds to 1200 nano seconds which repeats itself approximately every 1milisecond .
For design reasons I need to get the same pulse but at 10 or 20 volt in amplitude, without compromising the the pulse times .
My initial thoughts were using an schmitt-trigger as a buffer but It couldn't handle the high frequency of oscillation .
How can I create a pulse from 5V to 10V that lasts 100ns? Can I do this with a capacitor and a mosfet?
AI: You could consider just using a fast comparator such as a LM1711/1712
They will operate with a 10V or even 12V supply. At currents of 1mA or less you'll get almost the whole supply swing at the output.
Output rise and fall times are in the 2ns range with a 10pF load, and propagation delay is less than 5ns typically with 20mV overdrive.
A more Neanderthal method would be to drive the gate of a smallish MOSFET such as an 2N7002 hard with something like a fast CMOS inverting buffer and use a very low value of pullup resistor to 10V or 12V (something like 100 ohms, which would consume 100-120mA with your very low duty cycle.
Minimize loop inductance by keeping everything very compact and use very good supply bypassing right near the transistor/resistor.
|
H: Low power ESP8266 LDO or switching regulator
I'm going to use 1S Li-po, 3.7V. ESP8266 needs 3.3V and it takes 350mA peak on startup. When running it needs about 70mA.
The ESP task is to wake up, measure, connect to wifi and post data to server. It will wake up every 2 hours. I'm wondering which regulator is better for low power modules, LDO or switching. I know that LDO efficiency is a lot less efficient but they have lower quiescent current. On the other hand the switching offers better efficiency. Which one I should use?
AI: Good discussion already. I'll summarize the linear option as an answer.
A datasheet for the ESP8266 says that it can tolerate from \$3.0\:\textrm{V}\$ to \$3.6\:\textrm{V}\$.
A 1S Li-Po starts out almost \$4.2\:\textrm{V}\$ with low load and is nearly empty by about \$3.4\:\textrm{V}\$.
That leaves only about \$100\:\textrm{mV}\$ of headroom towards the end of battery life and that's probably not enough. It's almost certain that you'll need at least twice that much, using an LDO at \$70\:\textrm{mA}\$ and perhaps three times it at the initial \$350\:\textrm{mA}\$ you mentioned.
But it still may be possible to consider using a linear regulator, if you choose to operate at a lower voltage. One that is at or below \$3.1\:\textrm{V}\$. Sure. That's not much wiggle room. But at least it's non-zero. Still, you may also have to consider what else surrounds the ESP8266, too. And you've asserted the need for \$3.3\:\textrm{V}\$ but you haven't disclosed whether or not perhaps \$3.0\:\textrm{V}\$ or \$3.1\:\textrm{V}\$ could be successful. (Okay. I don't know of a fixed \$3.1\:\textrm{V}\$ linear LDO with low quiescent current. So probably \$3.0\:\textrm{V}\$.)
Given all of the above, all I can suggest regarding a linear LDO option is something like the LT1763. It may meet your needs. It's quiescent current is about \$30\:\mu\textrm{A}\$, too, and requires at most \$300\:\textrm{mV}\$ of overhead for your case. So perhaps.
Given a short search (you really should do your own), I did find this from TI: TPS783xx. The spec is below your peak startup current, but well above your operating current. And the quiescent current is extremely low, I believe.
|
H: Is there a way to control the PWM duty cycle of a 555 timer without using a variable resistor?
I need to control the duty cycle of a Pulse Width Modulation (PWM) circuit from a DC voltage input. To do this, I've been looking into 555-timer circuits - but every circuit I've come across involves controlling the duty cycle by using a variable resistor/changing resistor values (e.g. changing the value of Rb in the diaram shown below).
I've researched and thought about it a lot, but I can't find a way to use a certain input voltage to give a certain duty cycle in a PWM cicruit with the 555-timer, so any suggestions I may not have thought of or letting me know it can/can't be done would be appreciated.
Thanks
As a note: The requirement of controlling a PWM circuit duty cycle with an input voltage and not by other means is necessary for what I'm doing... I'm not trying to make my life difficult
Also, I forgot to mention I'm also not allowed to use microcontrollers!
AI: Yes – simply use your external DC voltage to bias the feedback voltage. Done!
(this means connecting it, with a appropriately sized resistor, to the threshold pin).
I'm not trying to make my life difficult
Well, still you're using a NE555 to generate a PWM. I'd simply go, get a cheap microcontroller with an ADC and a PWM unit (these start well below half a Euro), write ten lines of C firmware and be done. No analog circuitry you have to tune, no uncertainty and non-linearity of duty cycle as function of control voltage, just clean software in a microcontroller that doesn't need any external components but a decoupling capacitor for its power supply. Bonus: internal oscillators of microcontrollers these days would usually be much more accurate than a NE555.
Update better late then never: you mentioned you're not allowed to use microcontrollers. I know this will probably mean some learning overhead, but a CPLD-implemented PWM generator with either a resistor-ladder-based ADC implemented with pins and external resistor networks (so you don't have to solder a lot of resistors) or a cheap ADC IC would still be what I'd use. PWM is basically a pretty digital problem, so I'd go digital.
Another easy approach would simply use a quad Opamp IC: Opamp 1 & 2 to generate a triangle wave, opamp 3 to compare the instantaneous triangle voltage to your external DC voltage. Easy PWM, and gotten rid of the NE555.
|
H: Which is better from a Controls perspective, a Stepper or Servo Motor?
I'm designing a system to control the knob on an Air/Oxygen blender based off readings from a Pulse Oximeter. The idea to use a stepper to control the knob/retrofit an existing blender came from here.
I went with a Stepper motor for the initial prototype mainly because it doesn't require an encoder, and the speed and precision is more than enough for my requirements. I understand the differences in operation, power consumption, torque, etc and the only real argument for a servo over a stepper I see is heat. However this hasn't been an issue while using the driver circuit I selected. I'm about to graduate with an EE degree and am still learning how to use what I've learned to makes these types of design decisions.
AI: First off, servos and steppers both have pros and cons and tend to have slightly different use cases. This means that, in your case, it comes down to personal opinion (this is more of a yes/no, go/no go site)
Having said that. Steppers can be simpler to drive and can hold their position with no power (although their holding torque is much better with power in which case they're no better than servos), but have limited top speeds and tend to be a bit jittery - i.e. they don't have really smooth flowing movement (they're called STEPers for a reason). Servos can be really fast (1000's of rpm) but good ones can be quite pricey due to a much more complex control system, although good quality servos can have very fluid motion. But this depends on whether you mean RC servos or real proper AC Servomotors.
It's not that one is inherently better than the other, they're just different (like left and right hand drive). RC servos and un-encoder-ed steppers being more or less on par in terms of practicality.
As it sounds like this is a rough 'n' ready project, the distinction is likely a non-issue in your case, so it's going to come down to what you feel is appropriate, go with what you know works for as we engineers say: "Go with what you know" as "Good enough is almost always good enough"
|
H: help with circuit design
I have a problem and some of the components I think I need to solve the problem but I do not have the knowledge or understanding to put it all together and what I read online is just not connecting the dots for me.
Ultimately what I need to have happen is a PIR sensor detects movement and triggers a string of LEDs. Which seems like it should be simple enough. I can get the PIR sensor to light an individual LED but anything beyond that and the PIR is to weak.
The LEDs I am trying to light are in a ribbon format and I want them to run off of bare minimum 12v but preferably 18vdc as this is when they are at their absolute brightest. At 18vdc they draw about 2 amps of current.
The PIR is powered off of 6Vdc and outputs 3.3VDC with around 300uA.
I want to use a relay that I have which the coil can be pulled in at a little under 3 volts but needs around 100mA of current and the contacts could handle the 18vdc and 2 amps I want to run my lights.
My problem is how can I get more current from the output of my PIR to trigger the relay, being that I can't increase the voltage coming out of the sensor?
I have some transistors to work with but I don't have enough understanding to connect a proper circuit up and I believe I have been burning out some components needlessly just playing around with stuff.
My PIR is an HC-Sr501, my Relay is a JQC-3F (T73), the LEDs are a 12V RGB ribbon style, and I have available some standard NPN and PNP transistors BC547, BC557, 2N2222, and 2N3904. I also have a variety of common resistors and other standard components. Either a circuit somebody has already created or just some specific instructions of how the connections should be made in order to accomplish my goal would be appreciated.
I don't have a data sheet for the LEDs. This is what they are and all the specs I could get on them.
SUPERNIGHT (TM) 16.4FT 5M SMD 5050 Waterproof 300LEDs RGB Color Changing Flexible LED Strip Light
Voltage 12 volts
Wattage 60 watts
I don't have a problem running them at 12v if I have to and I understand it would be a better idea.
here is a data sheet to my PIR sensor http://www.datasheet-pdf.info/entry/HC-SR501-Datasheet-PDF .
Although it isn't marked as an HC-sr501 when I purchased it from a chinese supplier this is
what they claimed it to be.
I know you need real information from data sheets but I don't have much of that stuff.
I know what I can read from my DMM. The sensor puts out 3.3 volts as expected when triggered but the current is only 300 micro amps. my relay coil will react at about 2.8 volts but draws about 100 milliamps in order to do so. The LEDs are designed to operate on 12 volts but when I crank it up to 18 volts they are bright and although they will eventually burn out I have run them at 18volts for more than 24 continuous hours without fail, that doesn't matter so much to me anyways, even if I only supplied them with 12 volts that would be ok. I just can't get my PIR to trigger my relay to turn my leds on in the first place. There isn't enough current coming out of my PIR output to activate the coil.
AI: The below schematic demonstrates a circuit that will do this for you. When the PIR signal is low, the NPN transistor will be off, and no current will flow through the relay. When the PIR signal turns goes high, this causes the transistor to turn on, drawing current through the relay coil turning the relay on.
simulate this circuit – Schematic created using CircuitLab
Winny in the comments is correct, you should not use 18Vdc on the 12Vdc rated LEDs. While they may be brighter in the short term, this can lead to a few issues:
The LEDs will dim much faster over time than if you used 12Vdc.
There is a chance of overheating, and by extension, fire.
Now, you also say you have an RGB LED strip, not a simple one color strip. You will need to supply power to all three colors, not just one, if you want a white(ish) color out of them. (This is also assuming this strip is a common cathode model, datasheet/links haven't been provided)
EDIT:
To address @user2913869's concern that the sensor may not be able to drive a single NPN transistor, according to this website, and this datasheet, the actual output of that chip is approximately 10mA @ 5V supply voltage.
EDIT 2:
D6 is what is called a flyback diode. This is a necessary component when using relays (and other inductive loads). Inductive loads don't allow current to change immediately in a circuit. This diode allows current to flow back to the relay, rather than trying to push that current through the transistor, which can break the transistor. To put it simply, it allows the energy stored in the coil of the relay to dissipate without breaking the transistor.
|
H: What causes the uncertainty in the threshold voltage of a CMOS switch?
CMOS switches such as the DG419 have a threshold voltage - i.e. the voltage at which the output state is flipped - which is dependent upon the logic level and the positive supply rail. However, the exact threshold voltage at which the state will flip is undefined. For the DG419, the threshold occurs within a region spanning many volts, particularly for lower supply voltages:
What causes this uncertain region in switching? I guess it's not the material itself, but rather something else such as stray capacitance or source impedance.
AI: It's a CMOS input, so there are two transistors 'fighting' each other at the input threshold.
Variations in the semiconductor manufacturing process will result in different behaviors for one vs. the other, so the point where they switch will vary. That variation is accentuated at relatively low supply voltages where the transistors are barely working at all (what is 'relatively low' will be different for a high voltage analog switch like the DGxx series vs. a chip designed for 3.3V Vdd).
Chances are that under fixed and benign conditions (temperature and supply voltage) the threshold will stay pretty much fixed for any given sample of chip.
|
H: AD736 input amplitude for 32kHZ
I want to convert a sine voltage with frequency around 32kHz , what should i keep the amplitude of this signal to get the best conversion.
I looked up the datasheet , but the input range given is around 200mV, isn't this too low?
AI: Too low for what?
If the you read the third line of the first paragraph of the data sheet, it says 'use an input attenuator for larger input voltages' (or words to that effect).
It's not too low for a signal correctly scaled into the 200mV range by an input attenuator!
You can put any voltage between supply and ground into any pin without damage (see the absolute maximums table), but you won't be getting the specified conversion accuracy under those conditions.
If you carry on reading further into the datasheet (it never hurts to R all of the TFM before asking a question) you find that you can apply 300mV rms through the low impedance input with low voltage rails, and up to 1V rms when you use power rails of at least +/-5V. The peak transient is +/- 11V on the low impedance input when you use +/- 16.5V rails. Figure 4 page 7 has a nice graph of maximum input level versus supply voltage, and figure 6 gives error versus signal level.
If you are actually going to use the thing in anger, you need to pore over all of the 'additional error versus ...' graphs to see whether it will meet your accuracy needs.
|
H: MIDI out waveforms very different between two devices
I have two MIDI output devices: an Axiom49 keyboard and a Korg SQ-1 step sequencer. The Axiom49 drives a MIDI input circuit that I've built using an Arduino just fine. The Korg SQ-1, however, does not work with it at all.
The Axiom49's output is simply 31.25k baud RS-232, which conforms with the MIDI specification. The output on the Korg SQ-1, however, is totally different than I was expecting and does not work with my MIDI input circuit at all.
So, I hooked both of them up to an oscilloscope (across pins 4 and 5 of the MIDI 5-pin DIN connector, coming from the MIDI OUT plug on both devices) to see what their output looked like (screenshots below).
Also worth noting is that I was able to hook the Korg SQ-1 up to a friend's computer using a MIDI-to-USB cable and drive a software synth.
So, does anyone know why the waveforms between two MIDI devices could be so different?
AI: RS-232 and MIDI are completely different.
In particular, MIDI uses a current loop; the "space" state (logic 0) is defined as a current of (at least) 5 mA.
Depending on the MIDI output circuit, you cannot simply measure the voltage at a bare pin (or between two bare pins) of the MIDI connector.
Your Korg waveform looks as if you tried to measure a floating voltage, and got noise from something else. (And a sample rate of 80 kHz is much too low to capture the actual MIDI signal.)
You have to measure the current by, e.g., connecting a resistor between pins 4 and 5 and measuring the voltage drop over that.
|
H: Lm386 audio amplier - lyrics not audible
I am trying to amplify the music signal coming from a phone using an LM386 module for a 8ohm 3W speaker using a 5V supply,but when I connect it to the source,it amplifies a portion of the music,not all of it.The instrumental/background music can be heard,but not the lyrics(they are not audible).
Please help.
AI: You've got the input wired up wrong. What you should be doing is using two amplifiers; one for left and one for right but instead, you are taking the left and right signal (ignoring the common ground conenction) and feeding that to your amplifier. Because vocal and most bass stuff appears equally in left and right channels on most popular music, in effect, there is no net signal produced when you connect left to the input and right to the amplifier ground connection.
You get a similar effect on stereo speakers when one is wired in reverse - at certain points along the axis between the two speakers, bass sounds can dissapear. Many audio tools provide this feature too.
If you want to fix this try using two 1 kohm resistors to join left and right channels into one input and use the regular ground connection from your music source to connect to ground of your amplifier.
See this youtube video that explains what is happening when you invert one channel and add it to the other (which is basically what you are doing inadvertently).
|
H: How did scientists deal with electronics' problems before Kirchhoff and Ohm's laws?
Both physicists developed really powerful laws which still nowadays rule the electronic behavior of circuits.
These help us every day to solve problems, calculate circuit variables… but how did engineers do it before the said laws were discovered?
If alternative laws which would not be accepted nowadays were used before this, would this mean that the research done until the discovery of the laws was wrong? Did Kirchhoff and Ohm themselves rely on wrong theories to create 'the good one'?
AI: This is a bit like asking how Aztecs built cars without the wheel: they didn't.
There was a chain of invention by scientists in the early 1800s building off each others work. Prior to then there was only electrostatics: Benjamin Franklin rubbing insulators together and noting charged objects attract and repel. Leyden jars.
In 1800 Volta invented the battery or "pile". This allowed experiments with a constant source, rather than ephemeral electrostatic discharge. That led to Davy inventing the arc lamp, and Ohm in 1827 quantifying this electricity. Then Faraday's work on electromagnetism, allowing generators, dynamos and motors.
Engineers turning it into a "product" came later. Swan and Edison both invented the light bulb; Edison, Tesla and Westinghouse fought over distribution.
If alternative laws which would not be accepted nowadays were used
before this, would this mean that the research done until the
discovery of the laws was wrong? Did Kirchhoff and Ohm themselves rely
on wrong theories to create 'the good one'?
There's a little discussion of Kirchoff and Ohm here.
Kirchhoff's laws followed from applying Ohm's law but the way in which
he was able to generalise the results showed great mathematical
skills. At this stage Kirchhoff was unaware that Ohm's analogy between
the flow of heat and the flow of electricity, which formed the
accepted understanding of electrical currents at that time, led to an
incorrect understanding of electrical currents. Since no heat flowed
in a body at a uniform temperature, it was believed that a static
current could exist in a conductor. Kirchhoff's work would, a couple
of years later, lead to him to realise this error and to give a
correct understanding of how the theory of electric currents and
electrostatics should be combined.
Which suggests that the answer was yes - people were building off incorrect theory to some extent. In the case of Ohm, he was building off Fourier's work on heat conduction. Electrical conduction is similar but not exactly the same.
There isn't anything on quite the scale that "phlogiston" was in chemistry - a controversial popular theory that ultimately turned out to be wrong.
|
H: What is anti-reflection coating and in which application it is used?
I would like to know about anti-reflection coating and I am interested to know whether this can be used in antenna design also.
AI: So, anti-reflection coating is used in optics, for example, sunglasses, or lenses in microscopes to ... reduce reflections. It usually works by having a so-called thin-film on the material's surface that has a refractive index n chosen so that things that would be reflected on the air-lens interface are reflected back in a way that the way distance the reflected wave has to travel is an odd number of half wavelengths. That way, the incident wave and its reflection cancel out (interference).
For optics, that's a cool concept, and requires very accurate and fine manufacturing/coating technology, because light is in the ~800nm range, and thus, half a wavelength is very small.
The wavelengths of radio frequencies are much larger.
All antennas already work by matching their dimensions to the wavelength of the waves they want to receive or send, and not to others.
This question feels like you've been watching a documentary on nanomachines and how you can now produce little wheels in the nanometer scale, and now you're calling a car manufacturer and tell them to use this new technology called "wheel". Well, we already do.
|
H: what is current path in 6 winding transformer
The 6 winding transformer which i shown in figure is a 1:1:1:1:1:1 transformer.It is a schematic of wurth electronics flex transformer. If I apply voltage at one winding of primary what is the current path in secondary? and what is the applications of this type of 6 winding and equal ratio transformers?
AI: This is meant to be a flexible transformer that can do many things, depending on how you connect the windings. Each winding is magnetically coupled to the others. It is up to you how to connect windings in series or parallel or leave them open to get the ratios you want.
Secondary windings connected in series give you higher voltage at the same current. In parallel, they give you higher current at the same voltage. For primary windings, they have the opposite effect.
The advantage of a device like this is that you can stock one transformer and use it for different purposes in different designs. It also helps the manufacturer in that they don't have to make as many different transformers.
|
H: MOSFET: how to determine if a heatsink is required?
for a given mosfet, how do I know how much current can it handle without a heatsink?
I could just play it safe and add a heatsink + fan but the controller board am using doesn't have mounting holes for any pcb mount heatsink. It is fused but I just want to be sure.
So for example If I was using a mosfet as a switch for a 360w load and it's 0.31 degree c per watt then in 25degrees c air it'll be
0.31 x 360w = 111 degrees c!!!! based on a IXFH52N30P as a example
If for example I was stupid enough to not use a heatsink then it could termally runaway due to the mosfet rasing the amient temperature right?
AI: The power consumption of the load is completely irrelevant. The 400W power you're switching is not dissipated by the MOSFET. It is, by definition, dissipated by the load. The MOSFET dissipates much less power (hopefully).
In a switching (on/off) application, here is how it works:
Calculate the drain current ID that will go through the MOSFET. You can determine it by dividing your load power consumption (400W) by the load voltage (you didn't specify it). Or maybe it's directly given in the load specifications.
Check the gate voltage you'll drive the mosfet with. It depends on your controller board, so it should be specified in its documentation. It usually is from 5V to 12V. Maybe it depends on the voltage you're supplying to the controller board.
Check the datasheet of the MOSFET to get the VDS at the gate voltage and drain current you're using. In the IXFH52N30P datasheet, it is given in figure 3 (assuming 125°C - fig.1 is for 25°C, but we most likely be much hotter than this). For example, with a 10V gate voltage and 20A, it is about 2.5V.
The power the MOSFET (not the load) will dissipate is: VDS * ID. With the assumptions above, it makes about 50W (2.5V * 20A).
To check if you can go without heatsink, you would then get the RthJA (junction to ambient) thermal resistance value given in the datasheet (given in °C/W) and multiply it by the power dissipated. For the IXFH52N30P, RthJA is not given in the datasheet. It seems the engineers here assumed that the FET would only be used with a heatsink. Anyway, these kind of FET packages can't handle much more than a few watts without heatsink, and I guess you'll be above.
So, now, assuming you need a heastink, get the heatsink thermal resistance by applying this formula: (TJM - TA) / P - RthJC - RthCS. Basically, you compute the max total thermal resistance by dividing the temperature difference between "max junction temp" (150°C - given in datasheet) and ambient temp by the power. Then you subtract the junction-case and case-sink thermal resistances (both given in datasheet too) from the value obtained.
Choose a heatsink with a thermal resistance lower than this. The heatsinks thermal resistances should be given in their respective datasheets too (yes, even heatsinks have datasheets).
|
H: Why don't we get electrocuted when touching a pot sitting on an induction cooker?
As the title says, why does that not happen? As far as my understanding goes, an induction cooker is a coil of conductive wire through which AC is pumped. This generates an oscillating magnetic field around the coil, which in turn generates a DC current in the pot sitting on the cooker. Since the pot is not made of a superconductor, the current flowing through it heats it up, and that's where we get the electricity from.
Now this is where I get confused. First, yes, there's a current, but is there a potential difference between any points, or are the electrons just really excited? If there's a potential difference between any two points in the pot, then why does touching it not give us an electrical shock?
AI: The induced current in the pot (pan), is on the bottom of the pot where you are not likely to touch. In fact you would have to lift the pot in order to touch where the current is being induced.
By lifting the pan up and away from the induction coil, you will reduce the induced current in the pan (and reduced voltage, which is low already).
Very very little current would be induced in the side of the pot (where you can easily touch).
Comments made above correctly state that the induced voltage is low and that induced current is high (all relative of course).
"First, yes, there's a current, but is there a potential difference between any points"
You could measure a potential difference (voltage) by measuring on the bottom of the pot between two points that are on opposite sides of the pot bottom. The voltage will be very low (not going to address the value of voltage here).
Thus, if you were to able to touch the bottom of the pan in 2 places (difficult to do with pan sitting on induction heater), on opposite sides , you might get a mild shock.
And last,
"which in turn generates a DC current in the pot sitting on the cooker"
Just to set the record straight, AC is generated (induced) in the pot, Not DC
|
H: Calculating Field strength of a air core solenoid
I want to build a magnetizer for charging / magnetizing small alnico magnets for that I am thinking to discharge a cap into a air core solenoid.
The magnet to be charged would be kept within this solenoid.
The magnet would be of cylindrical geometry and 5mm dia and 10mm length, after searching on the web I have found that alnico needs a field strength of 3000 Oersted or around 239 kA/m.
I find that Oersted is related to A/mor Ampere/meter as both of these are units of field intensity but I don't know how to relate oersted to ampere-turns.
To generate this field intensity I would like to know how many turns to wind and how much ampere's to dump into the coil.
Any clues would be of great help.
AI: I don't know how to relate oersted to ampere-turns
Or, muliply the oersted value by 1000 and divide by 4\$\pi\$ to get amps per metre (aka ampere-turns per metre).
Any clues would be of great help
To get 3000 oersted requires a lot of turns and a lot of amps but I think you should be armed with the knowledge now.
Here's another hint - the H field is "per metre" and the length of your solenoid is 10mm so the amp-turns you need are 0.01 x 239,000 = 2390 ampere-turns so as a couple of examples, that's either 1 amp and 2390 turns (for every centimetre) or 10 A and 239 turns per cm.
I would also consider that the solenoid length should be three times the length of the material to be magnetized to ensure that the resulting flux density is constant along the length; at the open ends of a solenoid the flux can be significantly reduced compared to the centre.
|
H: Are special terminals required for connecting thermocouple wires to a PCB?
I'm currently trying to integrate a type K thermocouple into an electronics project. I intend on interfacing this thermocouple with the MAX31855 cold-junction compensated thermocouple-to-digital converter IC.
I am not an expert on thermocouples and the Seebeck effect, so I'm somewhat curious as to how I should actually land the thermocouple onto the PCB. Assume I'm using loose thermocouple wires.
Questions:
Can I use any type of terminal (say a simple through hole screw terminal) to land the wires onto the PCB? Or will these new junctions introduce errors?
Is the termination method dependent on the thermocouple type? E.g. use this terminal for type K, use this other terminal for type J, etc.
AI: Using a terminal block made with ordinary materials is quite sufficient for a relatively modest accuracy system as you're aiming for.
The cold junction compensation depends on the cold junction sensor (in this case, the chip itself) being at the same temperature as the two junctions where the thermocouple wire transitions to copper. In other words, all three should be isothermal- so you want to minimize gradients caused by dissipation on the PCB and by gradients caused by heat flowing down the wires. You can help this along greatly with ground planes or at least pours and by keeping anything that dissipates a lot of heat well away from the T/C block. Keep air currents away from the terminal block too. Of course you will put the chip as close as practical to the terminal block, physically as well as thermally.
There is no great difference between most sensors as far as this goes as most thermocouples are fairly linear (a couple percent) so 1°C error at the cold junction is around 1°C error in the temperature reading.
If the connection is out hanging in the breeze or is at an elevated (for example) temperature, it's better to use connectors that are made of thermocouple materials, and this is typically done for panel-mount connectors and inline connectors. They are usually color coded. In North America we use the ISA color codes, and type K (Chromel-Alumel) is yellow, type J (Iron-Constantan) is black. You could, for example, have a bulkhead K connector and connect that inside an enclosure to the PCB. You MUST use the proper thermocouple extension wire on the inside as well as outside in this example, and it MUST be connected the correct way (if you swap the polarity the error is actually doubled). Keep in mind that red = negative in North America T/C color codes.
|
H: How to close a circuit using a level
First off, let me apologize if this question is too basic for the community - I've searched quite a bit for a solution, but wasn't able to locate anything.
I'm a hobbyist working on a project that would require a circuit to be closed when an object is out of bounds on an inclinometer (basically a curved level) . I would like to use a rudimentary ball inclinometer (as pictured) over a more complex electronic solution.
I my goal is simple: close a circuit to activate a LED when there is more than a 5 degree bank - or alternatively, open a circuit when the ball leaves the +/- 5 degree range. I can't seem to locate a device that would accomplish this goal commerically and would like to build it for my own use. I'm stumped as to how to build this, though it seems as if it would be fairly simple.
My only other requirement is that the solution does not involve magnets, as it will be mounted on metal housing which could effect accuracy.
Thoughts?
AI: One obvious way to build such a thing would be to start with two metal (conductive) plates with the curved slots cut into them.
Then mount the plates with an insulator between them, and a ball-bearing sitting in the slot between them.
Insulate the slot in the range from 0 to 5 degrees inclination.
Finally, you need an electrical connection to each plate. When the ball bearing is on the insulated section, there will be no connection between the plates. When the plates are tilted enough that the ball bearing sits on the un-insulated portion, it will complete a connection between the plates.
|
H: How is the input not floating when there is a pull-down resistor?
I am wondering why the input to D2 will not be floating when the switch is open. If the switch is open, wouldn't it act like a mini antenna and provide a random input to the D2 pin?
AI: I guess you are getting the doubt because of confusion that the moving part of the switch is opened and this is making you think that it is floating and so D2 must also be floating. No no no. The moving part of the switch and so D2 are at zero volts.
D2 is input. And GND is zero volts. There is no current through R1. So no potential difference across R1. So the voltage at D2 is also zero when the switch is opened.
In other words, Voltage at D2 = Voltage across R1 + GND voltage = 0 + 0 = 0 Volts.
|
H: Soldering Station Extras?
I recently bought a cheap soldering hotter and solder iron station from E-bay. It came with lots of extras. Sadly it never came with any good tips, but it did come with these..
What are these for?
Thanks
Student
AI: These are replacement heater cartridges, from the number of wires likely with built in sensing.
Looks like the small one is for an iron, and the larger for a hot air tool or something else.
|
H: Why we use conjugate of current than the original phasor in the calculation of Complex Power i-e S=VI*
i have studied different books and sites for the above question but did not get the right concept that why the complex power encounter current conjugate not the original Phasor.if someone explain it with good example in simple words i would be very thankful...
AI: The voltage and current signals have an angle associated to them, better known as \$\theta_v\$ and \$\theta_i\$, respectively.
In terms of power, you want the phase difference between those two parameters, an angle we can call '\$\theta\$'. That is, you are looking for:
$$ \theta=\theta_v-\theta_i$$
If you were to find \$P=\text{VI}\$, where \$\text{V}\$ and \$\text{I}\$ have the form \$\text{a}+\text{bi}\$, you are implicitly finding $$ \theta=\theta_v+\theta_i$$
instead of the difference. This can be easily seen if you look at this in terms of Euler's identity.
Let's say that \$\text{V}\$ and \$\text{I}\$ now have the form \$|{\text{V}}|\angle\theta_v\$ and \$|\text{I}|\angle\theta_i\$.
If you now try to find \$P\$ as \$P=\text{VI}\$, you get
$$ P=|{\text{V}}||{\text{I}}|\angle(\theta_v+\theta_i)$$
Instead of the correct way:
$$ P=|{\text{V}}||{\text{I}}|\angle(\theta_v-\theta_i)$$
What makes you have the phase difference instead of the sum, is the conjugate of \$\text{I}\$ or \$\text{I}^*\$
When you find the conjugate the magnitude stays the same but the angle has opposite sign. So when you multiply the complex voltage and current, you are also subtracting \$\theta_v\$ and \$\theta_i\$.
Hopefully that clears things up.
|
H: Is it safe to hook up a device with a higher voltage battery through a regulator like this?
So using a 6v 26 Ah battery and a 5v regulator, am I safe to hook it up to a 5v by 2.5 A max power draw device? I think so, I'm still get familiar with EE concepts though. What complications could arise? What is efficiency like on regulators, I know they heat up, so the must lose power.
[from the O.P.'s comment] The regulator in questions is MC7805.
AI: Traditional LINEAR regulators indeed drop power by heating up. Newer-technology SWITCHING regulators only draw power for fractional amounts of time so that they produce very low heat compared to the old-fashioned ones.
Note, however that many (most?) regulators need some better margin to work with. You will need a rather good (better than average) regulator to produce a 5V output from a 6V input voltage. Most 5V regulators you will find are not capable of operating on inputs as low as 6V. Now you MIGHT find a very "low-dropout voltage" regulator that can work with only 1 volt of differential.
There are also switch-mode regulators which can handle "drop-out" voltage down practically to zero (slightly more than 5V source for 5V output.
|
H: Why boost converter efficiency increases at low D and load resistance?
This slide (page 19) analyzes the efficiency of nonideal boost converter. As you can see from the formula below, the efficiency of boost converter is dependent on duty cycle D and load resistance RL.
The smaller the duty cycle D and load resistance RL, the higher the efficiency.
I would like to understand intuitively how D and RL affect the efficiency.
For duty cycle, the smaller duty cycle, the less time that current flows through inductor resistance. So the efficiency will increase.
Is this right?
However, how to explain the similar for load resistance, why smaller load resistance results in higher efficiency?
AI: Current always flows through the inductor resistance. When the switch is in position 1, Vg only supplies power to RL and the inductor. When the switch is in position 2, Vg also supplies power to R. No energy is transferred from source to load in position 1, so position 1 is 0% efficient. Think about what happens with a 0% and a 100% duty cycle.
RL isn't the load resistance, it's the inductor resistance. Decreasing the inductor resistance obviously decreases the conduction losses.
|
H: MATLAB: Pole at 0.1 rad/s contributing to nothing on the Bode Plot - Why?
Consider the transfer function
$$G(s) = \dfrac{48000}{s(s+100)}$$
The Bode plot is given as
Now observe that the gain at frequency w = 1 Hz is 53.6244 dB. Completely reasonable, because this is exactly $$20\log10(48000/100) = 53.6244 dB$$
Now I change the system to:
$$G(s) = \dfrac{48000}{s(s+0.1)(s+100)}$$
Consider the Bode Plot
MATLAB is telling me that the gain at frequency w = 1 is 53.5812 dB. This is roughly unchanged. Wouldn't the pole at 0.1 = 10^-1 contribute to greater drop in gain than the previous system? I should expect the Bode plot to show full 20 dB lower at w = 1 compared to the the first system, so according to this logic 53.6 - 20 = 33.6 dB, instead of 53.5812 dB.
Can anyone explain this? Why isn't the pole at 0.1 contributing to anything on the magnitude plot?
Code:
G = zpk([],[0,-0.1,-100],48000)
bode(G)
grid on
More craziness:
$$G(s) = \dfrac{48000}{s(s+0.001)(s+0.1)(s+100)}$$
We have two poles before 10^0 = 1 Hz and contributing absolutely nothing to the magnitude plot!
AI: In the first calculation you correctly calculated the gain before the second pole to be 48000/100
On the second example the gain before the second pole is 48000/(100*0.1). So it is 20dB higher. This compensates with the higher attenuation of 20dB for the additional pole, so at w=1 the gain is unchanged.
To avoid these issues it is better to use a normalized notation where each pole is in the form (s/n + 1).
|
H: Why 3v3 should be connected to the center point of Ethernet termination?
I am designing an 10/100 Ethernet on my board and I can't understand why 3.3V should be connected to the center point of the termination.
The schematic checklist recommends it actually and I checked several design and datasheets even gigabit ones and somewhere it is used somewhere it doesn't. As I see it is not necessary because the centerpoint of the transformer is on 3.3V which can be pulled down by the ethernet chip in case of signalling (transmission) and it would require less current without it.
I will do this in the recommended way but I don't understand why which I hate very much. :)
On the other hand the receiver side would not require 3.3V at all right? (if I wouldn't use Auto-MDIX I mean)
Schematic checklist
Datasheet
I have found an interesting article in the topic from 2004. Unfortunately it doesn't explain this directly but provides and overview about line drivers of gigabit ethernet. (I think it is similar in 10/100 and this is a current-mode driver)
There is an image at the end of the article which actually shows this connection with dashed line but as I read it doesn't explain this.
Voltage-mode line drivers save on power
AI: IMO it's a pull-up resistor, TXP and TXN is just a open collector output that ties the line to GND level. The total current is now the current sum from both pull-ups. One goes directly from pull-up to ground, the second goes from pull-up trough a transformer and to ground.
The receive channel is the same, since it is auto-MDIX. Therefore you can use a cross over cable and the TX and RX channels got swapped automaticly.
EDIT:
I said IMO, there is no detailed description about the IC output. Here is the picture of CML output type. Do the 50 ohm pull-ups make sense now?
EDIT2:
The ethernet transceiver IC differs from manufacturer to manufacturer. It is necessary to use the template or demo board as recommended by the manufacturer. If some manufacturer uses centre tap tied on +Vcc and some else is floating, .... its due to the IC characteristics and can't be swapped or used an universal schematics that will fit all ICs and all magnetics.
|
H: Compensated Attenuator question
In the explanation, it explains that C2, C3, C4 are for frequency compensating the impedance of C1 which varies according to frequency.
Can someone explain how C2~C4 does it? and why the designer chose particular value for the three capacitors? It seems that he wanted to proportionally match the parallel connected resistor in terms of impedance, but I don't understand why.
Thanks
AI: Let's just think about the 10V position for now. The Thevenin resistance of the divider will be 90kOhms. This resistance will work with the equivalent input capacitance to from a low pass filter. In this case that will be around 177kHz. In order to increase the bandwidth of the scope, capacitance is added to the anttenuator to put back the high frequencies lost due to this low pass filter. In practice all or part of these capacitors are variable. To calibrate the anttenuator a square wave is applied. The variable capacitor can then be adjusted to go from a low pass to high pass characteristic. On the scope you would see a rounded or peaked waveform. The capacitor is adjusted to get the best square wave possible.
Here is a simulation;
It shows how the scope output will change if the capacior across the 900k resistor is 0.1pF, 15pF, or 30pF. The input is a 1V square wave. You can see the rounding and peaking as the capacitor changes values;
|
H: Basic N-MOSFET Circuit Simulation Results
I am confused about some features of the very basic N-MOSFET circuit shown below:
Which I also created in an online circuit simulator you can find here.
The problem is the VDS value, which the simulator reports to be about 11.99V and the voltage drop of the resistor 11.03mV.
I've found consistent results in all the 5 simulators I've tried, both online and offline.
I've created a real-world circuit using a N-MOSFET STP100NF03L-03 and I've found that resistor voltage drop is about 12V and VDS is near 0V.
So the the question: anybody can tell me why this difference?
EDIT
I have to apologize, I misreported the problem. The simulated circuit has a 12V power supply and a resistor 0.01 ohm, while in the real-world example I used 9V and a 100 ohm resistor.
The measured voltage drop of the resistor was about 9V, while VDS was about 0V. Now I realized that STP100NF03l-03 has a very low RDS while the simulation includes a generic N-MOSFET which values are not known, but likely RDS is generally far greater, even greater than the simulated resistor value, thus leading to a VDS greater than the voltage drop across the resistor.
AI: The STP100 has an on resistance of 0.0035 Ohms. So the real world was not driving a 0.01 Ohm load to near 0V. The best it should have said was about 3V. The transistor you have made in the model is not capable of the current needed to lower the voltage at the mosfet end to 0V. Change the Beta in your model to 1000 and you'll get there. So the problem is too small a transistor for too big of a load.
|
H: Measuring humidity with Honeywell 5030/5031 sensors
I'm trying to hook up Honeywell's 5030 humidity sensor (datasheet here). Figure 10 (on page 7) of the datasheet has a typical application circuit, only I'm not sure why they have also included a resistor between the output and ground pins. The resistor is not mentioned anywhere else in the datasheet (i.e. the humidity-voltage curve does not include a parameter for this resistor). Is the value of the resistor relevant? What is it supposed to do? Does it change the response curve at all?
I can see that the resistor is labelled 'minimum load', but I'm not sure what that means either. Sorry if this is blindingly obvious, I'm CS trained and jumping into embedded stuff so some of the electronics is a bit confusing :)
Leon
AI: The resistor is between the output and 0V of the supply. Minimum load implies it requires to be connected to a load to maintain a stable output, essentially it is there to prevent oscillation and false readings, as they are analog devices.
Normally you might connect the output to an ADC of a microcontroller or data acquisition system and read the voltage output.
Often inputs into digital system have very high impedance, like around >1Mohm so the minimum load is there to ensure the device will draw the correct current to work.
You then need to take that votage reading and apply the conversion formula which is given in the datasheet.
You might need to do a bit of algebra to the formula to get the actual humudity value. Also its a really good idea to have a temperature reading along with the humidity reading as there is a temperature factor involved to get a close to accurate reading.
I've had some experience with these particular sensors and they work quite nicely in normal temperature conditions. Just don't try and use them in very hot thermal chamber/oven type conditions or sub zero temperatures.
I hope this answers your question.
Edit: I will add that 65kohm is not a standard value, but 68kohm is, and I would say that would be close enough, by the time you have connected it to whatever you are connecting it to, it would below the 65kohm minimum load as lower resistance = more load ie. current.
|
H: Circuit analysis with resistance and diode
We just started learning about diodes and now I want to find the currents \$I_1\$, \$I_2\$ and \$I_3\$ in this circuit.
We should assume that the amount of voltage the diodes recieve is 0,7V each and that they got no resistance.
I worked this through:
For I1 I got: \begin{equation}I_1 =\frac{(U-2\cdot 0,7V)}{\frac{R_1\cdot R_2}{R_1 + R_2}}= 5,003\,mA\end{equation}
For I_3: \begin{equation}I_3 = \frac{U-2\cdot 0,7V}{R_1} = 2,268\,mA\end{equation}
And for I_2: \begin{equation}\frac{U-2\cdot 0,7V}{R_2} = 2,735\,mA\end{equation}
However, I am not sure if those values are correct.
AI: I won't answer this fully but you have to treat each path separately to get the correct answer. But heres 1 path you should be able to work out the rest yourself. We work on the assumption the diode forward voltage drop is 0.7V.
$$ I_2 = 0.7/6800 = 0.103mA $$
|
H: What is the "CV" unit for leakage current in a capacitor datasheet?
I've looked at a few leakage current specifications for electrolytic capacitors, and they all seem to specify the value as something like this:
I < 0.01 CV or 3 (μA) after 2 minutes, whichever is greater
Here are a few example datasheets: Panasonic, Multicomp, Nichicon, Rubycon.
Am I right in thinking that the leakage current is a product of capacitance and voltage, i.e. for a 100µF cap on a 5V supply I'd be looking at a leakage current of \$I = 0.01\times100µF\times5V=5\times10^{-6}A = 5µA\$.
Or is that CV unit something totally different?
Additionally, why the long time delay for this rating when a capacitor typically charges in seconds or less?
AI: The leakage spec- in this case 0.01CV (or 3\$\mu\$A) is the product of rated voltage and rated capacitance, not applied voltage. The 3\$\mu\$A, of course, means "whichever is higher" (aka "worse"). So if your cap is rated at 10V/100\$\mu\$F, leakage would be less than 10\$\mu\$A.
SP's rule #1 of data sheet interpretation is:
If a spec can be interpreted in two ways, and one is worse than the other, the worse one is the correct way.
The actual leakage of an electrolytic cap may be much less than the rated value or a bit less. Chances are a higher voltage rated capacitor will have lower leakage when operated at a much lower than rated voltage, but it is not guaranteed, nor will it necessarily last if the capacitor is continuously operated at lower than rated voltage.
The (relatively) long time is, of course, because the initial leakage may be quite a bit higher than spec and it may take some time to drop down to the guaranteed value. This is because the dielectric in an electrolytic cap is actually a very, very thin oxide layer on the etched aluminum plates and it can develop pinholes etc. that are anodized away when voltage is applied.
Here is what United Chemicon has to say about leakage:
Leakage Current (DCL)
The dielectric of a capacitor has a very high resistance which prevents the flow of DC current. However, there are some areas in the dielectric which allow a small amount of current to pass, called leakage current. The areas allowing current flow are due to very small foil impurity sites which are not homogeneous, and the dielectric formed over these impurities does not create a strong bond. When the capacitor is exposed to high DC voltages or high temperatures, these bonds break down and the leakage current increases. Leakage current is also determined by the following factors:
Capacitance value
Applied voltage versus rated voltage
Previous history
The leakage current is proportional to the capacitance and decreases as the applied voltage is reduced. If the capacitor has been at elevated temperatures without voltage applied for an extended time, some degradation of the oxide dielectric may
take place which will result in a higher leakage current. Usually this
damage will be repaired when voltage is reapplied
A strong 'forming' effect of this type is relatively uncommon with modern parts, and seemed to happen a lot more often in olden days when parts were sitting for some time before being used. Maybe the modern electrolyte is better controlled or more pure, or has preservative additives.
Edit: Note @Dave's comment that the units of the 0.01 parameter must be 1/s.
|
H: Is it wrong to assume the op-amp has the same voltage at its both terminals when deriving its non-inverting closed-loop gain?
I am referring to the approach taken in this video, at about 00:43. I know that the op-amp "tries" to balance the voltages at its both terminals but is it a completely correct approach to take V+ = V- when deriving the op-amp's closed-loop gain? As far as I know, in practice, V+ is never equal to V-. But from a theoretical point of view, would it always be correct to assume V+ = V- when calculating the op-amp's closed-loop gain?
AI: Yes. V+ = V- in cases whenever an OP-AMP gives out stable output and negative feedback dominated cases. In the real world, they are not equal. But almost equal. The difference between V+ and V- is almost zero and the gain of the OP-AMP is huge. But for a theoretical purpose, we just take it as zero and assume the gain of the OP-AMP is infinity.
If you want a better answer, use the formula V0 = A.(V+ - V-). In the video, V+ = Vin and V- = R1.V0/(R1 + RF) and then derive V0.
|
H: Why does my atmega32's malloc function returning me the pointer address in multiple of 4?
I am in a learing phase so I am no pro when it comes to understanding the memory and hence I am having a little trouble in understanding few things
The datasheet of atmega32 says that it comes with 2K of SRAM
How does this make 2k?
When I ran the following code in an infinite loop until it returns 0, the malloc function returned me with addresses in multiples of 4 for Atmega32.
p = (char*)malloc(sizeof(char));
From what I am understanding , the char is of 1 Byte hence it should be giving me the adjacent addrresses instead?
AI: For the first question:
There is 2kB of SRAM. There is an additional 32bytes of CPU registers, and a further 64bytes of I/O registers (things like the PORTA/DDRA/etc. registers).
Add these all up we get 2048+32+64 = 2144bytes of addressable memory. Convert that to hexadecimal you get 0x860 - or in address terms 0x000 to 0x85F.
The SRAM memory itself is the 2kB block from 0x060 to 0x85F.
For the second question:
malloc uses a so called "freelist" which maintains which chunks of memory are available to allocate or reallocated. This list is stored in the same SRAM as the memory you are allocating, and is updated on the fly as you use allocate memory.
Each entry in the freelist is a 16bit pointer, which points to the next entry forming a chain - each entry keeps track of where the next entry is - and a 16bit size which says how big the current block is.
As a result, each block of memory allocated using malloc must be at least 4 bytes in size in order to store the list entry. When you request a single byte of memory, it has to round that up to 4 bytes so that when that memory is deallocated, there is room for a list entry to be stored there.
Rather than go into exact details, the freelist approach is explained in the avr-libc documentation here. The following quote explains the need
The freelist itself is not maintained as a separate data structure, but rather by modifying the contents of the freed memory to contain pointers chaining the pieces together. That way, no additional memory is reqired to maintain this list except for a variable that keeps track of the lowest memory segment available for reallocation. Since both, a chain pointer and the size of the chunk need to be recorded in each chunk, the minimum chunk size on the freelist is four bytes
If you can't get your head around it (the process is not immediately obvious), you need not worry. The simple thing to remember is that each chunk you malloc will be at least 4 bytes in size, even if you only want 1.
Also worth noting after a bit of experimentation, the amount of memory allocated appears to always be the larger of either 4 bytes, or the requested amount + 2bytes. For example if you ask for 3 bytes, then 5 bytes will be allocated. If you ask for 8 bytes you will get 10.
As to why the extra 2 bytes are allocated, I am unsure. I cannot find reference to this in the avr-libc documentation.
|
H: Over current protection for a 1-cell battery
I have a one-cell battery which protection circuitry is based on a DW01-P (with dual mosfet 8205S).
The datasheet says on page 2 "Overcurrent detection voltage = 150mV". How do I translate this number into how much current (expressed in Ampere) that this one-cell lipo battery can discharge at maximum before it cuts off?
AI: You need to look in the data sheet for the DW01-P. One copy can be found here:
https://cdn.sparkfun.com/assets/learn_tutorials/2/5/1/DW01-P_DataSheet_V10.pdf
The device senses on the CS pin using a preset fixed voltage threshold. The discharge current limit is determined by the MOSFETs that you select. For quick reference here is the relevant text from the data sheet.
Selection of External Control MOSFET
Because the overcurrent protection voltage is preset, the threshold current for overcurrent detection is determined by the turn-on resistance of the charge and
discharge control MOSFETs. The turn-on resistance of the external
control MOSFETs can be determined by the equation:
RON = VOIP / (2 x IT)
(IT is the overcurrent threshold current). For example, if the
overcurrent threshold current IT is designed to be 3A, the turn-on
resistance of the external control MOSFET must be 25mΩ. Be aware that
turn-on resistance of the MOSFET changes with temperature variation
due to heat dissipation. It changes with the voltage between gate and
source as well. (Turn-on resistance of MOSFET increases as the
voltage between gate and source decreases). As the turn-on resistance
of the external MOSFET changes, the design of the overcurrent
threshold current changes accordingly.
|
H: USB to TTL converter: why doesn't TX need a voltage selector?
I plan to flash new firmware for a Sonoff ESP8266-based switch.
So I need a 3.3v USB to TTL for the programming. I see an inexpensive converter from China which has pinouts for
3.3v
TX
RX
GND
5v
My question: I understand the 3.3v and 5v pinouts as providing constant voltage for the load (device being programmed) with the power coming from the USB port. But why isn't a switch of some sort needed so the TX line can be set to either 3.3v or 5v output?
What I am missing here? Thanks.
AI: Why doesn't TX need a voltage selector?
Because it's a cheap, low-quality adapter that only supports one output level.
The Prolific PL2303, around which this adapter is based, can't output 5V signals. By default, it uses a 3.3V voltage standard, and will output at least 2.4V for high outputs. This will be high enough for most 5V TTL devices, but some devices may have trouble with it.
|
H: What current should I charge/discharge my 12V lead battery at?
Here are the specs listed on the battery:
AI: You can charge and discharge the battery over a wide range of currents but the information on the side of the battery indicates values that the manufacturer accepts as reasonable.
You can charge it with a constant voltage supply of between 14.4 and 15.0V with limited current that will ensure the current into a discharged battery does not exceed 6A.
You should disconnect the battery when the current falls below about 1Amp (1/20C) which should occur in about 5-10 hours if fully discharged.
The other mode of charging is 'Standby Use' where a constant voltage is applied to the battery and it will charge itself as necessary - this mode of operation is often used for standby power supplies rather than cycling charging then discharging (called cycling).
It is not recommended to charge this with a constant current because the voltage can easily go over the 15V level at the end of charge and this will cause the electrolyte to gas and be lost - there is no provision for topping up the electrolyte on this type of battery
You can use a lab supply for this but I recommend that you put a diode in series to avoid the battery discharging back into the supply if the power supply is unplugged or switched off.
The information on the battery also indicates the acceptable discharge currents and capacity - at the 2 Hr rate the capacity is 16AH - this means that if you discharge it with 8 Amps it will discharge in 2 hours. It will probably be acceptable to go up to ~2-3 times this value (i.e. 25 Amps) without damage although the battery will discharge more quickly. If you discharge more slowly you will get more total energy out of the battery, at 1 Amp it will take 20 hours to discharge.
There is no lower limit to discharge the battery but at very low rates you may lose more energy to self-discharge than powering the load - the battery will discharge all by itself in a few months.
I recommend that you learn about batteries from somewhere like Battery University.
Note - you will often see the 'C" terminology related to batteries - this is merely the capacity in Ampere-Hours (AH) expressed as a current, for example this battery has 20AH capacity. If you discharge it at 1/20C (0.05C) this means discharging it at 1Amp.
|
H: how do I change a Ultra High voltage AC currents frequency?
I am currently making a theoretical tabletop particle accelerator. and i need the materials in the accelorator to be resonating together at a set frequency (Whcih I can change to test speicific theories)
The key is High voltage. ---
The secondary requirement is AC current ---
The third is the frequency
I can step up household AC via a transformer up to 300KV AC
Since transformers are tuned typically at a set frequency I cant change the frequency before I step up the voltage , I have to do it afterwards
NOTE: the frequency needs to be changeable so that the different ferromagnetic/parametric and diamagnetic materials i am using have their own fundamental harmonic frequencies and what i am attempting to do is match 2 different materials fundamental frequencies be picking a higher frequency that resonates with both materials used.
Ok with that all explained - How do I change a Ultra High Voltage AC's frequency ...
Other materials lets say one is 15hz and the other is 20hz would resonate both at AC frequency 60hz...
Im testing a lot of materials . One of the most common frequency for a few materials is 85-90 Hz . Some higher frequencies are 300-340hz, lowest would be 10hz
AI: Mechanically drive a DC generator with a motor.
Excite the DC generator with your 10-350 hz AC.
Send the output through a variac to your HV transformer.
Vary your frequency with your exciter.
Vary your HV with a variac between the generator output and your HV transformer.
Generator output frequency should be independent of mechanical speed.
Be careful as insulators breakdown with higher voltages and higher frequencies can conduct through capacities.
|
H: What places, or where Can i get a AC transformer that will Increase voltage too 150KV + from household power
I am currently in need of locating a Transformer that will step up my household wall power to output in the ballpark between 150kv - 500kv.
Having a bit of a hard time
My project is a table top particle accelerator and the primary need for the device is Ultra High AC voltage
Current Amperes are very low and are not to important so eddy currents and heating hopefully will be manageable.
AI: I'd recommend three things:
Buy the book, "Exploring Quantum Physics through Hands-on Projects," by David Prutchi and his daughter Shanni Prutchi.
You contact the authors of said book immediately and ask for their thoughts about your ideas, thoughts, and wishes -- which to start I assume is a cyclotron with two dees.
Go to their web site, perhaps starting here: DIY 250 kV Power Supply. Read carefully through their book and their web site.
A cyclotron is pretty easy by today's standards and there's no reason you can't do it. But I don't think this is the place for a discussion about making one. Not unless you've shown an aptitude for it and are asking specific questions along the way.
Just be careful. You casually mentioned Pb (lead), saying you'd sometimes be using 45 cm thickness of it. While at 220 keV, lead is 12 times better stopping power than steel, at 2 MeV it's only 2.5 times better than steel. So don't get too cavalier and imagine that lead is some kind of panacea. It's still mostly vacuum space and you can easily generate gamma radiation that is very much smaller than an atomic diameter. The cross-section (measured in barns) of atoms varies depending the atom and energy of the gamma. You need to be fully aware of these details, and more.
A quick check can be made by looking up "radiographic equivalence factors," often shown in tables that include keV/MeV and materials.
You asked about what a cyclotron is used for. That made me pause a moment. But you can look up a cyclotron on Wiki. But they are used to make short-lived radioisotopes, some cancer treatments, and used to be used for particle physics experiments. Dirt easy to make, these days. Must be plans on the web. (There are also betatrons, linear accelerators, and well.. whatever you can think up, actually.)
Bottom line is that you need to understand ideas of atomic cross section, fully understand different kinds of radiation and how they are generated, and I'd recommend that you have appropriate instrumentation available to monitor what is going on. Not just a Geiger counter. (Which you should have several varieties of.) But also a scintillator capable of differentiating energies -- commonly using NaI(Tl) crystal and a photomultiplier. I designed and built a Geiger counter when I was 16 (and it worked, first time, on a sample of Autunite I had then [still do have that crystal].) So they are very, very easy to understand and design and use. A scintillator is a little more complex. But not that much more so. You should be relatively competent with at least those tools. (There are many more, but probably too expensive unless you are borrowing tools from a lab.)
You might also consider joining the Yahoo group called "Geiger Enthusiasts" which is a very active group and a very knowledgeable one. I'd definitely recommend conversing with some of those folks, as well.
Get advice, read, study, learn, test, validate, and enjoy. But don't do this alone. I contacted the designer (a physicist) of a Geiger tube I used (originally just to find out where I could buy his tube) and gained some advice, even as a mere teenager. I also spent days, and weeks, in the science library of the largest university in my State.
So mitigate your risks by getting good information and advice. And if possible, contact your local university and see if someone there may be willing to help (unless you are already involved in one right now -- in which case, why in the heck are you asking these questions here?)
|
H: Current Sensing with Differential Amplifier
Hello, I have to use this circuit as part of a school project so I'm limited to this design and these components. The voltage regulator works as anticipated however the differential amplifier does not. Here it amplifies, by a factor of 10, the voltage drop across the shunt resistor R8 which is ~28mV. I'm having two issues with this however circuit however.
The output of the LM324N on the simulation is 303mV but shouldn't it be 280mV since the gain is set to 10 by the 100k and 10k resistors ?
When I physically replicate the circuit the output voltage of the LM324N is 562mV with all the same parameters as the simulation but the voltage across the shunt is 27.2mV.
How can I solve both problems ? I think that it may have something to do with CMRR and the input offset voltage which according to the datasheet is typically 2mV. I'm also new to all of this and I don't completely understand CMRR
AI: Yes, your problem with the physical circuit is CMRR -- common mode rejection ratio. The CMRR of the opamp itself is pretty good, but you have used very low tolerance (5%) resistors, and that has given your circuit a very poor common mode rejection.
To illustrate what is going on, open up your LTspice simulation and change one of the 10K resistors to 10.5k (i.e, 5% error). You will see a ~500 mV change in the output.
The reason for this is that the gain of your positive and negative inputs are different. What you want to measure is G*(V2-V1). In reality you have (G2*V2-G1*V1). When the gains are not matched (because your resistors are not matched), you have a term that doesn't precisely cancel that is proportional to the voltage at the input terminals, rather than only the voltage difference. This is called the common mode.
It is common to use high precision resistors (0.01% or even better) when constructing differential amplifiers with high common mode rejection. Frequently people use instrumentation amplifier ICs that have laser-trimmed resistors on the chip.
What can you do to fix this with what you have? The simplest is to move the current shunt resistor to the "low" side of the load (i.e., between RL and ground). This will reduce the common mode signal, and thus the impact of the poor CMRR.
You should use the most precise resistors you can find. 1% are common and cheap, but if you don't have them available, you can just grab a handful of 5% resistors and an ohmmeter and find matching pairs. Selecting parts like this isn't as good as actual precision resistors, since low spec parts usually have worse drift and temperature coefficient, but it still helps.
|
H: How much AH should my battery has for my sound system (3 Amplifiers)?
I would like to know if I can mount my sound system in my car.
My current battery is 45AH, and the problem is that I can't go higher than a battery with 50AH, because there isn't space for it on the battery spot near engine.
My question, will I be able to use my sound system upgrading to a 50AH battery or it won't even be possible?
I am not planning to play music with the car turned-off by the way, so the car alternator will help right?
I don't know the capacity of my alternator, but the car is a Chevrolet Aveo Bi-Fuel 2011 (T250 Model).
My sound system specifications are the following:
Amplifier 1: Gravity PS1000 - RMS power: 2 x 125 Watt 4 ohm - Max. power: 2 x 500 Watt 4 ohm powering this Subwoofer: Climax Hazard 5000 - 500W RMS - 1000W Max Power
Amplifier 2 (next buy): L2.210 - Class A/B 2 Channel Amplifier 70W RMS x2 @4 ohm
Amplifier 3 (next buy): EV200.2 - Class A/B 200 W MAX Power, 38 W X 2 RMS @ 4 ohm
My calculation result in a total RMS of 466W (2x125 + 2x70 + 2x38)
And according to this rule Total RMS Wattage times 2 (Inverse of Amp Efficiency) divided by 13.8 Volts equals Current Draw in Amperes (explanation below) I will drain from my battery 68 A [ my calculation was ( (250+140+76) * 2 ) / 13,8 ]
So concluding my question, what will happen if I use a battery of 50AH in my car? supporting all this system draining 68 A?
Thank you so much for your help!
AI: The peak current you can draw from a battery is not directly related to its capacity in Ah. For instance, your existing 45Ah battery starts your car, and the starter motor probably draws north of 200A for a few seconds. If it does that, it will clearly deliver 68A peak.
So the real question is, how long will the battery and the alternator power the amplifiers at your listening level?
There are too many unknowns to give a real answer.
If the question is, 'I have 50Ah, how long can I deliver 68A for?', then the answer is 50/68 = about 45 minutes. However, that wasn't the question. Amongst the confounding factors are:
Your amplifier will not draw 68A continuously. It may do that at the peak, but it will usually be much less.
Your alternator will deliver some current so the battery will last longer ... unless it's a cold night and you're using headlights and all the electric heaters you can.
You would not want to drain a 12v car battery fully. Going regularly below 50% capacity shortens its life, and if you flatten it with music, you won't be able to start your car in the morning.
You mention in a comment that maybe you just wouldn't turn the system up loud. However, unless your electrics are instrumented, you won't know how loud you can set it.
My recommendations are
a) Install the system.
b) Plan for the continuous load to be supplied by the alternator, so the battery evens out the supply between traffic stops and loud songs.
c) Install a large 'audio capacitor' at the amplifier terminals, which evens out the load on the battery between the individual peak cycles of the loudest notes. This reduces the reliance of the amplifiers on very short fat cables to the battery, and the battery impedance.
d) Instrument the electrics. Install a voltmeter on the battery, and/or a current meter in series with it.
The current will give you an indication of whether you are charging or not, and by how much, but it won't tell you the state of the battery.
The voltage should be around 14v when the battery is fully charged. If it goes below 11v, turn the system off, you've really over-done it. If it's in the 12-14v range, you are using both battery and alternator, and the battery is not fully charged. I wouldn't want to see less than 13v shown on a voltmeter, especially towards the end of my journey when I had to start in the morning.
|
H: How to calculate Resonance Frequency of Audio Speaker?
Okay so I searched all over the Web for the resonance frequency of my car speaker so I can properly apply crossover filters, but didn't found any reference of the resonance frequency of this speaker model.
So how can I calculate it?
I have the following data:
Peak power: 150w
Rms power: 20w
Response Frequency: 90Hz-20kHz
Sensitivity: 91db/W/m
4 ohm
Thank you in advance!!
AI: You cannot CALCULATE Fs, (Resonant frequency of loudspeaker moving mass in free air). But you can MEASURE it relatively easily. You will need a power amp, a variable frequency source, and a DMM. THere are many resources online for measuring driver characteristics. Fs is one of the primary measurements for the popular Thiele/Small parameters. Here is one straightforward explanation:
Ref: https://sound-au.com/tsp.htm
|
H: Turns for saturating a toroid at a particular frequency
I need to know the winding turns required to saturate a toroid for a particular drive frequency.
The toroid I have is having the following dimensions
OD = 19 mm
Thickness = 8 mm
Effective Area (Ae) = 32 sq. mm
Al (nH) = 1900
Bsat = 0.49T
My driving signal would be
f = 100kHz
Duty = 50%
Drive Voltage = 15V
From my initial research I have found that this relates to the following equation
N = (V * t) / (B * Ae)
where
V = drive voltage
t = on time of the drive voltage
Based on the above I am getting turns of 4.8
I need to know whether this is the right equation, I mean is it actually giving be the turns to drive the toroid to saturation?
AI: It's unusual to specify a 50% duty cycle at the outset. This is kind of weird. Usually, you look at the input source situation and the output situation and work out the required duty cycle from there. The output is arranged so that the off-voltage and off-time of the inductor works out correctly. Starting with a duty cycle is an unusual approach.
The inductor volume is a useful metric (though not often directly considered in most write-ups.) You can compute it in your case (square wave) as:
$$l_e\cdot A_c \ge \frac{\mu_0\cdot\mu_r\cdot I_{peak}}{B_{max}^{~2}}\cdot V_{on}\cdot t_{on}$$
That can all be turned around to compute \$I_{peak}\$:
$$I_{peak} \le \frac{\left(l_e\cdot A_c\right)\cdot B_{max}^{~2}}{\mu_0\cdot\mu_r\cdot V_{on}\cdot t_{on}} $$
Plugging in your figures, I get:
$$I_{peak} \le \frac{\left(44.4\:\textrm{mm}\cdot 32\:\textrm{mm}^2\right)\cdot \left(0.49\:\textrm{T}\right)^{~2}}{\mu_0\cdot 2100 \cdot 15\:\textrm{V}\cdot 5\:\mu\textrm{s}} \approx 1.724\:\textrm{A}$$
And of course from your basic equation for an inductor, I can now compute:
$$L \ge \frac{V_{on}\cdot t_{on}}{I_{peak}} = \frac{15\:\textrm{V}\cdot 5\:\mu\textrm{s}}{1.724\:\textrm{A}}\approx 43.5\:\mu\textrm{H}$$
(I misspoke in a comment to you, earlier. I meant greater.)
From this, you can work out the windings needed on your toroid, of course. You have \$A_L\$.
The problem will now be that since you require 50% duty cycle, then you must provide for a reversed voltage across the inductor during the off-time that is at least as large as the applied voltage.
The reason for this is that the on-time volt-seconds (Webers) must be matched by off-time volt-seconds (except that the voltage polarity must be opposite.) It must be the case in each period that:
$$V_{on}\cdot t_{on} + V_{off}\cdot t_{off} = 0$$
Given enough cycles, there is no escaping that need. If there is even the smallest consistent deviation, each cycle, then it will build up and given enough cycle periods, walk itself so that it exceeds whatever limitations your core has (unless it is vacuum, which has no limitations -- with neutron stars perhaps demonstrating this fact.)
If \$t_{off}=t_{on}\$, then it must be the case that \$V_{off}=-V_{on}\$. Some circuits allow the inductor to find its own reverse voltage and provide for a long enough \$t_{off}\$ time so that in all circumstances it can return to zero. Then the inductor will automatically drop its own voltage back to zero, too, for whatever time remains in \$t_{off}\$ (until the next cycle starts.)
If \$\vert V_{off}\vert\ge \vert V_{on}\vert\$, then less time in the off period is required than you have provided. And that's fine if and only if the voltage across the inductor is zero for the remainder of the time. Not reversed. But exactly zero. Anything other than that will gradually accumulate Webers in one direction or the other until you exceed the core's ability to handle the flux (again, unless its vacuum.)
I started out talking about core volume. Yet far more often you will hear about core area. These are related, though. A Tesla is just flux divided by area. So 1 Weber divided by 1 Tesla comes out as area in \$m^2\$.
When you are winding a core, you are winding it around an area. So this is why the focus in most papers and documents discusses that concept a lot more. But to get a feel for magnetics, I think it's better to think about flux (Webers) than to think about flux density (Webers per meter squared.)
[Perhaps, just as it is often somewhat easier to think in terms of mass than to instead always force yourself into thinking of density. Density can be converted back to mass, of course. But it confuses a lot of equations and thinking if you aren't allowed to use mass, but always have to use density and volume, because now you have to keep two things in mind instead of one. Also, because of our biology, we have a more intuitive feel for the idea of "weight" (given gravity, proportional to mass) than we do of "density," which our biological sensors have no direct way to "observe."]
Sure, flux density is important. That's because matter acting as little magnetic dipoles that respond to an applied magnetic field does have its limitations. (In Tesla.) But flux is what must return to zero. Sure, you can also say that flux density must return to zero, too. Clearly, the area is non-zero, so if flux density goes to zero then flux also goes to zero. But again this forces you to imagine a slightly more complicated concept, the ratio of flux to area, instead of just focusing on the actual thing under consideration, which is just flux.
So if you have a material, such as some specific ferrite or powdered iron or whatever, it will have a limitation in flux density. Given a magnetic cross section (\$A_C\$), you can work out the allowable flux. But keep in mind that flux, itself, is only one of two orthogonal parts of the magnetic field. The total field energy isn't completely determined by just flux. It is determined by the magnetic flux AND the magnetic force, combined. You already know that the magnetic force is measured in amperes and the intensity (another paired dimensional unit thing) is the amperes per meter (H).
At this point you should be thinking: "Hmm. Flux times Force is Energy!! Wow! And Flux density times Force intensity then must be Energy per unit volume!! Incredible!!" Ah. See that 'volume' bit there, sneaking up on you??
Vacuum has no limitations here. But all matter does. And matter that can form into dipoles (by definition, a magnetic dipole opposes the applied field) will have some kind of limitations in the number of useful dipoles it can form up into, right? And this limitation is really throughout the entire material. Not just an area. But the entire VOLUME must exhibit this limitation. There's no reason to imagine that this is just a cross section behavior only. It's almost certainly a factor that affects all 3 dimensions!
So, now, if I know the energy I need stored, and if I know a particular material which can only support a certain 'energy per unit volume', then knowing the energy alone I can now compute the volume of the matter (material) I need to properly hold that energy and still be within its limitations of flux density and force intensity!
Another way of sweeping aside complications here is to imagine that energy MUST ONLY BE STORED in vacuum (which has no limitations) and that the magnetic dipoles formed up in matter are "short circuits" which cannot store energy of any kind (ideally; in practice, of course, it takes energy to cause them to rotate, which may also cause friction and heating and energy loss into the core, etc.) And that the value of \$\mu_r\$ is nothing more than a ratio of physical material volume to remaining magnetic vacuum volume in the matter itself. This is why the volume increases for a specific energy to be stored when you increase \$\mu_r\$. Nothing comes free. A point of a high value of \$\mu_r\$ (ignoring energy losses for the moment) is to concentrate flux lines and to keep them from spreading out into a huge volume of space around the inductor. To contain them, in short. You pay a price for that, which is the energy losses in a practical core and the limitations in energy storage given some volume of material to work with.
|
H: Mosfet operation wrt AC signals
I am trying to understand about the Mosfet operation.
It would be great if someone could help me get answers to these questions -
What happens to an AC signal (DC+ small AC signal) if it is applied to the gate of a MOS which is biased in Linear mode.
In the saturation mode, the output at the drain/source can be found using the small signal analysis mos model.
In the Linear mode, how do we analyse this? As the mos behaves as resistor here the output automatically should be DC+ AC component? What happens to the gain here?
From the transfer characteristics of, say CS amplifier, it is seen that the output at drain is small and dvo/dvi is small.
What would happen to the gain and other parameters for a circuit which is biased in saturation with an Ideal current source as seen in the image, for small signal analysis.
vo (node) DC value would be Vdd-I.Rd
Also with an assumption that device is huge (KW/L = huge), vs (node) DC component can be taken as -Vth.
What would be the AC component of voltages at these nodes? Will there be any gain here?
Also, If we have a big bypass cap at vs node, then the analysis would be the same as that for a CS amplifier. Is that right?
AI: In the linear region we can calculate the transconductance like in the saturation region.
$$
\frac{\partial I_D}{\partial V_{gs}} = \frac{\partial}{\partial V_{gs}} K^\prime \frac WL \left[ (V_{gs} - V_T)V_{ds} - V_{ds^2}/2\right] = K^\prime \frac WL V_{ds}
$$
Unfortunately the expression contains \$V_{ds}\$ which tells us that the transistor behaves like a resistor and the model is not very useful.
Nevertheless, we can calculate the gain which results in
\$A_v = K^\prime \frac WL V_{ds} R_{out} \$ so we can calculate it, if we make an assumption about \$V_{ds}\$. E.g. we could bias the transistor at \$V_{ds} = 100mV\$ and continue from there.
Since the current is fixed by the current source and the resistors are fixed as well, there will be no small signal gain. We have a gain of about 1 from Vsig to Vs, since the transistor acts as a source follower. Vo will stay constant.
|
H: sample input signals and check their values VHDL
I have 2 input signals - ID_1,ID_2 which sampled into id_vec.
LEDx_GRNn are output.
In this point, only one of a,b,c,d should be '1' and the others '0', which after should make only one led on and the others off.
For some reason all the leds are on so I'm guessing I do something wrong.
Waht am I missing?
ID_1,ID_2 have the constants values.
signal id_vec :std_logic_vector (1 downto 0);
signal flag :std_logic;
signal a:std_logic;
signal b:std_logic;
signal c:std_logic;
signal d :std_logic;
id_vec(0)<=ID_1;
id_vec(1)<=ID_2;
a <='1' when id_vec<="10" else '0';
b <='1' when id_vec<="00" else '0';
c <='1' when id_vec<="01" else '0';
d <='1' when id_vec<="11" else '0';
LED1_GRNn <= not (a);
LED2_GRNn <= not (b);
LED3_GRNn <= not (c);
LED4_GRNn <= not (d);
AI: It must be = and not <=
a <='1' when id_vec ="10" else '0';
b <='1' when id_vec ="00" else '0';
c <='1' when id_vec ="01" else '0';
d <='1' when id_vec ="11" else '0';
You want to compare if id is equal to 0 or 1 or 2 or 3. Not assign right?
After EDIT
<= also means LESS_THAN_EQUAL comparison which you don't want it to be for your case.
|
H: Cloning a USB signal - feed two USB hubs with one input port
I have a schematic for a 4 port USB hub (4 output ports, 1 input port). I'd like to make two of these hubs and connect them together, thus making a 7 or 8 port USB hub on a single PCB.
I know there are 7 port hub ICs, but I find them harder to use, plus I have this schematic already tested.
I can think of two ways to connect these 4 port hubs together.
First way, which I'm pretty sure will work, is to connect one of the hub A's output ports to hub B's input port.
This way however we get a total of 7 usable USB ports.
The second way, which I'm not sure if it will work, is to use only one input port, but clone its USB signals - they go both into hub A and hub B. This way we get a total of 8 usable ports.
But would that work? I know the differential signals are a bit tricky, so I'm not sure if cloning them like this would be acceptable.
AI: You can't duplicate the signal because, ignoring the electrical challenges, the protocol does not support it. The USB host would not be able to understand who it is talking to if two slave devices were talking back to it at the same time.
To put it another way, why do you think we have USB hub ICs in the first place? If you could just share the signals between multiple devices there would be no need for a hub.
If you have two devices (e.g. two hubs) which need to share the same host port, you simply need another 2-port (or more) hub IC to connect them. So using 3 of the 4-port hub ICs you could make your 8 usable ports - in fact you would have 10 usable ports, 2 spare on the first hub, and then 4 on each of the other hubs.
|
H: How to shift down a square wave at 0 - 5 V?
I have a square wave that oscillates at 500 Hz from 12 to 14 V. I would like it to oscillate at the same frequency from 0 to 5 V, so I could use to trigger i.e. a switch. Is there any circuit to do this (shift down the signal)?
It goes from 12 V (low level) to 14 V (high level). I would like it to shift it down so it will go from 0 V (low level) to 5 V (or higher from 5 V... I don't mind).
AI: @Olin's answer is a good one if you can live within the limitations which are:
High voltage supply for the comparator (or special comparator)
Duty cycle of input close to 50%
Some pulses will be missed when the signal is first applied.
It has the great advantage of being non-critical as to component values (it automagically adjusts to the average input voltage) and has a good noise immunity once it adjusts.
There are a couple other simple options. First, you could simply use a 2.5V reference and divide down the input with a voltage divider such that 13V gives you 2.5V out. Since you would have only +/-7% (minus whatever noise immunity you require) to play with, this makes for a relatively critical circuit.
simulate this circuit – Schematic created using CircuitLab
It will also stop working if the input voltages change by a small percentage.
A third option is a DC restoration type of circuit that AC couples the input. Again, a comparator is used.
simulate this circuit
In the case of my first circuit the signal level is faithfully preserved if the input stops switching. In the case of Olin's circuit, the input noise immunity drops to close to zero if the signal disappears so you could see random noise at the output. In the case of my 2nd circuit, the output will eventually drift to one state or the other (the state can be enforced by a resistor from the + input of the comparator to +2.5 or ground).
In all 3 cases you could add some hysteresis by feeding back a bit of the output voltage to the + input.
|
H: Power transformers?
So I'm interested in welding, and I've seen people create viable welding power supplies by putting microwave transformets in series - I would like to do this, but the question had arisen - how does replacing the secondary coil of the transformer with an insulated coil turn it into a power transformer?
Thank you for your time
AI: When you use a transformer to transfer electrical power from a source to a load, we call that a "power" transformer. The adjective is added to help understand the function of the particular transformer. For exemple, they also are "audio" transformer", "data" transformer, "gate driver" transformer, "isolation" transformer and many more depending on the specific application.
All types of transformer are made the same way. They all possess one primary and one secondary wound around a ferromagnetic core (laminated iron core for your microwave transformer). Both made of insulated wire. The use of the transformer for the application will determine if we "call" it a power transformer or another adjective.
|
H: Where is the negative connection of an oscilloscope?
I have found this circuit from http://vector.com/portal/medien/cmc/application_notes/AN-AND-1-106_Basic_CAN_Bit_Timing.pdf
Please can someone explain to me what is the negative connection of the oscilloscope : does it mean the GND ?
Thanks in advance :)
Pss the oscillo is Tektronic MSO 2024B: Mixed signal oscilloscope
AI: If your scope has math functions you can connect a second input to the negative line and choose something like 'A minus B' for your trace.
That is to say, you don't use a dedicated negative input. You use another normal input connected to the negative side of what you're measuring, and you go to math functions and look for the setting to subtract one input from the other and draw that on the screen.
|
H: In solving an op amp problem with two voltage sources, why can't we combine it?
Say we have a circuit like
Why can't we solve by doing $$V_{out} = -(V_{in} - V_{offset}) * (R_f / R)$$
We know that \$V_{out}\$ for a typical inverting amp (that doesn't have \$V_{offset}\$) is just \$-(V_{in}) * (R_f / R)\$, so why is \$-(V_{in} - V_{offset}) * (R_f / R)\$ for this diagram not correct?
My reasoning for \$-(V_{in} - V_{offset}) * (R_f / R)\$ is because we know \$V_{offset}\$ is the constant voltage for the positive side of the op-amp.
\$V_{in}\$ decreases until it reaches \$V_{offset}\$.
Thus, can't we just think of it equivalently as \$-(V_{in} - V_{offset}) * (R_f / R)\$?
AI: There is a way to combine the solutions for the the two voltage sources using the Superposition Principle, which works for ideal opamp problems because they are Linear Systems.
Here is the general procedure for N voltage sources: Short all of the voltage sources except for one. Find the output voltage and call it Vo1. Iterate through the voltage sources, shorting each voltages source, and solving for the output voltages Vo1 through VoN. The total output voltage of the original circuit is the sum of each solved output:
$$V_{out} = \sum_{i=1}^n Vo_i$$
@SpehroPefhany 's answer is the shorthand way to say all this, which is that each input has its own gain.
|
H: Calculating specs for power supply of array of high power LEDs
I doing an experiment with plants, for which I want to use commercially available LED lamps. They provide the light I need but the area they cover is too small. So I am trying to replicate a lamp of high power LEDs but I have straggle deciphering the necessary power supply, and I wonder if the specifications can be calculated?
The image below is the circuit I want to replicate. There are four arrays of 30 LEDs connected in series (orange outline in image; each LED is claimed to be 10W) and it includes two power supply (based on the image it appears each power supply is 50W). Each power supply feeds two of the four LED arrays (one circuit is shown in red, in the image below).
Using standard values for 10W LEDs: LED forward voltage=3.5v, LED current = 1.05Amps
For the sake of simplicity, I do calculations for each of the two independent circuits (i.e., two LED arrays and one power supply, red outline in the figure). So these are my calculations so far:
Expected voltage driver=Number LEDs * LED Voltage
Expected voltage driver=30*3.5
Expected voltage driver=105v
Expected power driver=Number LEDs * LED Voltage * LED current * Number of LED arrays
Expected power driver= 30*3.5*1.05*2
Expected power driver=220.5W!
Yet the only information in the power supply it says it is 50W. I wonder where my error is? The circuit works, so I wonder if perhaps the LEDs are lower wattage (e.g. 3W instead of 10W) or if this type of circuit uses a PWM so that only some LED's are ON at a given time?.
AI: LEDs work absolutely fine at below their rated power. So if it's a 50W current limited power supply, it will give as much power as it can, and the LEDs will glow at the corresponding brightness.
|
H: Power consumption of MOSFET
If I use a P channel MOSFET as a switch for a battery 9v 0.5Ah controlled by an external power source 12 volt will battery last its shelf time if MOSFET is by off?
Update: I dont mind the leakage of the external power supply.
AI: Look for the off state leakage current spec for the MOSFET you are using. Usually called Idss or zero gate voltage drain current. Typically it's in the range of a few uA, which isn't a lot of current draw but you would have to evaluate the leakage against the battery capacity to see if it will significantly impact the shelf life. The leakage can increase dramatically with temperature, so if high temperature storage is a possibility be sure to take that into account.
|
H: How do scanned interface inputs work?
I'm learning about different ways to interface physical switches to microcontrollers and I don't understand how scanning interfaces work.
I am told in the notes that the software from the microcontroller would scan each row at a time, in the following scanning pattern:
I dont really understand how it could work, and how the limitation is that it can only detect a maximum of two simultaneous switch closings.
AI: The trick with keyboard matrices like this is the fact that outputs can (on most microcontrollers) be in three possible states - HIGH, LOW, and High Impedance (Hi-Z).
When an output is in Hi-Z mode it is effectively disconnected.
So by setting three of the outputs to Hi-Z and the fourth output to LOW, and having the four inputs pulled up through resistors, pressing a button connects that input to a LOW signal.
In effect the currently activated (driven LOW) output becomes the ground connection for a standard pulled-up button.
By cycling through which output is playing the role of the ground you can have any number of buttons attached to your inputs.
It is simplest to think of each of the outputs as a switch to ground. Work with just one input for now, and you can get an idea of how it works:
simulate this circuit – Schematic created using CircuitLab
In that example outputs 1, 3 and 4 are Hi-Z. Output 2 is LOW. You can now see that pressing buttons 1, 3 or 4 will do nothing since the switch (output) connecting them to ground is open. Pressing button 2 will pull the input LOW, registering a button press.
Now just multiply that circuit 4 times, one for each input, and there's your matrix.
You can also turn the whole thing the other way up - use a pull-down resistor and drive each input HIGH instead of LOW - that way the inputs will register a HIGH for active. It makes no real difference which way around it's done. Pull-ups and driving outputs LOW is the most common way though, because MCUs often have built-in pullup resistors and the ability to select open drain outputs which switch between LOW and Hi-Z much easier.
|
H: Inverting Amplifier Generalized Gain Derivation
I understand that the derivation of an inverting amplifier's gain is generally treated as \$G = \dfrac{-R_2}{R_1} \$, and have seen many resources supporting this claim. I understand how the proof works by solving for the current flowing to virtual ground at equilibrium, and am satisfied with this expression. However, my EE textbook showed a different, more generalized version of the expression for gain,
\$ G = \dfrac{-A(1 - B)}{(1 + AB)}, \$
where A is the open-loop gain of the op-amp and \$B = \dfrac{R_1}{R_1 + R_2}. \$
In the limit of high open-loop gain, I understand how this expression simplifies to \$G = 1 - \dfrac{1}{B} = \dfrac{-R_2}{R_1}.\$ Can someone provide an explanation as to how this more generalized transfer function is derived, and why the general derivation that follows is inaccurate in some way for low open-loop gain?
Here is a sample of the derivation that I followed - what simplifying assumptions are made?
\$ I = \dfrac{V_{in}}{R_{1}} \$
\$ V_{out} = V_{in} + IR = V_{in} - V_{in} \times \dfrac{R_2+R_1}{R_1} = V_{in} \times \dfrac{-R_2}{R_1}\$
\$ \therefore \dfrac{V_{out}}{V_{in}} = \dfrac{-R_2}{R_1}\$
Follow-up: Does whatever correction needs to be made to the closed-loop gain of the op-amp affect the input resistance of the op-amp, or is it always \$R_1\$, regardless of the op-amp gain?
AI: This is your typical inverting opamp configuration:
simulate this circuit – Schematic created using CircuitLab
The one thing you know for sure is
$$V_o=A_{ol}(V^+-V^-) $$
and notice I am not even talking about negative feedback here. The output always follows that equation and what limits the output voltage are the rails or supply voltages (\$V_{cc}\$ and \$V_{ss}\$).
Now, back to the equation provided. In order to find the gain, you want to express it in terms of \$V_{in}\$ and \$V_o\$.
You can easily see that \$V^+=0\$. You can find \$V^-\$ in several different ways. For simplicity here is what \$V^-\$ is:
$$V^-=\frac{R_2}{R_1+R_2}V_{in}+\frac{R_1}{R_1+R_2}V_o $$
If you plug those values into the \$V_o=A_{ol}(V^+-V^-) \$ equation, you get:
$$V_o=-A_{ol}\bigg(\frac{R_2}{R_1+R_2}V_{in}+\frac{R_1}{R_1+R_2}V_o\bigg) $$
After some algebra, you can find \$\dfrac{V_o}{V_{in}}\$ to be:
$$\frac{V_o}{V_{in}}=-\dfrac{R_2}{R_1+\dfrac{R_1+R_2}{A_{ol}} }$$
So, in order for the gain to be approximately \$\dfrac{V_o}{V_{in}}\approx -\frac{R_2}{R_1}\$, the open loop gain has to be large enough so that the \$\dfrac{R_1+R_2}{A_{ol}}\$ term in the denominator is negligible.
|
H: What amp rating ESC do I need?
I am working on an electric longboard, retaining the original battery pack and brushless hub motors. I want to be able to control this with an Arduino. I want to know what size ESC I will need. The battery pack is 30V and the hub motors have a 1200W rating each. Will a 50Amp ESC be sufficient for each motor?
AI: This is why we have Ohm's Law: I = P/E
P = Power (Watts)
E = Electromotive Force (Volts)
I = Current (Amps)
You can calculate any of the factors given two factors:
I made an Ohm's Law calculator online here: http://www.rcrowley.com/eirp.htm
If you plug in 30V and 1200W, it shows that the current is 40A.
So your 50A ESC should be adequate.
|
H: Why is the current wrong after resistors in series with simple LED?
Setup
source --- 1.5V batteries x2 = 3V
LED --- 2V 35mA http://www.futurlec.com/LED/LED5R.shtml
Resistors --- CR1W 1W 10ohm +-5% x 4 = 40ohm
Description
I have your average "Light up a LED" experiment. I have a 3V source(worn out to 2.72 now) followed by 4x 10ohm resistors and a 2V 35mA LED.
Problem
3V... need to drop that 1V on the resistors and the current below 35mA... LED lights up but the reading is wrong! I get 2.08V after resistors which is good. But my current reading after resistors reads 58mA. Since current in series is said to be same for all circuit, it was supposed to be 1V/40ohm = 0.025A = 25mA(worn out now, so: 0.66/40 ~= 16mA). Why on earth is it 58mA?
I have tested the resistance between source and before LED(right after resistors), reads 40ohm.
Any ideas? I'm lost :(
AI: You're probably measuring the current wrong. Are you placing the meter in series, with the resistors and led?
If you are only measuring the current across the resistors, 2.7V across 40 ohms = 68 milliamps. But then you have to take into account that the battery has a high Equivalent Series Resistance. As the current rises, the voltage drops across the ESR, leading to a lower battery voltage, etc etc. That's the most possible situation.
You have to measure the current with the multimeter in series with the battery, resistors, and led.
|
H: Stability in control theory and electronics
In control theory, if the impulse response of a system dies out then the system is stable.
In electronics, Bode plot usually uses the see the gain and phase margins and determines the stability of a system.
Does the stability in the two fields mean the same? Also is there any relationship between them?
Update:
In control theory, stability is defined as a measure of the tendency of a system's response to return to zero after being disturbed. So does the definition is also applied in electronics (for example OpAmp) and how to test it (say OpAmp) using this definition?
AI: At first - two basic considerations:
The impulse response is a closed-loop test in the TIME domain (and can give you some rough "impression" regarding the degree of stability);
The BODE diagram is an analysis of the loop gain (loop open) in the FREQUENCY domain (and can give you some figures for phase and/or gain margin).
Hence, at first sight, both test are not related to each other. However, the term "stability" has the same meaning in both cases - and the mathematical tools of the system theory connect both domains to each other.
EDIT (UPDATE): Here is the desired answer to your update:
Regarding stabiliy there is, In principle, no difference between control systems and electronic (opamp based) applications. The DEFINITION of stability is in the TIME domain (BIBO: bounded input gives bounded output), however, the exact proof of stability properties (expressed in terms of stability margins) is conveniently done in the FREQUENCY domain (loop gain analysis). Note that this is one of the main reasons for introducing the frequency domain and the complex frequency variable s.
|
H: Can we replace relays with MOSFET?
My question is simple, why are we still using relays ? Can't we say that MOSFET are the future of relays?
AI: Can't we say that MOSFET are the future of relays?
You will probably never replace some radio frequency relays with MOSFETs because a MOSFET has several pF of drain-source capacitance and at 1 GHz that will continue to look pretty much like a closed contact. At 1 GHz, 100 pF has an impedance of 1.59 ohms - It wouldn't make much of an open-contact if you used a MOSFET.
Even at (say) audio frequencies this capacitance is problematic if you were considering using a MOSFET as a signal switch. At 10 kHz, 100 pF has an impedance of 159 kohms and this could still let through a significant signal into a high impedance amplifier.
|
H: ac model for buck-boost converter
This lecture shows how to create ac model for buck-boost converter.
What I am confused is the equivalent circuit model for inductor loop equation
on page 38.
As you can see from the model, there are two dependent voltage source (square boxes), \$D\hat{v}_g \$ and \$D'v \$, and one independent voltage source (circle symbol), \$(V_g -V)\hat{d} \$.
What is the basis to distinguish/know which one is dependent or independent source here?
For example, why is \$D\hat{v}_g \$ considered as dependent source? I thought it is an independent source because \$\hat{v}_g\$ is the input voltage and it is independent source. However, I am wrong here.
AI: In the AC modeling of converters, what is considered is the average over one period of a certain quantity (e.g. \$\langle v(t)\rangle=\frac{1}{T}\int_0^Tv(t)\mathrm{dt}\$), in order to maintain the slowly varying behaviour of electric quantities, while eliminating high frequency variations and ripples.
Using this approach to model the inductor in the circuit you showed, it is
\begin{equation}
\langle v_L(t)\rangle=d(t)\langle v_g(t)\rangle + [1-d(t)]\langle v(t)\rangle = L\frac{d}{dt}\langle i(t)\rangle
\end{equation}
where \$d(t)\$ is the duty cycle in the period that was considered, and the other quantities are named as per the circuit you provided.
If you now consider all quantities as being a DC value + an AC small variation/perturbation (e.g. \$\langle v_L(t)\rangle=V_L + \hat{v}_L(t)\$), you obtain
\begin{equation}
L\frac{d}{dt}[I_L + \hat{i}_L(t)]=[D + \hat{d}(t)][V + \hat{v}(t)] + [1-D-\hat{d}(t)][V+\hat{v}(t)].
\end{equation}
If you now compute all the products and neglect all DC (e.g. \$DV\$) and second order terms (e.g. \$\hat{d}(t)\hat{v}(t)\$), you obtain the very expression shown in the picture you uploaded (where \$D'=1-D\$).
Now, consider which quantities are you using to control your converter: you obviously use the duty-cycle, and \$\hat{d}(t)\$ is the perturbation on it that causes the circuit to steer away from the precise working point you wish to have; thus \$(V_g - V)\hat{d}(t)\$ is your independent source, since it is due to your control circuit.
What about the other two sources? They are the effect on the inductor of variations/perturbations happening on input and output voltage, which you do not control directly, since they are due to the power supply you are exploiting and to the correct work of your circuit, respectively.
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.