text
stringlengths 83
79.5k
|
|---|
H: Can you split track width among layers to equal one big trace?
Assume you want a trace to be able to carry 20 amps of current on 1oz copper with a temperature rise of 20. Using an online calculator this yields 12.60mm trace width.
My question is, can you split that width among bottom and top layers (6.30mm each) and stitch together those layers using vias? If so, is there any special considerations with this method?
AI: 1 - Yes. With attention to detail by both you and the board house, you can move a hundred amps with +/-10% sharing.
2a - The internal traces will run warmer than the external ones. Some online calculators address this.
2b - Pay attention to the vias; they have a large impact on how evenly the current is spread among the layers. Depending on the board stackup, you might have 2 oz copper traces but only 1 oz copper in the vias. If you want to get picky about it, a via is a very short trace whose width is equal to the hole circumference. Use the hole diameter minus one plated copper thickness as the diameter. This is equivalent to the average of the inside and outside circumferences of the copper tube.
Also, while it is good to spread out some vias along the trace run, the clusters of vias around the ends (or any high-current branches) have the biggest impact on current sharing.
|
H: IR emitter to keep TV on
We installed some TVs in the main floor of our facility. We have personnel here from 8am until 12am, and the TVs have been set to turn on at 7am and off at 1am. Sometime during the day, YouTubeTV asks "Are You Still There?", and the video is paused during this prompt. Suspecting the remotes are bound to get lost, they aren't left out for someone to acknowledge the prompt, and eventually the app stops and the TV screensaver takes over.
Can I record, on an oscilloscope, some key presses from the remote with a photoresistor, and play those back with a small circuit? We have sections of drop ceiling in our facility. I could place the circuitry above the ceiling tiles close enough to the TVs, and still maintain line of sight.
Pressing the up arrow shows the video progress bar. Without any interaction, that goes away after ten seconds or so. I think replaying the up arrow once an hour would be sufficient.
The IR blasters I've seen only replay incoming signals from a remote, not pre-recorded signals.
If a device like I describe already exists, can someone tell me the terminology to google?
AI: This is absolutely doable. In fact, depending on the TV, you might not even need to record the sequence because you can just look it up.
Lots of sites like THIS exist, that have databases of IR codes. All you need is a small µC to send out the pattern and the IR LED, resistor, and a transistor.
The circuit itself could be as simple as this:
simulate this circuit – Schematic created using CircuitLab
Just send the sequence out the I/O pin of your controller however often you need.
|
H: Where would a pure inductive or capacitive load be placed on a Smith chart, assuming the TL impedance is real?
I'm a new HAM operator, and just started learning about the amazing tool that is Smith charts; and it got me wondering what the edge case of a pure inductive or capacitive load would be? Since the normalized real part is 0, would they be plotted in the same spot?
AI: Forgive me for my knowledge of transmission lines and microwave circuits is very minuscule.
Wikipedia has a very good image of how the Smith Chart is organized (for impedance):
Looking at the imaginary vs. real axis, you can basically determine real and imaginary components of the impedance. If you had to cut horizontal line across the middle of this circle in half, you would see that top half would be a more inductive load and the bottom half being a more capacitive load.
Henceforth, using the picture above, if the reflection coefficient, \$\Gamma=j\$, it would mean that the transmission has a purely inductive load. Otherwise if the reflection coefficient, \$\Gamma=-j\$, it would indicate a purely capacitive load.
I'm not sure what you mean by "assuming the TL impedance is real". If the impedance was only real and not complex, it would mean your transmission line would be purely resistive with no indication of induction or capacitance.
Remember that your formula for impedance is:
\$z=r+jx\$,
where \$r\$ is your resistance, \$j\$ is your imaginary number \$\sqrt{-1}\$ and \$x\$ is your reactance which is your "resistance" of your capacitance or inductance, so to speak.
|
H: SATA power connector usage
I'm making a small device that I want to power by the sata power connector/ATX PSU. It needs 5v and 12v power supply. This device will use standard 15pin SATA power male connector to get the power from the PSU. According to the specs I found here, there are 3 5v and 3 12v pins, some of them are longer and used for precharging. My question - is it ok in the device to shorten all 5v together to get more power, and also the same for 12v and ground? I suppose so because on all my PSUs there's just one wire comes to the SATA power connector for each voltage, but is it standard and stated somewhere in the specs?
Also, there are power disable and Staggered spinup/activity pins - can they just be ignored?
Thank you.
AI: It's totally ok to wire all pins of the same voltage in parallel. That's how they're wired inside the connector anyway, so don't worry about it. The power disable pin is not used often on standard ATX power supplies, so don't worry about it unless you're plugging you device into some server appliance that might have a backplane that uses the power disable pin. Big server drive sleds use this pin, but standard PSUs don't. Same goes for staggered spinup, if you're connecting your device to a standard ATX power supply, just treat it like a GND and wire all the GNDs in parallel.
|
H: what is Vfb in this datasheet
Beginner here. I'm trying to read this datasheet, and I'm not sure what Vfb is or how it was obtained. See pages 19-20 in section 8.2. https://www.ti.com/lit/ds/symlink/lm3410.pdf
AI: Vfb is your feedback voltage, that's how the current driver IC "calculates" how much current is actually flowing through the LEDs it's trying to drive with a constant current.
The LM3410 will push the output voltage higher and higher until it reaches 190mV across the feedback resistor.
|
H: Why is 12V Common for PSU
Looking around at various PSUs and I see 12VDC show up pretty commonly, both for PSUs and motors they drive. Can anyone tell me some of the history and/or rationale behind this? E.g. is that a particularly efficient voltage to work with for some reason, do most of the more power-heavy motors commonly need that particular level, etc. New to power electronics.
AI: A short story
Long before solid-state electronics and vacuum tube electronics, lead acid batteries were and are the longest living battery technology. Being a multiple of 2V +/- 0.1 per cell or 2.36V on float charge or 14.2V, the 12V battery was not the 1st common voltage for cars. Initially it was 6V but as gas engines grew , they used 2x 6V batteries then became a single 12V pack for economical and performance value.
Meanwhile
Telecom used 48V (24 cells) to power ringing bells and powering millions of phones in local exchanges for landlines. They needed and still use 24 x 500 lb. 2V lead acid batteries designed for long life with ceramic and must all be balanced by monitoring specific gravity. This is the plain old telephone system (POTS).
Then Solid State Analog came along
Since lead acid batteries chemistry was chosen for best performance and operating at a float charge of 14.2V or more with tolerance, Automotive electronics rounded up to 15V for linear IC's requirements. So 15V and +/-15 for bipolar voltage is still used in datasheets specifications.
However for safe operating margins, 12V is still the most popular Analog voltage and is still used for wireless routers and all sorts of applications.
Rechargeable laptop voltages were all over the map for LiPo cell array voltages but eventually the industry settled on 19.2V as a "Universal" standard charger voltage which is converted for every different battery pack used inside.
Meanwhile for logic before and including TTL Logic IC's performed best 5V. But the datasheets will often specify then at the standard 10% tolerance of 4.5V or 5.5V and the same for 3.3V as 3.6V logic that works at 3V on Lithium cells.
Now personal computer CPU's operate below 1V for performance and efficiency advantages and only increase voltage on demand for power determined by the BIOS.
Who knows
While Lithium is most popular for portable applications, 12V lead acid in cars is still the most economical. My Mercedes SUV has 2 of them. One in the rear for auxillary use and the main under the rear passenger floorboard. There has been a lot of research to minimize copper costs and higher voltage car batteries have been considered but the cost of industry change must also be considered and the profits on 12V.
|
H: Advantages of using active inductors in ICs
I have seen that there exist many methods to realize equivalent inductors in IC through active components. For instance, through gyrators realized with Op-Amp.
This solution will require a lot of transistors. Which is the advantages of choosing it, and not a proper inductor? I know that in ICs it is very difficult to realize inductors with L greater than 20nH because of the lack of space in the substrate, while transistors are little, with lower parasitics and their realization pattern is "standardized". But I'd say it is convenient only if it allows to get high values of L, in general L>>20nH. Which is a typical value?
AI: The name gyrator is just a convenient term for an electronic circuit that replaces the functionality of a grounded inductor. There may be circuits called gyrators that can replace the functionality of a series inductor but these are much rarer and are probably explained away using conventional filter explanations anyway.
Which is a typical value?
But, if you want a really common example of a circuit that somewhat mimics an inductor, try thinking about a constant current source or sink. It mimics an inductor value of infinite henries and they are used all over the place, inside and outside analogue chips.
|
H: Limiting the inrush current
I have a transformer with several secondaries, and this is how I'm planing to control the inrush current:
simulate this circuit – Schematic created using CircuitLab
I'm using a resistor to limit the inrush current and MCU will energize the relays after a few seconds to bypass the resistors.
Can I use a single resistor and relay on the primary to limit the inrush current instead of using a relay for each secondary?
How can I activate the relays without an MCU, using only analog circuitry?
AI: Putting inrush protection on a transformer primary is a pretty standard way of doing things because it also handles primary magnetization current inrush problems too.
Avoiding an MCU means having a small timer circuit that, when timed out (having been initially triggered by the instance of power being applied), activates the relay as you show in your diagram.
Or, you could avoid all the complication and use a negative temperature coefficient varistor/thermistor in series with your primary winding. This is a pretty standard solution in many power supplies.
|
H: What is the collector current of this transistor circuit?
Knowing that:
$$V_{BE(on)}=0.7V, β=100, V_A=100V$$
I'd like to find the collector current of the following circuit:
It seems I will have to use the fact that the base emitter voltage is 0.7 volts but I have no idea how to do this. Using the equations for the collector current in the forward active region, saturation current Is is required, which we don't have in this case.
Another idea, is to try and find the voltage around the 1KΩ resistor which can be found easily if we know the voltage around the 15KΩ resistor. If we assume that the base current is very small,the two resistors on the left branch form a voltage divider and calculations become very easy. However, I want precise numbers so I can't make this assumption.
Maybe I could find the input resistance seen from the base and that would allow me to calculate the base voltage precisely. I know the Rin seen from the base but only for small signal models. I don't think I'm allowed to use it here, since we don't have "small signals".
$$R_{inB}=rπ+(β+1)R_E$$
AI: You say you want precise numbers (I think you mean accurate) and you are not willing to assume that \$I_B \ll I_C\$. At this point you should use a good SPICE simulator instead of hand calculations. However, the model parameters for the transistor will also be approximations so those results won't be 100% accurate either.
You could build the circuit and test it, as long as all of your test equipment is under calibration and traceable to NIST. However, there is a great deal of variation between transistors of the same part number, so your results will only be valid for that one single transistor and those particular four resistors. If you build another copy of the circuit you will get different results.
We are engineers here. Make the assumption about \$I_B\$ and get along with the project.
|
H: How does a load decide the amount of current drawn in a circuit?
I realise this question might seem similar to other questions asked on the site. But what I'm asking is actually completely different.
Suppose I have a 5ohm resistive load hooked up to a 10v battery. The current drawn by the load would be equal to the current through the circuit, right?(assuming the load is connected in series). Using ohm's law, the current through the circuit would hence be 2A.
Now,if I connect the same 5ohm resistive load to a 15V battery, the current through the circuit would be 3A, using ohm's law.
So doesn't the same load draw different currents when the supply voltage changes? If that's the case, why do we say "current drawn depends on the type of load connected"?
Moreover does the above reasoning of using ohm's law hold valid for any type of load? (i.e. L load, C load, RL load or a combination of all 3)
Edit: this highly upvotedsays that for a constant voltage supply, current drawn depends on load. So how does the load draw two different currents for two different voltages?
AI: If that's the case, why do we say "current drawn depends on the type of load connected"?
Usually we're dealing with a fixed / constant voltage supply. Most common examples would be domestic or industrial mains voltages or 12 V automotive systems. In both cases the voltage remains within certain tolerances so that means the other's have an inverse relationship, \$ I = \frac {V}{R} \$ or, since V is constant, \$ I \propto \frac {1}{R} \$.
Figure 1. Each of these various bulbs, connected to the same supply, will present a different effective resistance and will draw a current inversely proportional to their resistance. The higher the power consumption (watts) the lower the effective 'resistance'. (I use the word resistance cautiously here as LEDs are not resistances but it's getting harder to find images of incandescent lamps!) Image source: Banggood.
Moreover does the above reasoning of using ohm's law hold valid for any type of load? (i.e. L load, C load, RL load or a combination of all three).
No. Ohm's law is specific to resistance. We can extend the law to inductors and capacitors but we have to calculate impedances and use complex numbers.
Edit: this highly upvoted says that for a constant voltage supply, current drawn depends on load. So how does the load draw two different currents for two different voltages?
Because you've changed the voltage. There's no inconsistency there.
|
H: Can I safely ground myself by clipping an anti-static wristband to a half-inserted plug?
I'm having some trouble grounding myself for laptop repair. I cannot get the crocodile clip of my anti-static wristband in to my plug socket (I'm in the UK, so our sockets are a bit difficult) and I can't find a non-painted part of any of my radiators to clip myself to.
This has made me a bit desperate, but I've found a solution. Is it safe to get a normal plug, insert it about halfway in to a switched-off plug socket, and clip the crocodile clip of my anti-static wristband to the half-inserted ground prong of the plug?
AI: Bad idea. The pin may not be making contact internally in the socket. You also risk contacting the live pin on an older plug.
Figure 1. The earth pin of a UK 13A plugtop.
Instead, get a length of insulated wire, strip off about 6 mm and terminate it in the earth pin and run it through the cable clamp - maybe looping around it so that it grips the single core adequately.
Then refit the back of the mains plug and screw it closed, so the mains Live and Neutral pins are safely inaccessible.
Strip a cm of insulation from the other end and connect your wristband to that. Make sure that your wriststrap has a resistor in circuit to reduce the possibility of electric shock should you touch a live part.
|
H: Question about Vbe and operation regions of BJT
Updated/Corrected question:
Realizing what confuses me the most, I will make this more straight-forward. Say that a transistor works in the forward active region and it receives an ac input signal at its base.
Is there a chance that the change in the base voltage will decrease Vbe and force the transistor into the cut-off region? Is there any chance this might happen before first entering the saturation region?
Original post:
I'm soon taking an exam on transistors and I've just started my revision. After having a look at the conditions for each region of operation and some example problems, there is something I don't understand or that maybe I don't remember very well.
The condition for an npn bjt to be on is that the base emitter voltage is above 0.7 volts. From what I remember this is not an accurate value but let's say it is around there.
When we analyse the small signal model of a circuit , the values we calculate are superimposed on the dc values, right? So, when we apply an ac signal to the base of the bjt, the base emitter voltage changes.
So, my questions are :
Is there a chance that the ac input signal will take the bjt in the cut off region?
If yes, how do we check it?
If yes, when deciding on the swing of the output voltage are we more worried about entering saturation or cut-off region? I mean, which comes first as the swing increases? Does it depend on the operation point?
AI: Is there a chance that the change in the base voltage will decrease Vbe and force the transistor into the cut-off region?
Yes this is possible, and it would result in non-linear operation of your amplifier.
Is there any chance this might happen before first entering the saturation region?
Yes it's possible to (mis-)design an amplifier so this happens.
when deciding on the swing of the output voltage are we more worried about entering saturation or cut-off region?
You need to consider both when designing your amplifier.
Assuming we're talking about a common emitter amplifier, you need to be sure the amplifier can handle the complete range of input amplitudes desired without going into cut-off, and you need to be sure it can handle the complete range of output amplitudes without entering saturation.
|
H: A simple Arduino Uno HC-05 system doesn't work
I have an Arduino Uno, HC-05 module and an LED.
I want to connect the HC-05 to my phone and with an app be able to turn on and off the LED and at the same time sending a message from the Arduino to the Phone that the LED has been updated.
I have built this circuit: (But Rx is pin 12 and TX is pin 13 and with 1kΩ and 2kΩ resistors instead)
The HC-05 blinks properly (so I assume it's not broken).
Here's the Arduino code:
(I tried to put the code using the code sample button but it didn't work properly sorry)
Now, I have two problems with the system:
- bt.println(...) should send text to the phone but the phone side doesn't receive anything.
- Serial.println(LEDState) should print what the phone has sent to the Arduino and it does but it is all ASCII characters and are printed on new lines.
I believe the problems are connected with the baud rate but I tried using 9600, 38400 and both mixed but it doesn't work. I also tried reversing the places of Rx and Tx in the SoftwareSerial initialization.
AI: After a chat with OP to debug connections, here is how the problem was (somewhat) solved:
HC-05 should be connected:
VCC and GND to the Arduino's VCC and Ground respectively
TX of the Arduino (the designated TX pin; the second argument of the SoftwareSerial declaration) goes into a resistive divider of 1k:2k, and the output goes into the HC-05's RX pin
RX of Arduino straight into TX of HC-05
If hardware serial is used to talk directly to the HC-05 from the Serial Monitor (i.e. using it as a USB-to-TTL converter by bridging RESET to GROUND) then the RX and TX will be reversed on the Arduino - the Uno's RX pin goes into the resistive divider and Uno's TX goes directly to HC-05 Tx. If the Arduino is connecting using Serial (normal hardware serial used as Serial.read() and Serial.write() in code) then it is conventional (un-reversed) with pin 0 going into the TX of the HC-05 and pin 1 going into the voltage-divider.
AT commands should be issued with Newline set to "both NL and CR" - thats what the HC-05 expects and AT commands should be issued at 38400 baud.
It had become apparent that SoftwareSerial was the culprit here as connecting the HC05 to the hardware serial pins worked (pin 0 to HC-05 TX and pin 1 to HC-05 RX through voltage divider), for:
Talking directly to the HC-05 using the Arduino as a USB-to-TTL converter - at 9600 baud, messages from phone over Bluetooth can be sent and received from the Serial Monitor
Replacing the original code SoftwareSerial with just Serial, and including a character as a receive buffer and testing this character against '1' and '0' to turn on the LED also worked. See code below:
void setup() {
// Open serial communications and wait for port to open:
Serial.begin(9600);
while (!Serial) {
; // wait for serial port to connect. Needed for native USB port only
}
Serial.println("Serial link established.");
}
char recv;
void loop() { // run over and over
if (Serial.available()) {
recv = Serial.read();
if (recv == '1')
Serial.write("on"); //No newline produced here
else if (recv == '0')
Serial.write("off");
}
}
If you want to use Strings, then you have to make a character array and every time Serial is available, write to a new indexed character of that array and increment indexer - break this loop when '\n' (newline) is received and reset the indexer counting variable, for example. This is because Serial.read() returns a char at a time, so an entire string cannot just be read as the original post might suggest. This is also why Serial.println() printed the character one by one followed by a newline - println ends with a newline and Serial received one byte at a time.
OP's code was changed to utilise normal hardware serial for communication with the HC-05 which avoided the use of Software Serial. If anyone knows why Software Serial causes this, please comment why.
|
H: Parameters needed to identify the correct capacitor
In a traditional power switching supply PCB like this one,
(source of the image)
ceramic capacitors are sometimes present as discrete components. They appear as:
(source of the image)
I would like to correctly identify them, when no schematics is available, to replace them if/when they get damaged. In the case of the above image, these parameters can be obtained:
the capacitance 822, which, according to this answer, is: 82*10^2 pF = 8.2 nF;
the tolerance Z, which, according to this table, is -20 %, +80%;
the voltage rating 1KV, which is 1000 V.
(A letter can occasionally appear above these parameters, and I do not know its significance).
Is any ceramic capacitor, which has these same values, suitable to replace the above one, or other features must be taken into account when replacing such a component?
This question is related to a previous one.
AI: Is any ceramic capacitor, which has these same values, suitable to
replace the above one, or other features must be taken into account
when replacing such a component?
Be very careful here. If the capacitor is X or Y rated then swapping it with one that isn’t is asking for trouble. X and Y rated capacitors are specifically designed to fail safe thus, if they are used in filters across live AC terminals or AC to ground, you must replace with the correct type.
The good news is that every single X or Y rated capacitor has markings that are reasonably unambiguous in that respect: -
The ones above are Y rated.
These are X rated.
|
H: What differentiates Laptops from PCs in terms of power consumption?
Today I as a Computer Scientist stumbled about some strange facts.
Recently, I bought a new Macbook Model 16" inch. Today I tried to stress test it because I wondered if the 96 Watt charging cable could even keep up with this processor and GPU power consumption.
Running Cinebench showed, that running the program without charging cable, would completely overstress the battery. Running with the 96 Watt charging cable, however, the Macbook did not lose any more energy,
Therefore the whole computer must have run below or at exactly 96 Watt.
CPU: 2,6 GHz 6-Core Intel Core i7 (Turbo boost to 4,5 GHz)
GPU: AMD Radeon Pro 5300M 4 GB and Intel UHD Graphics 630 1536 MB
(And with the same charging cable, you can even power stronger Macbooks)
In comparison, I know that some Gaming PCs including mine can take up more than 500 Watts of Power. But they are not that much better in terms of performance. (they are better, but not proportionally to the power consumptions)
So, how exactly do manufacturers like Apple ensure low power consumption, while offering good computing ability? (Price does not matter)
Thank you!
AI: So, how exactly do manufacturers like Apple ensure low power consumption, while offering good computing ability?
By programming the firmware to cap long term average power consumption. A high end Intel 6 core processor can easily consume 150+ watts if allowed to run full out. Yours has a 45w TDP, which means after a certain delay, the processor voltage and frequency will be restricted to reduce power consumption and keep temperatures within reason. Since most laptop tasks are composed of quick bursts of activity followed by lower loads, most users never notice that the average performance is much lower than peak.
|
H: TTL Logic Gate Resistor Values
I've been teaching myself the basics of circuits for the last few months and have been enjoying going into chip schematics. I'm trying to understand why the resistor values are as they are. For instance, for this NAND gate:
Source: Big Code - Discrete TTL NAND Gate
How would one decide, if making this from scratch, to use 4kΩ, 1.6kΩ, and 130Ω resistors? I've noticed 130Ω to be common in the output stage of TTL and would love to understand why.
Please share any information or good book links for me, as well as offer any feedback on how I can improve my posts here. Thanks!
AI: How would one decide, if making this from scratch, to use 4kΩ, 1.6kΩ,
and 130Ω resistors?
The resistance values determine how much current can flow in the associated parts of the gate circuit. This then determines the input and output currents and operating speed of the gate.
R1 limits the current sourced by an input when pulled low. The previous gate must have enough strength to sink at least this much current (preferably a lot more to get a reasonable fanout). When all inputs are high R1 supplies current to the Base of Q2. This current must be sufficient to tun Q2 fully on.
R2 sets the current supplied to the Base of Q4 when Q2 is turned on, and also limits the current going into the Base of Q3 when Q2 is off.
R3 limits the current that can be drawn from the output when high.
R4 shunts some current away from Q4's Base to ensure that it turns off quickly when Q2 turns off.
Once you have chosen the desired input and output currents, the required resistor values can easily be calculated using Ohm's law and transistor characteristics (Vbe, Hfe etc.). However (as you might suspect) the values interact to some extent.
Choosing a higher value of R1 to get lower input current will require that the other resistances also be higher, so the maximum output current will be lower. With smaller currents flowing the internal parasitic capacitances of the transistors will take longer to charge and discharge, so the gate will have longer delays and a lower maximum operating speed.
|
H: How do Software Defined Radios have such big bandwidth?
I've noticed that the majority of "amateur" SDRs have ADCs with a sampling rate not higher than 20 MSps, but still the HackRF One for example can scan it's whole frequency range (1MHz to 6GHz) in the same time.
How does that happen?
If a SDR were to have a ADC with a 20MSps sampling rate, it could safely sample a signal with a maximum frequency of 20MHz (if quadrature, 10MHz if else). So basically it could work with signals from 0Hz to 20MHz and that range is shifted with the help of a down converter, but it still remains the same width. Do they work with the Nyquist zones or something, somehow reconstructing the spectrum for frequencies higher than the Nyquist frequency?
AI: can scan it's whole frequency range (1MHz to 6GHz) in the same time.
How does that happen?
Not at all.
The HackRF (and other SDRs) can just be tuned to arbitrary frequencies, giving you arbitrary coverage (in chunks of sample rate bandwidth at once), but not simultaneous coverage. I'm not quite sure where Mossmann claims that it would, so I think this might be a misunderstanding :)
|
H: Discharging all energy of one battery into another
I was wondering how discharging all the energy from one 18650 Lithium-Ion battery into another 18650 Lithium-Ion battery would be achieved. So far I’ve consulted using MOSFETs, DC to DC Converters and current sources/sinks without avail.
The whole circuit I’m attempting to achieve is a circuit that can fully charge battery 1, and then transfer all of the energy (less any losses) from battery 1 into battery 2.
Any direction or reference papers would be appreciated. Thank you.
Edit: Will update with more details shortly.
The end goal is to have a battery testing circuit that does not entirely dependent on the energy from a wall adapter. The circuit will also be testing for battery voltage, charging current and temperature among other things.
So far I have a discharging circuit to discharge both batteries and a charging circuit to charge one of the two the two batteries. The transferring of energy from battery 1 to battery 2 is the essential step I’m currently working on.
Thanks again for all the insight it’s has been a huge help.
AI: The way lithium-ion batteries work is by being at full charge at ~4.2V and "no charge" at ~3.0V. The limits on voltage is due to safety, because lithium-ion batteries are unstable outside of this range.
So given this information, and we know from electronics that the higher voltage-potential is the one that will control the way the energy flows, we can see that you would want the "charger battery" to stay at a higher voltage than 4.2V. Of course you aren't going to charge it to a higher voltage, but you can step up the voltage with a buck-boost converter, also called a step-up regulator. Then you could have the voltage at for example 5V, and it would charge the other battery.
If you want to charge the other way again, for whatever reason, then you can use the same step-up regulator and some switching-circuitry with mosfets to control which battery that is stepped up to 5V.
I have drawn a suggested schematic below. You want to research this more of course, but I think this would be at least close to what you want.
simulate this circuit – Schematic created using CircuitLab
My schematic does not include over-charge and under-charge protection. This is of course an important aspect you need to add to ensure you don't charge above 4.2V. That part of the design is up to you to include, but should be simple enough.
Sidenote: 3.7V stated on the batteries in the schematic is because this is the nominal voltage of lithium-ion batteries. They will charge up to 4.2V and discharge as much as you allow it.
|
H: Pin Layout Unknown - 5-way navigational switch
Completely at a loss, been searching for a datasheet or anything to help for a cou0ple hours... So I came to the only place where I can usually find help
Does anyone know what the pin layout of this 5-way navigational switch would be:
Here's the Link to where I purchased it: https://www.aliexpress.com/item/32815153642.html?spm=a2g0s.9042311.0.0.113f4c4dmzJuNZ
the only mark i see is letter D on the bottom right
Unfortunately my wife took out my battery from my multimeter, without using a screwdriver so I have none right now
AI: Thank you everyone
I figured it out the hard way (hooked up an Arduino + led) and tried each trial and error style here's what i found
Note the Bold "D" wit the arrow is the select button
|
H: Determining values to convert square wave to sine wave at low frequency
For a project I am working on, I have a 1Hz square wave coming from a real time clock, and need to convert it into a sinusoidal wave to pulse LEDs. After some googling and playing around with simulations I came up with the circuit below. The 1Hz square wave is connected at the net labelled SQUARE and then filtered:
My question now is how do I determine values for the passives to effectively filter out the higher frequencies and get the single sine wave I want? I got the values you see here from messing with a simulator, but I don't understand how to derive them analytically.
Thank you for your help.
AI: Square wave has odd harmonics 3f,5f,7f at levels of ~ -10dB,-14dB, -17dB ..
A 1st order filter is -6dB/octave and 3f is 2 octaves above f or -12dB more than the -10db harmonic @ 3f.
Triangular Current will not be much different to the eye than sinusoidal as the eye is logarithmic .
You might try something like this.
Then you can always try sine PWM control using a FET switch with current limiting R for each LED.
But the hardest thing it seems to learn is how to define a design spec (before attempting to design it) including the LED datasheets with inputs, Processess and Outputs in a point form Hierarchical format.
IBM used to call this, the HIPO format for design specs.
Simple, Elegant, Complete, Measurable and Verifiable.
If you graduate and do design/test work, you must learn to create Design Specs then Design Verification Tests ( DVT plans and reports) based on these simple specs and record it. This is how Japanese Engineers think and how I did it in Winnipeg and Silicon valley. in early 80's as well as an Aerospace Engineer in late 70's.
If it meets spec. It's perfect. But it has to be a good specs with above Design values, budget and design time. (always double estimated design time)
|
H: How to flip LTspice voltage controlled switch so that it opens at threshold voltage? (versus close)
I am trying to figure out how to tell the VSWITCH of the LTspice program to open after it crosses the threshold voltage and close when it is below (with hysteresis). Normal behavior is that with Vt and Vh defined, it'll close when the voltage is greater than (Vt + Vh) and open when the voltage is lower (Vt - Vh).
Unfortunately I can't use another component or edit my schematic since I am working off of a .lib file for an IC, and am attempting to edit the text commands so I can import it into MATLAB for Simscape simulations with physical components (torque, etc.). Basically, I want to ensure that my IC behavior is correct once I import it.
AI: From the LTSpice help file:
Symbol Names: SW
Syntax: Sxxx n1 n2 nc+ nc- [on,off]
Example:
S1 out 0 in 0 MySwitch
.model MySwitch SW(Ron=.1 Roff=1Meg Vt=0 Vh=-.5 Lser=10n Vser=.6)
The voltage between nodes nc+ and nc- controls the switch's impedance between nodes n1 and n2. A model card is required to define the behavior of the switch.
So if you switch the connections to the nc+ and nc- pins and invert (multiply by -1) the Vt value, you should get the effect you want.
|
H: Help Identifying a few Molex-style Connector?
I would like to find replacements for the connectors in the below photos. They're in an arcade game called Pump it Up (think the Korean version of Dance Dance Revolution), and connect the dance pads to the main cabinet. The cabinet was likely made in the early 2000s.
I'm only interested in finding the two plugs with the female pins (red in the below photos), but the mating connectors are shown below for reference. They just connect from wire to wire. I suspect they may have been made by Molex since I've found other connectors in the cabinet that I've identified and replaced with Molex-made parts, but they could've just as easily been made by anyone else.
Any help identifying those two red connectors would be a huge help, bonus points if you can point me towards the pins I would need to get for the connectors, too.
AI: What I did was estimate the pin pitch uing the ruler in the photo and then do a look-up on digikey. none of the parts had photos so I downloaded datasheets.
The 3x5 is Molex mini-fit Molex 0015311152, Molex 0015311151
Digikey has the part in the catalogue, but no stock
https://www.digikey.co.nz/product-detail/en/molex/0015311152/0015311152-ND/3122315
If the 2x3 is the same pitch it's probably mini-fit too.
Molex 0015311063, Molex 0015311062
Crimp pins, and tool available separately.
|
H: In a 3 phase inverter what is the maximum voltage across each switch
What is Vds value for switches used in the network
AI: Ideally the max voltage any of your switches could experience is equivalent to your DC source voltage (B2). In reality however, you would want your switches to be rated much higher then that. In real life inverters, the reasons to have higher rating for Vds are:
The rail voltage B2 could fluctuate when load rapidly changes. In many inverter designs like Variable frequency drives, the rail voltage maybe sourced from a boost converter maintaining a certain bus voltage. Transient spikes in voltages due to load fluctuations aren't entirely uncommon and could damaged your switches.
In real world, your load is not necessarily resistive, inductive loads would give your switches a hard time when you try to turn them off, especially when the load current is non zero. This could lead to rapid rise in Vds and I have seen devices frequently damaged due to this.
For the reasons above, you would want your Vds maximum to higher then B2, preferably twice B2 and you would in addition want to put in additional protection to save your switches from breakdown. The choice of protection would vary and this could be taken care of using an efficient gate drive circuitry or other means(like fast TVS diodes).
Hope this helps.
|
H: MOSFET for switching single LED
I try to design a cost-effective but still "reasonably engineered" way of driving a single white LED with a micocontroller with the following "specs":
For the single white LED no extra boost converter should be necessary
(as a side note: I found that most simple led boost converters only work from >=2LEDs in series, so even If I used such as boost converter I would need to use two LEDs even if i only need one)
The supply voltge is nominal 3.3V, but could be set higher to about 3.6V
The single white LED should be run on a rather small but fixed current, let´s say somewhere between 3 - 5mA (across different PCB charges)
The white LED is available in binned forward voltage groups and one group limits VF to be in between 2.8 and 3.2 volts (the unbinned range is up to 3.6V where I see no way of achieving what I want without a dedicated boost converter)
Due to various reasons I'd prefer not to drive/sink the LED current directly from the uC pin, since even if 5mA are small, the driving results in some "ground offset" on the uC pin which could disturb another pin).
This seems pretty simple but is challenging for me because after some number crunching I seem to miss something.
I assume the most simple form of driving a LED with constant current, i.e. a series resistance, the LED and a low-threshold MOSFET as low-side switch. As an example, I assume using the https://www.mouser.com/datasheet/2/302/NX3008NBKW-842194.pdf
To achieve a specific, constant led current (I assume 4mA as target) I think I need to use the MOSET operating in saturation(current not dependent on VDS but settable via VGS and I assume the VGS set by the uC is constant).
I assume the uC can drive the gate of the MOSFET with 3.0V. To operate in saturation I need to keep VDS > (VGS - Vth). If I understand correctly, in this case, I actually want the minimum Vgs,th to be actually rather high, so that the saturation condition is fulfilled and the VDS voltage (voltage drop across the MOSFET) is still low.
According to datasheet I see a Vth,max = 1.1V, and Vth,min of about 0.6V, so let´s take 0.5V. Would mean that VDS must be greater than VGS-Vth = 3.0-0.5 = 2.5V. And there starts the crisis. Even with a forward voltage binned between 2.8 and 3.2V that would mean I need a minimum VDD of 2.5+3.2V = 5.7V, not even accounted for a series resistor to the LED?
There must be obviously something wrong in my thinking. I am pretty sure of having seen single blue LEDs (which are to my knowledge the same as white LEDs in terms of forward voltage) driven with the described topology. Is my assumption wrong that I need the MOSFET in saturation for setting a constant current? Driving a single LED with a few mA can not be that complicated, can it? ;-)
AI: Why not use a charge pump?
V1 is a square wave source, for example the output of a PWM, can be buffered by a logic gate.
"X1" is a dual diode, make sure the two Vf in series reduce output voltage enouhgh so the LED is off when it needs to.
This is a somewhat constant current driver. If R2 and C2 are large enough to smooth the LED current well, then it will have a constant voltage output at VCC minus the two diode Vf. In this case R2 sets the current across LED Vf dispersion.
However if we remove C2 and set R2 so that C1 is discharged at each cycle then each cycle pumps an amount of charge equal to C1 times (Vcc minus one diode drop). Charge times frequency is current, so it makes a constant current drive. It's a bit more efficient than the one above, but the LED will blink at PWM frequency.
|
H: How do I read a simple touch sensor value from Arduino?
I suppose instead of connecting to the LED it will connect to A3 on the Arduino for example but I'm having a hard time making it work.
I don't know which point on the circuit should be connected to which pin on the Arduino. I seem to get values alternating between 950 and 1050.
My question is in general how this should be connected to the Arduino.
AI: Firstly, be cautious not to connect the two touch terminals together, it may damage the transistor.
The circuit can be manipulated in various ways to sense for a finger touch at the touch terminal. One of the ways can be to take a wire from the collector of the transistor and give it to the Arduino(make sure to use a lower voltage battery). When the transistor is off(no touch), the voltage at collector terminal should be at a higher voltage level and when transistor is on(touched), the voltage should be at a lower voltage level.
You can check for the typical values that occur at some of the points in the circuit using a multimeter and decide how you can use these points to give input to the microcontroller.
|
H: Open loop gain and Percent Overshoot Relationship
In the context of control theory, I found a text stating:
"Higher the loop gain of the system, larger is the percent overshoot"
Can anyone show that analytically?
Consider it to be a second order system.
Edit: Doing the Maths I figured out that for a Step input, percentage peak overshoot is independent of the magnitude of step size, and is related only to damping ratio, zeta:
AI: "Higher the loop gain of the system, larger is the percent overshoot"
Can anyone show that analytically? Consider it to be a second order
system.
My answer is based on assuming that a 2nd order system is modified with gain and put inside a control loop hence, the system can then be regarded as having “loop gain”. This then makes it a “control-system” as per the tag in the question and, the term “loop gain” then makes sense. Other answers may not have made this assumption: -
Starting with an open-loop, 2nd order, low pass system; it may have an output/input equation similar to this: -
$$H(s) = \dfrac{\omega_o^2}{s^2 + 2\zeta\omega_o s + \omega_o^2}$$
Where \$\omega_o\$ is the natural resonant frequency and \$\zeta\$ is the damping ratio.
This is modified to include a gain factor (\$G\$) and, inserted inside a unity gain closed loop with negative feedback. The new equation for H(s) becomes: -
$$H(s) = \dfrac{\dfrac{G\cdot\omega_o^2}{s^2 + 2\zeta\omega_o s + \omega_o^2}}{1 + \dfrac{G\cdot\omega_o^2}{s^2 + 2\zeta\omega_o s + \omega_o^2}}$$
$$= \dfrac{G\cdot\omega_o^2}{s^2 + 2\zeta\omega_o s + \omega_o^2 + G\cdot\omega_o^2}$$
$$= \dfrac{G\cdot\omega_o^2}{s^2 + 2\zeta\omega_o s + \omega_o^2 (1 + G)}$$
Notable is the \$1+G\$ term in the denominator because it modifies the old natural resonant frequency from \$\omega_o\$ to a new resonant frequency (\$\omega_n\$) that is \$\sqrt{1+G}\$ times bigger.
And, the important thing here is that this has the effect of lowering the damping ratio (in other words the old damping ratio is greater than the new damping ratio i.e.\$\zeta_o > \zeta_n\$). Hence, the terms in front of s (namely \$2\zeta\omega_o\$) relative to the new natural resonant frequency (\$\omega_n\$), become: -
$$\dfrac{2\zeta\omega_o}{\omega_n} \Rightarrow\dfrac{2\zeta\omega_o}{\omega_o\sqrt{1+G}} = \dfrac{2\zeta}{\sqrt{1+G}}$$
This, in effect, lowers the old damping ratio and creates more overshoot.
|
H: laptop batteries charging myths - Long life OR short life
I'm very conscious about the life of battery.
Some people say that don't keep plugged-in laptop all over the time and remove battery charging at 70% and again charge at 40% so your battery cells don't charge or discharge at their full position, and battery cells will keep at their optimal voltage rate.
VS
Some of says keep your battery plugged-in all the time so that your battery will not be used by laptop and battery's life cycles will not drain and important this is laptop will stop battery charging automatically when it reached to 100%
Which is true?
AI: The best report on this subject
https://batteryuniversity.com/learn/article/how_to_prolong_lithium_based_batteries
The lowest aging rate is known to be 0'C 66% SoC which is the approx. SoC charge level used by OEM's to ship/store LiPo batteries.
Depth of discharge %DoD, greatly affects total lifetime Ah capacity.
NMC LiPO4
100% DoD ~300 ~600
80% DoD ~400 ~900
60% DoD ~600 ~1,500
40% DoD ~1,000 ~3,000
20% DoD ~2,000 ~9,000
10% DoD ~6,000 ~15,000
Control depends on laptop OEM and bios + drivers. Lenovo has the best user options for float charge level and slow charge option prior to going mobile to get max capacity. Choose Longest life settings for 70% to 80% float SoC rather than 100%.
"A laptop battery could be prolonged by lowering the charge voltage when connected to the AC grid. To make this feature user-friendly, a device should feature a “Long Life” mode that keeps the battery at 4.05V/cell and offers a SoC of about 80 percent. One hour before traveling, the user requests the “Full Capacity” mode to bring the charge to 4.20V/cell"
Also FYI http://batteryuniversity.com/learn/article/why_mobile_phone_batteries_do_not_last_as_long_as_an_ev_battery
https://batteryuniversity.com/learn/article/rising_internal_resistance
https://www.researchgate.net/publication/303890624_Modeling_of_Lithium-Ion_Battery_Degradation_for_Cell_Life_Assessment
https://batteryuniversity.com/index.php/learn/article/bu_808b_what_causes_li_ion_to_die
|
H: Diode selection for step-down Converter
I have built quite some buck converters for ~3A@3-5V (~1MHz, 24V Input) and I always used schottky diodes due to their low forward voltage drop (and therefore low power dissipation during free wheeling). Now I naively replaced the converter‘s feedback divider to regulate an output voltage of 12V. After a short time the regulator failed and soon I found out that it was the shottky diode that was destroyed.
I think I understand the Problem: schottky diodes don‘t only have a low forward drop (which is desirable), but they also have a relevant reverse leakage current which rises significantly with die temperature. While that wasn’t a problem in the 3-5V versions, at a reverse voltage of 12V the power dissipation (12V*100mA=1.2W) is too high for the device. I replaced the diode with a common silicon diode for experiment and it failed as well, but (as I assume) for a different reason: the ~0.7V forward drop at 2A causes at least 1.4W of dissipation.
How do I choose a suitable diode for such a buck converter (high current and high voltage) where schottky diodes don‘t appear to be a good candidate?
Is there a class of diodes which has a low forward drop and low reverse leakage or should I use a bunch of schottky diodes to share the Power?
Someone told me I could use a transistor instead of a diode, but I have no idea about that; does it make sense?
Edit: this is the schottky diode that failed in my case
Edit2:
I'm using the TPS54240 regulator and I need the converter to provide at least 2A.
AI: When power loss is square of current at 4A, the differences between low RdsOn FETs and diodes becomes more significant if you want up to 94% efficiency while up to 80% @ 0.5A.
20W out with 6% loss best case = 1200 mW heat that must be dissipated.
This requires a Synchronous Buck converter with 50A Pch/Nch FETs and around 500 mm² 2x Cu fill PCB with about isolated 5 nodes to dissipate losses.
Your problems could be any component incl. PCB area and convection air. I can suggest LM3150MH with dual FETs using TI's design webench power designer or AD's similar tool but this depends on your investment of existing design and other demands.
edit
with added schematic & layout , but still missing critical parameters for Ptot = Pcon + Psw + Pgd + Pq and Rth*Ptot= ΔT 'C rise , DCR for each part, I don't see any obvious design electrical errors for 2.5A out a very nice compact layout, but a thermodynamic failure.
Consider new Diode retrofit with force air cooling with an 18mm 24V fan
or rip up and retry , with some useful experience gained to pay attention to Rth*Ptot +Tamb. (max) next time with better design specs.
Diode should look more like this. https://www.digikey.ca/product-detail/en/toshiba-semiconductor-and-storage/CMS04-TE12LQM/CMS04-TE12LQM-TR-ND/871534
And future board design should look more like this, using TI Webbench.
Note > 500 sq mm with above specs
This is a more efficient Buck converter for reducing diode losses.
Heat can cause thermal runaway issues with ferrite and higher peak currents results in thermal issues with diodes from lack of copper to release heat. Higher RdsOn also causes higher Coss and higher T causes more leakage in FETs and Sch. diodes
So SMD is nice and small but switches need to be chosen carefully to be efficient and resonate at zero valley current when turning off.
|
H: What is the use of 5V pins on Raspberry Pi GPIO?
On a Raspberry Pi, GPIO pins are 3V3 tolerant. And there are two 3V3 pins that we can use to play with GPIO (driving LEDs or sensing Buttons). But what is the purpose of the two 5V pins?
edit: I am talking about the pins specified in this official webpage - https://www.raspberrypi.org/documentation/usage/gpio/
AI: The Raspberry Pi board is powered off of 5V from the USB connector. The pins you see on the header can also power the Raspberry Pi in case you wanted to make a Hat that did something like provide battery power to the Pi. You could also make Hat's that use the 5V supply (sparingly) to power peripherals (like an Arduino Hat) and use level-shifters for communication back with the Pi.
5V is required for things like the HDMI, so it just makes sense to use 5V as the power input, and step it down to 3V3 for everything else. It also makes sense from a supply standpoint because phone chargers (that supply USB-5V) are cheap and in ready supply, so using one to power a super-cheap computer brings the total cost down to somebody who wants to get into it.
You can look at the Raspberry Pi "schematics" here although they are pretty reduced since they don't release full schematics.
|
H: ON Semiconductor diode - markings 751 KE
I can't seem to find this on the ON Semiconductor page. It's 4mm in length and about 3.5mm in width. Can you tell me how you figure it out, so I can learn?
AI: It could be Transient Voltage Suppressors (transil) with KE markings:
1SMB5.0AT3
http://www.s-manuals.com/pdf/datasheet/1/s/1smbxxat3_on.pdf
|
H: General question about safety-critical use of microcontrollers
I was wondering if microcontrollers that hobbyists use, such as the 8-bit Atmel AVR ATMega328 which is used in many of my small projects, can be used in sensitive things, for example a life-dependent application such as an aircraft of space shuttle?
I am asking out of curiousity: I am a student in school.
AI: There is a whole industry of reliability engineering which is how safety-critical electronics (and everything else) is made.
Amongst many other things, most chip manufacturers have special versions of their products which are specified for military, aerospace, medical and safety control purposes. This starts with simple things like extended temperature ranges. But the difference between these and ordinary products is mostly one of testing and certification, sometimes to the point where parts have individual serial numbers. You can imagine that such parts -- even if they run the identical code at identical speed -- are much more expensive that the kind for ordinary commercial products. As illustration, a given chip design might be rated with a particular clock rate, and perhaps the manufacturer tests 1 part every 10,000 to check it really does. They might sell you parts where each one has been numbered and tested and give you a certificate with it. The testing and tracking costs money. Another point of difference is doing something to make the part more reliable: such as using different packaging materials which have less background radiation, or radiation-hardening the device; all this prevents some of the causes of random behaviour of devices. The functional performance of these devices is unchanged, but you can see how they would be a better choice for safety-critical products.
You may have noticed at the bottom of the Atmel datasheet
SAFETY-CRITICAL, MILITARY, AND AUTOMOTIVE APPLICATIONS DISCLAIMER: Atmel products are not designed for and will not be used in connection with any applications where the failure of such products would reasonably be expected to result in significant personal injury or death (“Safety-Critical Applications”) without an Atmel officer's specific written consent. Safety-Critical Applications include, without limitation, life support devices and systems, equipment or systems for the operation of nuclear facilities and weapons systems. Atmel products are not designed nor intended for use in military or aerospace applications or environments unless specifically designated by Atmel as military-grade. Atmel products are not designed nor intended for use in automotive applications unless specifically designated by Atmel as automotive-grade.
This is how they tell you these are not specified for safety-critical applications. All manufacturers have this kind of thing on their datasheets.
The idea is that while the general specification of the part might make it suitable for a given application, it would be negligent to any particular chip in a safety-critical application without special testing. This is the case whether it's a piece of rope, a screw, or a CPU.
On top of this, there is many design practices, regulations, and certifications of design process which go to make safety-critical engineering. One place to start is the Wikipedia article Safety Engineering
Specifically about aircraft, here are a couple articles from Avionics Stack Exchange
https://aviation.stackexchange.com/a/52935 8080 in F-16 fighter jet
https://aviation.stackexchange.com/q/28009 CPUs in airliners
|
H: Minimum recommended LED drive current
I've stumbled upon a LED's datasheet that includes a strange recommendation. They want me to drive the LED at >5mA for reasons of "long term performance." I've never seen a similar thing anywhere else, what's the rationale behind it?
AI: The properties of aging on transparent substrate LEDs have been documented in research by many but are not handy to me the time of this writing. I have known this for about 10 yrs so Google or Microsoft Scholar ought to have some papers on this. I got mine from EOS by NDA.
Generally 10% of rated max current is needed to prevent some corrosion of the crystal electrode interface. But then with 50kh given as expected min MTBF this has been greatly improved by many brand name suppliers by quality controls to millions of hours if you want to use PWM dimming to 1% some of the time, that would be OK if 100% of the time was also experienced.
Heat generally has a positive aging effect initially only either by soldering or operating at rate temp for several hours or days. This can burn off leakage R and sometimes crystalline-metallic shorts. (dendrites)
|
H: What do CT numbers in SO-DIMM RAMs mean?
Here is a picture from the "crucial" website. I am trying to understand:
What is the difference between buying 2 separate 8GB sticks and one kit of two sticks.
What do the CT numbers mean and why are there two sets for each product?
AI: Those are manufacturer's codes for items you can purchase.
1) Difference is you get one box with two 8GB sticks or one box with one 8GB stick. Sticks are all identical. Sometimes buying a pack of 2 is cheaper than two packs of one.
2) You might want to ask the manufacturer about this, as it does not have anything to do with electrical engineering.
|
H: Steal 5v from USB Module in 3v Mode
My project includes an ESP8266 (runs at 3.3v) and a USB C-Serial breakout. The USB module can optionally run at either 5v or 3v, and there's are solderable jumper pads to make that switch.
Here's where it gets tricky. While the ESP8266 requires 3v, I also have a separate battery charging module that requires 5v in to charge. So I have two obvious options:
Set the USB to 5v mode. That requires the power to the ESP8266 use a voltage regulator, and it needs a logic level converter for the serial connection. That's two extra discrete components that can cost about $5
Keep the USB at 3v mode. The ESP8266 will have no trouble talking to the USB module, but I'll need a 5v step-up converter to charge the battery, which will also cost about $5.
However, a third less orthodox idea comes to mind. Is there a way to "steal" 5v from the USB module (even while it's in 3v mode) by soldering a wire to the jumper pad in the middle? Maybe it would also require a resistor so it doesn't drain all of the power? Ideas on any of these 3 possible solutions are welcome, I'm entirely new to prototyping electronics.
Edit: I did it! Here's a picture of the final product:
AI: I don't know who designed that board, but it has several questionable connections.
First, CH340C chip datasheet requires V3 pin connected to VCC if VCC is 3.3V. On the board it is permanently connected to decoupling capacitor, which corresponds to 5V configuration.
On the other hand, the LEDs are connected to 3.3V supply, so they will (maybe) operate in strange mode when the logic levels are 5V.
So basically, the schematics is wrong whether it is configured for 3.3V or 5V power.
I'd recommend leaving the jumper at 3.3V and soldering a wire to the other (5V) pad of the jumper to bring 5V VBUS out of it.
UPDATE:
Set the USB to 5v mode. That requires the power to the ESP8266 use a voltage
regulator...
I realized that the statement above bothered me before. You seem to be under impression that you don't need voltage regulator if USB breakout is set to 3.3V.
However if you power ESP8266 from lithium battery you MUST HAVE voltage regulator between the battery and ESP anyway. Fully charged LiPo can go up to 4.2V, way above 3.6V maxium for ESP.
Also note, that VCC on the USB breakout is an output, not input. It will put out 3.3V when jumper is set to 3.3, but you cannot use it to power ESP, because there is no way to connect battery to the input of on-board AP2112 voltage regulator. Too bad, as it can supply up to 600mA, plenty for both ESP and CH340C.
Well... technically there is a way. You can cut a trace between VBUS of USB jack and input of AP2112, taking care to leave C1 on regulator side and C5 on VBUS side. Then you can solder another wire to "IN" pin of AP2112 and connect it to battery output. Then power ESP from 3.3V VCC output.
So, if you feel adventurous, you can download Eagle files from SparkFun and figure out how to mod the USB breakout. Otherwise you need additional 3.3V regulator between LiPo and ESP.
|
H: Difference between the following two very basic circuits
What is the difference between the below two circuits?
I know the bottom circuit has 'ground', but what does that mean exactly? Why would someone want to construct the second circuit over the first? And, what would be an example of doing the bottom on a breadboard -- where does ground 'go'?
Finally, when doing the above in EveryCircuit, the first example doesn't even work, or allow current to flow through it. Why is this? (The same circuit works on my breadboard).
AI: There is no functional difference in reality. However, your simulation software is stupid and needs to do calculations, so you need to tell it which node you want to represent zero and will become the reference to which every other node is measured (remember that voltage is a potential DIFFERENCE).
The circuit with a GND connection has designated that node as zero volts, he node with which all other nodes are measured with respect to.
Note that physically making that GND connection to something else in reality doesn't make your circuit operate any differently, but does make it so your circuit doesn't "float" if you are using a battery source or some other isolated source. It anchors your circuit to a known potential and not doing so can become an issue for things like noise, shielding, EMI, and safety (arcing for high voltage things).
|
H: Best practice in transmitting sensor analog signal over longer distances (approx. 100') -- in cost sensitive products for microcontrollers
I'm looking for best practices for transmitting an analog sensor signal over long distances to a microcontroller such as the Arduino running at 3.3v or 5v. Keeping costs low is imperative.
SPI and I2C is very common in Arduino land, but it seems there just isn't much of a solution when sensors need to be more than several feet away from the microcontroller. But perhaps I2C or SPI will work with the right components and wiring. Any ideas?
4-20 mA current loops seems like a solid solution since current doesn't drop along distances as long as the supplied power stays consistently above the demand. But 4-20 mA systems of sensors, transmitters and receivers seem to be mainstays in the industrial world and virtually absent in low cost embedded systems.
A classic need is detecting temperature through a resistive based thermistor. Are their inexpensive transmitters that can be coupled with a thermistor to transmit a 4-20 mA current loop signal? And then an inexpensive receiver on the microcontroller side? I would think something like a MOSFET on the transmitter side would work, coupled with a resister on the embedded side.
Alas, being a software engineer, electronics is still a bit fuzzy to me.
UPDATE:
Should have specified more. 30' - 150' distances through potentially electrically noisy environments. I can go VERY low on bit rate: 10Khz. Even 1Khz if such a thing is viable. Putting in a low pass filter and/or software filters is not a big deal. I would like to be less than $10 per sensor (not including housing). But I would also like to stay away from a custom PCB board at the sensor if possible.
AI: The other solution can be I2C — there are extenders, allowing to create buses up to 100m (300ft). One of them is NXP P82B96. It is very useful in amateur constructions, because it is not expensive and exists in an easy-to-mount DIP-8 case too. Additionally it allows to translate logic levels to higher voltages (for example 12V). You can find some application notes on NXP pages, for example: https://www.nxp.com/docs/en/application-note/AN10658.pdf
There is also a possibility to build an opto-isolated bus using P82B96: https://www.nxp.com/docs/en/application-note/AN10364.pdf
Another possibility is P82B715, but it is probably less useful because the maximum bus length is about 50m (150ft) and there are some other limitations. There is this appnote: https://www.nxp.com/docs/en/data-sheet/P82B715.pdf
Using I2C is a very effective solution, because there are many cheap interface chips on the market (ADC, DAC, IO, temperature sensors etc.) so, from the sensor side of view, there is no need to use an MCU (and to create dedicated software for this MCU).
Personally, I built my home automation system using this solution (P82B96 with opto-isolation) and I can confirm that it works very well and is stable and reliable.
|
H: Does using "16's complement" for hex always yield the same result as two's comp for binary?
I'm sure it's obvious as day, but i'm having a hard time convincing myself with all these carries and conversions...
Given that i'm operating on the same number (say, 0xA451) in different representations(0x and 0b), is it always safe to use "16's complement" (not an actual term) to represent the opposite number instead of "going down" to binary and using two's complement?
I.e. I want to know what minus 0xA451 is so i do (0xFFFF - 0xA451) + 0x0001 = 0x5baf.
Else, i could have done 0xA451 = 1010 0100 0101 0001, flip them bits and add one:
0101 1011 1010 1110 +
0000 0000 0000 0001 =
0101 1011 1011 0000, which is indeed 0x5baf.
I would appreciate somebody confirming that this is always true.
AI: You can use whatever base you want, as long as you use the same limit to your number range. For example, here's how 16-bit two's complement would work with decimal numbers:
2s_comp = 0b10000000000000000 - number = 0x10000 - number = 65536 - number
Number Decimal 2s_Comp_Hex 2s_Comp_Dec
0xA451 42065 5BAF 23471
0xE227 57895 1DD9 7641
The method where you invert each bit and add one works for numbers of any length, but mathematically, this is what's going on -- the range of your numbers is limited, and you're doing addition modulo that range.
To subtract the first number (0xA451) from the second (0xE227):
(0xE227 + 0x5BAF) % 0x10000 = 0x3DD6 = 0xE227 - 0xA451
(57895 + 23471) % 65536 = 15830 = 57895 - 42065
You can see from converting hex to decimal that this is exactly the same calculation. The same thing happens with binary vs. hexadecimal.
Computers work in binary, so they use two's complement, but you can use the complement of any base. Here's an example showing three-digit ten's complement in both decimal and hexadecimal:
10s_comp = 1000 - number = 0x3E8 - number
Number Hex 10s_Comp_Dec 10s_Comp_Hex
284 0x11C 716 0x2CC
635 0x27B 365 0x16D
(635 + 716) % 1000 = 351 = 635 - 284
(0x27B + 0x2CC) % 0x3E8 = 0x15F = 0x27B - 0x11C
|
H: Current mirror: is there a name for this NPN-PNP pair, and how does it work?
I'm looking at the schematic for the MC1460 voltage regulator. At the top it has three current mirrors, each with an NPN-PNP pair. Is there a name for that transistor structure? These pairs look like Sziklai pairs except there is an emitter-base connection rather than an emitter-collector connection, and four connections in total.
I've experimented with the circuit in LTspice, but I'm not sure why they implemented it this way. It's more complex than other current mirrors I've looked at. The mirrored currents are sourced by the PNP collectors, but I'm not sure what current is controlling the mirror. There's a Zener-controlled current going to the bases of the NPN transistor, and a second current going through resistors to the emitter-collector connection. It appears that both of these currents affect the output. Any suggestions on how this current mirror works?
AI: I think they are just composite NPN - (lateral) PNP.
In the day (ca. 1970), the only real PNP transistors they could make with that bipolar process would have been lateral PNP types with miserably low hFE (maybe 1/10 of the NPNs).
This contemporary patent has some similar structures eg. Fig 5.
|
H: Wiring colors for 'intermediate' connections
It seems to that red is the color usually used to connect to the positive terminal and black is usually used to connect to the negative terminal. However, I have seen numerous other colors as 'intermediate' connections. What is a good coloring scheme to use when wiring various (basic) components into a breadboard. For example, here is an example of a circuit I've done below with red meaning direct connection to [+] and blue/black meaning direct connection to [-].
AI: Over my many years of engineering, I developed a colour scheme for myself that I thought was obvious. However, arguments with other engineers over 'correct' colours have disabused me of the notion that anything is obvious (witness the international mess over mains wiring colours).
FWIW, I use red as +ve, and usually +5v. Black is usually 0v. If I'm running a +3v as well, that tends to get orange (for 3!) or yellow (a sort of unusual red!)
When I need a -5v, it's blue.
When I need +/- higher voltages, like 10, 12 or 15v for opamp rails, then it's orange and purple. I shouldn't have to spell out which of those is +ve and -ve, it should be 'obvious' (is it?). The 0v wire to weave with those from the PSU is green.
'Unusual' supply voltages, and signal voltages get white or yellow, grey or pink.
That's all if I have the colours to hand. If not, I use whatever.
Mains wiring
Here in europe, we have brown as hot, blue as neutral, green/yellow stripes as earth. In the US, they use white, black and green. Other continents use other colours. A report a decade or two ago came up with 'the one true colour scheme' which had to be (a) obvious (b) colour-blind friendly (c) low-light friendly (d) different to all the other existing colour schemes, which is white for hot, grey for neutral and black for ground. I think it's obvious and a great idea, but it's not going to catch on, not least because it would electrocute 90% of americans.
|
H: Choosing the right MOSFET for 5V power relays
I am planning following circuit
PR_SIG: a 5V logic level coming from a uC
+5V: is powered trough a external power supply (max 10A)
Power_relay: 5V, needs 70mA to switch
My problem now is choosing the 'right' MOSFET. I used the search on Mouser using the parameters that come to mind:
- Vgs = 5V
- Id > 250mA (3x70mA when all relays are switched on)
- Another parameter is the power loss, but can also be neglected
With my research I found two candidates
PMF250XNEX,
DMG1012TQ-7
Comparing the two both should work as switch but I am not sure that I consider all necessary parameters, since there are ~450 different MOSFETS which apply to my search.
Do I need to look at more characteristics or can I choose the MOSFET using only Vgs and Id?
AI: Since relays are a lot slower than any MOSFET, your only concern are the DC parameters.
The MOSFET must be able to handle a \$V_{GS}\$ > 5 V
The MOSFET must be able to handle a \$I_D\$ > 250 mA
You also need to be sure that the MOSFET switches on properly when you apply \$V_{GS}\$ = 5 V.
When the MOSFET is switched on it needs to have a low enough resistance \$R_{DS,on}\$ so that it will not drop much voltage. At only 250 mA a value of less than 0.5 ohm should be OK. Many MOSFETs will have a much lower \$R_{DS,on}\$.
It is also OK to use a hugely overspecified MOSFET. For example, if you have some MOSFETs lying around that are large and can handle 20 A but their size isn't an issue, you could use them, I mean, that would work fine.
The two MOSFETs you found look quite expensive to me. To keep things cheap (and for hobby projects, not professional use) I look on Ebay and find the cheapest MOSFETs which can do the job. A very common and cheap type is the AO3400 which costs around 10 Euro cents (including shipping) if you buy 20 from China.
|
H: Power output neccesary to power multiple USB devices at once?
If I have a device that requires 5V and 2A to function, what kind of power supply would I need to power 4 of these devices at the same time to their full capacity?
Here is an example setup:
Power bank-Portable Charger 30000mAh 2USB Ports (Output 5.0V 2.0A)
USB Hub-4-Port USB 3.0 Hub
Device-Electric Cloth Heater Pad Heating Elemen
=====================================================================
Based on this battery, im pretty sure the portable battery must be able to at-least give off a total of 8 amps..
ZeroLemon ToughJuice 30000mAh 7A Output Portable Charger
t
But im willing to bet it needs more than that to maintain that sort of power for over two hours. I wonder if a USB hub can handle 4 amps in a single cable...
AI: You will be connecting all devices in parallel, so the voltage requirement stays constant at 5V. To get the needed current just sum up all device input currents.
If you have 4 devices with 2A each you need 8A. The power is V*I, in this case 40W.
If you want to power the devices from battery for a longer time, you obviously also have to keep in mind the capacity of the battery. 8A for 2h requires the battery to have at least 8A*2h = 16Ah = 16,000 mAh.
When choosing the equipment for this make sure all devices can handle the current you need. Check this question for more detail on power handling of USB devices.
Is this answering you question completely? Otherwise feel free to specify more precisley what you want to know.
|
H: Single Supply AC Gain Opamp With Unity DC Gain
I'm trying to build a AC amplifier that will amplify an input signal of around 0.4Vptp @ 160kHz to around 1.7Vptp. I'm also attempting to do this with a single supply (3.3V) opamp. I built the circuit depicted below.
simulate this circuit – Schematic created using CircuitLab
After building and testing this circuit I ran into some very weird behavior.
The voltage at the non inverting is suppose to be around 1.65V but during measurement it was around 0.5V. My first assumption was possible the input bias current was causing this. After checking the datasheet the bias current is only 0.5pA and with a 2M resistor that should only cause a change of around 2uV.
Despite the weird bias I was able to get some scope measurements (note: yellow trace is at the non inverting pin of the opamp and the blue trace is at the output:
Over time the output started to clip and eventually the output became flat (over a periods of a few minutes)
Once the output had flat lined I noticed the current being pulled by the opamp was around 50mA!!
I was hoping for some direction on this issue. I have no idea where to start.
AI: A basic problem with attaching oscilloscope leads to high impedance nodes is that the input impedance of the o-scope circuits are loading those nodes. O-scopes typically have an input impedance of around 1 Mohm and, with an effective impedance of 2 Mohms || 2 Mohms your bias voltage of 1.65 volts is going to approximately halve to around 0.8 volts. If your o-scope probes have the option of a 10:1 operation, the effective impedance becomes more like 10 Mohms so, this would be the preference.
Another observation: you do need to have a supply bypass capacitor but your circuit doesn't show one. Data sheet extract: -
Connect low-ESR, 0.1-µF ceramic bypass capacitors between each supply
pin and ground, placed as close to the device as possible. A single
bypass capacitor from V+ to ground is adequate for single-supply
applications
Next....
Once the output had flat lined I noticed the current being pulled by
the opamp was around 50mA!!
This may be a fault on your circuit pulling down the 3.3 volt supply rail - please check to see at what point the output started to flat-line and confirm if it coincided with the draw of 50 mA or whether the 50 mA built up gradually over the period. As a side note, the TLV9062 is capable of outputting 50 mA into a short circuit so maybe there was an excessive load on the output?
You say you are using a TLV9062 but some of those parts come with a shutdown pin and it's not clear from your question whether you are using one of those parts and, if you are, how you are wiring the shutdown pin.
To ensure optimal shutdown behavior, the SHDN pins should be driven
with valid logic signals. A valid logic low is defined as a voltage
between V– and V– + 0.2 V. A valid logic high is defined as a voltage
between V– + 1.2 V and V+. The shutdown pin must either be connected
to a valid high or a low voltage or driven, and not left as an open
circuit. There is no internal pull-up to enable the amplifier.
And, both shutdown pins need to be properly controlled to avoid problems even when only using one amplifier in the package. The other amplifier (if unused) should have applied sensible levels to its inputs as a precaution.
I was hoping for some direction on this issue. I have no idea where to
start.
Some direction given!
|
H: What are the max. overmold dimensions for USB-C plugs?
In contrast to earlier USB specs, the USB-C specification doesn't seem to define a maximum width and height of the overmold part of the plug (according to figure 3.3, there is just a length defined).
1) What is the idea behind not even giving a recommendation for this?
2) If I am to design a device with several USB-C receptables, what spacing should I use (horizontally and vertically) to make sure, that 95% of common plugs will fit in next to each other?
And yes, I did notice, that there are dimensions given for screwable USB-C plugs in this sepc, but those are obviously not relevant for normal plugs.
AI: See figure 3-11 (page 57,58)
Page 58 has the width at 12.35 max and the thickness at 6.5 max
the locations of sections A-A and C-C are given by the drawings on page 57
figure 3-3 has this information too. (note. it also covers multiple pages)
|
H: Using motor encoder with VESC
I want to use this BLDC motor on a project. In its datasheet, the encoder pin sequence is shown as follows.
I want to control this motor with a VESC. There is an overview of the board here. In it, the encoder connector is described as below
Can someone please tell me if I can connect the motor encoders to this connector? (Presumably the ABI) I want to measure the distance traveled at the end of the day.
AI: From your pictures, I have an opininon that this VESC can connect different type of encoders:
Hall type (HALL)
Quadrature incremental with zero index (ABI)
SPI encoder (AS5047)
Your encoder has embedded two types: quadrature AB no index, and hall. AB are quadrature incremental encoder output, while U,V,W are hall encoder outputs. From your pic, the U,V,W are going to H1, H2, H3. To not be confused with U,V,W motor power output.
Hall encoders are absolute type each sensor is placed at 120 degrees electrical angle. Quadrature can be arbitrary resolution, so you should somewhere configure the pulses per revolution and look in the manual if it can work without index (zero) signal (IMO no).
You have to watch out the specified voltage range of the encoders.
|
H: What is the effect of switching noise?
I am going to build a lab bench power supply using smps and LTC3780 step up and down regulator. Like a youtuber great scott built in a video which i will link below. Can such a bench power supply be used in laptop and mobile repair ?
What will be the effect of switching noise if I use that kind of supply in repair work ?
Video link https://youtu.be/wI-KYRdmx-E
AI: What will be the effect of switching noise if I use that kind of supply in repair work ?
That depends on what you will use the supply for when doing repair work.
Can the supply be used as a replacement for an AC adapter?
Probably yes, as an AC adapter is also a switched mode supply and will also have some ripple on its output voltage. Most laptops run from 19 V from an AC adapter, this voltage isn't used directly. There are many switching regulators inside the laptop as well which convert the 19 V to 5 V, 3.3V and 1.8 V for example.
Can the supply be used instead of a battery?
Usually yes unless the circuit relies on the low noise / ripple free output voltage of a battery. It depends on the device if this is the case. Cases where the supply ripple might be an issue are audio amplifiers, radio receivers and sensitive measurement equipment.
Even then, when using a supply which outputs a ripple on its output voltage, the device might still just work only with audible noise or measurement equipment might not perform as accurately as it should.
|
H: Will this 'radio receiver' work?
I am a complete beginner with radio and electronics in general.
I did some research on antenna designs and found that, with dipole antennas, at least from my understanding, the second antenna will act as a ground, balancing them. I have built the following circuit and it does not produce static, as I thought it would.
I would like to know:
If this circuit should at least produce static through the speakers in theory.
If not, how can I use a dipole antennae properly to do so.
Apologies if this question is really dumb, and the answer is obvious, but I've tried googling everything related to this and it seems like it should work.
AI: That circuit unfortunately wouldn't work even if the neighbour were an AM radio station. There's no closed circuit. The rectified AC voltage would charge the capacitance between the antenna branches and then the current would stop. The diode prevents discharging.
A working version of the minimum receiver has the diode in parallel with the speaker. The antenna is at the ends of the diode. It would work well if the speaker was sensitive enough having much more turns in the coil than ordinary speakers today have. Magnetic or piezoelectric headphones would be a good replacement for a speaker.
The next link shows how those sensitive old style magnetic headphones look: http://www.crystal-radio.eu/en2000ohm.htm There's also application guidance. Typically those headphones had so much wire in their coils that their DC resistance was 2000 Ohm.
FM reception would be impossible without parts which convert frequency to voltage. I guess the antenna is not selective enough, but if there's a strong AM station in the town something can well be heard with the fixed version. The antenna wires should be several meters long for substantial listening wolume even with headphones, but that depends on the distance and the power of the transmitter. A few watt strong 27 MHz CB transmitter with AM mode ON can generate within few meters so strong field that 50 cm long antenna wires catch a hearable signal.
Not asked:
The normal crystal receiver has a tunable resonance circuit because it makes possible to select between more than one transmitters. In addition it's resonance makes the voltage higher if there's not too much load. That reduces the attenuation caused by the voltage drop in the diode. If you try a crystal receiver, get a germanium diode. Its voltage drop is much lower than the drop in silicon diodes.
|
H: Fail safe led circuit
My question is about how to turn on a led before the GPIO pin has been initiated. When the IC has just been powered on and has not yet been initiated the GPIO pins remains in state HI-Z.
In Circuit-1 GPIO-P1 is pulled via a pull up resistor R7 HIGH and the led turns on before initiating the IC. once the IC has been initiated, GPIO -P1 is set to LOW to turn off the led. The problem with this circuit is that if GPIO-P1 is accidentally set to HIGH, it will short the led. (not current limiting resistor)
I believe I can solve this problem via circuit 2 and 3.
In circuit 2 I limmit the current through R2 in case of P1 is accedentially set to HIGH.
In circuit 3 The P mosfet is driven on via R5, lighting up the led, only when GPIO-P1 is set to high does the led turn off.
Out of interest would there be ANY reason at all to choose for circuit 3 over circuit 2? I see none.
simulate this circuit – Schematic created using CircuitLab
AI: Out of interest would there be ANY reason at all to choose for circuit
3 over circuit 2? I see none.
I would also consider circuit 4: -
Circuit 4 allows P1 to clamp the LED anode to about 0.7 volts. This, in effect will turn the LED off; in other words, P1 cannot drive "extra" current into the LED because of the direction of the diode; it can only shunt current away from the LED. This means that R7 can be chosen the deliver the maximum design current into the LED and P1 can only shunt this away.
In circuit 2, for instance, you can't deliver full design current into the LED except when IC is driving it.
So, I would consider circuit 4 because with circuit 2 you can never get full brightness of the LED unless P1 is activated. Also circuit 3 is slightly more costly than circuit 4. However, circuit 3 is more energy efficient if wishing to turn the LED off because R5 can be much higher in value than either R1, R2 or R7.
A less obvious advantage for circuit 3 is that pin P1 of the IC that drives the LED doesn't have to handle currents that are used to activate the LED. The current may be as high as 20 mA (for a standard red LED or even higher in some cases) and that current has to flow into P1 when deactivating the LED; this eats into the maximum current budget that can be handled by the IC (normally specified in the data sheets as the current that can be passed through the ground pin of the device).
So, there are other considerations.
Fail safe led circuit
I don't think any of the above can be called Fail Safe.
|
H: Why can't the two plates of a capacitor have unequal charge?
This is what was given in my textbook:
"The net charge on every component in the system is always zero. Thus no component can collect a net excess of charge, although some components can hold equal but opposite separated charges."
I can't quite understand why this is so. For example, what would happen if the two plates of a capacitor had unequal opposite charges?
AI: The assumption of charge neutrality is based on our notion of common electric circuits, where charge is free to move under the influence of electric fields.
If the amount of positive charge on one capacitor plate did not equal the amount of negative charge on the other plate then the electric field between the plates would attract and draw charge as needed to establish neutrality.
|
H: What is the name or number of this chip on this PCB?
We are trying to repair this pcb. We found that someone deleted the number or name of this chip which needs to be replaced anyone has idea ?
AI: Does it have two leads to the PCB plus the thermal lug - so 3 terminals total? If so, it could be a transistor of some sort.
EDIT 1:
OK, 5 (maybe 6?) terminals total. Thanks, I had trouble seeing that in the pictures. So it's probably not a transistor.
|
H: Is the BLDC motor and "brushless AC servomotor" same thing?
Are the terms BLDC motor and "brushless AC servomotor" exactly the same thing? I cannot find information so far whether there is difference.
For instance, at this site they are treated as if the same thing.
And at the moment I have this brushless motor where it is called at page 8 as "AC servo motor". And here is the photos of its stator and rotor:
stator:
rotor:
It seems very similar to BLDC but they don't call it as such and they call it brushless AC servo motor. What are the differences if there is any?
AI: BLDCs are a type of PM (permanent magnet) synchronous motors, as are AC synchronous motors without brushes. They are very, very, similar.
A BLDC always has armature magnets and not armature windings.
AC synchronous motors come in two types:
Permanent magnets on the rotor and windings on the stator. These requires no brushes because the permanent magnets are always placed on the rotor. If you placed the permanent magnets on the stator and windings on the rotor you would need brushes so no one does that.
Field windings replace the permanent magnets. This is where my
understanding gets a bit murky. In theory, you could have either the
field winding or the main winding on the stator with the other on the
rotor. But I think the field winding always goes on the rotor just
like the permanent magnets do.
The reason is because the field windings are DC so already needs
brushes anyways to rectify AC to DC. By sticking the field winding on
the rotor you can design the brushes to do double-duty to also
transmit current to the spinning rotor. If you placed the field
windings on the stator and the main winding on the rotor then you
would now need two sets of brushes: one to commutate AC to DC for the
field winding and another set of brushes to transmit power to the
main winding spinning on the rotor. Since the field winding just
replaces the permanent magnets, the brushes for the field winding are
lower power than those you would find in other motors that transmit
"load current" to the main winding.
You asked specifically about brushless though so if it is brushless, it must have magnets instead armature windings. That makes the two almost identical, only really differing in ratings and intended usage. (someone correct me if I'm wrong).
For comparison, a brushed DC motor has armature windings, while the field could be either a magnet or winding, which is the opposite which is interesting.
A BLDC motor actually does run on a form of AC when you think about it. It is just not expected to be run straight from an AC source, but a DC source with an inverter in between to synthesize the AC (though you could argue that's just a VFD and VFDs are used with AC motors all the time).
The BLDC motor is more likely to have a trapezoidal BEMF while the AC brushless motor is more likely to have a sinusoidal BEMF to optimize operation with the expected drive waveforms, but not always.
The "servo" part means it's not just a motor. It probably has a rotor position sensor like an encoder included and potentially an entire electronic motor driver and control package.
|
H: What does it mean when my car inverter's labeling says "Neutral Floating"? Is it safe to use?
I recently came across a fairly old car inverter. It plugs into a 12v DC lighter socket, and outputs 115 Vac at 60 Hz, and has a normal American outlet on the side of it (with a ground pin, which seems questionable) and is rated for 100W continuous or 150W peak. This tool could be very useful to me, mainly to run a laptop in the car (when someone else is driving, of course). The output is also non-sinusoidal, but, as far as I understand, that does not matter for switch mode power supplies, and all of the things that I want to power in the car run on switch mode power supplies.
More concerningly, however, it says "NEUTRAL FLOATING." My understanding of electrical engineering tells me that this means that the neutral wire isn't attached to anything, except maybe a capacitor. This may still work with lower current loads (perhaps up to 100W based on the labeling, although that seems high) due to the capacitance on the neutral wire, but it would also cause the device that I am using to start floating above 0V on its neutral or ground side, a dangerous situation. It seems, however, that such a dangerous device would not be legal to sell in the United States, where I am pretty sure that this was bought (it is very old, so I don't know exactly).
As such, it seems like I am misunderstanding what "NEUTRAL FLOATING" means. What does it actually mean? Is it safe to use my inverter, or do I need to buy another one?
AI: Many smaller inverters (and my Honda EU2000 generator) have "floating neutrals".
I don't think "floating" is quite the right term here, as the Neutral will alternately be connected to +170 V or Ground. What happens is that the inverter has a single high voltage supply (+170 V or so), and connects the Line to that supply on one half-cycle, and Neutral to that supply on the other half-cycle of the AC waveform.
A load connected between Line and Neutral will see an alternating voltage. As neither Line or Neutral should be exposed to the user, this technique should not present a hazard, and simplifies the inverter as it does not need to generate a negative high voltage supply.
|
H: Issues encountered when adding DC adapter to light powered by battery
I have an LED light bar powered by 6 AA batteries connected in series. There is also an RF control unit to turn the light on/off. The light eats through the batteries rather quickly thus I decided to experiment attaching a DC adapter to the light.
Due to the way the batteries are connected, I assumed that the light takes 9V to power, so to test my theory I grabbed a used 9V cell and it worked perfectly - even responds to the RF remote.
I then grabbed several DC adapters that output 9V with varying current and then began testing to see which one will work. First I tried one with 200mA and the lights turn on, but does not respond to the remote control. I then upped the current to 300mA with same result, but when I tried it with 600mA I heard the dreadful "buzz" as soon as I connected the circuit (did not even turn on the switch). As expected, I fried the RF receiver module in the light - yet the light still works. Even going back to the 9V battery does not allow the light to be operated by the RF remote.
So now I'm trying to find out what happened. If I have the voltage correct, shouldn't the worse that could happen is under-powering the light due to insufficient current? I'm assuming that even if I increased the current, there shouldn't be a problem since the resistors in the circuit should be limiting the current so there shouldn't be a risk of burning something out. The only thing I can come up with is that the adapter was putting out way more than 9V to fry the RF receiver circuit. Are there other possibilities about what could've went wrong?
AI: Possible reasons:
Many of those adapters are unregulated which means it might give 9 V when fully loaded but a higher voltage when lightly loaded.
You used an AC adapter. The alternating current alternates the polarity and that or may have caused damage.
You used one with polarity reversed. Check the label. Normally they are centre-pin positive.
You accidentally reversed the polarity.
|
H: voltage potential energy clearance?
1 coulomb = charge of 6241509074460762607.776 protons(fixed amount of energy passing thru certain point).
So 1 volt = 1 joule(energy) per 1 coulomb.
and 2 volts = 2 joule(energy) per 1 coulomb.
does that mean that more volts literally mean packing more energy(joule) per fixed amount of charge(coulomb)? and how?
AI: does that mean that more volts literally mean packing more energy(joule) per fixed amount of charge(coulomb)? and how?
Yes absolutely..
It's simple, think of a lake containing 10 million tons of water, located in a mountain 500 meters above sea level, it has some amount of potential energy associated with it. Now imagine the same lake, containing 10 M tons water, now located 1km above sea level, it has twice the amount of potential energy associated with it.
This analogy is very fitting, as voltage (or energy per unit charge) is often associated with distances, just like in the example with the lake and the distance located above sea level.
To give an example; let's say that you had a capacitor consisting of two 1m*1m metal plates separated by 1mm, between them would be some capacitance C, let's say you charged up C to 1v, and then separated the plates such that the new distance was 2mm, the capacitance would fall to 1/4 the original, and the charge would drop to 1/2, but the voltage would also double just like the potential energy contained in a lake above sea level as the function of its height, the net result of the capacity (and charge) decreasing and the voltage increasing is that the energy is conserved.
Another example; the unit of energy called an "electron volt" (eV) is the energy of one electron of charge, moved by one volt of potential. The way you can think of this is as if the charge was a mass in kg, and the voltage is a hight at which we drop the mass from in meters. In this example the potential energy on the mass from gravity is converted to kinetic energy, and eventually to heat, the same is (kind of) true in the example with voltage and current, the voltage is converted to motion of electrons (current) and eventually to heat (I'm simplifying things, I know there are more complex examples).
Positive and negative charges attract each other, and the larger the charges, and the further the distance, the more potential energy the two charges have. Masses attract each other, and the larger the masses, and the further the distance, the more potential energy the two masses have.
So you can think of charges like masses, and voltages as heights.
|
H: Voltage divider getting HOT!
The scenario went like this: I used to run this amp with +-20v DC, and it seemed to work fine except for a little bit of clipping. Now, I changed my mind to make it work with bigger voltage!!!
I've increased the voltage to +-50V DC. After powering the amp, a nice smell got out but it was working!!!
Now I didn't cut the power to experiment what actually happens. Funny isn't it?
Anyway, so, after playing the amp for around a minute things got out of control with a little burst!!
Therefore I've opened the amp and found that the supply decoupling caps got fired along with a few small transistors and one Output transistor from one channel only!!! My amp got 4 channels!!!
So, I took out the burned parts and replaced only the decoupling caps. Problem arised when I powered it after replacing the damaged parts. The divider resistors in the diagram (marked red) are getting hot and funny smell is coming again. Now, these resistors have been changed when I opened the amp because they got burned out too.
I've checked all the components connected to that divider network but failed to find any visual fault.
What can cause the resistors getting too hot?
Any suggestions would be greatly appreciated!!!
I've just replaced the 220 Ohms resistors with 1.2k.
Still it's getting hot!!!
I think 1.2k won't hurt according to the formula!!
AI: The resistors circled in your schematic are not potential dividers, they are potential dropping or voltage dropping resistors feeding voltage regulators formed by ZD1 and ZD2. If you feed them with 20 V and the Zeners are 15 V devices then the current through them will be given by \$ I = \frac {20 - 15}{220} = 22 \ \text {mA} \$ and the maximum power dissipated in them will be given by \$ P = VI = 5 \times 0.022 = 110 \ \text {mW} \$.
You have applied 50 V so \$ I = \frac {50 - 15}{220} = 160 \ \text {mA} \$ and the maximum power dissipated in them will be given by \$ P = VI = 35 \times 0.160 = 5.6 \ \text {W} \$. Note the drawing says the resistors are rated for 1 W.
Therefore I've opened the amp and found that the supply decoupling caps got fired ...
This shouldn't have come as a surprise to you. The engineer that designed the amplifier would have chosen components rated for a 20 V supply, not a 50 V supply.
What can cause the resistors getting too hot?
We've covered that.
Any suggestions would be greatly appreciated!
Yes. Gain understanding first. Learn how the circuits work, why the designer chose certain components, do the calculations and then test your modifications.
|
H: Can failure of an old electrolytic capacitor be precipitated by motion?
I have report of a circuit board failure. The board was 20+ years old, and included at least one electrolytic capacitor. That capacitor would then be beyond its expected useful life. The board was still functioning, until the moment someone moved the panel the board was attached to. At that point, the board immediately failed, and has not recovered. Component-level failure analysis is not possible in this situation.
It occurs to me that the electrolyte in the capacitors is liquid, and thus subject to motion, perhaps moreso if the capacitor is partly dried out. So if an electrolytic capacitor is old enough that it should be considered end-of-life, but is still functioning and energized, is it possible that moving the energized capacitor would precipitate its failure?
AI: It's very unlikely that an electrolytic capacitor failed as result of being moved. They generally either degrade gradually as the electrolyte dries out (greatly accelerated by high temperature) or they fail catastrophically and vent or bulge, particularly when abused or badly made.
Much more likely is some kind of mechanical damage, a solder joint or PCB cracking or a connection becoming dislodged. If it's an unpackaged board, the possibilities of mechanical damage are much greater- from missing configuration jumpers to surface-mount components completely sheared off.
And if the board is not enclosed (sometimes even if it is) there is the possibility of ESD damage, especially at this time of year in the Northern Hemisphere when humidity levels tend to be low.
|
H: Linear Voltage regulator becomes very Hot when connected to Nichrome wire heater
I'm trying to build a heating element using nichrome wire 24 gauge. I'm using a 12v linear voltage regulator (L7812CV) as a power adopter which is connected to a rectifier bridge providing 19v. When the power adopter is used with other devices it works well (e.g 12v Fans and Arduino) .But as i have tried to connect the Nichrome wire about 40cm length the heat sink connected to the regulator becomes very hot just in about a minute.Can someone tell me what's going wrong and how to solve it?
AI: Nichrome wire acts as a resistive load. Check the resistance of yours and I'm sure you'll find that it's pretty small--under 100 ohms, maybe even 1 ohm. At 12 ohms your circuit draws 1A through the linear regulator. You say that you're inputting 17v and outputting 12v in that regulator--this 5V drop means that it's losing 5W per amp in heat. With a small resistance load, that means that the linear regulator is releasing a lot of heat.
|
H: Turn on LED with the opening of a switch
Hi I’d like a circuit that turns on an LED when the switch opens. I’ve made this so far, but it’s drawing too much current in the “off mode” that it will just drain my battery. Any ideas? I’ve seen one with a mosfet, but I don’t have one of those, just transistors or a 555?
AI: Something like this should work. Tune R1 for LED on current. Tune R2 for your specific transistor, to get it into saturation.
Update: To save battery current, you can run the transistor not in saturation mode (make R2 even bigger). But you won't get good repeatability from circuit to circuit as the gain of transistors can vary wildly from part to part.
simulate this circuit – Schematic created using CircuitLab
|
H: SRAM: Purpose of Upper and Lower Byte Enable when Data Bus is greater than 8-bits?
What is the point of an upper byte and lower byte enable on an SRAM if the data bus is already 16-bits?
Is it related to multiplexing? I had always assumed that SRAMs were either designed with either multiplexed or dedicated data and address buses. Maybe I am wrong and they all have dedicated buses and how you wire them up determines whether or not they are multiplexed. Seems like you wouldn't save many pins or traces that way though. It would only save them MCU pins, but would not affect the SRAM pins or the number of traces running between the two ICs.
I am looking an an MRAM Datasheet, though this seems broadly applicable to all SRAMs:
https://www.everspin.com/family/mr5a16a?npath=258
AI: Upper and lower byte enables are needed when a processor with a 16-bit bus wants to do an açcess to modify just one half of the data RAM location.
For byte write operations this greatly increases performance over having to do a 16-bit read-modify-write operation.
|
H: Grounding for a battery
Given a 9V battery as the only voltage source for a circuit, where is the Ground usually put? Is there a standard, for example, that puts the negative terminal at 0 and the positive at +9?
Does it every vary, or are there usually hard-set rules for grounding?
AI: Is there a standard?
Not a universal one writing... but by popularity positive grounds and coaxial center negative have faded fast in the last 2 decades, due to user error.
Since ground only means a 0V reference, it does not matter locally, only when interacing. Zero Volts, 0V is always floating , unless connected to anything that is connected grid earth or PE, Protective Earth Gnd, which, coincidentally is also floating ;) in the Universe, but locally , it is our biggest common-mode 0V reference. Yet with trans-atlantic cables there can be a thousand volts difference. E.g. VGA monitors are PE grounded.
Like all voltage sources, even 0V has impedance so ground shift voltage depends in V(f)=I(f)*Z(f). It has both differential mode impedance (Zdm) and common mode (Zcm). So floating means at DC high R but at some frequency that Z can drop low by conductor length/gap ratio due to inherent capacitance.
Grid tower grounds need to be < 100 Ohms for lightning protection. But grid to home much less, so it is distributed.
|
H: Can I use any digital component to avoid 7805 killing my microcontroller directly when its ground is disconnected (hence 12v goes to uC directly)?
My idea is to have an intermediate cheap component that will blow-up before my uC if the voltage goes above 5v from 7805 (or even some fuse, but I am not sure if a fuse will blow-up with over-voltage like it blows-up with over-current). I have lost many Atmega328P chips because of 7805 accidentally supplying 12v to them. I have heard of SCR - is it the component I am looking for which blows-up with voltage?
Also, please tell me how to use a zener with 7805.
AI: One simple way is to use a fuse in conjunction with a relatively big Zener diode.
This is not perfect but cheap and simple and will work in most cases.
If the input voltage goes above the Zener voltage (in this case 5.4V) the Zener will short and the fuse will blow.
For the fuse, you can use a PTC resettable fuse
simulate this circuit – Schematic created using CircuitLab
A better way is to use a voltage clamp circuit instead of a zener.
Another solution is to use an LDO (or other types of regulator) for the input but you need to consider the voltage drop caused.
Overvoltage protection circuit (note this circuit will cause a voltage drop, LDO IMO is simpler):
https://circuitdigest.com/electronic-circuits/overvoltage-protection-circuit
|
H: How come distilled water hasn't replaced transformer oil as a high voltage insulator?
I've noticed that distilled water has an extremely high dielectric straight (65-70 MV/m).
With that considered, how come it isn't used in the place of transformer oil for 10-20 kV applications where the structure might not need to last multiple decades?
AI: Water is good for making coffee but in electronics over time, it causes rust , and rust particles cause static discharges which drastically reduces the breakdown voltage of the insulation.
Even epoxy coated stainless steel will eventually emit particles of rust thru, enough to degrade water to cause partial discharge which detonates particles and evaporates the medium. This can cascade like lightning to cause a massive explosion. The same can occur in oil, but unlike water TFMR oil has a Dk of only 2 and can support anti-oxidizer additives.
Worse yet, the polar properties of water make capacitance and lower the self resonant frequency. H20 has a dielectric constant Dk= 80 or about 20x worse than oil lowering the impedance of coils and self-resonant frequency.
Oil also has the same pristine breakdown voltage or BDV of 75k/mm or 75MV/m but that comes at cost of post-processed pressured , high temp moisture and particle removal to the parts per billion level so regulator transformer oil only must pass 25kV/mm and higher levels require in plant contamination removal equipment worth >$50k.
So if you don't mind the creapage effects in contaminated water, reducing the breakdown voltage from 70kV/mm to 1kV/mm go ahead and use it.
|
H: How do I learn OpAmp? Do I need to know the internal circuit? Are they works the same way?
I am learning OpAmp. I watched many good teachings on YouTube. Some teachings mentioned the OpAmp type/series he/she is explaining, such as uA741. But many other teachings without any explanation, like is in here and here. The fact, there are many type OpAmp mentioned, even in the CircuitLab which the software attached to this SE, there are many type of OpAmp included. Then my question:
How do I learn OpAmp? Can I just think that they work the same way?
Do I need to know the internal circuit and treat that every type is working specific based on their own design?
AI: There's an excellent free Ebook about opamps, it is called "Opamps for everyone" and you can download it here.
An Opamp is a device/circuit that tries to behave in a certain way, it amplifies the voltage difference between its inputs. Ideally an opamp would have an infinite gain and also work for every frequency and have no flaws at all. Obviously that's not possible so every opamp is a compromise.
The 741 was one of the first "usable" opamp ICs (a chip) and in its days (1970s to 1980s) it was very popular. We have moved on since then and have better technology to make better opamps. A limitation of the 741 is for example that it is useless at the low supply voltages we use today like 5 V. A 741 really needs a symmetric +/- 15 V supply.
I would not worry too much about what opamp is used as an example in teaching presentations. Usually the opamp is assumed to be "ideal" or at least good enough for the function that is explained.
The circuitry around the opamp, like a feedback network, stays the same whatever opamp you use. Focus on those circuits and how the opamp is used, not the model of the opamp as it doesn't matter much.
The internal circuit of an opamp is really only of interest to IC designers and experienced designers that can determine the limits of what the opamp can do depending on the circuit that is used. For example, I am an IC designer so I can look at the schematic of an opamp and see how close it can drive its output voltage to the supply rails from the architecture of the output stage (if that's an common emitter, common collector, common source etc...). However, in the specifications of the opamp there will also be information about this under "Output voltage range". So you don't need to know or look at the circuit, the information is also in a table. And that information is leading as it is guaranteed behavior. I could have missed something in the schematic or not understand it properly or the schematic could be incomplete (some manufacturers do that to protect their design).
|
H: Query on writing KVL equation for an RL circuit
Say I have the inductor in the following circuit with certain current already in the circuit flowing in the specified direction:
simulate this circuit – Schematic created using CircuitLab
If I had to write KVL for this circuit, would I write $$ L{di}/{dt}-iR=0$$ $$or$$ $$ L{di}/{dt}+iR=0$$
I always get super confused about what sign to take for the induction. I know by intuition(by the fact that the solution for the differential equation of (1) would be an exponent with positive power which is absurd) that it should be (2), but I always get stuck on that sign.
AI: I'd use KCL (not KVL) with the bottom node "grounded" and the top node as \$V\$:
$$\frac{V}{R}+\frac1{L}\int V\:\text{d}t=0\:\text{A}$$
Taking the derivative with respect to time:
$$\begin{align*}\frac1{R}\frac{\text{d}V}{\text{d}t}+\frac{V}{L}&=0\:\frac{\text{A}}{\text{s}}\\\\\frac{\text{d}V}{\text{d}t}+\frac{R}{L}\,V&=0\:\frac{\text{A}}{\text{s}}\end{align*}$$
The integrating factor is then \$\mu=e^{^{\int\frac{R}{L}\text{d}t}}=e^{^{\frac{R}{L}t}}\$ and the solution is:
$$\begin{align*}V&=e^{^{-\frac{R}{L}t}}\int0\:\text{A}\cdot e^{^{\frac{R}{L}t}}\:\text{d}t\\\\&=e^{^{-\frac{R}{L}t}}\left(0\:\text{V}+C_0\right)\\\\&=C_0\cdot e^{^{-\frac{R}{L}t}}\end{align*}$$
From initial conditions, we know that \$C_0=I_0\cdot R\$. So:
$$\begin{align*}V_t&=I_0\cdot R\cdot e^{^{-\frac{R}{L}t}}\end{align*}$$
Here, \$I_0\$ is the initial current in the specified direction.
Using your approach, let's start with a schematic:
simulate this circuit – Schematic created using CircuitLab
The above schematic will help you keep the signs straight, I think. Note that \$V=L\,\frac{\text{d} I}{\text{d} t}\$ where \$V\$ here is the difference between the top of \$L\$ and the bottom of \$L\$ when the current in \$L\$ is positive when it is down in direction. So in the above case, we find that \$\left(0\:\text{V}-V\right)=L\,\frac{\text{d} I}{\text{d} t}\$ or \$V=-L\,\frac{\text{d} I}{\text{d} t}\$. So,
$$\begin{align*}
0\:\text{V}-L\,\frac{\text{d} I}{\text{d} t} - I\cdot R&=0\:\text{V}\\\\
\frac{\text{d} I}{\text{d} t} + I\cdot \frac{R}{L}&=0\:\text{V}
\end{align*}$$
The integrating factor (again) is then \$\mu=e^{^{\int\frac{R}{L}\text{d}t}}=e^{^{\frac{R}{L}t}}\$ and the solution is:
$$\begin{align*}I&=e^{^{-\frac{R}{L}t}}\int0\:\text{V}\cdot e^{^{\frac{R}{L}t}}\:\text{d}t\\\\&=e^{^{-\frac{R}{L}t}}\left(0\:\text{A}+C_0\right)\\\\&=C_0\cdot e^{^{-\frac{R}{L}t}}\end{align*}$$
From initial conditions, we know that \$C_0=I_0\$. So:
$$\begin{align*}I_t&=I_0\cdot e^{^{-\frac{R}{L}t}}\end{align*}$$
Here again, \$I_0\$ is the initial current in the specified direction, with positive values being down.
I don't like this approach quite so much, in this case, because you have to "think more" about the signs. It works fine. It's just slightly more "twisting of the mind" to make sure everything is fine. The KCL approach is much easier to keep straight because all you do is make sure that you put the outgoing currents on the left and the incoming currents on the right (or visa versa.) And at least to me, that's easier to keep track of.
|
H: Wireless power transfer choosing of mosfets
I'm an ee student and for the group project we have chosen to make a Wireless power transfer. Our goal is to charge a laptop. As for the result we have to demonstrate charging a laptop, so considering modern laptops I think we need at least 60-80 watts of power to be transferred to the receiving coil.
So for the first step we need to create the driver for the primary coil which will creates an alternating current using DC voltage, for the purpose we intend to use a H-bridge and a mosfet driver and an arduino.
The purpose of using an arduino is to be able to vary the switching frequency.
Also we intend to use resonant mode instead of inductive mode because the resonant mode can transfer energy more than inductive.
As our syllabus doesn't cover any of this wireless power transfer, we are gathering information and learning of it.
Is the use of h-bridge is suitable for this purpose?
Input voltage of transferring is 12volts, on which factors we should choose the mosfets on the current rating.
We are also in a tight situation as we only have about 2 months to complete this.
Note:- the distance between two coils can be just few centimetres and the receiving coil won't be built to the laptop. For the demonstration purpose we will just use the secondary coil and laptop apart from each other.
Any help would be appreciated.
Regards
AI: Any help would be appreciated
The most significant recommendation I can give is to simulate the set up. This starts fairly easily with a transmit coil driven by a sinusoidal voltage source coupled to a receive coil that is resonant tuned. The coupling is dictated by the dimensions of the coils and this, in turn can be calculated by first understanding the top picture below (the transmitter): -
Previous answer where the diagram came from.
The receive coil can be assumed to be at point P and therefore it will collect a total flux of approximately B x Area and, knowing frequency of the field you can calculate induced voltage.
So develop some math that gives you induced voltage against the diameters of the coils. What you will always find is that bigger diameter coils makes the problem of distance easier.
Develop your coil size this way and then work out what the coupling factor is at the maximum distance you want.
Then go to a proper circuit simulator (I use micro-cap) and model the circuit using the coupling factor k. You will also need to add a bridge rectifier to the receiver along with smoothing capacitor and simulated load (representing the power consumed).
You can then see how much drive voltage is needed on your transmit coil to get the power output. You will also need to model coil losses (series resistance) for both transmit and receive ends. These will act against you.
At that point you will probably want to figure out how you will resonate your transmit coil and, if you are using a H bridge then series resonance is a likely candidate. Any driving circuit that produces "hard" voltages will need to be loaded by a series resonant circuit. A class A stage would be parallel resonant.
Practical advice: you might consider using ferrite plates behind the Rx coil to reduce the field travelling further and upsetting your laptop. And you will definitely not want your receive coil brought into close contact with your transmit coil because the total induced voltage might by hundreds of volts and blow up your charger circuits.
|
H: Metal enclosure grounding for audio project
Fellow engineers,
For various reasons I picked extruded aluminium enclosure for my current audio related project. All exposed metal connector chassis are connected to earth, which then goes to ground trough ceramic cap (100pF 10% 2kV 1206) for ESD protection.
Schematic:
Render:
My gut feeling says there might be artifacts (like spikes) and noise if a consumer touches the enclosure, which will result in audio distortion. How can I make sure my audio ground is totally protected against this scenario?
Note: I do not have a separate audio ground. All I did - carefully splitted ground plane to divide analog and digital parts, no components in between, directly connected to each other close to the LDO.
AI: How can I make sure my audio ground is totally protected against this
scenario?
The short answer is "with difficulty".
For instance, a static discharge (ESD) onto the enclosure will create a current pulse through to ground but your chassis is connected to real ground via an earth wire that has inductance. Inductance can be generally approximated as 1 uH/metre and so a fast discharging current can produce an instantaneous surge at the chassis of hundreds of volts with respect to real ground.
Now you might argue that your box provides equipotential protection to the electronics inside. Are you going to argue that?
Unfortunately, you have cable coming from the box and going somewhere else so, that hundred odd volt surge is going to try and pass currents down those cables via cable screen capacitance to real ground and, that current is also going to flow through your circuits.
But it could be worse than this if the far end of the cable screen is grounded (as often happens in audio). It's also made somewhat worse if the far end equipment's input circuit is "single-ended" and and hence is "unbalanced". This is why top audio gear always uses balanced circuits and cables.
So, the way to deal with this is to use balanced audio connectivity; in effect both conductors pass equal amounts of the surge current and, theoretically you have minimized the "audible" problem but, the current still flows through your circuits and this means you might need some form of protection like a TVS (or common mode chokes that will attempt to shut-down the path into the cables). Keeping the problem local is the key to not upsetting up-stream or down-stream circuits.
Not unheard of is lightning (as opposed to ESD) and this can raise local ground by several thousands of volts but how does other equipment survive? An example is ethernet - they use isolating magnetics on incoming and outgoing wires and some of these are rated to withstand up to 6 kV pules. The fact that balanced data systems are used is also more than just the icing on the cake.
A full schematic would be required to give more ideas.
|
H: When to factor duty cycle into power consumption calculations?
This is probably a simple question for most of you.
I am calculating the power consumption of a device. The load is driven using a square PWM signal. At what point do I factor in the duty cycle? For now let's assume a 50% duty cycle with a starting voltage of 10 V, resistance of 1 ohm (for simple maths).
Currently the calcs look like.
Effective Volts = 10 * 0.5 = 5 V.
Watts = 52 / 1 = 25 watts.
Instead should I be doing:
Watts = 102/ 1 = 100 watts
Effective watts = 100 * 0.5 = 50 watts
AI: In your first calculation you are computing the average voltage when in fact this should be the RMS voltage. In your second calculation you are correctly calculating the power (one half cycle) and dividing by two to correctly get average power.
So, taking the average power of 50 watts and multiplying by 1 ohm then taking the square root delivers RMS voltage because: -
$$V = \sqrt{P.R} = 7.071 V$$
But you can calculate RMS voltage by squaring each voltage, taking the mean then square rooting: -
$$\sqrt{\dfrac{10^2 + 0^2}{2}} = 7.071V$$
If the duty cycle were 25% then it would be: -
$$\sqrt{\dfrac{10^2 + 0^2 + 0^2 + 0^2}{4}} = 5V$$
Notice how I've split the latter solution into 4 distinct time slots.
|
H: Flyback Transformer Connection & Construction
I don't understand how the coils are wound in a flyback transformer.
Can someone post an image or explain.
Basically I can understand all the calculation needed to design the flyback transformer. However, when it comes to winding structure, it is really confusing.
Based on my view, different transformer maker sometime uses different transformer winding structure. Example such as separate winding & stacked winding. I totally dont understand the reason behind that.
Does anyone know what is the better way to understand the transformer winding connection & construction for below image?
This transformer is almost same as Flyback Transformer Design for the IRIS40xx Series
You can refer Fig.3
AI: To get transformer action, both the primary winding and the secondary winding must simply link the core.
The shape and relative disposition of those two windings are then arranged to meet other requirements. Flybacks can be used in a range of different circumstances, so you may be confused by seeing different arrangements for different purposes. The final design is always a compromise between several mutually exclusive aims.
If you want tight coupling, so low leakage inductance in the transformer, then you might want to interleave the windings. This reduces the kickback that the primary must handle.
If you want high voltage isolation between the two windings, then you might want to put one at one end, and the other at the other, at the expense of leakage inductance.
If you want the highest frequency of operation, then you would segment the windings, making each of several thin 'pancakes', rather than several layers. This reduces the self capacitance of each winding, thus raising the self resonant frequency (SRF).
Segmenting the high voltage winding is also good for improving the withstanding voltage across the winding. Segmentation of course reduces the percentage of copper filling the winding window, increasing copper losses.
If you do wind in layers, then winding each layer in the same direction, with a quick return between layers, gives better insulation performance and better SRF than winding to and fro on alternate layers.
|
H: How can I fix EMC problem over my PCB?
Here is my PCB drawing that couldn't fix EMC problem. There are peaks at 120-130 MHz at radiated measurement. Are there anyone who have any opinion to fix this?
Note: My circuit behave as led driver which drives at 500 kHz and output current is nearly 210 mA. R14 and R13 set output currents, these two resistors have 100 mA voltage drop and we can set output current by changing these resistors.
At input stage I have EMC filter but there is still something I couldn't suppress.
AI: Because this is a switching device, care must be taken in proper decoupling of the input.
You have a wrong order of your input caps C15 and C13. You must connect C15 (100nF) as close as possible between Vin and Gnd inputs of the switcher.
Just try it out, you will see much difference in EMI measurement.
BTW: In my experience, this is one of the top 3 reasons/causes why designs fail at EMI tests.
|
H: Issues with this USB ESD protection circuit?
I've used this ESD circuit on four boards for hobby projects. I have had no issues on two boards. On two others, the TVS diode for VBUS conducted when the board was plugged into a USB port, causing an overcurrent condition and destroying the ESD suppressor (the host is grounding the USB shield). The two working boards have probably just not seen a transient on connection.
Is there a better design for this? Should there be a resistor between the suppressor and shield ground? I have not seen this in the reference designs I see published by companies that make TVS diodes and worry this could slow transient response or prevent conduction.
The ESD suppressor in the circuit is a TI ESDS314DBVR, which is four TVS diodes in one package tied to a common ground. Digikey link https://www.digikey.com/product-detail/en/texas-instruments/ESDS314DBVR/296-ESDS314DBVRCT-ND/9745477.
The PTC is a LittleFuse rated for the power requirements of the board.
AI: From the datasheet:
Breakdown voltage, IO to GND is the critical parameter. Some parts may conduct at 4.5V, shorting the 5V VBUS to ground and other may not. This is not a suitable part for VBUS ESD protection.
|
H: Driving a 74HC245 in one direction with 3.3V when Vcc is 5V
I have a 5V device that accepts parallel data in. That data is buffered by a 74HC245 IC (TI datasheet, NXP datasheet) where the Dir pin is hard wired to drive in one direction. I can't change/modify this device hardware-wise.
To date I have been driving this device using an ATMega328P, running at 5V, with no problems.
I now have to switch to a 3.3V microcontroller for other reasons out of my control.
I am hoping that I don't have to add logic level adjustment circuitry in-between the microcontroller PCB and the device being controlled. Reading the datasheet of the 74HC245, the input high threshold is 2.4V(typ.) to 3.15V when Vcc is 4.5V. Being that the 74HC245 IC is supplied by 5V, I am concerned that 3.3V logic out of the microcontroller is going to be right on the limit.
Will I be able to get away without any logic level adjustment circuitry?
AI: This will not work reliably. If you look at the TI datasheet you will see that \$V_{IH}\$ is specified at \$3.15\$V when \$V_{CC} = 4.5\$V, and it is clearly a strong function of \$V_{CC}\$. If we assume that \$V_{IH}\$ scales linearly then it would be \$3.675\$V when your power supply voltage is \$5.25\$V, which could be the case if your voltage regulator tolerance is 5%. In fact, this arrangement won't work is the supply voltage is above about \$4.7\$V.
Furthermore, if the supply voltage for the microcontroller is \$3.3\$V that doesn't mean that the logic signals will be right at \$3.3\$V. Check \$V_{OH}\$ for the microcontroller.
|
H: Virtual ground in positive feedback op amp
In negative feedback Op amp Virtual ground condition is satisfied but in positive feedback it doesn’t why?
AI: OP Amps, by circuit design, are linear ONLY when the output is not saturated which means if the input differential is >>0 there is zero linear gain. ( ie no virtual ground, Vin NOT =0 ).
As soon as the input >> 0 OR the output goes to the rail, the closed loop gain must (by definition ) = ZERO (0)
Av(f) = linear gain = ΔVo/ΔVin ( at any frequency including DC)
Positive Feedback is also called a latching comparator with hysteresis, designed to ignore input pk=pk noise and have only one output transition until signal change exceeds hysteresis levels.
Positive feedback with low phase margin just causes ringing and still has linear gain, but if output cannot null the input circuit offset error. The inherent bias offset error is still called a virtual ground.
Comparators also have linear gain like logic but only "during" the transition, then Zero gain at any logic level "0,1" after that. It is this clipped level that gives logic such a high SNR and low Error rate. The logic levels are defined so that with good layout and decoupling these "grey area" noise margins ensure very high SNR and thus near zero errors. This can be computed based on thermal noise but often inductive and crosstalk noise is the real threat to logic errors not Thermal random noise..
|
H: Review of my LED spinner Schematic
After a few months of questions and modifications to my design. I have what I believe to be a LED Spinner that can change the speed of the spinning LED. Before I send this design off to get printed on a board, I wanted to see if I could get it reviewed by the magnificent people that roam this forum.
A quick overview: I have a 1.2288MHz crystal feeding into a 74HC4040 to bring the frequency down to 300Hz, then that will be fed into a 74LS390 to bring it down to 60Hz, 30Hz and 15Hz. those frequencies will then be fed into a 74LS126 so that only one frequency gets fed into the CD4017 that controls the LEDs. Most of this design was made possible with the help of user peufeu in a previous question: Link
The power coming in is from 2 series fenix 4800mAh 3.7V batteries pushed through a linear regulator down to 5V. Then a circuit that I must say came from user Dave_Tweed in my previous question:Link to make it so my switch will turn the circuit completely off when put in position 5. this switch will feen into the 74LS126 to choose at what speed the LEDs spin. I added a pull down resistor on each line here since I figured leaving it open might not go well.
Here is the schematic:
If needed, I can post the datasheets for all the ICs in the schematic.
I just want to know if I did everything correctly, and if not, what did I do wrong and why is it wrong?
Thank you
PS: Thanks in advance for the correction edit.
AI: At least it is missing all bulk and bypass capacitors. The voltage regulator might be unstable. You also have chips gulping current at their respective frequencies, max being 1.2 MHz spikes, so any inductance and resistance in the supply wiring will cause voltage sags during each current spike, so chip supply voltages can have too much ripple for it to work at all.
|
H: Short-circuit in CAN lines
I have a device with a SNx5HVD251 CAN transceiver. I opened the datashett and read the following parameter:
IOS(SS) - Short-circuit steady-state output current
Below are two images of the datasheet that I put together here:
I didn't understood the relevance of this parameter or what it means.
1)For example, if I have VCANL at 12 V, CANH Open . It means that the current over CANL line will be at maximum of 200mA ? Or, it means that CANL line can support at maximum of 200mA ?
2)The CAN line will be passing through a relay. (The relay will not switch during communication, but the CAN line need to pass through it). So, how relevant is to take this value of current into account when I need to specify my componentes ? A relay with less than 200mA is much more cheaper, so I need to know the relevance of this parameter.
AI: Those are worst case short circuit currents. If something breaks and the CANL pin gets directly connected to 12V supply voltage, maximum of 200mA can flow when it tries to drive it low. And the other case, if CANH gets connected to -7V, max 200mA can flow. Wires must handle at least this current. Surely most small signal relays can switch 1 or 2 amps, 200mA is nothing.
|
H: What is this IC "30=70H"?
I'm hoping to identify the IC on this Philips Hue lightstrip controller in order to replace it.
The IC blew after a barrel connector head used had its polarity reversed, so 24 V went in the wrong way and melted pin 5 (top row, middle pin on the IC in the picture).
Suggestions on how to identify what it is would be greatly appreciated.
AI: It's a Richtek RT6208GE. It's a switching power supply (buck regulator) chip (obvious from observing the PCB layout).
I disagree with DKN.. worth trying a replacement in this case for a $1.50 part. I give it better than even odds.
|
H: How to clamp opamp output to +-12v when supply is +-15v
I have the following circuit. It's purpose is to take a 0-3.3v signal from an MCU and provide a square, -12v to 12v wave across R3, R4 and either R5 or R6. The idea is to read +-9v when only R4 is connected, +-6v when R4 and R5 are connected, and +-3v when R4 and R6 are connected. The peak-to-peak voltage of the square wave are reduced depending on the configuration of R4/5/6. I have no control over the values of R4, R5 and R6. My problem is that, considering the +/-15v supply, the output of the opamp is somewhere around +/-13.7v. This offsets the values at the "Output" point by a few volts and is out of spec.
I thought of using a rail-to-rail amplifier with a +/-12v supply, but even the rail-to-rail amps can't truly swing to the rails. Also, I'd have to re-work a whole bunch of other subcircuits to work with +/-12v or add a whole other AC/DC converter and I'm not willing to do either.
I then thought of using two back-to-back zeners right after R3, but that would only clip the max voltage, all of the other points would still be too high/low.
Then, I thought of using simple voltage divider for the power inputs of the amp, but that would add 4 resistors to the board and I'd rather avoid that if possible.
In the schematic, assume SW2 has another unconnected pole so that having only R4 in series with R3 is an option.
simulate this circuit – Schematic created using CircuitLab
Can anyone think of how I can limit the output of the opamp to +/-12v and maintain the 9-6-3v ratios with the regards to the output?
AI: You're using an op-amp as a comparator, which isn't all that great an approach.
Consider replacing the op-amp with an analog switch such as an ADG1419.
You can either reduce the supplies to +/-12V or replace R3 with two resistors that give you the Thevenin equivalent. eg. Ra = 1137\$\Omega\$ series and Rb = 4550\$\Omega\$ to ground.
Ra || Rb = 910\$\Omega\$
15V * Rb/(Ra+Rb) = 12V
If you insist on using a quad op-amp you could add a series resistor and back-to-back zeners before the 910 ohm resistor, but you'll need an op-amp that swings a lot closer to the rails under load than the TL081, and the tolerance on the voltage will not be great. Another option would be a Schottky bridge and a single zener or a TL431 with resistors to set the clamping voltage (at least that would be very symmetrical).
|
H: Transistor operating point problem doesn't make sense?
Find the operating point of the transistors T1 and T2.
I've been given this problem and when I try to apply Thevenin's theorem on resistors R2, R3 and then try to calculate \$I_{C1}\$, I get \$I_{C1} = 0\$ and have no idea why.
simulate this circuit – Schematic created using CircuitLab
Sorry for the not so good explanation I'm new in this community and new in the electronics domain, it isn't really my strength. Any help will be appreciated.
Applying thevenin's on R2, R3, we get:
simulate this circuit
If we take the loop with V1, R1, T1, V2, using K2:
$$Vcc - 0 = U_{EB}+I_{B1}*R1+ V1$$
$$I_{c1}/\beta_0 = (Vcc - V1)/R01.$$
With \$\beta_0 = 100\$ and \$U_{EB} = 0.6V\$ and approximating V1 to 9.6 we get I_c1 = 0. And it's really weird.
AI: This problem isn't a good candidate for hand solution.
Because the resistor divider formed by R2 and R3 is going to have about \$(10\ V)\frac{0.6}{9.3+0.6}\approx 0.6\ V\$ across R2, and thus there is about 0.6 V across the b-e junction of T1.
That means that T1 is only very weakly turned on, and our basic rules for hand calculations don't handle this situation well. If we have a rule that the b-e junction must be -0.7 V (negative because it's a PNP) to turn the transistor on, then we think this transistor is in cut off mode. Or if we think the b-e junction needs -0.6 V to turn on then we see that the transistor is just barely turned on, but there's nothing to tell us how much current is flowing out of the base.
Even if we use the more accurate 9.393 V you calculated for the Thevenin equivalent voltage of the 10 V source and resistor divider, there's nothing to say that the b-e voltage of the transistor will be exactly -0.6 or -0.7 under these bias conditions. It might be -0.65 V or -0.55 V, and given the small difference between the Thevenin source output and the base voltage in the circuit, this will cause substantial errors in the calculation result.
The best way to solve this circuit would be to plug it in to a simulator (with reasonably chosen transistor models). Or to adjust the resistor values to be sure the transistor is either fully in active mode or fully cut off.
|
H: Plotting values into the graph
I recently found this op amp exercise and solved it using superposition .The result was UA
I first need to draw U1 and U2 and then based on UA draw the output result.Stuck doing this and now way to figure it out if I'm doing this right so I came here for some needed help
This is my attempt to just plot U1 and U2 without then trying to plot the output UA.
Asking for some advice if my approach of graphing this is right. U1 seems to be right I'm just not sure about U2
AI: I can tell you nothing about what these functions mean with regard to op-amps, but the following should help you get the graphs you're after.
You go about this by first understanding the notation:
U1 is a linear function on t, Û1 is essentially a scalar constant to adjust the amplitude, and T1 is a scalar to adjust the slope. The function is defined over a T1 wide period centred on t=0, ie, from -T1/2 to +T1/2.
U2 is a sinusoidal function on t, expressed as a cosine so it is also centred on t=0, so expressed as cosine. Û2 scales the amplitude to 2V. f2 is the frequency: 2π is the constant to convert cycles to radians.
UA is simply a linear combination of those two functions.
The two easiest ways to graph this are use either a graphing program/website or a spreadsheet.
Graphing Website
You might find the graphing site https://www.geobra.org/graphing helpful. I use it for quick graphs of this kind. There are plenty of graphing packages and sites available.
First I folded up the constants and multiplied everything by 1,000,000 just for convenience, giving T1=20 and f2 as 0.1. Then you create the three functions by typing them in to the box on the left (by the three coloured dots). You might have to read the manual to see how to type in the functions, but it's easier than it looks. The program then just draws the graphs for you. If you click on the graph lines or coloured dots you can see various values on the graph (eg points of inflection).
Spreadsheet
Having noticed that we go from -10μs to +10μs, we construct a table in a spreadsheet with a least a hundred rows, which will give us a detailed enough graph. I pick 0.1μs just for convenience, which means we will have 201 rows. Then you make a column for t and each function; then you make a graph of it. The scatter plot function expects each row to have the horizontal coordinate and several vertical, just what's needed. You adjust scales and grids until it's easy to read.
|
H: Are bypass diodes in low voltage solar panels with parallel cells really needed?
I have 5 solar cells, ~3W ~3.3v 900mA each, a total ~15W, connected in parallel.
Considering the low voltage and current involved, do I still need a diode attached to each cell?
Will I still risk damage if one or more cell are shadowed or will I just risk power drop?
AI: It probably won't damage them.
Any light absorbed by a solar cell will either be converted to heat or electricity. If the electricity generated is not drawn off then it will be absorbed by the intrinsic diode in the cell and converted to heat.
The worst case is with no load and one panel shaded on all cells but one. Because the shaded cells in that panel are producing a lower voltage, the non-shaded one gets a greater proportion of the voltage produced by the other panels, as well as having to absorb about double the current (the current it is producing plus the current pushed into that panel from the other panels). In a 6 cell panel that one cell will have to dissipate about 2.25 times more power from generated current than it would normally. Thus that cell will become a 'hot spot' in the panel.
Sounds bad, but the cell already dissipates ~75% of the incident light anyway, so total dissipation only increases by ~25% which won't heat it up much more. In a large panel it might be a problem, but small panels generally have a higher surface area to volume ratio so they can dissipate heat better.
Without blocking diodes any shaded panels will pull the voltage down a bit, but under load the effect is small. Diodes also drop voltage, so the loaded voltage will probably be lower with them than without. Suitable Schottky diodes (eg. 1N5820, rated for 3A max) could drop 0.3~0.4V at full current, which might be unacceptable on a system expecting 3.3V.
|
H: Wrong ground reference producing wrong output on ADS1115
I am currently working on a project, which involves measuring current in a DC circuit and storing that generated data on an arduino microcontroller.
I decided to go with the ADS1115 and a shunt, which is located in series on the low-side of the circuit I want to measure. It is a 0.001Ω shunt, which results in a voltage drop of 1mV per amp.
I have attached a schematics of the wiring (Note that there are multiple other devices connected in parallel on the arduino, which all have very little power consumption; I left them out for better reading)
The issue I'm having are the results the ADS1115 is giving me:
When no voltage is applied to the left circuit, i get a single-ended reading of about -20, which in my application is unacceptable. When voltage is applied and the load draws current, i also get a reading, which is -20 bits off.
The ADS1115's supply voltage is around 4.98V and I am not sure if the problem lies in all the other components that are attached to the arduino's circuit.
I am not an electrical engineer but a software developer, so please take that into consideration when commenting on my schematics and/or wrong wiring. :)
Any help is appreciated!
AI: Unlike other devices in ADS111x series, ADS1115 includes source MUX that selects which of the input channels will be used as inputs for differential amplifier.
Since you are measuring shunt voltage referenced to the ground, you have to select GND as one of those inputs. See "9.3.1 Multiplexer" diagram in the datasheet. This is done by setting bits 14:12 in the Config Register to 0b100 (for A0 input).
Also, many ADC chips have "zero offset", which can be accounted for by calibration. I don't know if that is the case with this particular chip.
|
H: Identify chip on 2S BMS / Balance card
I picked up a bunch of Chinese-made 2S BMS / Balance boards. As is the norm with inexpensive Chinese boards, they came with essentially NO documentation aside from how to hook the board to the batteries and load.
I've identified all of the devices on the board except for the chip that controls the BMS function. This is a SOT23-6 with the letters "PAYG" on top.
I'm hoping that someone can identify this chip and point me towards a datasheet for same.
AI: PAY is Seiko S-8242AAY-M6T2GZ battery protection IC. The last letter G is related to lot number.
Here is PDF (see table on page 23 for part code): http://img.hqew.com/file/Others/1090000-1099999/1098111/Electronic/201251816354378771.pdf
And there is the list of codes, which contains PAY: http://www.dl7avf.info/charts/smdcode/cpa.html#TOC
|
H: type I phase detector - analog vs digital inputs
Section 13.13.2.A of the Art of Electronics (3e) describes a type-I phase detector. It is explained that the analog version of this circuit uses a multiplier and the digital version uses an xor gate. Additionally, this quote seems to indicate that the xor gate is a digital "version" of an analog multiplier:
The type I phase detector is applicable to either
analog- or digital-input signals, and performs a simple multiplication of the inputs. For digital signals this is just an exclusive-OR gate...
I fail to see how an xor gate gives the same result as an analog multiplier with digital inputs. Here's the digital circuit and corresponding voltage output as a function of phase.
How this circuit works and why it functions as a phase detector make sense to me. However, I don't see what the analogous analog circuit would be. For instance, replacing the xor gate with a mixer produces the following circuit:
If I assume that the RC lowpass filter produces a perfect average of the mixer output over a period (and for simplicity both input signals have identical frequency), then the relation between \$V_{out}\$ and \$\phi\$ is shown in the following plot.
I guess I could invert the output (e.g. amplifier with gain=-1) to get a roughly analogous result. Is this what is done? If not, how can I implement an analog type-I phase detector (presumably with a mixer)? Did I misinterpret the phrasing used, and what was really meant was that a mixer is used as part of the phase detector but is not actually analogous to the xor gate?
Something else I've missed?
AI: About logic XOR gates acting as a multiplier....
It is easy to confuse logic states of an XOR. One makes the mistake of assigning logic low (a zero) to a multiplier's equivalent of zero volts. That's wrong.
A logic low is assigned to some negative voltage, while a logic high is assigned to some positive voltage when considering it as a multiplier. Now the logic equivalent to multiplication is apparent:
+1 x +1 = +1
+1 x -1 = -1
-1 x +1 = -1
-1 x -1 = +1
It is striking to look at the transistor arrangement inside an XOR gate of non-saturating emitter-coupled-logic chips. The arrangement bears a striking resemblance to a Gilbert-cell-mixer, which is often used as a four-quadrant analog multiplier. Shown below is the internals of ECL type MC1674 from Motorola, a triple XNOR done in MECL III...intended for use in PLL applications :
There are DC offsets to contend with when using XOR, XNOR logic gates as multipliers, since most logic gates don't provide symmetrical input/output voltages about zero volts.
For this logic chip, Vcc on pin 1 is often considered ground while Vee on pin 8 is -5.2V.
Logic 1 output on pin 15 (inputs on 7,11) is about -0.885V
Logic 0 output on pin 15 (inputs on 7,11) is about -1.735V
The mid-point of output on pin 15 (about -1.31V) would correspond to nearly zero volts in an analog multiplier.
|
H: Paralleled voltage sources error in LTspice
I have the following LTspice simulation schematic.
(It's a simplified version, just to demonstrate my dificulty). The LTspice simulator returns a paralleled voltage sources error even when the sources V5 and V6 are clearly not in parallel. Is there any walkaround that avoids putting internal series resistances with the sources? Why is this error coming up at all? I'm using the most recent version of LTspice.
AI: Look at your node naming. That “GGL” on both will short them or draw an invisible line between them. It’s intended to make your schematic less crowded and human readable, but if you take a look in your netlist which any SPICE uses behind the scenes, they are shorted.
Solution: remove the “GGL” statement for one of the nodes or just rename it.
|
H: Why is the current mirrors output unequal?
I made a LTspice simulation of a pmos current mirror. The current source should model a photodiode (without parasitic capacity). The W/L ratio is the same for both Mosfets. I expected the plot of the output current (red) to be the same as the input current (blue). I thought it will rise in a linear manner as the input current (the sweep is linear).
Now I see that the output current is logarithmic (see plot: red curve).
Can someone explain this difference?
AI: Your power supply is only 1 volt so you are operating the mosfets in the linear part of their characteristic. Try increasing the supply to about 10 volts.
|
H: What is your go-to prototyping connector?
I have completed a few projects - sensors and an ADC circuit - and I find myself often connecting discrete boards together by tediously soldering wires to break-away headers like those below.
I worry about poor connections and the time it takes is excessive.
What are some good, flexible go-to prototyping connectors? I am considering getting some RJ45 connectors and using old cat 5 cable, but they aren't 0.1" spacing friendly, which makes prototyping with them harder. (Do 0.1" compatible RJ45 connectors exist? Could I rig these up anyway?)
I realize this is highly dependent on the application but I was hoping to get a solution for the general case. Or maybe soldering ribbon cable to headers is the way to go.
For a little more information, my current projects have involved carrying a ~2v signal ~1m to an ADC, carrying 1MHz SPI ~0.5m between an ADC and a PI, as well as carrying around 3-5V power and ground - so nothing terribly fancy.
Update: I went with IDC connectors (2x5 pin and 2x20 pin) and I am happy with this decision.
AI: Why solder the wires to the header pins when you can use flat cable with IDC connectors?
No soldering at all, compatible with 0.1 inch spacing, and the cable and connectors are easy to assemble (and cheap as well).
|
H: Why are there diodes going to D+ on this alternator/generator?
Above is a schematic of an alternator/generator. I understand the rectification going to B+, but I don't understand why there are 3 diodes going to D+.
The way I understand the setup, D+ measures the voltage at the battery, and the delta between that and the preset voltage in the regulator drives the field to increase/decrease the output of the generator. I know why you need the diodes to charge the batteries, but why would you put the rectified current on the D+ sensor line? Seems like that would defeat the purpose of the sensor.
Maybe I'm completely misreading the diagram, if so, please point out my error.
AI: You have a misconception about what the regulator of a car alternator does. It doesn't regulate the voltage of the battery, and so, it doesn't measure the voltage of the battery either.
The regulator is there so the voltage delivered to the battery to charge it is independent from the speed of the drive shaft.
The regulator does this (simply said) by controlling the field winding of the alternator. That's the winding on the rotor, in your schematic that one on the right. This field winding needs some power, and it has to be independent from the battery, so the speed the alternator runs at can be measured.
That's why these three leftmost diodes are there. They provide an additional DC output which is not filtered to "pure DC" by the battery. Instead, it still has an AC part with a frequency dependent on the drive speed.
|
H: What is affecting the power supply current rating?
My question is generally specific on how to determine the current rating for a power supply? does the current is always constant? or it's according to the load of the circuit? Also does the maximum current rating of the power supply is determined according to the power rating of the components used in the circuit?
What will happen if I connect an 4 Amp LED to a transformer based 10 Amp power supply of the same voltage rating?
AI: I'm assuming you're talking about a AC-DC power supply.
Ratings:
The main ratings on an AC supply that you can't usually exceed are:
Input voltage (and frequency): Don't give it higher than this voltage on the input.
Output current: Don't draw more than this amount of current from the output.
Exceeding either of those two will likely break something.
There are couple other basic ratings that a power supply has, but they relate to how the supply will behave if your adhere to the above:
Output voltage: At the specified input voltage, this will be the output voltage
Input current: At the maximum specified output current, this will be the maximum input current
Since your LED draws 4 amps, you are just fine connecting it to a 10 amp supply.
How they are determined:
Primarily by two things:
Heat and voltage the parts can take before breaking.
The supply behaving as specified.
Whenever you draw current through a wire, it heats up since some energy is lost. A transformer is two or more giant coils of wire wound around some kind of core, so the wires inside the transformer heat up with current. At some point, the temperature of the transformer will get hot enough that the insulation between the wires stops working as designed and your transformer shorts out between the windings. This causes the wiring ratio to change and the output voltage of the transformer to drop (or rise!!), often far enough so that the DC side of the supply stops working. The maximum current rating of a transformer is the current that can be drawn through it without heating it up to the point where its life is reduced. There are other factors as well in addition to winding resistance (for one, the core of the transformer has a maximum amount of magnetic flux it can contain before it stops behaving as expected), but heating is generally a good indicator if something is being pushed too hard.
For the input voltage rating, there is a point where the transformer will start shorting out because the input voltage is so high that it breaks down the insulation between the individual coils on the same winding or the insulation between the input (primary) and output (secondary) winding. This voltage is known as the "isolation rating" or "breakdown voltage" and most properly certified transformers will have a breakdown voltage of at least a thousand volts for a short period of time. However, that doesn't mean that you could put several thousand volts on the transformer and expect it to work. The magnetic properties of the transformer see to that.
As for the DC part of the supply, on a basic level it consists of a diode, capacitor, and possibly some semiconductors. Each of these components also has a voltage and current rating that needs to be respected. The exact ratings are determined through a variety of ways, but again, they will heat up when this is exceeded. Putting too much voltage into the supply could cause too much voltage to be put into the DC components. Drawing too much current from the output could exceed the current ratings of these components. Certain components, such as capacitors, are rated to operate for a certain number of hours at some temperature (105C, 85C, etc). The output current rating of the supply is also calculated to prolong the life of these kinds of components so that they will hopefully last as long as the customer expects them to. This is also a place where corners are cut, since the power supply might get just warm enough so that the useful lifespan of the parts is shortened, but it will work good enough for some time and use cheaper parts.
In reality, perhaps you could draw more current through a supply than it has specified, or you could put more input voltage to it. However, the supply wasn't designed to work within those conditions and could misbehave. The specifications on the supply also reflect the designer's intentions for how the supply is to be used.
|
H: Power line communication by inducing a signal into a wire
I would like to make a small board with a simple MCU that just repeatedly sends a message via a modem on one end of a power line wire (14awg or so). On the other end will be a similar device for reading that message. The only requirement is not interfering with/cutting the wire or isolation in any way. There will not be access to a plug or any way to directly tap into the line like PLC home networks do. The device will be powered from an external source (3.3 or 5V).
I've done some reading on this topic and the ST7540 seems to be a good option, in my case I would not need to filter out the 50/60Hz because there would be no electricity on the line. I am mainly uncertain about the induction, possibly using some kind of coil around the power line. What I basically want is to use the power line wire as an antenna between two modules (approx 3m or 10ft apart). I can't use full wireless (315MHz for example) for various reasons.
What is the best way to inductively transmit data (only one direction is needed, tx to rx) over a power wire that has no voltage/current and what kind of parts would be required? Are there modems specifically made for something like this? I remember reading about something similar being used in under-water applications in the past.
Hypothetical example: There's a 2-lead mains wire permanently plugged into an outlet in my room. The wire leads to the next room and it powers a lamp with a switch (relay). I can't go into the next room. I want to induce a modulated signal into the mains line by means of a transceiver/MCU while the lamp is off (no voltage/current in the mains wire), the signal goes through the mains wire, which acts as an antenna. There's a receiver/MCU on the mains line in the other room, which reads the signal inductively and turns on the relay, lighting the lamp. I do not want to physically manipulate the mains wires in any way, but also can't use a regular RF link (I know this could easily be done wirelessly).
AI: Consider PPM....pulse position modulation.
Couple pairs of spikes into the hot-wire. For ZERO, place the spikes 5 uS apart.
For ONE, place the spikes 10uS apart.
Send the bits every 100uS.
Or, to have a very narrow-band datalink, so you can narrow-band filter to reject more noise, for ZERO, send +spike,+spike, For ONE, send +spike,-spike.
Examine some Fourier signals and pick your spacing and modulation.
Here are useful waveforms and spectra from Signal Wave Explorer. I used the JBStage (canned example) with trapezoidal input (50%, 50%, no plateaus)
Notice how narrow-band (centered at 200KHz) the output spectrum is.
By the way, the powerline can easily kill you. Are you thinking about transformers wrapped about the HOT wire?
|
H: How can I set up a transmission line in ADS?
I'm in the palette section with TLines-Ideal. I'm confused about what all these options mean, for example "terminal transmission line" Where is the part for me to draw the transmission line and specify a characteristic impedance, phase constant, etc.
If you can point me to some sources, that'd be great. I couldn't find much about transmission lines on ADS online and everything else was confusing to me.
AI: There are a lot of different transmission lines in ADS. There are ideal lines, coaxial cable models, printed circuit board traces, coupled lines, etc. For an ideal line, the simplest is TLIN, which just has Z, and an electrical length that is described as a phase shift at a specified frequency. The 2-Terminal transmission line only has two places to connect a wire. The 2-Terminal model doesn't mention if there is a ground plane or a coaxial shield. The 4-Terminal ideal line adds extra connections for a ground plane or coax shield. For physically realistic models, you have to define a substrate, dielectric constants, trace widths, and so forth.
Another ideal transmission line is the TLIND element, which has a characteristic impedance and a delay in seconds.
After you place the TLIN or TLIND component, double click it to bring up the form. The Help button on the form takes you to the documentation.
The online documentation is at http://edadocs.software.keysight.com/display/support/Knowledge+Center
You need to be a registered user to get to the online manuals.
Find good materials available without registering on Youtube. There are a lot of ADS videos there, with many created by Keysight. Perhaps "Getting Started with ADS" would be helpful.
Full disclosure: I don't speak for Keysight Technologies, even though I work there. I don't think anyone from Keysight regularly visits this StackExchange site, and this is usually not a great source of ADS help.
|
H: How can an ISP program an AVR over SPI if there is no code on the device?
I am trying to understand how an ISP can write machine code into a microcontroller's memory.
In the case of AVR microcontrollers (such as the ATMega), you can use a special in-circuit programmer that will "engage" the target microcontroller over an SPI bus. (I am getting this information from this document)
What I'm wondering is this: if I had a fresh AVR with no prior programs written to it, how could it possibly "listen" to the SPI bus? Doesn't there need to be a program running?
AI: A fresh AVR, while it has no code programmed into the user space, doesn't mean it has no capability.
You'll notice that the RESET line is used as part of the ISP protocol. While the RESET line obviously does what you'd expect for most of the chip, it also enables the mechanism that handles the ISP. It also clearly delineates 'normal' operation from ISP operation, you are in one, or the other. You'll also notice there's a handshake of 'magic bytes' for the programmer to establish that this mechanism is working.
Whether this extra mechanism does all the work to handle protocol and control the memory busses, or whether it just flips an address line to allow the main CPU to run ISP code from a ROM (sounds more sensible?), doesn't really matter from the programmer point of view.
|
H: Level shifting a clock signal
I need to step-up a clock signal of 8 MHz from 3.3V to 5V, I was wondering if this type of level shifter would work :
The BSD840N as rise and fall time fast enough for my application but I'm not sure I'm thinking this the way it should be thought. Is it correct to assume it will work in such a setup ?
AI: The rising edges of your signal will be slowed down by the RC filter formed by the pullup and any parasitic capacitances in your circuit. Assuming those are 10 pF, you get a time constant of 100 ns, which is way too slow for an 8 MHz square wave.
You could try to build a level shifter that actively drives both low and high signals with multiple FETs, but it would be easier to just use a level-shifting logic chip.
There are level shifters with two supply voltages (e.g., 74LVC1T45), but 3.3 V signals are TTL compatible, so you can use any 5 V buffer with a TTL-compatible input, such as the (SN)74AHCT1G125/MC74VHC1GT125/TC7SET125.
|
H: Input and Output Impedance parameters of two port network
I am reading about two port network and I come to about Z-parametrs
$$\begin{bmatrix} \mathbf{V_1} \\ \mathbf{V_2} \end{bmatrix}=\begin{bmatrix} \mathbf{Z_{11}} & \mathbf{Z_{12}} \\ \mathbf{Z_{21}} & \mathbf{Z_{22}} \end{bmatrix} .\begin{bmatrix} \mathbf{I_1} \\ \mathbf{I_2} \end{bmatrix} $$
but then it introduces the concept of input and output impedance which is $$Z_{in} = Z_{11}-\frac{Z_{12}Z_{21}}{Z_{22}+Z_L}$$and also $$Z_{out} = Z_{22}-\frac{Z_{12}Z_{21}}{Z_{11}+Z_S}$$
Where \$Z_L\$ is load impedance and \$Z_S\$ is source impedance.
I don't know how writer concluded these impedance. Please provide me the explanation.
AI: For the input impedance, expand the matrix equation and then use the output condition: \$Z_L=-\large \frac{V_2}{I_2}\$
Similar approach for the output impedance.
|
H: LTC6993-1 Circuit not working
I'm having a problem with simple pulse generator circuit using LTC6993-1 (datasheet).
Circuit is not starting to pulse... This is my schematic and PCB layout:
I have checked the voltage levels and all of them are fine... I'm running it from 3V supply.
I don't know what I managed to get wrong with this simple circuit... Any ideas?
AI: The circuit works as expected. The problem is probably your understanding of what a "Monostable Pulse Generator (One Shot)" is.
If TRIG is set to a fixed level you will never see any output.
The output goes high only if a rising edge is detected at TRIG and then it stays for a given time. That's what a monostable (one shot) does.
EDIT:
Maybe the confusion came from the fact that this IC contains an oscillator to implement the one-shot functionality by a digital counter.
|
H: Does a 555 astable dissipate more power at higher frequencies?
I'm planning on running a circuit on a 9V battery and it uses a 555 in the astable mode to generate pulses.
I want this circuit to be as efficient as possible, and not have to replace the battery very often if it can be avoided.
Looking at the 555 datasheet, the max power dissipation is 1180mW (page 4) -- but is this lower at smaller frequencies? Or is the chip designed to dissipate the same power at any frequency?
I can't find any graphs on the datasheet (p6-7) regarding power dissipation with frequency, which is why I'm asking :/
AI: Yes it does.
Reducing the oscillation frequency \$f_{osc}\$ will have an impact on your circuit power consumption, because you'll be charging and discharging the capacitor less frequently, thus reducing consumption.
If you also want to reduce the current drawn by the 555 itself, then the first thing you should look after is lowering the supply voltage \$V_{cc}\$. The efficient way to do this is with a switching regulator to bring down 9V to 5V.
A lower \$V_{cc}\$ has the extra benefit of reducing the amount of charge required for the capacitor to achieve \$\frac{2}{3}V_{cc}\$, thus reducing the power consumption. You can have the best of both worlds and combine this with the reduction of \$f_{osc}\$.
EDIT:
As Jim Dearden has correctly pointed out in his comments:
Using the CMOS version of 555 will help in reducing the consumption of the IC.
On top of that, the CMOS version can work with higher values of the timing resistor than the bipolar version, thus making possible to reduce the value of the timing capacitor while keeping the same time constant. A lower valued capacitor stores less charge, thus consuming less current while charging up.
Double win for the CMOS version, which adds these benefits on top of those already achieved by reducing \$V_{cc}\$ and \$f_{osc}\$.
Note: selecting the highest possible value for the timing resistor in order to lower the value of the timing capacitor will always yield a reduction in current consumption, regardless of the version of the 555. So it should be considered for any design in which consumption is a concern.
|
H: Simple Diagram Question
I am following a trimming tutorial that requires me to replace the U5 transistor with the U10 transistor, and I need some clarification.
When it says to cut the trace on the fifth image, is that represented by the point he chipped off at the bottom right of the transistor's traces?
Am I supposed to bridge the three bottom pins together in the sixth image?
AI: The images are a little confusing, but I don't think removing the bottom-right pad was the intention or is necessary since C6 is removed and there are no vias. Colouring the trace red (solid and not showing a cut) also doesn't help, but the meaning is that is the trace you need to cut. It's a very thin trace so it should be easy and not require the butchering that was done to the board in the image. Actually the 6th image gives a better view I think of how the trace was cut.
I don't know what that IC is so can't comment on whether those pins should be shorted together, but I would assume that they should be based on the photo, even though it's not explicitly stated.
|
H: How useful is C for working as embedded engineer?
I have finished a course of basic C programming and I wonder is C useful/powerful for working as embedded engineer, should I spend a lot of time to master it or should I move to a higher programming language?
And if it is useful, as I want to work as embedded engineer, can you advise me what aspects should I learn/practice more about C (like pointer or array, driver...) so it can become a useful skill for my future?
AI: In my experience, almost all embedded programming is done in C. I find knowledge of bitwise manipulation is the main thing that is important compared to C used for programming other types of applications.
C is considered very useful and powerful for embedded work because it is low level enough that it compiles easily to machine language, and yet is still high level enough to be readable, easily understood by many other developers (if you do it well), and give access to abstract data types.
|
H: Connect PCIe x1 port to an ASIC PCIe x4
I want to connect a microcontroller with a PCIe x1 port to an ASIC with PCIe x4 interface. It is not possible to connect microcontroller to lane 0 of ASIC, and it should be connected to other lanes (like what depicted in picture). is it possible to connect lanes this way and work with microcontroller PCIe x1 mode? In general, could we connect PCIe lanes not in numbered order?
ASIC PCIe is gen 1 compatible with PCIe Base Rev 3.0 and supports x1, x2 and x4 modes.
It would be great if link some documents.
AI: In terms of connecting a x4 endpoint to a x1 host, that is perfectly doable. It is required as per section 4.2.4.10.1 of the PCIe V3.0 Base Specification that:
The ability for a xN Port to form a xN Link as well as a x1 Link (where N can be 32, 16, 12, 8, 4, 2, and 1) is required.
Emphasis mine
Therefore you will have no problem connecting Lane 0 of your ASIC to the MCU.
However, the question is whether or not you can connect to any lane other than Lane 0 on the ASIC. The answer to your question can be answered as nothing more than "maybe" due to lack of any specific documentation on the devices. From the same section of the standard:
Support for Lane reversal is optional.
• If implemented, Lane reversal must be done for both the Transmitter and Receiver of a given Port for a multi-Lane Link.
• An example of Lane reversal consists of Lane 0 of an Upstream Port attached to Lane N-1 of a Downstream Port where either the Downstream or Upstream device may reverse the Lane order to configure a xN Link.
Emphasis mine
Some PCIe devices support lane reversal, which typically allows Lane 0 to be mapped to Lane N-1, keeping all lanes in order (e.g. 0,1,2,3 connect to 3,2,1,0). However this is not required by the specification so is completely device dependent.
If your ASIC supports lane reversal, then you can connect your MCU to Lane 3 of the ASIC. However if the ASIC doesn't implement the optional feature, then connecting the MCU to anything other than Lane 0 will fail.
|
H: Deducing sinusoidal expression for V
I have the question "An alternating voltage V has a periodic time of 20ms and a maximum value of 200v. When time t = 0, V = -75 volts.
Deduce a sinusoidal expression for V and sketch one cycle of the voltage showing important point."
So I know that:
time t = 20ms
Max value = 200V
angular value = \$2{\pi}f\$, f = 1/T = 1/20 = 0.05.
Therefore angular value = 0.314
But the solutions say that the expression should be:
V = 200sin(10\$\pi\$t - 0.384) V
I'm not sure how this is achieved. I understand the maximum value which is given as 200 but not how the rest was achieved.
AI: \$V(t) = \hat{V}Sin(\omega t + \theta) \$
So we know \$\hat{V}\$ = 200
We know T = 20ms -> f = 50Hz -> \$\omega = 100\pi \$
if V(t) = -75 when t=0
\$V(0) = -75 = 200Sin(\theta) \$
\$\theta = Sin^{-1}(-75/200) = - 0.384\$
main expression = \$V(t) = 200Sin(100\pi t- 0.384) \$
|
H: How to figure out if a 25V capacitor can replace 50V in this circuit
I'm following this video Astable 555 timer - 8-bit computer clock - part 1 where he builds this circuit:
simulate this circuit – Schematic created using CircuitLab
I just realized that the 1000uF capacitor I have is 25V and not 50V as the one he uses in the video. I suspect that 25V will do but I don't know for sure.
I've tested the voltage where the capacitor would have been in the circuit on my multimeter and that reads 55.3mV which is much less that 25V but maybe the 555 timer can change that?
Is this a valid way to figure out the maximum voltage the circuit will apply to the capacitor?
AI: The best way to know for certain would be to measure as you did (although you said 'where it would have been' which might suggest you had it removed). In this case since the highest voltage source is 5V, and there is no boost converter topology present to increase the voltage (you would need an inductor to do that), the cap would never see more than 5V. A 25V rated cap would be perfectly fine.
In general, if you rely on measurement though, be sure that you check it under all possible operating conditions. Ideally you should understand the circuit well enough to know what potential voltages (and current) may be present at any given component.
EDIT: If you plan to take measurements (of a non-steady-state circuit such as this) to confirm your expectations of peak voltage on a cap, it's probably best to use a scope if possible, since meters won't reliably capture brief pulses or transients.
|
H: Is it possible to connect USB printer to Atmega8
I have a usb thermal printer, that I want to control using Hex commands in its datasheet from cheap micro-controllers. But to do that I need to create USB host in my board and it requires more than just Atmega8. Instead If I just throw usb packet, which has my command, from controller to printer, would it work?
AI: USB is a particularly complex and timing-sensitive protocol which is not conducive to bit-banging on a microcontroller. Even attempting to manually implement the USB state machine in an FPGA from the ground up would be a significant amount of work, involving deep knowledge of the USB specification and practical implementation quirks.
You'll need an external USB host IC, or a microcontroller with inbuilt USB host support.
|
H: At what current and voltage should I run my TEC/Peltier modules for max cooling effect?
I have 8 x TEC modules each with specs 12V, 6A, and I have a power supply of 12V, 10A.
If I connect all them in series, I know current stays the same and voltage increases.
But what I'm unable to understand is that will be voltage divided equally among the modules?
Also, will the cooling effect decrease with low voltage input? What's the relation of cooling effect with voltage and current?
To run all the modules with full effect, what kind of power supply will I need?
Currently, when I run two modules with my power supply, I don't get temp. lower than 10°C. What could be the problem?
AI: If you put the modules in series the current will drop.
Read the datasheet for your device, here's one for the TEC1-12706
If you put 8 in series each one will have approximately 1.5V and it's likely you will draw less than 500 mA current. Notice that operation at this level does not even appear in the graphs.
If you connected 2 in series then you'd have approximately 6 V per device, and you would have something like 2-3 A flowing, so could connect at least three pairs to your power supply, but might not be able to connect 4 pairs without exceeding your supply current rating.
You are giving up a lot of cold potential in this configuration since each device is limited to about 18 W and you could only use 3 series connected pairs on your power supply. This gives about 54 W total.
For 8 devices operating at 12 V you'd expect to be able to get somewhere near 72 W per device for a total of about 576 W from the devices.
Your current power supply is not rated to run 2 of these modules in parallel. To test out your heatsinking I'd suggest you connect only 1 device, and sort out you hot side heat dissipation, then replicated that 8 times larger.
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.