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H: Current flowing inside voltage source or battery
Current is the flow of charges (many say electrons).
Do these charges flow just between the terminals of the source/battery (terminal to terminal) ?
Or
These charges flow through/inside the source/battery as well?
What i know is that the positive terminal is the lack of electrons as compared to the negative terminal which is the excess of electron.
Current flows when electrons flow from the negative terminal of the battery to the positive terminal of the battery/source?
My question is what happens when the electrons reach from the negative terminal to the positive terminal? Are electrons then "absorbed" by the positive charges or they continue to flow through the source/battery to move in cycles?
AI: Current is the flow of charges (many say electrons).
It's electrons in metals, but can be any other charge carriers in other media. In batteries, the charge carriers are positive and negative ions.
When an electron arrives from the external circuit at the positive terminal and heads off into the battery, it will either release a negative ion into the electrolyte solution, or turn a positive ion arriving from the solution into a neutral species. When it neutralises a positive ion, it's getting 'absorbed by the positive charges', but of course those charges have traveled across the electrolyte. In the event of sending a negative ion off, you could see that as the electron 'hitching a ride' across the battery. Different battery chemistries may employ all of one, the other, or both polarities of ions. |
H: Difference between SVM and sinusoidal PWM in BLDC
In this answer Answer to question regarding difference between 6-step and FOC, the author said:
"Likewise if you have gone to the effort to implement a FOC you might as well implement a SVM block to maximise the utilisation of your DClink voltage (sinusoidal PWM is only 50%, SVM is 86%)".
What i fail to understand is, how is SVM different from sinusoidal PWM. I thought SVM is just sinusoidal PWM on all 3 phases.
What is the difference between SVM and sinusoidal PWM when used with BLDC motors , and how does SVM utilise the DCLink voltage better?
AI: A BLDC machine is part of the PMSM family. From a machine perspective a BLDC machine will produce higher torque and also higher efficiencies at the expense of higher torque ripple (the commutation dips are large!).
*NOTE: IPM and PMSRM push the efficiency and performance higher for PMSM topologies. These are excited with sinusoidal waveforms
Since BLAC and BLDC machines are PMSM you can excite either with either methods (BLAC with quazi-squarewave and BLDC with sinusoidal). You will gain some of the benefits with crossing their drive topology but not equally when compared to a machine (for the same framesize) optimised for the specific backEMF profile
How is SVM different from sinusoidal PWM. I thought SVM is just
sinusoidal PWM on all 3 phases.
What is the difference between SVM and sinusoidal PWM when used with
BLDC motors , and how does SVM utilise the DCLink voltage better?
I am highlighting that one part specifically because of the opening paragraphs. How you generate the PWM is agnostic to the type of motor being used. "when used with BLDC motors" is a red herring.
Sinusoidal PWM and SVPWM both produce sinusoidal currents,both compare a reference voltage signal to a triangular carrier (for PWM generation). The difference is in how they produce this. The key difference is: SPWM operates on a per-phase basis independently while SVPWM works on all three at once.
Both start off with a sinusoidal phase demand reference (either due to the output of an inverse Clarke-Park transform or via some other source)
This is where there is a divergence in methods. For S-PWM, these references are directly compared to the triangular carrier. As a result it can only ever produce an output phase voltage that of 0.5Vdc (when compared to the neutral of the motor) because the inverter will switch about a virtual midpoint.
Even if you were to include a single DCLink capacitor, the "virtual midpoint" would be that of the neutral and equally the star of the motor is a "virtual midpoint". SPWM can therefore only produce +0.5Vdc and -0.5Vdc. At some point you will run out of voltage, voltage needed to push current into the machine.
So how is SVPWM different? Well because it works on the 3-phase vectors as a whole and tries to synthesis all at once (rather than independently), the resultant voltage waveforms are different. How these are generated (the exact timing, the resultant line-line, the mathematical equations) are an implementation specific point but the result are voltage demands seen below. These are directly compared with the triangle wave carrier to create the needed PWM
So what is so special about these voltage waveforms? Their shape presents a root-locus which utilises more of the available DC-link. but how? The common mode voltage swing
As mentioned previously, S-PWM is centred around the midpoint of the DClink, with SVPWM, this midpoint moves (due to the dips, the same dips if 3rd harmonic injection was used). Because there is a resultant shift in the CM-voltage, a higher potential difference can be used on a cycle-by-cycle basis, thus extending the potential voltage available to generate line-line voltages.
The downside is the starpoint of your motor now has a common-mode component, which causes some serious considerations when you are dealing with EMI.
Finally the reason I stated Likewise if you have gone to the effort to implement a FOC you might as well implement a SVM block to maximise the utilisation of your DClink voltage was the beauty of quazi-squarewave control is it is simple. If you are wanting to go the route of FOC, this will produce sinus outputs with a fair amount of computational overhead. A systematic decision must have been made to utilise FOC (performance possibly) and thus to utilise SVPWM is a tiny additional computational overhead ontop of FOC
--edit--
Adding visual overlay of the reference voltage waveforms for
Sinusoidal
SV-PWM
3rd harmonic injection |
H: Does NRZ-I have a synchronization problem?
I'm new to EE, just have a question on synchronization.
I was reading a book which says:
NRZ-I has no problem with sequences of 1s, but has problem with sequences of 0s.
It certainly has a synchronization problem with sequences of 0s. But for sequences of 1s, it is still possible to have a synchronization problem.
For example, if a sender sends 1111 and the receiver's clock is twice as fast as the sender's clock, the receiver will interpret the pattern as 101010.
Isn't that a synchronization problem?
AI: For example, if a sender sends 1111 and the receiver's clock is twice as fast as the sender's clock, the receiver will interpret the pattern as 101010.
Isn't that a synchronization problem?
Not in this context, no. When you design a synchronous communications link like this you assume that the uncorrected receiver clock is pretty close to the transmitter clock. "Pretty close" can vary (and you design the communications protocol with an allowable mismatch in mind), but \$\pm 10\%\$ is easy to achieve with RC circuits, and \$0.1\%\$ is easy with crystal-controlled circuits.
Part of a system designer's task would be to assess the tradeoff between protocols that allow quicker synchronization at the expense of a closer initial match between receiver and transmitter (and hence more expensive receivers and transmitters) vs. the opposite. |
H: What would be an LDOs output if the input is its rated ouput
I have an LDO TLV733 that output 3.3 volts. I need that the rail would always be 3.3v because anything below that the slaves data pins might get damaged because the host signal is 3.3v. The input voltage can range from 3.3 volts to 5 volts. Anything beyond 3.75volts is guaranteed to output 3.3volts on this regulator but i am not so confident voltages below it.
How do i compute the voltage output of the LDO compared to its input. In the datasheet i could not find a graph of its input output relationship.
AI: The parameter you are looking for is known as the Dropout Voltage - that is the minimum required difference between the input and desired output voltage in order for the regulator to maintain the correct output voltage.
This parameter is highly dependent on the load current, so you will often find a graph showing dropout voltage vs current. You simply need to add on the nominal output voltage to this value to get the corresponding minimum input voltage.
For your part, an example curve is shown on the very first page of the datasheet you linked to.
It should also be noted however the curve will also be dependent on temperature as well, so you will need to check the datasheet for further information regarding temperature effects. There are more complete graphs later in the datasheet showing dropout vs current for different temperatures for different output voltage settings. For example: |
H: How standard relays mitigate the problem of electrical arc?
I disassembled (tore apart?) an old 812H-1C-C relay to see what's inside, and found an expected piece of metal which moves between two other pieces of metal when activated by an electromagnet.
I was however surprised that there is nothing special to prevent arcing, and the contacts were rather small, compared to the contacts of a 10A 250VAC wall socket, for instance. Despite the fact that the relay was used a lot, there was no traces of contact wear.
What's the trick? Why arcing is such an issue with wall sockets or any other connector, but is not an issue inside a relay rated 16A 277VAC? Is it because the relay is sealed, and filled with non-conductive gas which somehow prevents an electrical arc?
AI: Relays at full capacity will spark a bit and generally wear out after a relatively small number of operations, perhaps 100,000 though some are rated for only 10,000-50,000 if the load is heavy and/or somewhat inductive or has a surge. They will, however, generally be extremely reliable during that lifetime- they tend to run quite cool and are robust against surges in voltage and current, unlike semiconductor switches.
If no current is flowing through the contacts (or minimal) they can easily last for millions of operations.
The relay contact material and the speed of opening/closing is what minimizes the arcing. Different metals and alloys are used for different purposes, optimized for the best possible life at reasonable cost. |
H: Why is the verification of Thevenin's theorem more accurate than that of Superposition theorem?
In my first course on Electrical Engineering, we have an experiment in the Electrical lab that asks us to verify the Thevenin theorem and the Superposition theorem and cross-check our results with our calculations on paper.
Here are the circuit diagrams, straight from my lab manual.
Thevenin's theorem:
Superposition theorem:
Question:
I was asked, during my post-lab viva, to explain why the errors for the Superposition theorem were greater than those of Thevenin's. I answered saying that both theorems were dependent on the linearity of the circuits for their functioning, and the increased current in superposition theorem was heating up the rheostats and promoting non-linear behavior. However, she was dissatisfied. She said there was something more important that I was missing. I couldn't figure it out. This was months ago, and I still wonder about it but haven't been able to come up with an alternative explanation.
TL;DR
Why is Thevenin's theorem more accurate than the superposition theorem, experimentally?
Note: Please forgive the incorrect labeling of the DPDTs in the first schematic. I am redrawing them.
AI: Why is Thevenin's theorem more accurate than the superposition theorem, experimentally?
The last word in this question is the most important one.
To verify the superposition theorem you measure your circuit voltages and currents several times. Each measurement implies a measurement error. The error is generally a number of fixed digits and a percentage.
Other errors are also introduced: your measurement equipment is not necessarily linear - the devices introduce possibly non linear impedances.
Every measurement will also include thermal noise. You will add this up in superposition, but measure it only once for a global measurement.
So once you add up all these errors in superposition, your total error will be bigger than when you make a single measurement. |
H: balanced to unbalanced audio switch
I have an audio source input, that I am putting through the circuit below to create a balanced audio signal.
There will be one of these for each channel, left and right.
I have been asked to provide a "balanced or unbalanced switch", so that a user can flip the switch and have either balanced or unbalanced audio out.
But, as far as I can tell, if you want unbalanced audio from this set up, you would just take the audio from pins 1 and 2, and dont connect pin 3. This gives you an unbalanced audio output.
If I added an extra stage to combine the balanced audio to unbalanced, then I would just be converting back to the original input signal.
Or am I missing something here?
Is there some way to add an "balanced or unbalanced switch"?
AI: Firstly, that's a differential output, not balanced.
A true balanced output would couple interference on one leg to the other, improving common mode rejection at the receiver.
To test : short either leg to ground. If the amplitude on the other leg doubles, it's balanced. This has two consequences :
the differential amplitude doesn't change due to a short on one leg
You could unbalance the output simply by shorting one leg to ground.
However, differential may be good enough for your purpose, and it's not unusual outside pro audio.
As you say, hot + ground comprises an unbalanced signal.
But if you want to make an unbalanced output on pins 2 and 3, you could simply disconnect teh cross-coupling resistor from the -ve leg, leaving that amplifier as a voltage follower with 0V input.
The trivial way to generate a balanced output is to use a transformer; the secondary provides a fully floating balanced output.
For an example of a transformerless balanced output driver, see the SSM2142 which can be replicated with opamps and matched resistors if you like.
Some further discussion in this Q&A: attracted quite a few answers, of which mine was only one.
And also this one |
H: Why adding capacitor to a 10x passive oscilloscope probe
I'm trying to understanding input impedance and its circuitry of a simple, passive oscilloscope 10x probe. From reading Input Impedance of an Oscilloscope and the video EEVblog #453 - Mysteries of x1 Oscilloscope Probes Revealed, I don't understanding why we add these capacitors in the probe circuitry and make things more complicated. The point is to add enough impedance to a passive probe to minimize loading effect of the circuit being measured. A big enough resistor will do and the with just resistors, voltage divider works the same for both AD and DC source regardless of frequencies. If there are intrinsic reactances in the wiring, then can't we just add a big enough resistor in series with those to make any reactance negligible?
AI: I don't understanding why we add these capacitors in the probe circuitry and make things more complicated.
They make things far less complicated. The low-pass filter that would otherwise occur is eliminated.
The point is to add enough impedance to a passive probe to minimize loading effect of the circuit being measured. A big enough resistor will do and the with just resistors, ...
No. Increasing the probe resistance decreases the high-pass cut-off as it will be proportional to \$ \frac 1 {RC} \$.
... voltage divider works the same for both AD and DC source regardless of frequencies.
No. You have omitted the effect of the scope's input capacitance.
If there are intrinsic reactances in the wiring, then can't we just add a big enough resistor in series with those to make any reactance negligible?
Increasing the resistance decreases the signal available to the 'scope to the point that you won't be able to get high enough resolution and ADC noise will become a problem.
Figure 1. From Introduction to oscilloscope probes.
The trick is to use two potential dividers; one resistive and one capacitive. Since we've got a ratio of 9:1 with the resistive divider we need to do the same with the capacitive divider. Remember that the capacitors impedance is given by \$ Z_C = \frac 1 {2\pi fC} \$ so
$$ \frac {R_1}{R_2} = \frac 9 1 = \frac {Z_{C1}}{Z_{C2}} = \frac {\frac 1 {2\pi fC_1}} {\frac 1 {2\pi fC_2}} = \frac {C_2}{C_1}$$
From this we get \$ C_2 = 9C_1 \$. With the values shown in Figure 1 we can just about achieve this if C1 is 8 pF and CCOMP is wound up to the max to give us 72 pF total (although this model omits the cable capacitance so we have room to spare). |
H: how to transform 120V AC to 5V DC 5A load using flyback transformer circuit? Setting frequency - low cost, low footprint - change load capacity
i am using a flyback transformer circuit to keep the footprint small and cost efficient. Its been suggested to me to keep the frequency high so the transformer can remain small.
how is the frequency set in a flyback converter circuit?
Which components are different when transforming a load of 400mA or transforming a load of 5A? (i am using 5A, though most of what i see on Google is people looking to transform to 400mA for 30-60V DC)
is this something i can find load measurements on my components for, if so, which components measurements should be reviewed when changing the load capacity to a greater amperage?
this is my application .. 120V AC -> flyback transformer circuit -> 5V DC 5A power draw
the others i have seen are 30V DC -> flyback transformer circuit -> 5V DC 400mA power draw
the power draw is much greater in my circuit, i am looking to find what changes there are.
the flyback converter circuit i am designing is going to change AC to DC also. I think (don't know, looking for confirmation on) a typical flyback transformer changes DC to AC and then back again in it's typical application. So it does not matter whether it is AC or DC on its first input?
i am going to use a tool called WEBENCH - its available from Texas Instruments https://webench.ti.com/ . suggested to me by @user4578 on the post that lead to this
Which buck converter can be used for 120 V AC to 5 V AC rated at 5 amps? - on that post it was decided to use an isolated transformer circuit to step down from 120V AC instead of the buck converter i had originally intended.
AI: how is the frequency set in a flyback converter circuit?
The frequency is set by the switcher IC. To find out the details you must look at the datasheet for specific chips. Its going to be different for every one of them.
Sometimes its a fixed frequency determined by the chip design and you cant change it.
Sometimes you add an external component (resistor or capacitor) to
set the frequency.
Sometimes there is a pin that can select from a set of frequencies by
applying a voltage.
Sometimes you use a separate oscillator to set the frequency.
As far ad determining what a reasonable frequency is...
Somewhere in the 100KHz to 1MHz range is pretty typical, but higher and lower is not impossible.
It depends on your transformer. The transformer will probably only
be rated at a specific frequency, or range of frequencies. Using
it outside of whats specified you don't have any guarantee on
performance. So use what the datasheet says if you can.
Your switcher IC will also have some frequency limitations that are
stated in the datasheet.
Higher frequencies can use smaller transformers but have higher
switching and core losses. so its a trade.
Which components are different when transforming a load of 400mA or
transforming a load of 5A? (i am using 5A, though most of what i see
on Google is people looking to transform to 400mA for 30-60V DC)
All other things being equal the transformer is going to be 12.5 X larger for 5A vs 400mA.
Wurth electronics has a great tool called "Red Expert" that can help with transformer selection. They have a whole section in the tool to select fly-back transformers for offline switchers.
https://redexpert.we-online.com/redexpert/
is this something i can find load measurements on my components for,
if so, which components measurements should be reviewed when changing
the load capacity to a greater amperage?
When increasing the amperage, all the power path components need to be reviewed.
Rectifier Didoes, wattage rating increases proportional to current.
Transformer, volume increases proportional to output power
Switcher IC, need to check that it can handle current levels.
Switching transistors (if any)
Bulk holdup capacitors probably increase in size proportional to output power level.
the flyback converter circuit i am designing is going to change AC to
DC also. I think (don't know, looking for confirmation on) a typical
flyback transformer changes DC to AC and then back again in it's
typical application. So it does not matter whether it is AC or DC on
its first input?
Some converter designs work with either AC or DC input, some only on AC. There is no universal answer here. Usually if there is a full bridge rectifier at the front end of the design, then there is a good chance it will work with DC also. |
H: Variation in collector current if we change the collector resistor in a common emitter transistor
If we increase the collector resistance then how will the collector current change?
I understand the circuit behavior if i neglect early effect but when i try to include it that is allow Ic to vary with Vce i get lost because everything changes and it's as if i lose the grip on all variables.
P.S I'm a self studying person solely dependent on internet and books, so I hope that this question isn't blocked because it may look like a homework problem.
Thanking you in advance.
AI: Yasir Sadiq...here is my short explaantion:
When modifying the collector resistor (without changing the input signal) the collector voltage - and with it - the voltage Vce changes. For example, a reduction of Rc will increase the voltage VCE across the resistor.
This voltage increase will increase the depletion zone of the reverse-biased B-C pn junction and consequently reduces the width of the remaining base region associated with the forward-biased B-E pn junction. As the result, the E-field within this region will rise (constant B-E voltage) and a larger portion of the emitted charged carriers will arrive at the collector (Ic rises somewhat).
This effect is called "base-width modulation" and is the cause of the Early effect. |
H: PWM Three Phases with PIC
I am trying to program a PIC18F4550 to control a 3 Phase BLDC motor.
I have doubts about how to make the program, since I have seen many examples that what they do is use the TIMER to generate the 3 shifted PWM signals.
And I don't know why the PWM module that this PIC has is not used to do this process.
Is it not possible to generate 3 shifted signals with the PIC18F4550?
AI: The PWM module in the PIC18 that produces 3-phase PWM is not called a Timer. It is called the Enhanced Capture/Compare PWM module (EECP). It is not a timer, but uses the one of the other timers.
There seems to be conflicting info about when the ECCP's features (even if the ECCP itself might not be) are present on 28-pin devices: |
H: Can switching adaptor perform constant current mode?
I want to ask about wall adapter (switching) with 2.5mm jack
let's assumed its label as 9V 2A.
I saw many tutorial even some product use normal adapter (2.5mm jack) with BMS module to charge Li-ion cells. (for this function BMS just over voltage protect and balance for each cell) So BMS module it's self don't have function to control charging current.
Q1: What happen when load draw current more than 2A? (Will it perform constant current mode?)
Q2: Is it safe to use wall adapter to charge battery? (Can we consider current on label as charge current?)
AI: Q1: What happen when load draw current more than 2A? (Will it preform constant current mode?)
It could go into current limit. It might not and depending on the (lack of quality) could overheat and go on fire.
Q2: Is it safe to use wall adapter to charge battery? (Can we consider current on label as charge current?)
No. A charging circuit to suit the battery chemistry and capacity must be used to do this safely. It may be possible to find one that uses your 5 V power-supply as the mains to 5 V DC converter. |
H: Calculate resistance of coil/inductor made of PCB traces
This is to be a NFC coil/inductor of 1.1uH according do ST's eDesignSuite tool, with 7 turns.
The drawing above follows these parameters:
ST's tool give inductance value as 1.1uH, but also I need to know the resistance of entire track, in ohms.
I have measured the antenna total lenght on the layout software and it is 418 mm long.
So I have these parameters:
0.25mm width
418mm length
0.5 oz (17.5 micrometers) copper thickness
Then I entered these value of this calculator: https://www.allaboutcircuits.com/tools/trace-resistance-calculator/
And the result given by the site is 1.66 Ohms. Can I trust this value? I need the know correct value of resistance to calculate the components of RF portion of NFC filter and matching circuits.
The board will have 0.5 oz copper thickness, I can guarantee this.
Help please. Regards.
AI: Resistivity of Annealed copper: 1.72x10-8 Ωm
Resistivity is defined as:
\$ \rho = R\cdot \frac{A}{l}\$
therefore \$ R = \frac{\rho \cdot l}{A} \$
your length is 0.418m your area is \$0.25\,\mathrm{mm}\times 17.5\,\mu \mathrm{m}\$
therefore your resistance = \$ \frac{ 1.72\times10^{-8} \,\Omega \mathrm{m} \,\times\, 0.418\, \mathrm{m}}{ 0.25\,\mathrm{mm} \,\times\, 17.5\,\mu \mathrm{m}} = 1.64\,\Omega\$ at 20C |
H: Can cells get burned in EPROM by programming twice?
I have ordered my monthly electronic components dosage online, due to the covid-19 disease... in my dosage and for the first time I got an UV EPROM and started a personal challenge of programming it myself without following any tutorials because 1. I didn't find any tutorials on my UV EPROM and 2. I wanted to understand the datasheet for the first time, as you might know I am in secondary school not yet entered a college, yet I love electronic engineering and ask questions, search for courses etc... but anyways I have successfully followed the datasheet and programed my EPROM using 12V and pulsing the correct pulse time which was a great achievement for me !!! Yet I wanted to program another address but unfortunately as my setup was so crude, no switches or anything just generic hanging jumper wires on a breadboard, I forgot to change the address and programed again that address with new data bits, I quickly shutdown everything and when I toggle read mode on that memory address all bits were zero ! My PROM is "M27C256B" and I don't know if I have burned this address or no ..... I tried programming another address and it worked, yet I am now very nervous about the bits I reprogrammed unintentionally again.
In short, if an address on an EPROM is programed twice with different data bits, will that address get burned forever? Or just the EPROM need to get UV cleared?(I couldn't test as I didn't have enough amount of UV LEDs to successfully try clearing the PROM)
Thanks in advance.
AI: The MC27C256 and similar UV EPROM cells are at a logic '1' when bulk erased by exposure to UV light, and a cell can be electrically written to a logic '0'. Re-writing a cell that is already '0' to another '0' has no effect, and attempting to re-write a '0' cell to a '1' will be unsuccessful. Once a cell has been set to logic '0' it remains '0' until the entire EPROM is erased to all '1's.
By the way, the same silicon chip that is in the UV EPROM may also be sold at a lower price in plastic package as OTP (One Time Programmable); it's the same chip except it can't be erased because of the lower-cost packaging. The little window on the top of the UV EPROM package is made of quartz crystal, and it has to be encased in a package that has the same temperature coefficient of expansion, so plastic is out and metal is out, it has to be ceramic. The cost difference is significant.
Back in the day, we used UV EEPROMs for firmware development, and once the firmware was pretty stable we'd switch to the cheaper OTP for building up a larger batch of prototypes or building production boards.
The normal cycle involved plugging the EPROM into a device programmer to write the memory image, then cover the window with an opaque removable paper label (both to identify the firmware version and protect against stray UV light possibly eroding the image). Then unplug from the device programmer and plug into the target board for testing. More changes needed? Then put another EPROM into the device programmer, write image, label it, unplug the old EPROM and plug in the new EPROM into the target board. Then scrape off the sticky label from the old EPROM, clean off the window, and put it in the UV eraser like stacking dirty dishes in the dishwasher, until there are enough to run a full cycle. The UV bulb has a limited lifetime, so you want to erase as many EPROMs as you can with each 20 minute cycle.
All this changed for the better once Electrically Erasable (EEPROM) became widely available; no more need to pull EPROMs from the sockets and bake for 20 minutes in the UV eraser bin. Eliminated the need for the UV bulbs, the sockets themselves and the plugging and unplugging, so developers were willing to pay a premium for EEPROM. |
H: Why are inverter modules extremely oversized in VFDs?
I was checking a faulty Schneider VFD (ATV312H037N4) which is rated for a nominal output current of 1.5 A. Turns out the rectifier bridge of the inverter module (FP10R12YT3) blew up and after reading the datasheet I was surprised to find out that the IGBTs are rated for a nominal current of 10 A and the rectifier diodes are rated for 1600 V and 25 A. After checking some other drives from different brands I noticed the same thing.
I get that oversizing of components is normal for this kind of application but this seems a little extreme. Why would they do this? What's keeping this VFD from running a much larger motor that's closer to the ratings of the inverter module?
All the heavy lifting is done by the module so I don't see the reason to why the VFD is rated so low.
AI: Those current ratings only consider conduction loss ratings. But there are also switching losses. Even then, sit down and calculate the temperature. It is never anywhere near the rating.
There are also startup currents and peak currents where the heat generated by \$I^2R\$ losses is produced faster than the heat can spread and equillibriate on the transistor substrate causing hot spots.
There are also all sorts of nasty voltage transients.
I was monitoring VFD currents the other day on a scope. For a 1HP, 230V 3-phase motor, startup currents on a single phase were repetitive 18-30A bursts lasting miliseconds while the motor was spooling up. The nameplate current was 2.68A. |
H: XTR116 4-20 mA Transmitter Measuring Problem
I have created the following circuit:
On the DAC side, output in the range of 0-4.095 V, VREF = 4.095 and VCC_ISO = 5 volts. Totally normal. But XTR116, I cannot control less than 16 mA current. Between 16-20 mA ıs normal. But I have never seen less than 16 mA. Vcompliance > +7.5. (So no problem.) I can not figure out. Anybody have an idea?
Original Circuit: https://medium.com/electronza/4-20ma-current-loop-arduino-tutorial-part-i-hardware-bb50935da42e
AI: You cannot draw unlimited current from the regulator in the XTR116 and still reach the lower limit of < 4mA. It's just a linear regulator, after all.
The XTR116 draws 0.3mA itself, so you have perhaps 3-3.5mA to play with. I suspect you are drawing much more than that. |
H: Phase relationships in AC circuits
Is it wrong to use the mnemonics "ELI the ICE" [ From my understand means if its inductive current LAGS voltage and if its capacitive current LEADS] in understanding the phase relationship of a Parallel RLC circuit? because I've just read in a book that its the opposite because given a parallel LC circuit,if current in the inductor is greater than current in the capacitor the current lags or if XC > XL current lags.
AI: The mnemonics work fine and are for AC steady-state (phasor) conditions. They are restricted to the individual elements, not the circuit as a whole. So, when you say "ELI", it refers to the current into the inductor with respect to the voltage across the inductor.
In the phasor diagram to the right i arbitrarily align the current to the 0 degree axis. Steady-state, positive frequency phasors rotate counterclockwise. So, you can see the voltage is leading the current by 90 deg (it is also correct to say that it is lagging the current by 270 deg but we normally prefer to use the smaller angles when describing and that fits the mnemonic).
The circuit below is an inductor in parallel with a capacitor, fed from an ideal AC voltage source. The top-right phasor diagram shows both element's current phasor referenced to the voltage across them (VAC). The current coming from the source is the sum of these two (KCL). In my example the capacitive current is larger (light blue). When you add those two together you end up with a net resultant that is still capacitive (the bottom right phasor diagram). So, as far as the source is concerned, the load just looks like a capacitor. This is what @Reroute was describing concerning power factor correction.
When you introduce resistance you can now end up with a resultant current from the source that is somewhere between completely capacitive and completely inductive. |
H: Why CMOS turns on when input voltage is zero?
When the input voltage is high, Q1 is off and Q2 is on. In this case, the shorted Q2 pulls the output voltage down to ground. On the other hand, when the input voltage is low, Q1 is on and Q2 is off.
Now, the shorted Q1 pulls the output voltage up to VDD.
BUT I read that in CMOS, the high is +VDD and the low is 0. How does Q1 turns on when input voltage is zero(won't it only turn on when Vin is negative) ?
AI: Q1 is a p-channel MOSFET (aka "PMOS").
It turns "on" when the potential difference between its gate and source (\$V_{gs}\$) is negative.
But where is its source connected? It's not connected to ground (in fact, none of its terminals are connected to ground). It's connected to Vdd.
So what turns Q1 on is having the input voltage (its gate voltage) brought substantially below Vdd.
Since it has no connection to ground it doesn't "care" what the gate voltage is in relation to ground (i.e. whether it's positive or negative). |
H: Using a LiPo battery or different battery for outside project
First I am sorry if this is the wrong section to post this, I have been searching for a clear answer and haven't found one yet.
I have a project that I want to keep outside. I have a solar panel, charging circuit and a Lipo battery. It works great, though I have read a lot about how LiPo batteries can fail and catch fire.
I am wondering if Lipo is the way to go for something that will stay outside for years. It could get cold and hot from the sun :/
Should I redesign to use a lead acid? Would it be safer???
I know that was a lot, If this is the wrong place for this please let me know and I can remove this post,
If this is the right place, thank you so much for the assistance.s
AI: LiPo / LiIon batteries are almost always safe when charged and discharged within specs.
To get usefully longer lifetimes
Try to charge to less than the usual 4.2V (say 4.1 or even 4.0 V)
This gives longer and much longer cycle lives and whole of life capacity at cost of some per cycle capacity.
Do not discharge under say 3.2V.
Terminate CV phase of charging at high current - say CC/2.
Be sure CCCV charging is done correctly and NEVER "float" at 4.2V once charging is complete.
With solar you may have variable rate charging. This is acceptable as long as CC rate is within spec. Some cells allow C/1 and some C/2 rates. C/2 is less stressful.
LiFePO4 is better again for longevity and safety.
Look at manufacturer's temperature limits and observe them.
How cold does it get. Sub zero temperatures can cause issues if specs not met.
Ventilate to ensure summer temperatures are as low as reasonable.
How long is "years"? |
H: Idle supply current at frequency or vcc
I'm trying to figure out how much power I can expect my ATTiny84 to use in idle mode and I'm not sure which of the two graphs from the datasheet I should use, or why there are two in the first place. I intend to run it at a clock speed of 1MHz and off of a 3V battery. Also, if I choose to run it using an 8MHz crystal with CKDIV8 set (so the clock speed is 1MHz), should I read the graph at 8MHz or 1MHz?
AI: The top image is when using an external clock source and does not include any current used by that external clock source
The bottom image is when using the internal RC clock source at 1MHz, using that circuitry draws some power,
If your using the internal clock source, you would be using the bottom graph. |
H: Opamp input bias current and current sources
I have a current source connected to a load and I want to measure the voltage at the output of the source, and therefore I would like to buffer this voltage.
In this configuration, I am under the impression that the opamp will always draw the same amount of current from the current source regardless of what current it is set to output. (Eg an opamp with a bias current of 1uA buffering the output of a current source producing 1000uA will result in 999uA going to the load. If the current source produces 2000uA, 1999uA goes to the load.)
If this is the case, it seems like I don't need to necessarily seek out an opamp with a low bias current, but I could simply offset the current source by the bias current to ensure that my load gets the desired current.
Is this thought process correct? Or would using a very low bias current opamp still be of significant benefit here?
AI: The input bias current may be relatively constant, but it will tend to change with temperature, input voltage and supply voltage, and will generally not be well specified to begin with.
Take an example, such as the MC33078. At 25°C input bias current is typically 300nA but may be as high as 750nA (no minimum is given). Over temperature, it could be as high as 800nA. Pretty wide range- but you could measure it easily at room temperature anyway and calibrate it out. Remember that's only with one particular set of power supply voltages, one temperature, and one common mode and output voltage.
Now, how does it behave with the other parameters.. (typically, no worst case figures are offered):
Changes with supply voltage, but if you have regulated supply voltage it's not too bad.
Not great, changes about 50% with temperature.
Okay here the current changes with voltage, meaning it appears like a resistance
in parallel with your current source output (degrading it). It typically changes 100nA between -5V and 5V, which means it acts like a 100M\$\Omega\$ resistor. Of course it could be better or worse than that.
So it really depends on your accuracy requirements, and there is no substitute for a thorough analysis if accuracy is important.
JFET or MOSFET op-amps tend to have much lower bias currents, so even if they vary a lot it won't make much difference to a relatively large current like 1uA.
That all is at DC. At relatively high frequencies like your 20kHz, other (possibly much larger) error effects can come into play. |
H: Solar Panel-Photocell Wiring
I am trying to help my dad with a project of his where were trying to build a lighthouse powered by solar panel in which it spins a motor at night and charges battery during the day. This is a DC application. We have the front end, being charging the battery all figured out. However, the backend, being controlling motor and having motor only turn at night, is the hard part. We have a controller for controlling the speed of the motor working with no issues. It is the last part that we need help with, which is the photocell sensor light switch. We figured that we need to place the photocell between the battery and the speed controller. The photo sensor switch says it works for AC and DC application and says AC/DC on the device we received. The directions that came with it state this for working with DC application, "black line connect to input power (-) negative. White line connect to input power (+) positive and load device (+) positive. Red line connect to load device (-) negative." So I believe line is meaning wire and load device will be the speed controller with power being my battery. I have tried it using the directions and doesn't work. I looked on comments on the amazon page where we brought the products (got 2 of them) and the comment said that black wire is hot wire with red wire goes to load and then white wire is ground. That didn't work either. I have waited at least a minute with light shining directly at the sensor and it doesn't turn off the motor to simulate day. Here is a link to my diagram with both directions based and amazon comment based that I tried: https://i.stack.imgur.com/abZwW.jpg
Here is a link to the sensor on amazon: https://www.amazon.com/gp/product/B01M1O3C1V/ref=ppx_yo_dt_b_asin_title_o00_s00?ie=UTF8&psc=1
Here is a link to the speed controller on amazon:
https://www.amazon.com/gp/product/B00F839VNQ/ref=ppx_yo_dt_b_asin_title_o02_s00?ie=UTF8&psc=1
here is a link to the motor: https://www.amazon.com/gp/product/B07FKKWSFJ/ref=ppx_yo_dt_b_asin_title_o08_s00?ie=UTF8&psc=1
The battery is a car battery rated for 12 volts and 23 amp max. The speed controller is rated for 6v-90v at a max of 15 amp. The sensor is rated for 12 volts at 10 amp max. We need the speed controller to slow the speed of the motor as it is way to fast for those of you wondering. The motor is rated for 12 volts at 150 watts max. I do not know if I have fried the sensor or not, but I do not think so because I do not smell any smoke or anything and it is NOT black to indicate frying it. I do have the second one I can use, but I prefer not to use it until I know for sure what I did wrong in case my current one is damaged, if it is.
So my question is what is wrong with my diagram or why isn't it working? Thanks in advance.
AI: Your diagram is wrong.
Based on the labels on the photo-switch in this image.
Connect battery negative, controller P- and photoswitch white wire together
Connect Battery positive to photoswitch black wire
Connect photoswitch red wire to controller P+
If that does not work the photoswitch may be dead. |
H: Mounting a heavy PCB in standing position
Due to space restriction I want to mount the PCB vertically and secure it with M3 screws to the box bottom like this:
simulate this circuit – Schematic created using CircuitLab
It's a power source board with a lot of big capacitors and is heavy, I'm not sure about doing it. are the capacitors going to fall off after a while? is it OK to mount a heavy PCB vertically?
AI: The solution for this would be some rigid support rails that run up the side of the board, and ideally anchor it top and bottom to prevent how much it can flex, The top of the board is where most of the weight is, so I would strongly recommend you look for any way to make sure its secured some way at the top.
if you cannot, you will need to make up with thicker support rails, usually rectangular profile screwed to the board at the corners at minimum, more usually top bottom and center
In addition to this glues similar to the craft hot glue are used to secure each capacitor to its neighbors (a small amount in the gaps) this way they cannot work harden off the PCB |
H: What's the difference between similarly rated DC motors with different nominal voltages?
I'm looking at a datasheet that describes several versions of the "same" DC motor. The versions are mostly similar (e.g. same size, no-load speed...) except for the nominal voltage, no-load current and nominal torque. What's the internal differences between these versions (your guess)? and what are the pros and cons of choosing one over the other?
AI: For mechanically similar motors, the difference is the number of turns and the diameter of the wire on the rotor. One motor might use 200 turns of 1 mm2 wire, another wound for half the nominal voltage may use 100 turns of 2 mm2 wire. This means they have the same number of Ampere.Turns when running at the same power. They'll produce exactly the same heat from the rotor.They produce different motors simply for the convenience of being able to run them from different nominal voltages.
However ... Wire is only available in certain sizes, there may not be exactly the 'right' size available for the next voltage. The ratio of copper diameter to insulation thickness will not necessarily remain constant. Thicker wire is difficult to pull tight round the poles, so you may get fewer turns than the ratio suggests than with thinner wire. You can only use an integer number of turns. The air resistance of different wire thickness will be slightly different. All these differences mean there will be subtle, or even not so subtle, differences in the motor performance, especially with motors at the extreme end of the range.
All the motors in the datasheet have nominally the same speed, torque and power, with the exception of the torque of the 6 V motor, which is lower. Maybe this is due to the thicker wire meaning fewer turns on the rotor?
While the mechanical, rotor, heat are all nominally the same, the commutator will be running at a high current in the lower voltage motors. This may mean they need uprated or wider brushes, or that the brushes and sliding contacts don't last as long. The commutator in the higher voltage motor is subject to a higher voltage, as the rotor has a higher inductance. This may mean they need bigger gaps, or higher resistance brushes, or that the edges of the commutator don't last as long. |
H: Modify LED circuit with a MOSFET
I have this existing LED circuit with 2 LED strips with built in resisitors and 1 switch. Normally 1 LED strip is always on and when the switch is closed the other LED strip also turns on. Power Source is a Sealed Lead Acid battery 6Ah and the switch is a regular rocker switch.
I want to modify this circuit so that an MCU can control the LED strip which was wired to be always on in the existing circuit. So I have made this new circuit diagram which I want to verify is correct or not. In the new circuit diagram (+) power from the closed switch goes to a tiny buck converter which steps it down to 5V for the MCU (ATTiny13a). (- ground) of the second LED strip goes to the drain of an N channel Logic level MOSFET which has it's gate connected to the digital out pin of the MCU. The function of the MCU is to make the LED strip blink every other time the switch is closed. The programming for the MCU will make use of the EEPROM to achieve this so that it knows (remembers) when the switch was closed last time.
In short, the intended function of this circuit goes like this
Switch Open - Both LED strips OFF
Switch Closed - First LED strip ON, Second LED strip OFF
Switch Open - Both LED strips OFF
Switch Closed - First LED strip ON, Second LED strip blinking
... and this function repeats
Note: I want the second LED strip to stay OFF even if the MCU isn't powered.
So I wanted to know if my circuit was correct or is there something wrong? like the way MOSFET is connected to the MCU?. Also is it possible for me to switch the (+) of the second LED strip with the MOSFET instead of switching it's (-) ground?
AI: If you took more time thinking about drawing your circuit (extremely important) you'd be able to unclog your thoughts and see that the addition of a 1 Mohm resistor (purple) would be a good idea: -
Note: I want the second LED strip to stay OFF even if the MCU isn't
powered.
The 1 Mohm resistor is there to protect the MOSFET and achieve the above should the ATTiny become disconnected. It's optional of course.
It's still not drawn perfectly but is a lot better and folk might have a chance of understanding it better.
Also is it possible for me to switch the (+) of the second LED strip
with the MOSFET instead of switching it's (-) ground?
You could use a high-side PMOSFET and low side NMOSFET to do that like this: -
Picture from here. |
H: What board should I use to control my 3D printer
I'm trying to build a simple experimental FDM 3D printer as a hobby.
As I said it's a simple printer with only basic functionalities such as receiving the slicer file to print from a slicer software from PC and sending back the print status to the PC.
Following are other functionality of the printer:.
Sends the X and Y coordinates of the printhead and Z cordinate of print platform to the PC.
Sends the temperature of the cartridge heater to the PC with the help of a temperature probe.
Send the number of layers printed currently.
Now in order to achieve the above mentioned functionality and control the printer via the PC, what microcontroller motherboard is ideal to use? Is using Arduino Uno ideal enough? Or should I use Arduino Mega or any Rasbery Pi board.
AI: There doesn't seem to be anything different in your application to any of the 3D extruder-based printer applications. All this has been sorted out many, many times already.
Most use an Arduino-based board with built-in stepper drivers and temperature control interface. Marlin firmware, for example, allows monitoring of position and temperature on the printer's LCD so that you don't have to leave the laptop connected - although you can if you want to.
simulate this circuit – Schematic created using CircuitLab
The slicer software doesn't generate a "slicer file", it generates CNC g-code which can be transferred by SD card to the printer which reads and executes line by line. Most slicer software can drive the printer over USB or Ethernet but that ties up the laptop for the duration of the print. |
H: Clip a PCB against a 3 mm plastic protection
Is there a way to clip the PCB against the 3mm plastic protection surface? The image is below.
I need that it stays in the position shown in the image, with the connectors side sticking out.
That PCB is just for the example, it will be a new PCB so I can change its form or placement to solder some kind of clip to the PCB that would just clip into the 3mm plastic protection.
AI: Google search: Double-sided tape PCB pillars
You can get these in various sizes to suit different heights and different holes: -
They have double sided tape on the bottom so they just fix in place and you should be good to go. Here's a different type: -
And a range of heights: -
You can even get them that work with a screw too: -
There are plenty of options and quite a few suppliers: - |
H: Trying to understand this Class D amp (using IR2113) circuit
I understand the concept of the Class D amp and this circuit, but in this particular case there is one part I just don't get.
What I don't understand is how the low side of the signal should work at the power stage. In case of the high side as I understand when the MOSFET is on, the current can flow from VCC to the load. But at the low side I don't understand how could current flow through the load and the low side MOSFET.
Could you please clarify it? Thanks!
Schematic: https://easyeda.com/omerzaid/New_Project-e38fb4c8918946298058a93ec7ef210a
AI: If you dig up the original Great Scott video you'll see that he has put a 1000uF capacitor in series with the speaker. It can be seen at 5:37 in this video: https://www.youtube.com/watch?v=3dQjIeYoIdM
It stops the DC component of the output signal. Without it the speaker would get continuous DC about half of the DC supply voltage. It could still work somehow but the dissipation in the speaker would be unacceptable at least for me.
I have said it previously in other cases: Do not believe YouTube entertainers. Believe even less those who build their show on some YouTube video. You must get to know the things by understanding them. Your question here shows you have already thought it. |
H: ST ST25R95 - NFC chip - Impedance of transmission lines
I will supply the chip with 3.3V, and the datasheet of ST25R95 IC shows it has a ZOUT of 27R at this condition.
There is a circuit between NFC antenna/coil/inductor and the IC, the LC filter and matching circuit:
My doubt is about the black lines shown on picture above, the transmission lines. Should they have 50 ohms? or 27 ohms? I think it should have standard 50 ohms but I would like to have sure about that.
Here is the datasheet if you need: https://www.st.com/resource/en/datasheet/st25r95.pdf
Regards.
EDIT:
There is a distance of around 10mm between the transmitter pads and the antenna:
EDIT 2:
Calculation of microstrip with AppCad of Avago Technologies
AI: My doubt is about the black lines shown on picture above, the
transmission lines. Should they have 50 ohms? or 27 ohms? I think it
should have standard 50 ohms but I would like to have sure about that.
The device operates at 13.56 MHz and the wavelength of 13.56 MHz is 22 metres in free space and about 15 metres on a typical PCB. So, the golden rule is that if your tracks are longer than about one-tenth the wavelength then you should start to consider using transmission line calculations.
Are the tracks likely to be close to anything like 1.5 metres long is the question you should consider. If not then don't worry about matching because it won't make a difference practically.
I've used 13.56 MHz because the signals are largely sinewave at 13.56 MHz. If it were a clock of 13.56 MHz then I'd be considering t-line math if the track length was more than 20 cm. |
H: My oscilloscope tells me I vibrate at 60Hz. How?
As I live in North America, my initial hypothesis is that I'm picking the 60Hz from my house's power line.
But, to my understanding, my body can't be a 60Hz antenna (1/4 wave antenna length for 60Hz is ~1250 km.)
I feel like my understanding of "indirect" coupling is incomplete.
Could somebody help me?
AI: You're picking up near field effects, which is a different mechanism than normal electromagnetic radiation.
See also: Designing 60hz antenna to detect electric field |
H: Activate 3 or more PWM signals in PIC
I am doing a PWM with the PIC18f4550 and it managed to do it.
But only activate an output by varying the value of CCP1CON.
What I want to achieve is to have three PWM outputs and then be able to offset them to control a BLDC motor.
I attach the part of the code where I initialize the PWM:
public override PWM_Initialize ()
{
PR2 = ("% # x", (_XTAL_FREQ / (PWM_freq * 4 * TMR2PRESCALE)) - 1);
CCPR1L = 0x1F;
TRISCbits.RC1 = 0;
TRISCbits.RC2 = 0;
TRISDbits.RD5 = 0;
TRISDbits.RD6 = 0;
TRISDbits.RD7 = 0;
T2CON = 0x03;
//CCP1CON = 0b01001100;
CCP1CON = 0b11001100;
TMR2 = 0;
T2CONbits.TMR2ON = 1;
}
Attached image of the idea I have of the driver for the BLDC
AI: If you have the 40-pin PIC18 then CCP1 has been replaced by the ECCP and you shouldn't have any problems.
Just enable Quad PWM mode and you will have 4 PWM channels: P1A, P1B, P1C, and P1D.
That's not entirely ideal though since for a BLDC you want 6 PWM signals since you have 6 transistors and you want to be able to control the dead-time between them.
That means that you have to forego complimentary PWM since you can't have independent control of all 6 switches for dead-time control or signal inversion which would be required if you are using NMOSFETs on both high-side and low-side.
You will probably have to PWM just high-side switches and use regular logic signals to just turn on the low-side switches during the commutation cycle with no PWM.
If you have a high-side bootstrap capacitor, this might need tweaking, especially during startup where RPM is lower so commutation cycles are longer. The reason is that the bootstrap cap needs to be constantly refreshed and therefore requires the low-side switch to turn on regularly enough to recharge the cap and you won't be able to do that regularly during a commutation cycle without complimentary PWM. The easiest thing to do is probably just to oversize your bootstrap capacitor so it can sustain the switching required throughout an entire commutation period without needing to be refreshed. After the commutation cycle is over the low-side switch for that half-bridge will get a chance to turn on during the commutation cycle for another phase so you can recharge the bootstrap cap. |
H: Am I calculating my shunt FET power dissipation correctly?
simulate this circuit – Schematic created using CircuitLab
I'm designing an LED driver circuit that requires very fast PWM switching at a high resolution (10bit at 30khz requiring a minimum pulse width ~30ns). In order to achieve the pulse timing required, I'm using a FET to shunt the LED driving current to ground during the "off" periods. I'm trying to determine the worst-case power dissipation for the FET, so I can plan heat mitigation strategies when designing my board.
One of the FETs I am considering is the BUK9M52-40E, because it works on logic-level voltages and on/off times in the single ns range. It is not a power MOSFET, however, and does not appear to have a substantial heat sink build in, so I would presumably need to rely on copper pours and thermal vias for heat management.
I will be driving 8 LEDs in series at 24V at a constant average current of 360mA with a peak current of 540mA.
The FET has a maximum RDSon of 52mO at 5V, though I hope to drive it with a 3.3V signal, which from what I can determine would result in an RDSon of closer to 100mO. So given a constant average current of 360mA, and assuming 100mO of resistance, (0.36A x 0.01O) the voltage across the drain-source will be 0.0036V. So the power dissipation in the FET will be 0.0036V x 0.36A = 0.001296W.
This number seems... really low. And given that the FET is rated to 31W, I assume that this is well within its capabilities, and I could probably get away with very little thermal management, and potentially a much smaller FET. Am I correct in this assumption?
AI: Power dissipation is \$I^2R\$ so 12mW, plus switching losses. I didn’t check how realistic your 100m\$\Omega\$ looks. It apparently is not guaranteed, so you’re rolling the dice.
Note that average current is not as important as the average current squared (which is larger than the square of the average current if it’s switching at all).
Note that the switching speeds are specified with 1A peak gate current, so they are assuming a good gate driver. There is a lot of Miller capacitance in your situation due to the large change in Vds. |
H: Is this the right convention for SPI modes?
The table shown here, under "Clock Polarity and Clock Phase" doesn't agree with my conclusions from table for SPI modes on Wikipedia.
I'd like to distinguish between whether the one on the first like is another convention, or is it just just that 11 and 10 are mistankely swapped, I'm working on a school project and the document provided is from the first website.
AI: There is no SPI standard, so the Clock Polarity (CPOL) and Clock Phase (CPHA) selection bits do not need to map to modes 0-3 in any standard way. Sometimes the mode selection bits have different names so what you read about modes for one device does not apply to another device.
The point is, when you want to communicate SPI between two chips, you need to look at the timing diagrams between the chips and select matching setting to know the correct mode, instead of just looking at mode numbers. |
H: which voltage regulator to use for 12v 2A?
I have looked into using 12v linear voltage regulators, coming down from 16V, i need 2 amps and the regulator is only rated for 1.5A (7812 linear voltage regulator)
i am looking for a solution for getting 2A through the regulator.
my fan is 1.68A rated at 12V and i'd like a little more than that.
can a buck do it? Other linear voltage regulators rated for more amperage?
Thanks!
AI: A linear regulator would waste 8W which is quite a bit, so a buck regulator would be much better.
A buck regulator module based on Xlsemi's XL4016 would do the job nicely with plenty of margin, you can find inexpensive modules on eBay, Aliexpress etc. |
H: Solar powered mini water pump: how to turn it on/off automatically
I have acquired a mini water pump (5V 1.0A) and several solar panels (each: 5V 200mA). What I try to achieve is to pump water from a container above a vertical hydroponic system and let it flow down back to the container while watering the plants inside. I want to pump the water only for ca. 3-5 minutes each hour during the day. Pumping non-stop during the whole day is not good for plants and unnecessarily wear out the pump. What is the simplest/cheapest way to interrupt the pump most of the time and power it only for 3-5 minutes during the daylight?
AI: You don’t really need multiple solar panels, I would add a small battery, Perhaps lead acid for durability, and a solar charge controller to keep it at the proper charge level. You should be able to get all of this off-the-shelf.
After that, just get an Arduino or a PI zero, and a relay shield or hat, the code to start once an hour run for a few minutes and then put the processor to sleep is relatively straight forward. And, although this is overkill and simpler circuitry could be designed to do this, it would require more experience that what you seem to have, it’s not necessary unless you are designing a sellable product, and it would save you a lot of aggravation later on. Particularly if you feel compelled to modify the behavior in the future.
What you would need if you want to put together a circuit from scratch:
A solar charge controller. An IC such as this one, its purpose is to optimize power extraction from the solar panel.
A battery or super capacitor. To store the necessary power to drive the pump and the rest of the circuitry.
a small switching regulator for the circuitry power.
A power switch, relay, or transistor to drive the pump. You might want to regulate the voltage to it unless you can directly power it from the battery. You can do this via simple pen and an LC filter, ourjust a simple switching regulator.
A reasonably stable oscillator. A cheap circuit would just use an RC, a less cheap one a ceramic resonator, but a crystal oscillator takes all the guess work out of it.
A set of counters and flip flops to put together a state machine with the desired duty cycle of 3mins per hour (a purely analog oscillator with such a duty cycle would be very hard to pull off). This is where I would use a timer counter IC, which are being discontinued.
Such a system would be very hard to modify once put together unless the changes are designed from the beginning.
Or for less hassle and money you can use a simple 8-pin microcontroller (e.g., a PIC or a tinyAVR) with its built-in RC oscillator and program it to generate the duty cycle, the pump PWM, and even a heartbeat LED so you know it’s working, and go to sleep to conserve power. The same thing you would do with an Arduino, harder to program, but with no unnecessary parts and perhaps more satisfying. |
H: What does a High Impedance input mean for an ADC
The ADC ADS131M08 list a feature:
Can someone elaborate what this feature implies? Does this mean i can place a 300 Kohm resistor in series with the inputs? There is even a reference on page 88 with 900Kohm for measuring 220 VAC.
Because if it is it would be great because i can increase the overvoltage the pin can handle just as disscussed in this article. Would i suffer some aliasing using high value resistors??
AI: Nothing to do with aliasing. I won't go into what aliasing is but it is so far removed from the issue that you should go do some reading to understand what aliasing actually is before ever suspecting or worrying about aliasing in anything.
It's telling you how much the input will load down a signal source you feed to it, and adding a 300K effectively increases the impedance of the signal source and would utterly demolish the ADC's reading of the signal.
simulate this circuit – Schematic created using CircuitLab
Voltage divider giver: \$V_{load} = \frac{R_{load}}{R_{load}+R_{source}}V_{source}\$
Obviously you want \$V_{load} = V_{source}\$ which requires
\$input.impedance.of.load >> output.impedance.of.source\$
so as much voltage as possible appears across load and not lost in the source via voltage divider.
So do you see the problem if you make \$R_{source}\$ significant relative to \$R_{load}\$? It would result in something akin to drinking a sizeable portion of the wine barrel in order to sample the taste.
If you want to protect the ADC with a large series resistance, put a buffer in front of the ADC and diode clamp the input to the buffer. The buffer will have much higher input resistance so can tolerate a larger (but still small relative to the buffer) input resistance and not load down the signal as much. Simultaneously, the buffer has a low output impedance to work well with your ADC's input impedance.
simulate this circuit
The diodes, whether external or internal (for ESD protection), clamp the voltage to within on forward diode voltage drop of the power rails and are what actually protect the pin. The resistor in turn protects the diodes from frying by limiting the current through the diodes and dropping the extra voltage across the resistor rather than the diodes (which are basically short circuits when conducting). |
H: Summing amplifier design
I'm learning about op amps and I have the following problem. I need to make a circuit using one op amp that give me the following output (Vi input, Vo output)
I know that the output will be like
$$
v_o = 4\,v_i+6
$$
So I assume that I can make this work with an additive non-inverter amplifier, where one of the inputs will be the 6 V offset. I think that a configuration like this is what I need
I found that the equation for the output in this case will be this
$$
v_o=\left(\frac{R_f}{R_s}+1\right)\left(\frac{1}{R_1}+\frac{1}{R_2}\right)\left(\frac{v_i}{R_1}+\frac{V_2}{R_2} \right)
$$
And then I played around with the resistors but I didn't reach anywhere. I dont know what I am missing
AI: Opamps amplify AC signals and DC signals alike. In this case, you know you want a gain of 4, so look up an op amp summation circuit: one input will be your AC signal centered around 0V and the other input will be a DC offset BEFORE the amplification gain. So in your case, I'd use a voltage divider to get a 1.5 V input level and sum that with your AC signal. Once it's amplified you'll have your 4 Vp2p + 6 VDC.
Edit: Also regarding the voltage divider, make sure that the resistor values you're using to generate your 1.5 V DC offset are an order of magnitude lower than your addition stage resistors, you don't want your 1.5 VDC offset to be too high of an impedance that it gets dragged around by what I assume is a low impedance 1 Vp2p input sine wave source. In other words, if your voltage divider resistors form a 1 kOhm impedance voltage source, I would suggest that your summing resistors on the opamp input should be at least 10 kOhm. |
H: How resolve "logic" is an unknown type error in Vivado synthesis?
the verilog code at
output logic PIPE_PCLK;
AI: logic and bit are added in SystemVerilog 2012.
You need to tell Vivado the file is in SystemVerilog. It's documented
in chapter 8 of UG901 - Vivado Design Suite User Guide: Synthesis and
Appendix B of UG900 - Vivado Design Suite User Guide: Logic Simulation.
Targeting SystemVerilog for a Specific File
By default, the Vivado synthesis tool compiles *.v files with the
Verilog 2005 syntax and *.sv files with the SystemVerilog syntax. To
target SystemVerilog for a specific *.v file in the Vivado IDE,
right-click the file, and select Source Node Properties. In the Source
File Properties window, change the File Type to SystemVerilog, and
click OK.
Tcl Command to Set Properties
Alternatively, you can use the following Tcl command in the Tcl
Console:
set_property file_type SystemVerilog [get_files <filename>.v]
Depending on your taste, you may want the suffix to be .sv since it's
SystemVerilog. If you are creating the file from Vivado, you should
select SystemVerilog instead of Verilog.
You may also want to make sure the file's library property is not
empty. It usually defaults to xil_defaultlib which works. But on
some occasions (I don't know the exact conditions for this to happen),
it defaults to empty, and you get a strange error like "Incorrect
project file syntax" when you try to simulate it. It's next to the
"Type" property in the GUI. |
H: Boost Converter Stuck at Around 80 Volts
I'm building a boost converter that utilizes a 555 timer switching at 60 kHz at a 96% duty cycle which drives an IRF3205 MOSFET to power a hand wound inductor (calculated to have ~330 uH) with a 6A10 diode and a 400 V, 80 uF capacitor at 5 volts. My goal is to achieve 360 volts.
The problem I'm having is that the circuit consistently maxes out at around 80 volts and attempting to increase the voltage e.g increasing the inductance of the inductor by adding an iron rod, adding another inductor in series, increasing the duty cycle of the 555 timer and even increasing the voltage has no significant effect on the voltage. I can only lower the voltage using these methods.
I think it may be because of the diode limiting the current as replacing the 6A10 diode with a higher resistance one seems to decrease the voltage to ~60 and further attempts to mess with that also fail.
I do not have any more diodes that I can use and so I am wondering if it is current available to the inductors which is limiting the output before I buy any more diodes.
AI: The problem I'm having is that the circuit consistently maxes out at
around 80 volts
Consider that the IRF3205 might be a limiting factor: -
Then consider that the 6A10 diode will be next in-line for limiting the voltage: - |
H: 'true' balanced audio?
I was working on the below circuit, to create a balanced audio output.
But have been told this creates a differential audio output, not a 'true' balanced audio output.
"A true balanced output would couple interference on one leg to the other, improving common mode rejection at the receiver. Short either leg to ground. If the amplitude on the other leg doubles, it's balanced."
So my question is how to create this 'true' balanced audio?!
Everything I see online from hours of searching all show some variation of the above circuit, with opamps to create a signal and an inverse signal.
Can anyone point me to a circuit that would create a 'true' balanced audio output?
AI: "Balanced" generally means that the two wires that carry the (audio) signal have the same driving impedance at the source and the same receive impedance to ground. This means that both wires are affected equally by EMI when transporting the signal - if EMI comes along and "hits" the cable then both wires will be "interfered with" the same. This then allows a differential receiver to reject that interference.
It doesn't mean necessarily that the driven signal is differential - it needn't be in order to create a balanced signal but, quite often is because identical drivers are used at the transmitter in order to get a balanced impedance drive situation and making one output the inverse of the other gains another 6 dB signal to noise ratio.
But a balanced driver can provide 0 volts provided it is sourced through an identical resistor or driving impedance: -
Image from my answer here.
But have been told this creates a differential audio output, not a
'true' balanced audio output.
Your circuit driver looks balanced to me. It happens to be differential as well so that's an added 6 dB SNR.
Short either leg to ground. If the amplitude on the other leg doubles,
it's balanced."
No, that isn't a true definition of balanced in my book but, it doesn't exclude a circuit such as that (i.e. a transformer isolating output) from being balanced.
Wikipedia definition of a balanced line (and I agree with it) |
H: How is the current measured in this setup?
I came across this article where they measure the current called leakage current, but they are not using an ammeter or DMM.
I couldn't understand what(which device) they use to measure the current. What is the way of measuring current and what device is used in this case?
AI: They are using source-measure units (SMU). These are devices that combine voltage, current, charge, power, and energy, measurment with a variable power supply. |
H: How can current be in different phases in the same wire at the same time?
We know that either in ideal or practical transformer in no-load condition, the exciting current has two components Ic(core loss component) and Im(magnetizing component). Now since the same current will flow through the winding, how can the two current be out of phase as we learned the current goes out phase when the current is divided into paths, and there is the same voltage between two distinct terminals. You can explain this in an equivalent circuit but in the actual circuit the current division is not possible as there is just single wire winding.
Also, the same case is with the compensation current that appears in the full load condition. The two currents are out of phase in the same wire.
How is this possible? Is it just a theoretical concept just to make it easy for understanding, or am I missing something here?
AI: currents Ic and Im are components we derived for our mathematical calculations purpose,resultant of those both currents will flow through the winding.and for clear understanding read about vectors. |
H: 4-20 mA output from Arduino for VFD
I'm a beginner with electronics, so everything has to be explained in simple terms to me.
I have an Arduino with a program which provides a value via the Analog output, so that is anything from 0-5 V.
On the other side, I have a VFD with an IO card which allows it to connect to a 4-20 mA current loop, to vary the speed of the motor accordingly.
How do I convert from, let's say, 0-5 V output (PWM) from Arduino, to the 4-20 mA in a somewhat cheap and safe way (I don't want to destroy the VFD [and possibly the Arduino])?
Currently, I am using the circuit below:
The problem is that the simulation on "Multisim" is giving accurate results. However practically, the circuit did not give the accurate results.
For example:
When V = 5 V output (from Arduino analog), I = 20 mA (V=IR) --> which is correct
But, when V = 3.3 V output (from Arduino analog), I = 16 mA (V=IR)--> which is incorrect, as it should be 13 mA
Will this circuit work practically
AI: It will only work practically if the supply ground is isolated and an op-amp is used that is "single supply" such as an LM358. You could use the LM741 if you added a negative supply, such as -5V but that would be a lot of trouble to use an obsolete part. Using a more negative supply than -6V or so would expose Q2 to damage if the inverse Vbe rating is not high enough.
So you feed 5V from the PWM for 20mA (100% PWM) and 20% PWM (1V) for 4mA.
Maximum load resistance is about (12V-5.2V)/0.02A = 340 ohms to allow a full 20mA out. |
H: Is the current, Ids, through a MOSFET truly 0 when in the cutoff region?
I was testing a circuit using this circuit simulator and found that the nMOS transistor had a current of a few nanoamps running through it when it was in the "off" state. This was unexpected because I was taught (and found online) that the current through a MOSFET in the cutoff region was 0.
After a lot of thinking, I've decided that it seems odd that it would ever be exactly 0 since everything about a MOSFET varies in a continuous fashion. In the same way that there is no real point that the MOSFET changes from linear to saturation (as in, it does not suddenly switch to saturation as soon as , it's more of a transitional phase), it seems reasonable to assume that there is no real "off" state of a MOSFET, rather just a very low current linear region. But if that was true, surely I'd be able to find an equation which models the current in the cut-off region. I thought the linear region equation would work, but the results I get from that do not match those of the simulation.
The reason this is an issue for me is that it means the voltage output is dependent on the value of the resistor when in the off state.
AI: It's all about \$I_{DSS}\$ so pick up a data sheet for a MOSFET and look up that term. It'll tell you something like: -
It's the drain current that flows when the gate and source are connected i.e. \$V_{GS}\$ = 0 volts. It's usually specified for a high-ish drain source voltage of maybe 10 volts to hundreds of volts.
Basically it's the leakage current of the channel. |
H: Mutual induction
I need to find the mutual induction between two coils.
simulate this circuit – Schematic created using CircuitLab
The circuit look more or less like the one I have linked. I don't know how to calculate the mutual induction between the two coils. Does anyone have any experience they would like to share? I would appriciate it very much!
AI: Does anyone have any experience they would like to share. I would
appreciate it very much!
For simplicity, I'm keeping this answer based around a 1:1 transformer where the primary and secondary inductances are both 13.5 μH (as per the question).
100% coupling, no-load
If the coils are 100% coupled, the mutual inductance is \$\sqrt{L1\cdot L2}\$ and, the output voltage is the same as the input. However, if the output voltage is only 50% of the input voltage, then you can be sure that the coupling coefficient is half. This is because only half the flux produced by the primary couples with the secondary and, as per Faraday's law of induction, the induced voltage will be half. So, the unloaded transfer voltage ratio is a good measure of "\$k\$" when the transformer is a 1:1 type.
Adding a load, coupling less than 100%
When we factor in \$k\$, the mutual inductance formula becomes \$k\sqrt{L1\cdot L2}\$: -
When you add a load of high value you can usually assume that the transfer ratio defines the coupling coefficient \$k\$ but, be careful. With (say) an apparent 50% transfer ratio and a load that varies you can become mistaken. For instance, here's a 50% coupled transformer with a load that varies between 1 ohm and 100 ohm: -
As you can see, if you had a 1 ohm load resistor and your measured the transfer ratio at 100 kHz you'd see an attenuation of nearly 24 dB. At low frequencies (say 1 kHz) the attenuation is 6 dB (50%) so there's no problem measuring at 1 kHz but at higher frequencies you can misdirect yourself into assuming the wrong coupling coefficient.
Equivalent circuit
The reason why this happens because when the coupling is only 50%, leakage inductances become present and these, along with the load resistor, form a potential divider like this: -
In the above circuit, I've halved the coupled inductances to 6.75 μH and made k = 1 then I've added the leakage inductances as new inductors L3 and L4. This produces exactly the same result as my original circuit up above.
Now, the mutual inductance could be taken as \$\sqrt{6.75 \text{ μH}\cdot 6.75\text{ μH}}\$ which equals 6.75 uH or, could be taken as this: -
$$0.5\cdot\sqrt{13.5\text{ μH}\cdot 13.5\text{ μH}} = 6.75\text{ μH}$$
And, finally, the equivalent circuit not using a transformer is this: - |
H: Decode/Analyse the following UART signals
I have sent a 2 and a 9 using UART. The baud rate is 9600, there is no parity, and it has 8 data bits.
As we know 9 is represented as 1001 in binary. However, this is what I get from my UART signal:
A 2 is represented as 0010 but this is what I get from my UART signal:
I know I have to pay attention to the start bit, but other than that, I have a hard time seing the logic in this signal.
NOTE:
The signal works, but I just need help analysing it.
AI: Let's annotate your first scope shot. When we do this take note that the lowest order bit of the binary value is transmitted first. Conventional notation in this field is to order the bits in a byte like this:
[Bit 7][Bit 6][Bit 5][Bit 4][Bit 3][Bit 2][Bit 1][Bit 0]
So when I annotate and then evaluate the bit positions are flipped from the positions shown on the annotated scope shots.
The bit pattern represented there is 0b00111001.
That is the same as 0x39.
0x39 is the ASCII code for a '9' character.
Let's do the same for your second scope shot.
Here the bit pattern is represented as 0b00110010.
That is same as 0x32.
0x32 is the ASCII code for the '2' character. |
H: Specific transistor for voltage regulating?
In 2SD1047 NPN transistor datasheet described the application for power supplies and the description is:
The device is a NPN transistor manufactured using new BiT-LA (Bipolar
transistor for linear amplifier) technology. The resulting transistor
shows good gain linearity behavior.
How do we know which transistors used the "BiT-LA technology"? is it described somewhere in the datasheet?
Does it really make a difference between this transistor and for example a general purpose transistor like 2N3055 (I know the packages are different but what about the performance)?
Which specifications makes a transistor ideal for using it as main voltage regulating transistor in power supplies?
AI: Given the flatness of output I_V collector characteristics will affect linearity and power-supply-rejection ability, I'd expect the Early Voltage to be a key parameter for audio power devices.
And the size of the high_current contacts down into the silicon may also matter; this requires slightly larger metallization which requires more silicon. Another way to do this is with additional bond_wires, which definitely increases cost (slightly).
==================================
https://en.wikipedia.org/wiki/Early_effect
A large EARLY VOLTAGE tells the designer that particular bipolar transistor is a more constant current source as the Vce varies. |
H: USB 3: keyed connector protocol adherence necessary?
The USB 2 specification presents the concept of the "Keyed Connector Protocol" in chapter 6 section 2 (6.2). Basically, as I understand it, this limits the use of type A receptacles to downstream outputs form USB hosts or hubs. This is how it is meant to be:
+-------------------+ +----------+
| | | |
| Type A Receptacle +------> USB Host |
| | | |
+-------------------+ +----------+
I could not find the topic being picked up in any newer specification like USB 3.2. Can I build a USB device with a type A receptacle or is this not compliant? This is how I wanna use it:
+-------------------+ +----------+
| | | |
| Type A Receptacle +------> USB Dev. |
| | | |
+-------------------+ +----------+
AI: Can I build a USB device with a type A receptacle or is this not
compliant?
You can make a device with Type-A receptacle, but it won't be compliant to any USB specifications. Some early designs of portable HDD enclosures did this, steering confusion among users.
The main reason is that classic USB is a host-centric architecture, so there should be a host that supplies VBUS, and devices that consume VBUS (and/or use VBUS as a signal of "cable connect"). If you make a device, it must be a recipient of VBUS. The USB Type-A connector does not have any extra pins to swap port roles. More, you will need a Type-A-to-Type-A cable, so there will be a possibility that some user will connect this (illegal) cable between two classic hosts, creating VBUS source conflict between two systems and possible damage to ports.
However, there is a rise in demand for devices (mostly mobile) with necessity of having dual role USB ports (formerly known as "OTG", a meaningless term). So the industry is moving towards the dual-role functionality and truly universal connector, and the means to provide this is the USB Type-C connector. The legacy Type-A and Type-B connectors are systematically depreciated across the set of USB specifications. I would strongly advise to forget the idea of using Type-A on a device, and seriously consider Type-C instead. |
H: Change antenna so bluetooth beacon works in all directions equally well?
I am looking at using a long range bluetooth LE beacon on a project. The beacon comes with a SMA interface whip antenna that has the following radiation characteristics. It is over my head to clearly understand the report but I think it is indicating that the antenna has weak zones at the ~0, ~60, ~120, ~180, ~240, ~300 off of vertical.
First off, is this correct?
If so then this has some downsides to my project. This beacon will be mounted in an large ball and will rotate with the ball so the orientation of the antenna could end up in positions were it has a very short range.
Second question: Is there another style of antenna I could use that would be more uniform in all directions? I assume it would mean sacrificing the maximum distance it would be readable? However if I can get even half of the stated 4000m distance then it this could still work quite well for me.
In case you are wondering about the final application: The ball is actually a 3ft diameter 20 sided die that is an art piece which is free to roam around a festival site. The purpose of the beacon is to allow us to track it down periodically to make sure it's okay and to do any maintenance.
Thank you for any clarity you can provide.
AI: Yeah looks like there's some nulls like you identified so the received signal would be weaker if one of those nulls fell on your receiver.
The age-old antenna that might fit your needs is a half-wavelength dipole antenna (2.15 dB greater directionality than a perfect isotropic source). It's a linearly polarized omnidirectional antenna, but there are nulls above and below in line with the radiating members so it's not perfect (radiation pattern looks like a donut).
Alternatively (and way cooler), since your source orientation will be unknown, you could use a circularly polarized antenna! There are many premade ones since it's popular with the remote control community. Specifically, you could use a cloverleaf antenna pioneered by the legendary IBCrazy (1.2dBi supposedly) and a transmit/receive pair would only cost you around $10. It still has a null directly above and below the antenna but is pretty darn close to omnidirectional.
The benefit of circular polarization is that:
1) It will work at any orientation of the receiver relative to the transmitter. Theoretically, with a 90 degree rotation between a horizontally polarized and vertically polarized antenna (due to your ball's orientation for example) there would be zero signal received. In reality there will always be some reflection or cos(theta) power that makes it, but you could easily lose reception.
2) It will reject multipath better in a concrete jungle since the rotation of the circularly polarized signal flips on each bounce (right and left handed rotations, like corkscrews through the air). Circular polarized antennas reject the opposite polarity at -3dB when it receives the wrong rotation. This will manifest as an increased range since your receiver won't be swamped by garbage reflections (although the gaussian distribution of BLE already helps with that).
So to summarize, if you have a base station with a fixed antenna I'd recommend a circularly polarized antenna setup. If your receiver is mobile, you can get away with a linear polarized dipole but remember to try holding it at different angles if you lose signal. Also, if you generally know the area your ball will be located in, you should consider using a higher gain receive antenna to give you a longer range. In the RC community you can easily find high gain antennas at your frequency: regardless of the source polarization you could use a helical antenna since it will still receive a linearly polarized signal (if you stick with your dipole) but there will still be some loss.
The ultimate solution would be two transmit antennas hooked to a splitter and then oriented to cover each other's nulls. However, you'll get an insertion loss (less transmitted power aka less range) and now you're poorer... But at least you know where the giant ball is.
http://www.antenna-theory.com/antennas/dipole.php |
H: Feedback regarding isolation practices on a 5V/Mains PCB
I am currently designing a PCB for a project including PID-control of a heater. The heater runs on mains (220V) and is controlled using zero-crossing switching to regulate the power and minimize EMI and THD.
I already designed multiple PCBs , also ones with mains, but never low and high voltages on the same board. By now I have e prototype-design, where I made the following safety considerations:
strict separation (HV,leftside - LV, right side, see line on top silk layer)
connection HV-LV only via optotransistor/-triac (U1 and U2) and a switching converter U6 including a isolation transformer
further protection with MOV(R6), fuse (F1) and use of a snubberless triac Q1
copper planes of LV part start only after the last everything is converted to LV
Since I do not really have any space constraints I designed everythig with a lot of clearence (grid is 1mm) and used 2mm traces for the AC-powerlines. The PCB is then mounted in a metal box connected to PE or in one out of isolation material.
My main concern is that there are three JST plugs on the LV side (U4,7,8) that connect to rotary encoders, screens,etc, that are accessible to the user. While I am confident about the isolation on the PCB I still wanted to get second takes on the design, especially regarding the connections to the user-accesible features. Are there additional considerations to make in this case?
Any feedback regarding any part of the design is appreciated!
AI: I have the following feedback on your layout, roughly ranked by importance:
Your PE trace is far too thin. In the case of a live-earth short downstream of your board, you run the risk of vaporizing this trace before the fuse or a breaker / GFCI trips. You want mains traces as short and as fat as possible; of these PE is the most important for safety.
You could increase the clearance between the different mains traces. HV - LV isolation is top priority, but L-PE and N-PE are also important, and if you can manage, L-N clearance is also good. I would aim for at least 4mm clearance wherever feasible. I realize that's not possible everywhere, but there is room for improvement.
Consider making a slot in the PCB below U1 and U2, between the HV and the LV side. This further increases the creepage between HV and LV.
Keep mains traces as short and fat as possible; to keep resistance low. In case of a fault, you want a low resistance so a breaker/fuse trips decisively. Use top and bottom traces if appropriate. Your L trace is a bit too long for my taste. Keeping traces short also helps with EMI. I would prioritize keeping the traces "on the power path" (between P1, P2, F1, Q1, R6) over those "on the signal path" (dealing with R1, R4, R5, U1, U2).
Clearance between the pins of Q1 is particularly low. I like using a footprint with the pads further apart, and spreading the triac's legs out before soldering. See e.g. this answer.
Are you using huge footprints for R1 and R5 for clearance and power dissipation? You can also use several smaller resistors in series to achieve that. Use similar value resistors, assume voltage distributes equally over them so you can add their voltage ratings, and derate generously. Like a factor 3 or so. |
H: Calculating power consumption in portable air conditioning unit
Apologies for this likely very simple/stupid question.
I have a portable AC (air conditioning) unit that's supposed to be "energy efficient". On the back of the unit, it tells me some details about the unit's power consumption:
Voltage: 115 V-60 Hz
Current consumption: 11.4 A
Cooling capacity: 4100 W (14000 BTU/h)
So here's where I'm a little confused. I see the 4.1 kW and I'm a little confused, because my limited understanding of electricity would be that the power consumption shouldn't exceed the voltage multiplied the current (in this case, 1311 W). Does this mean that my unit is drawing 1.3 kW/h of electricity, or 4.1 kW/h? If it's the former, how does it output more than 3x the power that it draws in?
Also, I was wondering if my power consumption remains constant with this unit plugged in (whether at 1.3 kW/h or 4.1 kW/h), or if it will fluctuate based on the settings? In other words, are the numbers listed on the unit supposed to represent how much power it will draw in no matter what, or does that represent the maximum amount of power it will draw (meaning that it could draw less if I, for example, didn't need as much cooling)? Apologies for my limited understanding here, but just want to make sure I can get a roughly-accurate estimate for my upcoming power bill.
AI: Cooling capacity can exceed power consumption because it's moving heat from one place to another, not creating it.
Those ratings don't actually tell you the power consumption, but it should be less than 1311W, because 1311 is the VA (volts * amperes) and we don't know the power factor.
If you want to know the actual power consumption you can measure it with a device like a Kill a Watt or maybe there are additional specifications available.
The actual power consumption will vary with the settings and the operating conditions (temperature and humidity). The A/C unit turns on and off to try to maintain the temperature. If it is off all the time it draws nothing or maybe a tiny bit for the remote control etc., if it runs all the time it uses the maximum. |
H: Height adjustment of PCBs in increments of 0.05 mm
I have a 1.55 mm thick FR4 PCB with a thru-hole sensor on it. I was told these 1.55 mm are more of a suggestion than a number you can rely on. In fact the tolerances of FR4 PCBs are around 10 percent. This means that a 1.55 mm thick FR4 PCB can have around 0.1 mm of thickness variation across production batches. In my case there is a thru-hole sensor on this 1.55 mm thick PCB and the 0.1 mm of thickness variation has caused issues (the PCB is currently mounted on a piece of ABS plastic with four screws).
I cannot solve this issue by adjusting the sensor and thus adjusting the height of the PCB is my only option. I have looked into board spacers and standoffs as a way to negate this 0.1 mm in thickness variation, but could not find anything small enough. I have also thought of using thin plastic washers or shims in combination with small screws, but this seems like a crude method.
Are there sophisticated methods to height-adjust PCBs in steps as fine as 0.05 mm?
AI: Years ago I visited a local PCB fabricator to see how they actually make boards. Process starts with the two innermost copper foil layers, then additional layers are glued on (with epoxy). The whole sandwich is finally run through a roller press machine that sets the final thickness. For controlled-impedance boards, the PCB vendor does a test run, measures the impedance, and then adjusts the width of the traces to compensate for the actual FR4 batch electrical permeability and the actual layer thickness.
So no, I don't think the PCB vendor can control the PCB thickness with that fine a resolution. If you do find a PCB vendor that claims they have that much control of the initial dimensions at room temperature, it's probably going to cost a steep premium.
But even if the initial dimensions were exactly what you require, the material's thermal coefficient of expansion causes mechanical strain in the material itself (expand/contract, warp/fill, bow/twist, etc.) Usually these effects are considered negligible, but if you're concerned about a mere 0.05mm dimensional error, then you will have serious problems as the temperature changes. In civil engineering, every bridge must have a roller or other expansion joint, or else the bridge will fail. Same principle here. If there's no place for thermal expansion strain to go, then the design will fail at its weakest point -- apparently that's what's causing the issue with your sensor. (You don't say what kind of sensor it is, but most sensors have some amount of unwanted sensitivity to temperature and mechanical stress.)
I would recommmend spliting the PCB design into two smaller PCB assemblies. Use a very small PCB for the sensor and any support components that needs to be within 10mm of the sensor, and use a ribbon cable or a Flat Flex Cable (FFC) to connect the sensor PCB to the main PCB. The main PCB mounts to the box with screws, and the sensor PCB mounts with only the sensor itself -- the sensor PCB is supported by the sensor. This way, the sensor-only board ends up being small enough that dimensional stability effects are negligible, and any mechanical stresses from the box + main PCB will be absorbed by the cable.
If that's not a workable idea due to space constraints, maybe borrow a page from the LCD playbook and use Zebra strip instead of cable. See Can zebra strips (elastomeric connectors) be cut to size? -- note that this works by using compression force to hold the two boards together (sensor PCB is playing the role of the LCD glass in this scenario), and the Zebra strip elastomer is where all the mechanical strain goes. |
H: Radioactivity counter
I am designing a Geiger counter which measures radioactivity. I am planning to this using the photoelectric effect.
The band gap of insulators is pretty big and only gamma particles are energetic enough to jump an electron from the valence band to the conduction band.
Where can I find a photo-resistor made of material with a band gap of 30+ eV? Can it be ordered online or is this a special case and it needs to be ordered from a factory?
AI: There are apps (with Raspberry Pi and similar devices) for this. I'm not sure on their bad gap level, but I'd presume the very widely.
First Sensor provides some high quality photo-diodes for this purpose.
first-sensor.com/cms/upload/datasheets/gamma-ray-detection.pdf
Aside, I don't think it's correct to say "Geiger counter" because it's not using a Geiger–Müller tube. "Rad detector" is more apt :) |
H: Maximum AND minimum duty cycle in SPICE converter simulation
I am modeling a Push-Pull Converter in LTSpice and implemented this for a simple voltage feedback PWM control. This works great and I have added a bit on the end that allows for a maximum duty cycle control, as shown:
What I'm noticing is that when the duty cycle gets set to something low by the feedback control - under 10% - it starts to drive the two switches unequally during periods, e.g. one will be on for 40% and the other on for 7%. I tried putting another voltage source set to 10% pulses connected to each control line via a diode, but LTSpice did not like that.
Is there a better way to implement both maximum and minimum duty cycle control?
Alternatively, if someone can tell me why the switches are being driven unequally that would also be helpful.
EDIT: I've included a screen grab of exactly what I'm talking about:
Sometimes the feedback will drive one switch for 40% and the other for ~7% in the same period and that is not acceptable. I am looking for a way to have the control say "this period the duty cycle will be 10%" for example, and force BOTH switch driving signals to be that duty cycle for the whole period. I have been trying something with the feedback signal as a reset for an SRFlop and "carving out" the off-time, but it's not working correctly.
AI: I'm not sure why you chose to have two separate paths driven by the same signals, but if you mean this to be synchronously driven, then you have to mingle the two. You had the solution in the linked answer: simply use the complementary outputs of the Schmitt trigger. If you need dead-time, use the solution from this answer, for example. As for the limits, you could go the current-mode way, but you can also use a limit on the signal to be compared with the ramp: a simple (custom) diode. For example, here's a synchronous buck converter:
The output is awful, though correct, because the whole thing is a concoction of the sorts: it works just so you know it can work, no more. D3 is the limiting diode (the symbol doesn't matter, but it suggests bi-directionality), with the .model above, dlim. Since the ramp goes from 0 to 1 V, limiting the PWM to whithin 5% and 95% means limiting the signal from 50 mV to 0.95 V. And since a normal diode's reverse voltage would cause the signal to go negative, I have use a negative voltage for Vrev. It's not a solution that is encouraged, even if LTspice allows it, but in this case, why not? If in doubt, or if it causes hiccups to the simulator, use the [SpecialFunctions]/OTA, it's a better solution. For this case, the OTA's settings would be vhigh=0.95 vlow=0.05 isrc=0.95 isink=-0.05 rout=1, and this would make the limiting to be tamed by a smooth tanh().
For a push-pull configuration, this would take care of both the minimum and maximum allowed levels from the error amplifier (V(in)), and the distribution of pulses for the switches:
The dflop must have a delay, otherwise the solver will abort with errors (because of the direct feedback). Other delays or taus are also for aid in convergence. A7 is the OTA, as proposed above, but with the linear flag, to avoid unnecessary nonlinearities due to the tanh() limiting. All the vhigh greater than 1 are just for easier viewing when plotting. Note that T is 5 μs for a 100 kHz switching frequency.
As a bonus, A2+C1+G1 form a ramp generator with sync pulses. These will be a bit wonky (they go to -95 V here), but in terms of logic, the gates receving this signal only care about ref, or vt+vh. V(saw) can replace V(ramp), but the frequency may need tweaking. 0.97 seems a better fit for 100 kHz, but it varies, and I suspect the timestep to be of a major influence, as well as tau and internal parasitics. |
H: Using a slider potentiometer with an Arduino
I'm new to electronics and in the process of building a Joystick using an Arduino Leonardo. I am reusing parts from an old Joystick from the 90s https://www.amazon.com/F-15E-HAWK-Tactical-Control-Joystick-Pc/dp/B0012N0EOQ.
This is my sketch to test:
const int X_AXIS_PIN = 0;
const int Y_AXIS_PIN = 1;
int x = 512; // variable to store the value coming from the sensor
int y = 512; // variable to store the value coming from the sensor
void setup() {
Serial.begin(9600);
}
void loop() {
x = analogRead(X_AXIS_PIN);
y = analogRead(Y_AXIS_PIN);
Serial.print("X: ");
Serial.print(x);
Serial.print(", ");
Serial.print("Y: ");
Serial.print(y);
Serial.println();
delay(500);
}
So far, I have hooked up a Joystick component I bought separately (not from old Joystick) to my Arduino to learn the general concepts. And it works well!
GND to GND
+5V to 5V
VRx to A0
VRy to A1
SW not connected
The code prints the correct X and Y values depending on the position of the Joystick.
Now to my problem: Instead of using a Joystick component, I'm trying to re-use the old slider potentiometers from the Joystick, so I don't have to construct a new Joystick rack. The potentiometers seem to be rated at 120k ohm (isn't that a lot?) and look like this:
Rack and potentiometers:
Close-up view:
However, I'm struggling getting it to work with my Arduino.
First question: The joystick component itself does not seem be anything else than two potentiometers, isn't it? Why does the Joystick (or any other normal potentiometer) need three pins connected (GND, 5v, and Out), but the sliding potentiometer that I took off the Joystick only had 2 connected? From Googling, I learned that 3 pins make the potentiometer a voltage divider and 2 pins make it a variable resistor. But I don't fully understand the difference. Don't both reduce the current?
Second question I tried testing the potentiometers to find out if they work at all. I set my multimeter to 200k and probing each of the cables coming off the potentiometer, the resistence would move between 0 and 134k as I move the slider. I also put one in series with an LED and a 220ohm resistor. The LED would dim but not gradually. It would only dim very slightly between 0-99% and be at full brightness when the slider is at 100%. Is that because the 120k potentiometer is too strong? Why would the resistance still not change gradually?
Third Question I tried hooking the potentiometer to the Arduino and see if it generates any data as I move the slider. I connected one end to GND and the other to A0. It doesn't generate any data. Is that because it is missing the third cable?
Fourth question As the potentiometer has three pins on each side, I was wondering if it simply needs a third pin connected. However, none of the pins are labeled so I tried all 6 different combinations to find out which combination makes the potentiometer working as expected (regulating voltage between 0 and 5V) This is how I tested it:
GND to pot (1st pin)
5V to pot (2nd pin)
Multimeter set to DC and 20V (COM + Ohm, V, mA)
Multimeter - to GND
Multimeter + to pot (3rd pin)
I found two combinations where the multimeter would read constant 4.99 (5) volts, no matter where I move the slider. All the other combinations don't read anything, no matter how I move the slider. Did I test this right?
AI: For the answers to my questions, check @JustMe's answer. But I'm also writing one that explains how I got everything to work.
After trying out different pins and frying one pot, I found the right combination of pins and got it hooked up and working with my Arduino :) Please correct me if made wrong conclusions anywhere.
How potentiometers work on the inside
I couldn't find anything on the internet on how 6-pin potentiometers are wired up. But this video really helped me understand how a potentiometer works on the inside: https://www.youtube.com/watch?v=1G11DbaUlec
It seems like some pots have 6 pins because they are essentially two pots in one. These are called dual-gang.
To test which pin is which
Don't test with Arduino, test with separate battery (e.g. 9V) to be safe!
First, move slider to the center.
Set Multimeter to Ohm, to the next higher (200k) than the maximum resistance (120k) of my pot
Check which two pins have a fixed resistance (which will be the maximum
resistance value), i.e. the resistance does not change when moving the slider. These two will be + and -. According to the
video, it seems like the polarity doesn't matter for those two pins
(please confirm).
Move back slider to center, release one probe from one of the previous pins (keep other probe on pin) and try to find the 3rd pin. It will be the pin where it shows a resistance value lower than the maximum value. It should change when you move the slider. The 3rd pin will be the wiper!
Verify functionality
Next, we are going to confirm that the voltage division works (since Arduino doesn't measure resistance, it measures voltage).
Set Multimeter to voltage (e.g. 20)
Connect the pot's + and - (the two with fixed resistance) to battery (e.g. 9V)
Connect multimeter - to ground (e.g. battery), multimeter + to wiper
As you move the slider, the voltage should change from 0 to 9V. Also make sure, the whole thing doesn't get warm, or even worse, goes up in smoke like mine.
Connect to Arduino
Now, you should be able to connect it to the Arduino like any other pot. - goes to ground, + goes to 5V, wiper goes to your analog pin (A0 in my case). |
H: Hidden wire detector using BJT transistors
I made a hidden wire detector from this tutorial. It works, but I'm not quite understand why. Here is the circuit:
I used 2SC1815 transistors and the value of R was 1K0. My antenna was made of 300 mm length copper wire with diameter 0.3 mm. I've measured its inductance - it is about 2 uH.
First of all, I've measured the base-emitter junction voltage of the first transistor - its amplitude was about 80 mV. I believed that BJT transistors require about 0.7 V to open, but the circuit was able to detected the wire. How is that possible? The base-emitter junction voltage was 10 times lower, so I thought the transistor should stay closed.
The second question might be silly, but I was not able to find the answer. I realize that electromagnetic field develops the voltage across the antenna. But the antenna is connected to a circuit with only one side. So, which voltage do we use as input? My suggestion is that voltage may only occur when the resistance is high and the field does not need to develop a high current to maintain such voltage. So, maybe the equivalent circuit actually includes the Ra resistance, which is as high as the resistance of open circuit? See the circuit below:
If I'm right, the input resistance of the circuit must be very high too. So maybe this is the key for my first question and the transistor does not open at all? But in such case how on Earth this circuit can work?
AI: A better way to look at this is that your mains voltage is hundreds of volts wrt ground, and only a small capacitive coupling will drive enough current through the base to illuminate the LED. There is plenty of voltage compared to the few hundred mV required to forward bias the base sufficiently.
simulate this circuit – Schematic created using CircuitLab
Here you can see that only 1pF of coupling is sufficient to allow strong current pulses through the LED. C1 represents the near-field capacitive coupling to the mains wire. C2 represents the user's body coupling to earth.
Only a few tens of nA are required at the base, thanks to the Darlington triple transistor's high current gain. |
H: NFC - design of inductor/"antenna" made of PCB traces
I did a layout of a NFC PCB inductor/"antenna"(I know it is not one exactly) drawn only on top side copper (nothing drawn on other layers at inductor area), and I wonder if there will be a considerable loss of working distance range on the opposite side due the PCB FR4 core itself is an obstacle... I need to read tags to both sides of the board at approximated distances.
I will share some screenshots.
Schematic of RF side, based on ST ST25R95, 230mW output power, QFN32:
Rant is 1.65 ohms and was calculated HERE
Lant is 1.09uH and came from ST's online tool eDesign Suite
This is my current drawing for the PCB coil/inductor. 10.5 x 26mm, 7 turns.
I wonder if I need (if it is suitable) to continue the "antenna" drawing on the bottom layer, and sum the inductances and resistances? (Or some other equation to calculate the inductance).
Also I wonder if I do that (antenna drawn on both sides) if I could reach more working distance (between the reader circuit and the tags) than doing coil only on top side.
If I need to continue the coil on bottom side also, do you have any recommendation about how to draw a good PCB inductor? So that I can reach the best distance range the chip can offer (it is 0.23W output power only, ST's ST25R95).
Regards.
EDIT 1
4 turns x 2 layer NFC coil.
TURN x mm:
1: 11.2 + 25.7 + 10.2 + 24.9 +
2: 9.4 + 24.1 + 8.6 + 23.3 +
3: 7.8 + 22.5 + 7 + 21.7 +
4: 6.2 + 20.9 + 5.4 + 20.1 + (2.7)
side = 251.7mm
total (x2) = 503mm
PARAMETERS of inductor traces:
witdh = 0.3mm
length = 503mm
copper tickness = 18um (1/2 oz final copper thickness)
According to HERE, Rant = 1.66 ohms
So Then I have new input values, Lant = 1.68uH and Rant = 1.66 ohms to input on the ST25R95 calculator spreadsheet. Now I can easily recalculate the NFC circuit, I mean the caps and indutors of matching and filter circuits.
I wonder if I should consider the capacitance created between the traces of spposite sides, they are on exact same place, they just are in opposite board sides. PCB is 1.6mm FR4. dk (dielectric constant) can consider as 4.5.
Single side inductance
new inductor drawing.
Rant value
Coil "pins"
AI: Also I wonder if I do that (antenna drawn on both sides) if I could
reach more working distance (between the reader circuit and the tags)
than doing coil only on top side.
and
If I need to continue the coil on bottom side also, do you have any
recommendation about how to draw a good PCB inductor?
For near-field magnetic coupling it would be better if the coil was split between top and bottom copper layers with half the turns on the top layer and half on the bottom stacked above each other. That will allow you to use slightly fewer turns overall and get a slightly better coil Q factor. That will increase range slightly and, because coils are top and bottom you would get equal range performance from either side.
Designing coils to couple magnetism some distance requires the best amount of mutual coupling you can get between the windings of the transmit coil. Designs that are spiral are not the best but, given that you might have only one PCB layer to work with then it has to be a spiral but take note; the lower radius windings of the spiral are largely superfluous to the energy transmitted to the receiver and so, if you had (say) four layers, the optimum would be to put stacked turns on each of the four layers (just like a regular electromagnet).
So, I would look for a calculator that allows you to design the right inductance using 2 or more layers of the PCB. |
H: What is the purpose of this disk over a Samtec connector?
With beautification in mind I decided to attach 3D-STEP models to elements on my PCB.
One of them is a connector from Samtec. They do provide a STEP models for a download, but it was immediately suspicious to me, that the preview had this weird disc over the connector. At that moment I thought it was some kind of glitch and proceeded downloading STEP model regardless.
See the image below:
Attaching the downloaded STEP model I was surprised by this disc again. Turns out, it was also exported into STEP model along with the connector. Unfortunately, this serves bad for the purpose of "beautification" :)
Questions are:
What is the purpose of this disk?
Is there any way to download a STEP model of Samtec connector without this disk?
For a reference, the connector is QSH-030-01-F-D-A-K
AI: This hasn't been explicitly mentioned yet, but you got a disc because you asked for one.
It's a Polyimide film pad, and specified by the last code in your part: -K. The purpose of this pad is so the vacuum pick-and-place machine has something to hold the component on. It needs a flat surface big enough for the vacuum head, as already shown by Jeroen3 and DKNguyen. For manual manufacturing, you don't need the film pad. But some automated lines can't pick up the parts if there's no flat surface provided.
QSH-030-01-F-D-A-K has the pad, QSH-030-01-F-D-A does not.
030-01 says how many positions and the lead style. The other codes:
-F: 3u plating
-D: double rows
-A: alignment pins
-K: Polyimide film pad
See the QSH configurator. Find your new, padless STP files there too.
Polyimide (also known under the brandname Kapton) is stable for a wide temperature range and a good electrical isolator, making it ideal for this purpose. |
H: Which diodes are used in a high amperage rectifying circuit? (4007 are 1A rated) and other components
I recently designed a power supply that was good for about 700mA power and used these diodes
https://www.diodes.com/assets/Datasheets/ds28002.pdf They are 4007 diodes. As I am checking the datasheets for my 5A power supply I'm noticing that they are good for up to 1A so I was wondering which diodes are commonly used with high amperage 50 Hz / 60 Hz power supplies? What other components can I expect to change with this change in amperage? Looking forward to your responses
EDIT: this is a standard 50Hz / 60Hz power supply .. buck converter becuase it's high amperage.
That is what i'm referrring to in this question, as replacement for 1N4007 diode.
I am also making a different 5A power supply.. which in this case requires Schottky diodes rated for high frequency and 3-4A, in a full bridge rectifier
thanks
AI: 1N540x is a common 3A diode, so with a full bridge you can get 6A. Above a few amperes you should consider Schottky diodes, at least at lower supply voltages.
You can also get inexpensive packaged bridges with ~17A diodes rated for 35A that can be directly bolted to a heatsink. |
H: USB Type C swap A/B rows for easier routing
I'm routing a USB-C receptacle on a carrier board for a System on Module. The location and orientation of the receptacle and the SoM headers are dictated by the design of the device.
I noticed that if I would swap the signals running from the SoM to the A and B rows of the USB-C receptacle, routing would be substantially easier. As the USB-C port design is mirrored, one would assume that would works just as well.
But I'm afraid that somehow the bult-in logic of USB-C where it resolves cable orientation would somehow cause trouble in my swapped layout. Could I trick the system by swapping ALL signals except CC1 and CC2? Would DisplayPort still work?
If I just leave it as it is, I will have to have differential signals cross-over, as well as have P/N swaps, wheras in the swapped situation, I can simply route almost everything in more or less straight lines.
-edit-
Below is an image of the supposed routing if I make an exact copy of the schematics of the development carrier board that my SoM supplier has made avilable to me, shown within the physical constraints of my carrier board layout. I think everybody would agree that if I rotate the pins on rows A and B of the type-c connector, so (SS)TX1 becomes (SS)TX2 and vice versa and (SS)RX1 becomes (SS)RX2 vv., routing becomes much easier. And then I haven't even shown the amount of P/N reversal that would have to be applied in the below case.
Any way i guess the comments that have already been made reassure me I can go ahead and rotate A and B and don't do anything weird with the CC1 and CC2 pins, because I understand now that I can just see this rotation as exactly the same as a normal connector rotation.
AI: Let's say you connected the receptacle exactly as it is shown on your diagram. Then you attached the cable and marked corresponding pins on both plug and receptacle, A1..A12 in top row, B12..B1 in the second row.
Now, if you disconnect the cable, rotate it 180° and plug it back, pin marked A1 on the plug will be connected to B1 of the receptacle. Pin marked A2 to B2 and so on.
So, if you now swap the signal traces the same way, i.e. A1 <-> B1, A2 <-> B2 ... then the end result will be electrically identical to the original connection. Your SoM will not see any difference, but the routing will be more convenient.
Note, that the same result can be achieved by rotating SoM in place, but since you said it is fixed the swapping of the traces is a way to go. |
H: Matrix has a voltage drop on a coil that should not be receiving any current
I am making a matrix of electromagnets (each with about 120 Ohms resistance) and I am using a circuit like this for the columns (low side of the switch):
simulate this circuit – Schematic created using CircuitLab
For the high side (rows of the matrix) here is what I am using:
simulate this circuit
:
I have a test 2x2 matrix hooked up right now (with two rows and two columns as shown) When I power 0,0 (i.e. signals on col 0 and row 0 are high, the others low) I still get about a +4V drop on the coil at (0,1), which is definitely not what I expected. Does anyone know why? My guess is because of the lack of diodes but since the 4V is in the + direction I'm not too sure. Another guess is that my resistor values are wrong, but to be honest I'm not sure at all how to figure that out.
EDIT: After further probing, the coil at (0,1) also turns on and has a voltage reading of about 3.8V, as well as the coil at (1,1) with a voltage drop of 3.8V
AI: If you don't use diodes to steer the current then you will have multiple paths of current flow through all the electromagnets. Yes, there will be one electromagnet that receives a direct connection but, because that row is "grounding" several other electromagnets, current from the active column driver can flow to several series electromagnets and back to the grounded row. Just draw it out and use a crayon or something similar to plot the current flow. Like this: -
The purple electromagnet is the one that is directly driven and the light blue electromagnets are an example of series electromagnets that will also take current flow. I've only shown one path that the extra current will flow but there are many other paths. You need diodes to prevent this from happening. |
H: Rocker style light switch: do I need an isolated power supply
I'm building a Wi-Fi controllable light switch that will use rocker style (up/down) plastic buttons. My plan is to have all the electronics enclosed behind a metal gang plate (the part that screws onto the gang box) in it's own plastic enclosure.
I would like to have plastic 'legs' on the back of the rocker switch (for up/down) that protrude into small holes in the single metal gang plate so that the plastic legs push tactile switches on the inside PCB, mounted behind the metal plate in the enclosed area. The depth of the holes is roughly 1.25 cm.
Some of the sensing electronics I'd like to keep mains referenced, so a non-isolated power supply would be preferable. I would like to have this UL certified eventually.
My question is simple. Do I need an isolated power supply in this case? The buttons will only ever be pressed by the plastic 'legs', not touched by a finger and everything else is enclosed.
Don't get me wrong, I have experience designing switching mode isolated power supplies in the past. But am not too sure of what I'll need in terms of certification for this specific case.
Any help / clarification / expertise is appreciated. Thanks.
AI: From the point of view of users' safety, during manual operation, a Wi-Fi controlled switch is no different from a standard switch. Hence an isolated power supply would not be required. Isolation would also entail an increase in size and cost.
You would also need to factor in the requirement of a 2- wire switch as a replacement for a standard 2-wire switch (without neutral wire). The standby current of a 2-wire Wi-Fi switch would not light up a high wattage incandescent lamp but could cause a dull glow in low wattage LED bulbs.
A metal gang plate could hinder Wi-Fi access. |
H: Do I need to connect both GND and both VCC on this addressible LED?
I'm working with APA102-2020 addressable RGB LEDs (datasheet here). Each component has two GND pads and two VCC pads. Testing with a multimeter, I find that each pair of pads is connected (ie GND-GND and VCC-VCC). In the PCB I'm designing, should I connect both GND and both VCC pads, or do I only need to connect one of each? Why/why not?
The datasheet doesn't show these pads, but the datasheet is known to be inaccurate in other ways so I'm not sure if I should fully trust it.
I only know what they are from the KiCAD footprint.
The LEDs look like this:
AI: Yes, you should connect them.
A good datasheet would tell you that, but hey...
More connection points means better electrical contact (lower impedance overall), better mechanical strength and thermal management, there really is no advantage in not connecting them. |
H: PIC16F887 PWM multiple output
I am trying to design a control for a BLDC motor with a PIC16F887.
I have managed to activate a PWM output and be able to vary it with a potentiometer.
The point is, I need to get 6 PWM signals out.
When I change the CCP1CON I can vary the PWM output but it is always one.
I attach the part of my code where I configure the PWM.
void PWM_Initialize()
{
PR2 = ("%#x", (_XTAL_FREQ/(PWM_freq*4*TMR2PRESCALE)) - 1);
CCPR1L = 0x1F;
RC1 = 0;
RC2 = 0;
T2CON = 0x03;
CCP1CONbits.CCP1M = 0b1100;
CCP1CONbits.P1M = 0b00;
TMR2 = 0;
T2CONbits.TMR2ON = 1; //Timer 2 ON
}
How can I activate multiple PWM outputs?
AI: Your Controller has got only 2 PWM outputs. The PIC16F1783 could maybe help you. |
H: High Voltage applied to TVS diode
I am using the below input circuit which has this SM6T36CAY 36V TVS diode.
The input voltage range is 12-36V and the output is the same only dropping the diode forward voltage.
Typical load current is around 80mA.
My case :
I applied around 130V at the input side by mistake which resulted in the magic smoke.
After turning off the power supply and adjusting it to the nominal voltage range, the typical current draw increased from the usual 80mA to 500mA.
Upon debugging, I checked the individual power rails. This 12-36V is converted to 5V by the LDO which is fed to the microcontroller. I found that the 5V rail was showing short (continuity) and therefore, I replaced the microcontroller thinking that it might be damaged. But After replacing the microcontroller also, same result of 500mA current draw.
Later, I checked the resistance across the TVS diode. It was showing some 10-50Ohms only.
After changing the TVS Diode, the current draw reduced to 80mA nominal.
I thought TVS was to handle transient voltages. And I applied that 130V for some 30seconds.
My questions :
TVS diodes are not capable of handling some hundreds of volts for 30-45seconds but they can handle kV range for few micro or milliseconds. Why is that?
And why did the current draw was high till I replaced the TVS? The TVS was a short circuit impedance path for the current?
If the TVS diode was the only problem, why did my LDO output (5V) also show a short circuit?
AI: TVS diodes are not capable of handling some hundreds of volts for
30-45seconds but they can handle kV range for few micro or
milliseconds. Why is that?
As with all use examples of TVS diodes, they are trying to fight-off a surge of current. They cannot fight against a solid voltage source because if you applied one there would be theoretically tens of thousands of amps. So, the voltage source they "fight-against" is not a true voltage source because it has series impedance. It doesn't matter whether the surge is indirect lightning or ESD - there is a series impedance that limits the current and the TVS takes this current and clamps the voltage to a peak or limiting value.
However, it will only do this for a certain length of time before the internal temperature of the junction gets too hot and it melts.
Surges are limited in time to a few hundred micro seconds so why should a TVS diode be designed to fight-off a surge that lasted several seconds?
And why did the current draw was high till I replaced the TVS? The TVS
was a short circuit impedance path for the current?
Because the internals of the TVS melted and it turned into a good conductive blob of semiconductor material.
If the TVS diode was the only problem, why did my LDO output (5V) also
show a short circuit?
Look at the data sheet: -
The TVS might have allowed a voltage as high as 49.9 volts or maybe 64.3 volts when you applied the 130 volt supply. It might even of allowed a higher voltage - it all depends on how much current the 130 volt supply can muster.
Is your LDO capable of withstanding 50 volts?
How hot will it get
The device weighs 0.11 grams and lets say it is subjected to a terminal voltage of 50 volts at 12 amps for 10 seconds. The power is 600 watts and the energy is 6,000 joules. Now consider a material that has a really good specific heat. I'm thinking water and water has a specific heat of 4.186 joules per gram per degC. In other words, if you apply a heat of 4.186 joules to 1 gram of water it will warm 1 degC.
So, with 600 joules applied, 1 gram of water will warm 143.3 degC and 0.11 gram of water would warm by 1303 degC.
Do you see why a small 0.11 gram TVS diode will turn to a molten blob in a few seconds. Once it's got past 600 degC it's done for. The specific heat of silicon os only 0.7 joules per gram per degree Celsius so it would likely melt in less than a second. |
H: A basic question about RF multiplexer practical use
Below shows an RF switch(multiplexer):
So what I understand is that the same RF input is routed to multiple outputs. This can be I guess used to couple the same RF signal to many different equipment such as spectrum analyser, oscilloscope ect.
But lets say the scenario is such that we have many signals from several devices which we can couple these as inputs to a device under test's input. I tried to draw it as follows:
Now as in my above example, we can select which analog input will be coupled to the ADC. I named it as MUX.
So in such a case are RF switches used as well? Or only analog multiplexers are used? If RF multiplexer can also be used, what could be the reason?
(is this called switch matrix?)
AI: You can regard the HMC252 as a multiplexer if you want: -
So in such a case are RF switches used as well?
No, you wouldn't use RF switches like the one above for regular low speed applications because regular low speed applications don't need inputs terminating in 50 ohms when disabled. Look at the picture above and note that when any of the 6 RF inputs are not used, those pins are internally connected to 50 ohms.
So what I understand is that the same RF input is routed to multiple
outputs.
Or multiple RF inputs selectively coupled to a single output. They operate bidirectionally.
is this called switch matrix?
It's not an unreasonable description if produced by a marketing guy but I wouldn't use the word matrix because it's misleading technically. |
H: How bootstrap capacitor voltage is added to source voltage?
Circuit diagram below is a high side n-channel mosfet with bootstrap(while C1 is discharging)
It says that after charging of bootstrap capacitor:
The upper optocoupler opt1 gets ON for 1s and lower optocoupler Opt2 remains OFF for 1s. The capacitor C1 now tries to maintain the 12V across it and this raises the source voltage to 12V. This makes the diode D1 reverse biased as its cathode voltage is now 24V for maintaining the 12V across the capacitor. The capacitor C1 now starts discharging through upper optocoupler Opt1 and the gate of the MOSFET Q1 develops 24V.
BUT I cannot understand how the 12 volts from the source is added to the 12 volts of the charged capacitor. I mean, at the time of discharging, the capacitor is not in series with the 12 volts source. (it seems to me that the capacitor and source is in parallel to gate of mosfet)
I think the current does added up but not the voltage. How so the source voltage is added to capacitor voltage to become 24V?
source: https://www.engineersgarage.com/contributions/driving-high-side-mosfet-using-bootstrap-circuitry-part-17-17/
AI: The optocouplers are just your turn-on and turn-off switches.
You require a bootstrap capacitor because you are trying to drive an N-channel Mosfet on the high side, and therefore you will need a voltage between Gate and Source of the mosfet which is greater than its threshold value.
When the mosfet is on you will be sending 12V to the load and therefore the node between the source and the load is at ~12V so to keep the mosfet on you need a voltage above 12V.
Notice how the capacitor (negative terminal) is at the node between the load and the source of the mosfet.
When the mosfet is off, lower optocoupler on, the capacitor will charge to 12V approx through the load and partially through the optocoupler.
When you turn on the mosfet, upper optocoupler on.
a. the capacitor will provide a voltage to the gate with respect to the source above the threshold value.
b. The mosfet will turn on and the node at the source of the mosfet will climb to ~12 volts
c. The voltage between the gate and the source of the mosfet will not change because it is provided by the capacitor, but if you were to observe the voltage referenced to gnd, the voltage will climb whatever the voltage drop on the load provides as "offset".If you measure the gate voltage referenced to ground you are measuring two things : Vgs + Vload. For the mosfet to stay on, it only requires that Vgs be above its threshold value, the capacitor provides this "bootstaped"/"floating" voltage source.
d. So with respect to gnd, you will have initially a value above 12V, idealy 12V + 12V, which will discharge slowly through the resistances R3 and R4 and the reverse leakage current of the diode. If switching is fast, you will not even notice any discharge.
You turn off the mosfet and the mosfet source node voltage goes back down to Gnd and the bootstrap capacitor is charged again whatever charge it lost.
Cycle repeats. |
H: Find constants of ideal op amp circuit problem
Alright, I'm currently learning more about circuit analysis and op-amps using an electrical engineering and measurement book; the problem is that most questions don't have answers, and I don't really understand what they want me to do during this problem.
I've done some OP-amps examples before, but I've always been given the Uin value for that, which I don't have here, so don't really know how to obtain my desired values.
"Uout = A*Uin + B (where Uin is the sensor output voltage), figure out A and B" is the question in the book, and I don't really understand how to obtain the "A" and "B" value. Usually you can just take the two resistor values such as R1/RF, and multiply by the voltage source which in this case is 0; however taking (1.5k/4k)*0 would just result in zero. And even if I did manage to find Uout, I still don't understand how I would get the A and B constants from it.
AI: You are trying to find the transfer function, not the actual voltages.
To figure that out, find the gain of the path from Uin to Uout.
U1 has a gain of \$ \frac {Rf} {R1} +1\$ which yields a non-inverting gain of 3.67
U2 has a gain (from U1) of -1 and so does the final amplifier, so the total gain (from Uin) is 3.67.
Then we take Uin2 which goes to a summing junction and the total gain to Uout from Uin2 is +1 (from -1 * -1).
So the transfer function is Uout = (Uin * 3.67 ) + Uin2 which is (Uin * 3.67) + 1V
Therefore A = 3.67 and B = 1 |
H: Understanding the Base voltage of PNP when driving NPN transistor is cut off and there is no clear DC bias voltage
I am studying the Art of Electronics and trying to apply what I have learned to a motor driver circuit from an MITx intro to control systems course (https://www.edx.org/course/introduction-control-system-design-first-mitx-6-302-0x). I have a good understanding of the behavior when the driving NPN transistor is ON (pwmIn is 5V) but I am having trouble understanding the reason why the voltage at the base of the Power PNP is 5V (demonstrated on the scope) when the NPN transistor is cut off (pwmIn = 0). Most of the circuits in AoE had well defined DC bias voltages at the transistor base and in this case it seems the base is "floating."
Can someone please explain how best to understand the base voltage of the PNP when there is no base current.
AI: If the NPN transistor is cut off, then it doesn't allow any current to flow into its collector.
Therefore there is no current flowing through the base-emitter junction of the PNP transistor.
But the I-V relationship of the b-e junction is similar to that of a diode. The only way current is zero is when the voltage is also zero.
So the b-e voltage must also be zero in this case.
Another way to look at it is the b-e junction acts as a non-linear resistor, which is able to pull up the base voltage to equal the emitter voltage if no current is flowing through the junction. |
H: What is the correct pin spacing for this footprint?
I'm trying to import an IC footprint into CAD software and can't seem to nail down the exact dimension for pin spacing. The two supporting documents seem to conflict with each other (Land Patterns here, and Package Drawings here).
In the drawings document the pin spacing in the Y-axis is listed as 0.55mm whereas if you calculate the pin spacing from the land patterns it works out to be 0.50625mm (4.05mm/8 pins == 0.50625mm/pin).
If I try to go with the 0.55mm spacing the pins seem to be out of bounds of the package size and the spacing seem to contradict with the D1 dimension (5.0 BSC).
What is the correct dimension I need? And is BSC 0.55mm, 1.1mm or something else?
Curse you Maxim and your vague schematics! (Only their drawings, their technical specs are great)
AI: I think the confusion lies in the 0.55mm measurement. This is not the pitch between every pin on each side of the IC, but rather the distance between the top row and the pins below.
The distance between the pins on each side is given by the measurement e, which is 0.5mm.
I agree, this is a particularly confusing dimensional drawing. |
H: Proof of linearity of a transistor
When we deal with a transistor we break it up into two parts,
1 The DC bias.
2 The small signal perturbations.
Then we add the currents/voltages from the both the analysis to get the total response.
But what is the proof that we can do this procedure?
As an example say I've a DC source V and Ac source W.
When I apply V at VCB I get x as a response.
Then I use the small signal model and find that when I apply a small voltage W I get a response y. The total response is then x+y.
However isn't it possible that if i apply V+W together I would end up with a different response?
AI: As an example say I've a DC source V and Ac source W. When I apply V at VCB I get x as a response. Then I use the small signal model and find that when I apply a small voltage W I get a response y. The total response is then x+y. However isn't it possible that if i apply V+W together I would end up with a different response?
Of course if you apply V+W together you end up with a different response (NOT x+y). This is because the circuit is nonlinear so superposition doesn't apply. Separating the dc (bias) and ac (small-signal) response is NOT superposition, though it resembles superposition superficially. What you're actually doing in this case is applying a first-order linear approximation. Do you remember something like this in high-school calculus:
https://steemit.com/steemiteducation/@masterwu/estimating-2-001-5-with-differentials-and-linear-approximations
In the image, x is your dc bias voltage and Δx is your ac signal. The true response is f(x+Δx) (nonlinear), but you can approximate the response as f(x) + f'(x)Δx, which is the dc response, f(x), plus the linearized ac response f'(x)Δx. The term f'(x) represents the small-signal gain and is of course dependent on x, the bias point.
Remember that small-signal analysis gives approximate results that are 'close enough' if the excitation signal is kept small enough. Superposition applies only to linear circuits and gives exact results. If you want a proof for why superposition works, that's a whole nother answer. |
H: Identify component (diode?) based on photo
I am currently fixing a Akai MPK Mini mk2 midi controller with a loose USB port.
In addition to the USB, it is missing one part labeled as D1. It can be seen in the attached picture circled in red. I think the text on the component reads 62S? Any idea on what component it could be? All help is appreciated!
AI: NOTE Looking at more pictures, this is definitely incorrect, but left here in case anyone finds a similar component :)
My best guess (not totally convinced) is it's an ST USBLC6-2SC6 low capacitance ESD protection device.
Reasoning:
Near the USB connector, so probably a transient protector (as per comments)
Marking is almost correct (from datasheet, it should be 62SC6, but it may be truncated)
Pin 2 looks like it's connected to ground, pin 5 looks to be going (a long way!) to a decoupling cap.
My only doubt is the D- pins (3, 4) do not appear to be connected in that picture, but the vias may be hidden. |
H: Series internal resistance of Power Supply
I have this power supply - GPC-3030D
There are 3 independent channels. 2x (30V/3A) and 1x (5V/3A)
To find the internal resistance of the voltage source channels, its just 30V/3A = 10Ohms and 5V/3A = 1.66Ohms.
Is it correct?
AI: You can estimate the effective internal resistance of the outputs from the load regulation specification.
For the fixed 5 V output, it's given as <10 mV.
If the output voltage drops by 10 mV for a change in output current of 3 A, then the output resistance is \$\frac{10 mV}{ 3 A}\$ <= 3 mΩ. |
H: Transistor base voltage and current
I want to use a transistor to control a 5v power source but Im confused on how to control the base pin.
How much voltage can I put on the base pin? I've seen tutorials which use 5v with a 1k resistor, but doesnt this exceed the saturation voltage of 0.7v?
Also if I were to use a voltage divider circuit to lower the voltage, would this provide enough current to turn on the transistor?
Im using a tip120 transistor. Thanks for any help
AI: When you're using a transistor as a switch, the typical way to turn it on is to apply a suitable current into the base pin, usually from a voltage source through a resistor. The current will then develop about 0.7 V across the base emitter diode junction. Most transistors specify a maximum base current.
In the particular case of the TIP120, this is a darlington transistor, consisting internally of two transistors. The 'base' to emitter voltage will typically be 1.4 V.
The absolute maximum base current for the TIP120 is given as 120 mA in the data sheet. However, as the minimum gain is given as 1000, and the maximum collector current as 3 A, the base current needed to turn it fully on shouldn't need to be much more than 3 mA. This 3 mA to 120 mA window gives you a huge range to hit with your voltage source and series resistor. |
H: Simple 300-400VDC/100A power supply for EV charging from 400V 3-phase/32A
I would like to purchase or build a simple AC-DC power supply, with 400V/32A 3-phase input, and DC output adjustable (by a microcontroller which I would add) between 300-400VDC and upto 100A (not exceeding 22kW total). The purpose would be to charge an electric vehicle via the fast charge port (CCS or ChaDeMo), which provides direct DC access to the vehicle's battery (usually around 350VDC and 15-70kWh depending on car type). Such power supplies are really expensive (I haven't found any under 5000$) and I have a hard time understanding exactly why they are so expensive. It doesn't need to be isolated, since all parts are extremely well protected from touching. The output voltage is pretty close to the input voltage, and never higher, so shouldn't a rather simple PWM regulation be enough, no transformers, no big voltage differences? (I know such a power supply is not enough to charge an EV, the microcontroller part needs to communicate with the car via CAN (ChaDeMo) or Powerline (CCS), and multiple safety circuits must be implemented, but I already know how to do that, what's missing is the power supply part).
AI: Power factor correction is the hidden biggy with any powerful 3 phase AC to DC converter. It’s simply not the case that the rather elegant 3 phase rectifier circuit can be piggy backed with a boosting power factor correction circuit. This will not work. This is because the elegant 3 phase rectifier (used for many tens of years if not a hundred years) just doesn’t allow it to happen. See this for an explanation.
This won't work: -
Because the current in each phase falls to zero while its respective voltage is still very high: -
The only viable alternative is to regard the 3 phases as 3 individual supplies and have power factor correction replicated for each. Then there is the requirement to maintain load balance so, there has to be significant cross checking between each phase and finally, each of the 3 boost converter's outputs has to be managed so that it can be adequately merged with the other two boosted DC voltages: -
There is a lot of overhead here and I reckon power factor correction alone is about 50% of the BoM cost for the circuits.
I’m not saying that $5k is a justifiable price to pay of course but given the market and trends and hidden complexity of PFC, it doesn't surprise me at all. |
H: Use 2 of the 8 cables of an existing ethernet connection
There is an Ethernet cable that runs from a server room to an office. That cable is connected to a router. There are times when we need to reset that router and I will like to do that remotely. There is only one Ethernet cable that runs to the office I wish there was one more.
I will like to place an arduino or raspberry pi at the office that will be responsible for turning off and on the router with a relay.
I need to send a signal from the server room to the arduino located at the office.
I have tried using the NRFL01 and other radio signals but the distance is to far and the signal is lost because there is a lot of walls in between.
I want to use 2 of the existing Ethernet cables just like a Poe switch uses some off the wires to inject power. Instead of injecting power I will like to insert a signal. What cables can I use of the internet cable. I know very little about networking. Does it matter if it's a cat5/cat6? Can I do something like that?
AI: If it is a CAT5 or CAT6 twisted pair cable, and it's used by 10baseT or 100baseT Ethernet (max 100 Mbit Ethernet), then the existing connection is only using pins 1,2,3 and 6 (usually the green and orange pairs) of the RJ45 plug. You can use the other 2 pairs (brown and blue, on the remaining 4 pins) as you like, even for a complete second Ethernet line, although they will interfere with each other (which, depending on the shielding, the speed and the length, may or may not make a difference.
There are probably better solutions for your application (like TCP/IP controlled relays), but this answers your question. |
H: LTspice counter component how does it works?
I find this component on LTspice but I have some troubles to know how it works. It asks me to enter 2 parameters. I could understand that It needs to know the "divide" number but what is the second parameter ? I supposed one of the outputs is the complementary of the other
Thank you very much,
Have a nice day !
AI: Right click the component and in the field "Value" indicate "cycles=10". Change the value 10 to the desired cycles.
Obviously, connect a clock signal (voltage source as pulse) to the Clk input and connect GND to the pin in the left-bottom corner. Phi1 and Phi2 are outputs. |
H: Ideal inductor AC current LTspice
I am simulating an ideal inductor in LTSpice connected to a 1Vpk 50Hz sine wave and 1 ohm resistor.
The current through the inductor seems really low, at peak being around 3.2mA
As the current through the inductor should be the integral of the voltage multiplied by the inductance it seems that this should be higher. So why is it this low?
AI: 1H is a rather large inductor, and 1V is a rather low voltage, so it seems reasonable to me.
The impedance of the inductor alone is \$2\pi fL\$ or about 314 ohms at 50Hz, so R1 isn't doing much and the current will be about 3.2mA.
Remember, the magnitude of the total impedance of an inductive and resistive component in series is \$\sqrt{X_L^2+R^2}\$ so unless the two components are fairly close in magnitude the larger one tends to dominate. |
H: Using SPST Switch as Button using Capacitor, which type?
Before I start, know that I'm still very much a beginner at electrical engineering. I'm building a switch box to be used for PC simulators using mainly (non-momentary) SPST switches and a USB encoder.
However, most games/sims can't use these kind of switches as pulses, as they are on-off rather than pulse-pulse.
I could use one of the dual pulse encoders sold by "Desktop Aviator," but According to this article, you can also use a capacitor to create the same pulse effect whenever the switch is toggled. But, it only says "47uF." Which voltage though? Would any work? The pulse should be under one second. A̶g̶a̶i̶n Obviously, I'm very new to this. Thanks!
AI: The voltage rating of the chosen capacitor just needs to be greater than whatever logic voltage you are working at (e.g. 5 volt). Using a 100V capacitor in a low voltage circuit won’t do any harm. |
H: Why do the television manufacturers (Samsung, Sony, LG, so on) exceed the voltage capacity on LED strips (backlight) on TVs?
I don't understand why manufacturers always exceed voltage to backlight TV (LED strips.) For example, I'm fixing a TV with some LEDs burned out.
This is the backlight:
There are two LED strips. Each strips has 9 LEDs, and each LED consumes 3 volts. All LEDs and strips are in series.
It would be: 3 volts * 9 LEDs * 2 strips = 3 volts * 18 LEDs (2 strips are in series) = 54 volts.
Just 54 volts is enough to "on" the backlights.
Why does the SMPS of the TV deliver 133 volts D when I test the voltage to the backlight?
It's 80 volts above capacity of all LEDs together. Why don't the LEDs burn out instantaneously?
AI: Isn't 133V the open-circuit voltage? When LEDs are working, voltage should be lower.
LEDs are driven by constant current sources, it's far better than using voltage regulation, because LEDs have widely varying forward voltages and they tend to have negative temperature coefficient :
Overheating reduces voltage which raises current with a constant voltage regulator which leads to even more overheating until one LED in the chain die.
Here is for example the datasheet of a LED driver used in a LG TV I've repaired :
https://fscdn.rohm.com/en/products/databook/datasheet/ic/power/led_driver/bd9285f-e.pdf
"Current mode DCDC converter" |
H: Choosing a zener for dynamic Rload variations
I am attempting to simplify a schematic by using a zener diode as a regulator for a small radio. At peak, the radio will consume about 50mA @ 3.3V, and in its low power sleep state, virtually nothing (a couple uA). In a previous experiment, where a Zener diode with 1mA Iz and 400mW Pd was used, the voltage would shoot up far above 3.3V. After doing some of the analysis, something is still not quite clicking as to whether or not I would experience voltage fluctuations now while using this particular zener (20mA Iz, 0.5W).
If my voltages at Vz are basically 4.13V and 12V absent the diode on the high and low spectrum of the load draw, respectively, and both would trigger the zener ON, what are some things to keep in mind that might cause this zener to deregulate as mentioned? As far as I know, I'm currently looking at ~200mW dissipation at radio idle, but could come up short of Iz by about 12mA, providing only 8mA of available current when the radio is maxed out.
Thanks
AI: If you want something that just works, use a linear voltage regulator. They're available for pennies. If you want less power dissipation, use a switching regulator.
If you want to use your circuit, the things to ensure are that the value of Rs is small enough to provide enough current at 3.3V, and physically large enough to dissipate all the heat that it generates, ~450mW by my calculation (12v-3.3v/50mA). Your zener also needs to be able to dissipate the required power, which will be ~160mW (3.3v*50mA). Your voltage regulation probably wont be great, but your radio can probably deal with it (although if you post what the radio actually is that would help). You should put a capacitor across the output of your voltage regulator.
Overall, 600-700mW is quite a lot of power to be continuously dissipating in a small space. You need to account for the temperature rise that this will cause. The issue with this circuit, compared to a linear (series) regulator, is that it always consumes the same amount of power, which is the maximum amount. A series linear regulator would be a much better fit for this application, as presumably the 50mA draw doesn't happen much of the time (although even if it happens 99% of the time, a series regulator is better).
Try your circuit out here. |
H: Switch 12V loads with MCU and BDW93C darlington or IRF630N
Task is to switch 2 loads with an MCU (3.3V 10mA max) and few IRF630N, BDW93C and sample pack of S901x transistors (of switch relevant components).
12V loads, likely important context for question:
0.3A 3-pin fan, want speed control, research points to low frequency PWM method
2A XL6009 based boost converter with enable pin, which in turn feeds 24V ultrasonic mist driver (picture 3). Duty cycle around 20s/on 300s/off so I figured keeping converter always connected and switch 24V is not optimal, please point if wrong
According to my poor understanding of datasheet, IRF630 would barely open at 3.3V so additional transistor is needed to run it, so I tried darlington first and am very surprised at observed voltage. Open state: -1.1V closed: short burst to 18V, then steady 12V as wanted. Don't have oscilloscope yet, measured with cheap digital multimeter.
Question 1: please explain negative and burst voltage on load connector. Is it bad? If yes, how to fix it?
Question 2: please advice how to properly switch described loads with listed components.
simulate this circuit – Schematic created using CircuitLab
Just in case I messed up wiring, here is real photo. Long "bus" wire on the right is common. BDWs' pin 1 soldered with limiting resistor to GPIO wire. BDWs' pin 2 go to their load (-). BDWs' pin 3 connected to common.
Photo of load2, XL6009 based DC-DC converter with enable pin and its load, ultrasonic fog device.
AI: Q1) I would guess the cause is current taking a while to pass trough the high input impedance of the voltmeter, try connecting a resistor in parallel to see if results are more stable.
Fan: It only draws 0.3A and so the BDW93C would work just fine, as the fan has 3 pins, you switch the power and ground pins and leave the 3rd pin unconnected (pin for speed sensing)
Boost converter: this boost converter cant turn off completely, only stop boosting voltage. You can use an N-channel MOSFET controlled using a drive circuit.
The reason for needing a drive circuit is that the MOSFET barely turns on with 3.3V, one circuit you can use is this:
simulate this circuit – Schematic created using CircuitLab
This has 1 problem, setting the MCU pin high will turn off the converter and setting it low will turn on the converter. Also if the mcu is not active, the MCU pin will float and the converter will turn on.
One way to fix this is:
simulate this circuit
This circuit uses both a PNP and an NPN however, it will prevent the MOSFET from turning on if the MCU isn't driving its output. |
H: NPN BJT NOR gate to drive LED. Base resistance problem
Here's a circuit I'm trying to get working :
I have two digital signals (B and C in the diagram) coming from a CD40174 flipflop.
I use these to light up two LED's.
I also use these as the input for a transistor based NOR gate the output of which I call A.
I also need to be able to drive an LED from A
The problem is that if I calculate R11 (the base resistor for the LED driving transistor) correctly (about 5K) the voltage at A is significantly less than 9V.
I am using A elsewhere and it needs to be as near 9V as possible. Also I'd like to draw as little current as possible from the flipflop outputs.
If I raise R11 to 100K I get a nicer, very nearly 9V at A, but the current through the LED (on pin 8) is significantly less than 20mA
Can someone tell me how to resolve this?
AI: If I understand your requirements correctly, you can just connect the pin 6/7/8 transistor as an emitter follower instead of common-emitter.
simulate this circuit – Schematic created using CircuitLab
This will only put ~7-8 V on the anode of the LED, so if you need this LED to be equally bright as the others in your circuit, you may have to adjust the R1 and/or R4 values down a bit. |
H: Heinner OMX-600 blender motor not turning
I have this blender that won't spin when switched on, the leds turn on tough.
I managed to take it apart:https://i.stack.imgur.com/wqbRY.jpg
From the mains power (!caution is used all times!) we have:
on line going to a micro-switch that I shunted (it's purpuse is to make sure the bowl is secure on top of the blender).
a main PCB that has a few resistors, cap (0.1uF), triac (BTB06) , diode (DB3), turn knob ( has 7 states: off, CCW - pulse (has spring and turns back to off if let go) + 5 speeds (all CW) )
a secondary PCB that has 3 LEDs, 2 resistors and an IC that seems to read MDD M7
motor
My theoretical and practical background is limited to low voltage DC circuits mostly but I do understand the dangers of working with mains voltage and would like to further expand my knowledge in this area.
I was thinking of first ruling out that the motor is not faulty. Where should I probe with my DMM on it? Should I be measuring continuity(resistance) along the coils or try to measure AC with it ON as so to make sure it's getting power? ( Yes, dangerous if not careful)
Is there an practical way to test that the triac or diode ( based on the datasheet I see it's not a simple zener diode ) is ok?
Taking parts off the PCB is not an issue if there is a test rig I can create to make sure components are not faulty. I also have an oscope if there is a need for it.
Thanks!
Later edit:
I was lucky and managed to get locally a DB3 diac and a BTB12 triac (i'm guessing the higher amp rating shouldn't influence negatively). After soldering them on there was no change, no motor spin.
I managed to get my hand on an identical, working blender. I then cut the wires from it's controller and hooked it up to my motor and there was a small buzzing from the PCB but no motor spin (I only had it on for max 2 sec as i didn't want to burn anything ). When I hooked up the controller that I am reparing to the working blenders motor it spun up and worked normally. This way I was able to pinpoint that it's my motor that is faulty.
After a little research I found out that it's a universal motor and thus can be powered with my DC bench supply (something i'm more familiar with :P).
If I hooked my bench supply's leads (polarity didn't matter) to:
White lead and brush guide closest to me ( 1 and 2 pic ) the motor would spin.
White lead and brush guide furthest to me it would arc (short).
Blue lead and either brush guide nothing would happen.
Pictures: https://i.stack.imgur.com/C0bjV.jpg
Am I safe to assume that the stator winding that is on the blue side lead has failed?
If so, what are my options?
Thanks
AI: I do understand the dangers of working with mains voltage
and would like to further expand my knowledge in this area.
Please go to school for this. Making a mistake with mains is deadly.
Part of your schooling would be training so that you can preserve your life.
Proceeding without schooling is like Russian Roulette --
it's just a matter of time until you make a mistake,
and feel the lightning bolt. It is wise to learn from the wise.
Seek, value, and retain wisdom! It will preserve your life!
And don't forget to listen to wisdom, once you find it!
I was thinking of first ruling out that the motor is not faulty.
Where should I probe with my DMM on it?
Should I be measuring continuity(resistance) along the coils
This is a good idea. It looks like your rotor may be burned, as it
is darker, which is not a good sign, and indicates an area to
look for further problems, like short-circuits, or open windings.
The resistance reading should all be similar for symmetrical windings.
A winding that looks like the others, but has a significantly lower
or higher resistance is likely one part of the winding shorted to
another, or open altogether.
or try to measure AC with it ON as so to make
sure it's getting power? ( Yes, dangerous if not careful)
Please do not do this! There are other ways to make sure
that it's getting power. Just measure the resistance.
It should be fairly low. But keep the power off while working! PLEASE!
Only turn the power on to test for success.
And make sure to discharge any dangerous capacitors each time you unplug,
and before you work on it. (Yes, it's better to do it twice and be safe.)
Is there an practical way to test that the triac
or diode ( based on the datasheet I see it's not a simple zener diode ) is ok?
This I don't know. But the triac is cheap, so order a few and
swap them in. For less than 2 usd, here's one: "BTB06-600CRG - TRIAC, 6A, 600V, TO-220AB" Order at least 4. If one winding is shorted,
it might work for a little bit and then burn out the triac,
so you may literally burn through a bunch of them.
You may also look at the brushes (spring loaded pieces of graphite)
and make sure that they are contacting properly,
and conducting electricity.
EDIT:
Be careful when you're testing whether the motor works, to secure the motor, as it will "jump" and could damage or injure itself, you, or something else when it does. |
H: How does HFI (high frequency injection) work with BLDC motors
How does HFI work with BLDC motors.
In this video: VESC HFI: Sensorless position tracking at zero speed he explains how HFI works, however i don't get how it can work with motors that have trapezoid back-EMF such as the one he is using.
The code belongs to the author of the VESC project which is an open source firmware implementing FOC with HFI. I tried understanding the code, however i couldn't given the level of complexity.
Can someone explain how HFI is implemented and how it works with trapezoid back-EMF BLDC motors?
VESC github repo
AI: I skimmed through Google materials and there are two steps.
First it measures the impedance of the coils by sending in current at a particular D-Q angle (I think). Then it changes the angle and compares the two impedances. In other words, it's measuring inductance or the strength of the two flux paths through the motor. This lets you get the rotor position, more or less.
https://www.controleng.com/articles/understanding-permanent-magnet-motors/
But you can see from the graph above that it is symmetrical so there is an ambiguity which needs to be resolved. Unless perfectly centered, the
presence of the PM field will bias the ferromagnetic domains in the
iron core so it saturates in one direction earlier than the other. You can ramp up current in both directions and measure when it saturates to differentiate whether you are near a N or S pole. I'm not sure quite sure when this is done though.
Pretty cool. I'll have to read up on it more tommorrow. |
H: How to make hex file for attiny402 microcontroller using winavr
I have written a code in the 'Programmer's Notepad' of winavr. But when I went to make the hex file, I didn't find the attiny402 microcontroller in 'mfile'.
Can anyone help me? Now, what should I do? it is very urgent for me.
thank you
AI: Winavr has not been updated since 2010; the ATtiny402 was released in 2018. According to en.wikipedia.org/wiki/ATtiny_microcontroller_comparison_chart the ATtiny402 requires gcc arch id avrxmega3. So winavr will not be able to work. Check whether you have a new enough version of gcc to support ATtiny402 (avrxmega3), the version installed with winavr is probably out of date. |
H: LM7805 Voltage Regulator stops working when input voltage increases
I have the following circuit:
simulate this circuit – Schematic created using CircuitLab
I designed the PCB, had it printed and now I am testing it.
When I power it up the following happens:
When V1 stays below 27.8V circa, everything works fine. The WeMos board powers up, its led flashes as expected. Output voltage of the 7805 is 5V stable.
When V1 passes the 27.8V, the output voltage of the 7805 drops to 1V circa, shutting down the board. The transistor base voltage oscillates around 0.7-0.8V, causing the relay to open/close very quickly.
If V1 goes below 27.8 volts again the board powers up again and the relay stops flickering.
I tried to remove the WeMos board and put a resistor instead (with Q1 base floating), and in this case everything works perfectly fine, the 7805 doesn't switch off.
I am not an expert in troubeshooting circuits, so I'm asking for help.
What could be the problem? How can I start troubleshooting? Could it be the 7805 oscillates?
Edit: I installed a small aluminium Heatsink on the 7805 (TO-220 package). I calculated a maximum current consumption of 70 mA (there are some leds and another IC using the 5V), for a total power dissipation of circa 2W
AI: This is almost certainly because of thermal shut-down. These kind of old linear regulators perform poorly when they have to go from such relatively high voltages down to 5V - there will be lots of heat no matter the load connected to it.
When thermal shut-down occurs, the output voltage from the IC drops just as you describe. Then when the part cools off, it goes back up again. This leads to oscillation.
You will absolutely need a heat sink, and you could fiddle around with various heat sinks in an attempt to stall the thermal shut down to higher input voltage levels. A less expensive and less problematic solution is to use a LM317 first, to take the voltage down to 15V or so, then place a 7805 "in series" after that.
This is how it was done "back in the days" when these parts were new and it is still ok for hobbyist solutions (or if you actually need heating to drive out moisture), but it isn't how professional designs are done any longer. You should strongly consider using a switching buck regulator instead.
Also, I don't understand your relay circuit. It says 5V relay, yet you feed it the 27V-something voltage. I'm not sure what those 4 resistors in parallel are supposed to do, current limiter? Looks very fishy, did you think they form a voltage divider somehow? You will need a 24VDC coil relay, and even then that coil won't be happy if you stray too far from the nominal voltage. |
H: Exceed recommended power supply range on analog switch
I would like to use an analog switch in my design (NLAS5223C).
The recommended power supply range is up to 4.5V, and the absolute maximum rating is up to 7.0V (relevant specs attached below.)
I'm wondering, if it safe to use this device with the power supply rail at 5V.
I'm sure the official response from On Semiconductor would be that I shouldn't use this device above 4.5V, but I'm wondering what is the effect/risk of exceeding this limit, especially in this particular case of 0.5V. I've also posed this question to On Semi - no response yet, but I'll update here if they do get back to me.
I'm also curious what's happening to the device internally, and if there's a way to quantify the effect / risk.
Note: I intend on passing analog signals from 0-5V through the device. This device does have internal ESD diodes, but I'm assuming that the clamping diodes are relative to the power supply rail, and therefore shouldn't impact the analog signal.
AI: The obvious answer is no. That's what data sheets are for.
However, it's tantalising to have an absolute max above what you want to run it at, isn't it.
The fact that the analogue inputs give a voltage specification with respect to rail confirms that the protection diodes are to the rails.
A serious problem is hinted at by the fact the digital absolute max is not with respect to the rails, but to an absolute voltage of 5.5 V. This is still above 5 V, isn't it? But it's a lot closer than the 7 V for the rails, and suggests there are lower breakdown devices inside. The same applies to TTL, and there the recommendation is not to tie inputs to VCC.
If you're a professional, or intending to sell this stuff, then the answer still has to be an unequivocal no. If you're an amateur, making one, and its failure is not going to inconvenience you beyond making another, then why not try it? You'll probably get away with it. Its failure mode is unlikely to be bursting into flames.
If you're building it for a friend, then you're putting your reputation on the line, is that a thing worth risking? Don't post it either as a 'good' circuit, it isn't.
The manufacturer is rating this switch with their reputation in mind. They don't want to be known for 'fragile' products, so they will have built in some margin, which includes users sometimes having a bit of over-voltage on their rails. How much of that margin do you want to knowingly use up? You might want to put more effort into making sure the power supply doesn't spike. It would definitely be worth trimming a bit off the digital control voltage, or restricting its input current, as is the recommendation with TTL.
There are other analogue switches rated for the voltages you want, though this one has a rather nice low on channel resistance. Is that the reason for it? If so, consider designing with FETs, and you'll get even better on channel resistance, though at the expense of space and component count. Unfortunately, really low Ron often goes with low voltage, and a lot of designs today are targetted at single cell LiPo products, like phones. |
H: Why do people learn electronics? Need some practical examples
Examples of why one would to learn electronics are:
1) He works (or intend to) as an electrical engineer
2) He likes (or needs to) repair electronic equipment
One thing I find it difficult to understand is that some people build circuits just for fun. And there is nothing wrong with that. But are there any practical reasons why one would build that circuit? instead of buying one? or at least buy a chip that's going to make the circuit far more easier to build. In woodworking for example, one can build his own table instead of buying it. And it's not only a fun to build, but the table itself is useful and might be more robust and cost less.
So my question is mainly for people who do build circuits. Are there any practical reasons to build these circuits that I'm missing?
AI: Sometimes it is not the end goal of the circuit your building it for, but to understand the hows and why's of it all, usually starting blind from the outside you have no understand of why things are done a certain way, or what constraints are in place, maybe what things can be swapped out for others
Think of it like baking, some people cook from scratch to learn all this, while others just buy the completed item, usually for less, some people enjoy the process far more than the end result
In combination to this, the new skills you learn on this project your doing for fun can make things cheaper and easier for later projects you do need resolved as you have a prior knowledge base
Other times are that the item you are after does not exist, or is in an undesirable form / unrealistic price. This is where many of my pet projects end up, e.g. want to log 300 thermocouples to 0.05C and feed that straight into python via a USB link? good bloody luck going commercial, as a hobby project it only cost $250, and I learned enough that if I did it again, it would be cheaper still. |
H: Unable to generate AM wave using opamp log antiog multiplier
I was trying to make a op amp based AM modulator. I used a log anti-log multiplier to generate the below waveform.
This is the output I am getting as from this circuit
What I intended to generate
AI: Your circuit idea might be flawed and I can't understand what your waveforms are in relationship to your circuit AND, your inputs into R4 and R5 don't appear to be sourced from anywhere so, start simple using the simplest form of AM circuit i.e. a diode like this: -
That will produce these waveforms: -
All the detail should be in the schematic above but, just in case: -
Carrier is 1 MHz sinewave 5 volt p-p superimposed on an offset of 2 volts DC
Modulation is 10 kHz 2.5 volt p-p with 1.25 volt DC offset
Output band-pass filter (L1 and C1) tuned to about 1 MHz.
I'm using a triangle wave modulation so that I can see it's fairly linear quite easily. Here's what sinewave modulation looks like: -
Start here and build on that. With a bit of circuit invention it can be made to be a full modulator with suppressed carrier if that is what you ultimately aim for. |
H: Difference between DC polarity symbols
On some devices, the DC polarity symbol uses diamonds or squares for the poles instead of circles. Do these two styles represent any electrical of mechanical difference, or is it purely cosmetic?
AI: The purpose is always to see the polarity of the output pins, everything else is just visual taste. |
H: Wall Voltage in 3Phase
I have a relatively simple question on the grid voltage. When we say the voltage from the wall is 220 Vrms is that the line-line voltage or is that the phase voltage. ie are the two output lines from a socket both hot (in which case each line’s magnitude would be 220/sqrt(3) Vrms) or is one hot and the other neutral?
AI: When we say the voltage from the wall is 220 Vrms is that the
line-line voltage or is that the phase voltage
Usually, the output winding at the substation that feeds domestic buildings is a Y formation. This means it produces three phase voltages and one neutral. What should be wired to a wall socket is one phase voltage (live) and the neutral.
But there are variations and what I've described above is typical for UK outlets.
So, if the phase voltage is 230 volts (UK), the line (or phase to phase) voltage is \$\sqrt3\$ higher at 398 volts. If the phase voltage is 240 volts then the line voltage is 415 volts. |
H: Should an unused VCC pin with a higher voltage be left unconnected?
I'm trying to integrate an ESP32-PICO-KIT/V4.1. into a small project.
The documentation for the PICO power supply says this:
There are three mutually exclusive ways to provide power to the board:
1) Micro USB port, default power supply
2) 5V / GND header pins
3) 3V3 / GND header pins
Warning
The power supply must be provided using one and only one of the
options above, otherwise the board and/or the power supply source can
be damaged.
I have chosen to power the board with a 3.3V supply coming from a linear regulator. The PICO board has two pins for 3.3V, and I have connected each one to the 3.3V Vout leg of the regulator. I've also connected a 0.1uF capacitor next to each 3.3V pin.
My question is about the 5V pin on the PICO. It will not be used. Should I leave it unconnected, tie it with a large pull down to ground, or do something else? It's not an input pin so I'm not sure what to do.
Thank you!
AI: You don't need to do anything with the 5V pin, just leave it open, but you might want to consider what would happen if someone plugged a USB cable into the connector. From the official schematic:
There is an SS14 1A Schottky diode between the USB and the AMS1117 input, so the 3.3V cannot backfeed the USB, but the USB could backfeed whatever is connected to that 3.3V supply.
It's not clear from the AMS1117 datasheet but perhaps the 3.3V backfeeds the EXT_5V node and causes the LED to illuminate (albeit more dimly than if a 5V supply or USB was connected) (that would be my guess). |
H: What does it mean to send CAN messages at 10Hz or 100Hz? Is this configurable on Linux SocketCAN
I understand the meaning of message transfer speed, i.e. a theoretical maximum of 1Mbit/second, and the 8Mbit/second in the CAN FD.
That's essentially the rate at which bits are travelling on the copper for a lack of better description.
What I don't understand is the 10Hz and 100Hz that I see in some links like this. I looked at the Linux kernel docs for SocketCAN where it talks about the device properties, such as Clock Frequency.
It seems that this message rate is somehow configurable but I am not sure how to go about that.
AI: This is usually called the broadcast rate, how often you send that message on the bus, e.g. how fast a vehicle is moving should be sent fairly often, but the coolant temperature will not change very often, so can be sent far less frequently. |
H: Battery charger mAh rating - Can it be ignored if battery type is the same?
I have used a brand of NiMH batteries for years with a charger from the same manufacturer. After reading up on batteries I found the following statement on battery university All about chargers
The Ah rating of a battery can be marginally different than specified. Charging a larger battery will take a bit longer than a smaller pack and vice versa. Do not charge if the Ah rating deviates too much (more than 25 percent).
After checking the rating of my charger I found it to indicate 1.2V 500mA. The batteries I have been recharging for years rate at 1.2V 900mAh. So is the charger I have used not suited for the batteries which are from the same manufacturer?
AI: Don't mix up mA and mAh! One is (charging-)current, the other one is overall charge.
mAh = mA x hours, so if your charger charges a battery with 500mA for 1 hour, the overall charge transfered is 500mAh.
To charge the battery up to the full 900mAh will take \$ \frac{900mAh}{500mA}= 1.8~ hours\$. |
H: How to analize an ideal diode circuit with sinusoidal sources and storage elements (inductors and capacitors)?
In my first Electronics classes, we learned how to analyze a circuit with AC sources and ideal diodes but with only resistive elements. In such cases, the currents and voltages respond instantly to inputs (I mean, there's no lag). What we did was that, during the positive half cycle of the AC voltage source, we assumed the current was exiting from its positive terminal, and from there we imagined how the current would divide in each node as it traveled through the branches. I think this is the common way taught. However as soon as a capacitive or inductive element is added, currents and voltages can have lag, and a transient response exists. Therefore, in the positive half cycle of the AC voltage source, the current doesn't necessarily exits from its positive terminal at all instants during the half positive cycle. The \$i\$-\$v\$ characteristics of an ideal voltage source is \$v=v_s(t)\$ and \$i=\text{any value}\$.
My question is how can we analyze/solve a circuit with ideal diodes, AC sources, resistive elements, and energy-storage elements (L, C), in order to analytically obtain the expression for a voltage or current? I wasn't taught about that. My goal is to obtain an exact expression for a voltage or current. While I know using the ideal diode will actually give an approximate value, I'm asking for that model to simplify calculations. As an example, consider the following circuit, where each diode is ideal, \$R = 1 \text{ } \Omega\$, \$L = 1 \text{ H}\$, and \$v_s(t) = 12 \sin {2 \pi t} \text{ V}\$ (\$f = 1 \text{ Hz}\$), and the inductor is initially discharged. Let's suppose we want to solve for the inductor's current.
The current through the inductor, choosing the reference direction from node 2 to 4, and the current through D1, are the following according to LTspice.
Proposed solution
What I'm thinking to do is the following. In BJT circuits, in order to find its operating region, we initially assume the BJT is in active region, therefore we substitute it with its DC equivalent circuit in active region, then solve the circuit and compare if the assumption is correct. If it's correct, the analysis is done; if it's incorrect, we substitute it with the DC equivalent circuit in saturation region or cut-off region. Here I explain a bit more about this method.
So, I suppose that for diode circuits, we can also make an assumption and prove if it's correct or wrong. This answer describes the procedure I have in mind. However, since the source is AC and not DC, I think the solution we obtain will be valid only for half a cycle. So, we can not obtain the complete response for all \$t\$, but only up to the cycle we analyze. Is that correct?
AI: This is the reason simulators were created, because even with the diode, only, it gets complicated.
Consider the case of a diode as a half-wave rectifier, driving an RL load (i.e. your picture without D2). For a mathematical analysis, the diode should be ideal. This would mean that for the 1st half of the period the diode is shorted out, and the circuit disconnected for the 2nd half. But since there is a reactive element in there, the current will not stop when the input voltage would drop to zero. Then, the inductor current will tend to flow the other way, forward-biasing the diode.
But first, let's analyze the circuit without the diode: a simple series RL. The equations would be:
$$L\frac{\mathrm{d}i(t)}{\mathrm{d}t}+R\,i(t)=0$$
with the solution:
$$i(t)=i(0)e^{-{L\over R}t}\tag{1}$$
To solve for \$i(t)\$:
$$\begin{align}
Z&=\sqrt{R^2+\omega^2L^2} \\
L\frac{\mathrm{d}i(t)}{\mathrm{d}t}+R\,i(t)&=V\sin(\omega t) \\
i_{\mathrm{steady}}(t)&={V\over Z}\sin(\omega t-\phi)\tag{2} \\
\phi&=\arctan{{\omega L\over R}} \\
i(0)&={V\over Z}\sin(\phi)
\end{align}$$
So the expression of the total current would be \$(1)\$ plus \$(2)\$:
$$i(t)={V\over Z}\left[\sin(\omega t-\phi)+\sin{\phi}e^{-{R\over L}t}\right]\tag{3}$$
Plotting, side by side, the calculated current next to a SPICE simulation, they would agree:
If the diode is to be considered in the circuit, then the previous equation would only be valid for the first half of the period, plus the part where the diode is forward-biased by the induced voltage. At that point, the current is zero until the 2nd period starts, when the cycle continues. Shown below are the waveforms for the simple RL and for the RL+D:
So, until now, even with the diode, things look like they could be solved analytically fairly easy. But if you insert the D2 from your picture then the things complicate. Until now, there was only a part when the diode was on, and when it was off, and these were two states which could be separated and "stiched" to give the desired waveform. Now, with two diodes, there are 4 states, made by D1 on/off and D2 on/off. In each of these there are different things happening, each influencing the next:
in the 1st half of the period, the current rises according to expression \$(3)\$.
then, the induced voltage forward biases D2, not D1, which conducts until the next period, but it starts by continuing where D1 ended.
D1 again starts to conduct, but this time there are no more zero initial conditions, which means that the waveforms for the two previous states determine the values of the initial conditions for each of the next part.
etc
What remains, though, is the transient solution, \$(1)\$, which is visible in the slowly rising average value of the sum of the two currents, and the solution for the current through D1 (I(R1), green), but only for the first half of the period -- it can be seen that the waveforms coincide in that timespan.
The current through D2 can also be derived (similar to \$(3)\$) and calculated, but, as mentioned, the initial conditions are always changing, until steady-state is reached. So each next half-period has distinct solutions with initial conditions based on the previous half-periods.
At this point, even if I could derive the other formula (the same way, but the voltage has a displacement), I'd rather not because I'm hoping you can see why the simulators are used at this point. And, if you think the simulator decomposes the circuit into formulas, I'd have to dissappoint you, because it does nothing but compose a matrix of voltages, currents, resistances, conductances, what have you (simulator specific), then solves this numerically. It has no idea about complex operators, phasors, and the likes. It simply crunches numbers until a convergence is reached, at which point it declares the result satisfactory. |
H: What's the purpose of a drill symbol?
I use OrCAD environment for designing PCBs but I've never thought about what is the purpose of a Drill Symbol in padstack creator. The diameter is set in the Drill bookmark and other parameters like antipads or thermal pads in Design Layers. I can see it in the layout file. What is that used for?
AI: They're used to create a drill chart, which some PCB manufacturers use to double-check things like the match-up of the CNC drill files and the Gerbers.
I seldom-to-never send it with the Gerbers and drill file. In my experience, not much goes wrong on that account and I doubt they bother looking at it. From this site: |
H: Emitter-Follower Class A amplifier
I was trying to solve a problem from SEDRA-SMITH book. I got stuck so I refer to the solution. When Studying how the author solved the problem I got even more confused.
Let's refer first to the given vbe and ic in the question. As per the explanation in the earlier chapter of the book these givens are instantaneous values.
Here's the first part of the solution. This is only the part I need to understand.
the part of the solution that I don't quite understand is the equation for vbe1.
I suppose this is a DC value. An uppercase letter is used for the symbol and the subscript. But if that is so, the equation does not agree with the equation for the instantaneous vbe that is
Following the above setup the equation for vbe1 should be something like this
One other thing that confuses me is the use of 4.8 mA in the second term of the VBE1 equation. Isn't that supposed to be Is? How is Ie equal to Is? From what I understand Is is a very small quantity. Why is Ie used in this case?
AI: Simply, they use a Large-Signal analysis.
The small-signal analysis should only be used to find voltage gain only.
As for the \$V_{BE}\$ equation. It's strange that you never see this equation before in the book? Especially that is it in chapter 12. But ok.
First, think you need to do is to rearrange the Shockley equation and solve it for \$V_{BE}\$.
$$I_C = I_S \times e^{\frac{V_{BE}}{V_T}}$$
$$\frac{I_C}{I_S} = e^{\frac{V_{BE}}{V_T}}$$
$$\ln\left(\frac{I_C}{I_S}\right) =\frac{V_{BE}}{V_T}$$
$$V_T\ln\left(\frac{I_C}{I_S}\right) =V_{BE}$$
$$V_{BE} = V_T\ln\left(\frac{I_C}{I_S}\right)$$
Now let's define \$ΔV_{BE}\$ as the difference between the actual \$V_{BE1}\$ voltage and some reference \$V_{BE2}\$ voltage for a different collector current.
$$ΔV_{BE} = V_{BE1} - V_{BE2} = V_T\ln\left(\frac{I_{C1}}{I_S}\right) - V_T\ln\left(\frac{I_{C2}}{I_S}\right)=V_T \left(\ln\frac{I_{C1}}{I_S}- \ln\frac{I_{C2}}{I_S} \right) $$
$$ΔV_{BE} = V_T \ln\left(\frac{I_{C1}}{I_S} \times \frac{I_S}{I_{C2}}\right)=V_T \ln\left(\frac{I_{C1}}{I_{C2}}\right)$$
$$ΔV_{BE} = V_T \ln\left(\frac{I_{C1}}{I_{C2}}\right)$$
Therefore, if we know the \$V_{BE1}\$ at \$I_{C1} = 1mA\$ we can calculate the \$V_{BE2}\$ value at different current \$I_{C2}=4.8mA\$.
\$V_{BE2} = V_{BE1} + ΔV_{BE} \$
The end of the story. |
H: Design a MOD-N counter using LS293 circuits
This is the internal scheme of the LS293 circuit.
I understand that the CP0 frequency will be divided by 2 and the CP1 will be divided by 8, but I don't understand how to use this circuit to build a mod-n counter.
AI: you have access to the reset pin. exploit that, as part of state transition table |
H: MOS current mirror region of operation
The mosfet M1 is clearly in saturation as gate and drain terminals are shorted. Why we take M2 also in saturation? I know Vgs1 = Vgs2 and they are identical MOS but i cant justify why M2 should be in saturation, as drain voltage of M2 can be any value..
AI: According to Sedra/Smith text (because textbooks do a better job at explaining things than I do), Microelectronic Circuits, on the subject of basic MOSFET current mirroring:
Now consider transistor \$Q_2\$ (or \$M_2\$ from your picture): It has the same \$V_{GS}\$ as \$Q_1\$... (from Figure 8.1) For proper operation, the output terminal, that is, the drain of \$Q_2\$, must be connected to a circuit that ensures that \$Q_2\$ operates in saturation... To ensure that \$Q_2\$ is saturated, the circuit to which the drain of \$Q_2\$ is to be connected must establish a drain voltage \$V_o\$ that satisfies the relationship \$V_o \geq V_{GS}-V_{tn}\$.
– Microelectronic Circuits ed. 7 by Sedra and Smith
So in order to have a proper current mirror, we have to assume that the M2 is in saturation. Drain voltage for M2 can be arbitrary (to an extent) with that condition of \$V_o \geq V_{GS}-V_{tn}\$. |
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