How to prevent current spikes - Page 1
Forum section: electronics / beginners

author: LoveLaika
author_rank: Frequent Contributor
timestamp: August 25, 2022, 03:30:35 am
content: My colleague asked me a question about an experiment, and I didn't know how to answer, so I wanted to ask for some advice here. The experiment consists of sending an AC voltage source (~12 Vrms, relatively low voltage) to some kind of material sample. The voltage source is fixed, but they seem to plan to adjust the current going through the material to observe characteristics of it somehow. I don't know about the specifics; that's outside my expertise.The question I was asked is how to prevent current spikes/sudden changes in current as the current output increases. In my mind, I'm picturing the setup as a voltage source in series with the material. You can think of it as a resistor, but as voltage is applied to it, the resistance of the material will behave nonlinearly (drop a lot as the sample heats up). So, as a result, they want to prevent sudden large fluctuations in current so as to not destroy the sample. Their idea was to have light bulbs in series given that bulbs have a non-linear voltage response which can help maintain a stable current. I guess I can see that, but I never actually thought about how incandescent light bulbs work. Shame on me for never thinking about it.Back to the problem, aside from possibly stringing light bulbs together, what other simple solutions would work, if you had to put some circuit element in series to reduce current spikes? My first thought was an inductor and how it can't change current instantaneously. I also did some searching, and I found thissitethat showcased a RC parallel circuit in series with AC-line voltage. It says a capacitor in series would limit the current, but wouldn't it act as a short at high frequencies? Assuming 60 Hz, the impedance would be pretty high, so it's the impedance that's limiting current? These are all ideas I'm just looking at off the top of my head, but what other simple solutions would be good here?

author: Ice-Tea
author_rank: Super Contributor
timestamp: August 25, 2022, 06:36:46 am
content: Using a bench power supply with the current limit set appropriately isn't suitable?

author: Robert763
author_rank: Super Contributor
timestamp: August 25, 2022, 07:52:30 am
content: Quote from: Ice-Tea on August 25, 2022, 06:36:46 amAs the voltage source is AC, Not a typical bench supply is not suitable.One simple trick is to use a LM317 and a single resistor in constant current mode combinded with a bridge rectifier.For a 1.2 Ohm resistor current is limited to 1A use 1N4001 diodes or a > 1A bridge rectifier. (Z is the test item)Robert.

author: Ice-Tea
author_rank: Super Contributor
timestamp: August 25, 2022, 08:10:25 am
content: Quote from: Robert763 on August 25, 2022, 07:52:30 amIf you're looking for me: I'll be the one in the corner with his foot in his mouth.

author: tggzzz
author_rank: Super Contributor
timestamp: August 25, 2022, 08:31:14 am
content: The OP's question is too poorly constrained to be answered. The OP is aware of that, which is a good start.Start by ensuring this isn't an "XY problem".Before bothering to consider solutions, answer these questions and any others that spring to mindfrequencycurrent (we know ~12V)reactancewhat do the problem owners regard as the source of a "spike"; indicate any relevant physical characteristicswhat do the problem owners regard as a "spike"; draw a waveform picturecharacterise what would be anacceptable"spike"

author: MrAl
author_rank: Super Contributor
timestamp: August 25, 2022, 10:48:16 am
content: Hello,Capacitor in series with the load probably wont work because they only limit current with pure AC and that is not that case when you first turn the power on. There is an exponential part you dont see talked about much in AC theory but in transient theory it's always something to know.There is one exception and that is if you can turn the AC input source up SLOWLY and then you might see the limiting effect only,but if the load has an unusual response the exponential part of the capacitor response will again appear.The exponential part kicks in whenever the level of the AC changes, such as from 0vac to 5vac, or from 5vac to 10vac, etc. Amazingly, if that was a step in voltage amplitude the capacitor would not limit current at all, at least not for a short time. So i dont believe you want to use a capacitor.Incandescent bulbs work a little differently. Their resistance increases with current, so they do limit current. You have to be a little careful though because they do allow about 10 times the current flow when cold. So for example a 120vac 100 watt light bulb might have 10 Ohms resistance when cold and 100 Ohm when hot. They have been used in such a manner however, and i actually used them myself to test a CRT television one time. It was all i had on hand at the time so i had to do it.Another idea is a variac. You can turn up the voltage slowly with that and watch the current change. The only time this wont work is if you expect any negative resistance, such as with a fluorescent light bulb. You could actually harm the light bulb like that.A resistor helps here. If you think the max current should be 1 amp and you have a 12vac voltage source, then if you use a 12 Ohm resistor you cant get more than 1 amp, at first. You would then have to switch to a lower value to get the full voltage to the load once you know it does not draw excessive current. You may go down to 6 Ohms, then 3 Ohms, etc.There are also simple bipolar transistor current limiting circuits that are very simple, you could use that with a bridge rectifier, but with that or the LM317 idea you will drop a lot of voltage, as much as 2.5 volts.Another key point is you have to have some idea what "too much" current is for a device you are going to test. If the device requires 1/2 amp and you dont know it and you set the limit to 1 amp, it might blow the device anyway. I dont think there is any way around this unless you do not work with anything with a negative resistance as part of it's response. If you dont know the minimum resistance you cant know what current it will draw worst case.

author: mikerj
author_rank: Super Contributor
timestamp: August 25, 2022, 11:25:17 am
content: Quote from: MrAl on August 25, 2022, 10:48:16 amMore to the point, the impedance of the capacitor willreduceif there is a fast transient "spike", the opposite of the desired behaviour. An inductor seems like it may give the desired behaviour, but depending on current it may be impractically large or expensive.

author: Terry Bites
author_rank: Super Contributor
timestamp: August 25, 2022, 11:59:28 am
content: A series indcutor limits the rate of change of current.The indcutor stores energy and acts like a flywheel.The choice of value depends on the phase shift and attenuation of the supply voltage you can work arround.If you do this then want to measaure the current and voltage at the sample end not the psu end.If your sample is very capcacitive, then you may encounter ringing. You can mitigate this with a resistor in parallel with the indcutor.The current handling and internal series resitance of the indcutor need be taken into account.

author: MrAl
author_rank: Super Contributor
timestamp: August 25, 2022, 06:33:52 pm
content: Quote from: mikerj on August 25, 2022, 11:25:17 amHi,Not sure what you mean you only quoted the first line of that post.Second and third paragraph quoted:There is one exception and that is if you can turn the AC input source up SLOWLY and then you might see the limiting effect only,but if the load has an unusual response the exponential part of the capacitor response will again appear.The exponential part kicks in whenever the level of the AC changes, such as from 0vac to 5vac, or from 5vac to 10vac, etc. Amazingly, if that was a step in voltage amplitude the capacitor would not limit current at all, at least not for a short time. So i dont believe you want to use a capacitor.

author: MrAl
author_rank: Super Contributor
timestamp: August 25, 2022, 06:40:17 pm
content: Hello again,An inductor in series is also not a good idea because the value would most likely have to change with every new circuit to be tested. Inductor values are usually chosen based on a circuit that is well known up to that point they are not usually just randomly thrown into a circuit.That's also because they only limit the current for a short time, usually, although it depends on the value and the current levels.For these things usually an active current limit is employed that will always limit current and does not depend on time like a capacitor or inductor and is something that can be considered almost general purpose that would work with all loads up to it's design specs.

author: LoveLaika
author_rank: Frequent Contributor
timestamp: August 25, 2022, 08:11:49 pm
content: Thanks for all the replies. I really appreciate all the responses. Clearly, I didn't have enough information on hand. I was just spitballing ideas, but then I asked my colleague more about the experiment, what he's done so far, etc. Asking lots of questions has helped give a better overview of the whole thing.So, first off, he has previously used a DC source for this. From what I understood, he used a DC supply in constant-current mode to pass a current to the load/sample, using the voltage knob to pre-set a fixed upper-voltage for safety. He starts off at 0 current, and then slowly ramps up the current output by turning up the knob slowly, stopping for a moment before incrementing again in steps. As he waits, he waits for the temperature and resistance of the sample adjusts (seeing the sample resistance drop and the voltage across the sample be reduced). The resistance would drop, but because the power supply is in constant current mode, the sample will only draw as much current as the supply is set to. At least that's the jist of it. This is done slowly as he heats up the sample, so the sample won't be exposed to too high of a current or voltage as its resistance drops. In some ways, I get the idea behind it, but it's hard to think about when you envision the load as a constant, fixed value.Now, he wants to do the same with an AC voltage source. To my understanding, he wants to limit the voltage source and the current source, same way as he did with the DC supply, only this time, with AC.So, following tggzzz's comments, here's what parameters I'm able to gather:To start off, he wants to have a max voltage of 8Vrms and a max current of 1.3 Arms. This may change, but this gives a good place to start off.With regards to samples to pass current through, he says that it can range from 2-50 ohms for doped samples, and from 2 to 200kOhms for undoped samples. As the sample heats up, the resistance drops. For the doped sample, he observed (V=5 V, I=0.5 A) at room temperature and (V=2.8 V, I=0.8 A) after heating the sample to 500 degrees Celcius.Frequency is okay at 60 Hz.A sudden jump in current by 0.1 amps can cause the sample to rise drastically in temperature and destroy it.So, given how the resistance changes non-linearly, I think it's safe to say that he wants some kind of current control with a low-voltage AC power supply. Forget putting an inductor or capacitor in series. This seems more complicated than I originally thought. Though, I think Robert763 brought up a good idea: using something like the LM317 as a constant-current source mode to limit the AC current. I ran a quick demo in LTspice showcasing the differences in constant-current mode with a DC and AC source. So, as the load resistance changes, the load draws more and more current while keeping the load voltage more or less the same. Once it draws more current than what the circuit is set for, due to the resistor in front of the LM317's output, it stays fixed at that current, and the load voltage drops accordingly. I think this may work with a bit of modification. For instance, adding a potentiometer in series with the resistor at the LM317's output would allow for a finer control and achieve the same constant-current that was achieved with the DC source, though I'm not sure if having the voltage be constant is okay. With the DC power supply, voltage can vary in constant-current mode, but using the LM317 in a constant-current configuration fixes the output voltage.For the AC input voltage, I was thinking that a VARIAC would work here. We can connect it to wall-power and just slowly adjust the input AC voltage to whatever voltage we need at the output.

author: LoveLaika
author_rank: Frequent Contributor
timestamp: August 25, 2022, 08:19:38 pm
content: Robert763, thanks for your reply. I asked my colleague for more info about the experiment, and I wrote up a reply below detailing what I've found out. Thanks for your circuit. Since I got a better understanding of what is required, I think this might be what he is looking for. Previously, he used a DC-source in constant current mode, and now I think he wants to do the same for AC. Your circuit does seem to do that, although the output voltage is fixed and doesn't change as the current draw changes. I'll have to check if that is okay. The sample will naturally draw as much current as it can given its resistance, but if we slowly ramp up the current limit and it always hits that limit, then the voltage will drop rather than stay constant.Nevertheless, thank you for pointing me in a good direction.

author: David Hess
author_rank: Super Contributor
timestamp: August 25, 2022, 10:03:31 pm
content: The described problem is the same one dealt with when driving gas discharge lamps, including fluorescent lamps, because of their negative differential resistance. The solution is to "ballast" the voltage source so that it has a higher output impedance, or implement a current source instead.Power supplies are typically considered to be voltage sources, with an adjustable current limit, but nothing prevents designing one to primarily act as a current source with a high impedance output and voltage limit. This applies equally well to an AC output.A practical high performance design might use a transconductance (current) output plus an additional voltage limit loop. Some curve tracers operate this way.

author: Benta
author_rank: Super Contributor
timestamp: August 25, 2022, 10:13:44 pm
content: I'd use an audio power amplifier for this (using a signal generator as source). Plenty of those around, we're talking 25...40 W here.Enough drive for the load, and designing a current feedback mechanism is not hard.

author: LoveLaika
author_rank: Frequent Contributor
timestamp: August 25, 2022, 10:25:05 pm
content: tggzzz, thanks for your reply. Yeah, I only started off the problem with the info at hand, but I asked more questions about the experiment and the specifications that he require. I did my best to answer the questions that you posted, and I also wrote up a description as best as I could about how the sample was experimented on with DC voltage. Previously, it seems that he used a bench top DC supply with a set voltage, and stepped up the current in certain increments as he heated up his sample. Admittedly, it's kind of hard to envision this experiment, ramping up current while the sample's resistance changes at the same time. It sounds like a recipe for disaster. Nevertheless, it seems he wants to do the same with an AC source.Through Robert763's simple trick, it seems that it would allow for current-limiting of an AC source, with the max AC voltage being limited by the diode voltage drops and the AC source itself. Frequency doesn't seem to be a problem here, though I'm concerned that this would not behave exactly as it would when using a DC source.

author: LoveLaika
author_rank: Frequent Contributor
timestamp: August 25, 2022, 10:37:48 pm
content: Thanks for your reply. I'm unfamiliar with the examples that you brought up, but I'm rather glad to see that this isn't new territory. What do you mean by ballasting a voltage source? I'm only barely familiar with it in name only through use in fluorescent lamps, but I've never used it. It seems that it's just a simple ballast resistor? A resistor that changes as current flowing through it changes? I can see that being one solution along with that trick circuit from Robert763. I'll have to see if there are any SPICE models of a ballast resistor and see if it will work in theory.

author: tggzzz
author_rank: Super Contributor
timestamp: August 25, 2022, 11:00:29 pm
content: Quote from: LoveLaika on August 25, 2022, 10:25:05 pmYou're welcome; it is pleasing to be able to help someone help themselves.From your description, it sounds like the interest is in an AC signal superimpoposed on a DC value, such that the instantaneous current is never less than zero.Given that the AC frequency is low, you have many options for controlling it, possibly including a commercial PSU that can be configured by a PC or Arduino or RPi via a USB or Ethernet or GPIB interface. The combination of fast (but expensive) implementation, plus controllability might be advantageous.

author: David Hess
author_rank: Super Contributor
timestamp: August 25, 2022, 11:01:07 pm
content: Quote from: LoveLaika on August 25, 2022, 10:37:48 pmA resistor works fine as a ballast but has the disadvantage of dissipating power. In older circuit designs it was common to use high voltage and a series resistor to make a current source.Gas discharge lamp power supplies produce AC and use capacitive or inductive ballasts in series with the load for high efficiency. Sometimes a series LC circuit is used as a ballast.I am not really suggesting that a ballast element is the solution to your problem, but it might be part of it. An inductor placed in series with a transconductance (current output) power supply increases the impedance at higher frequencies where the feedback loop controlling the current might have problems responding quickly enough. In the same way, a large capacitor placed across the output of a voltage source lowers it impedance at high frequencies.

author: David Hess
author_rank: Super Contributor
timestamp: August 25, 2022, 11:03:31 pm
content: Quote from: tggzzz on August 25, 2022, 11:00:29 pmSome commercial 4-quadrant power supplies allow for an input signal to control the output voltage, essentially operating as an not-bridged audio power amplifier, but offhand I am not aware of any which can operate as a controlled current source.

author: Ian.M
author_rank: Super Contributor
timestamp: August 25, 2022, 11:48:02 pm
content: If square wave AC is acceptable, reversing the sample cell's connection to the CC/CV PSU previously used at an appropriate frequency may be sufficient. Use a MOSFET H-bridge, controlled by a toggle flip-flop, clocked by a signal generator, set to 120Hz for 60Hz squarewave out.

author: LoveLaika
author_rank: Frequent Contributor
timestamp: August 26, 2022, 04:02:36 am
content: I can see how that can act as a current source, though I also see many risks with that approach. I can't say for sure whether or not we need a lot or a little impedance going to the output. I think maybe what we really need is control so as to not burn out our sample during testing. In that sense, I'll need to work up a demo to understand how my colleague wants to control the power supply.

author: LoveLaika
author_rank: Frequent Contributor
timestamp: August 26, 2022, 04:09:04 am
content: I appreciate the reply and the options you suggested. To be honest, thats something I havent considered, whether a DC bias is necessary or not. Right now, it feels like I'm getting all sorts of suggestions (some simple, and some complicated) when I may not need that much. I think my colleague wasn't looking for any fancy solutions. He was told about the lamp idea, asked me what I thought of it, and asked about some other ideas I may know of when it comes to current controlling an AC source. From all the replies, using lamps to limit the current seems out of the question. I think I'll have to create a mock-demo to get a better feel of what he wants.

author: LoveLaika
author_rank: Frequent Contributor
timestamp: August 26, 2022, 01:34:26 pm
content: To give a bit more information about the whole thing, I was told about why they needed an AC source in the first place. I don't think I articulated it well enough, so I put it in a comment here, about why they needed it and their initial plans of accomplishing it in the first place.So, they typically send a current through a metal sample to do some experiments. However, when applying DC current through the sample, it will cause oxygen vacancy, whereas AC current won't. That's why they need an AC-source; their goal is to drive current through the sample to heat it up, in this case, AC current.What they have is a VARIAC (0-130 VAC, 60 Hz). Their initial idea was to connect in series with light bulbs to limit the current along with a potentiometer to fine-tune the power.

author: tggzzz
author_rank: Super Contributor
timestamp: August 26, 2022, 03:25:25 pm
content: Quote from: LoveLaika on August 26, 2022, 01:34:26 pmIncandescantlight bulbs are of course a non-linear resistance. Experimentation would be required to establish whether that hindered or improved controllability. I doubt other types of light bulb would work.Perhaps it is worth considering a simple high power resistor bolted to a heatshink. Or maybe a signal generator driving an (audio) amplifier. At 60Hz, the signal generator could even be a PC's sound card, thus providing automatic stepping/control of power levels.

author: LoveLaika
author_rank: Frequent Contributor
timestamp: August 26, 2022, 07:14:35 pm
content: Thinking about it, a resistor would simply reduce the current going to the load; it doesn't necessarily cap the current going to the load. To simplify things, what he wants is to control the current coming from an AC voltage source. Boy, I wish this was clear the first time around. People don't know exactly what they want, so all I can do is show them demos and ask if this is what they're looking for.I demo'ed Robert763's circuit in a simulator, but their complaint is that it cuts off the current (when the current reaches the pre-set current limit or goes over it, the regulator will cut it off at that limit, like how an op-amp output will saturate to +/-VCC if it tries to go any higher or lower than its power supplies; the output gets clipped). What they want is to limit the amplitude of the AC current output rather than have it clip. I think I get what they want, but how to do it? Is it even possible? You don't know how much current will be drawn from the load, but what they want sounds like dynamically adjusting the current limit based on the current being drawn...dynamically adjust it such that you never have clipping in your AC-current. Does that make sense?

author: LoveLaika
author_rank: Frequent Contributor
timestamp: August 26, 2022, 08:59:07 pm
content: Thanks for the suggestion. Using an H-bridge, I can kind of see how to get a rough AC voltage, but if using a DC source for it, how does it work when it comes to current control?

author: MrAl
author_rank: Super Contributor
timestamp: August 27, 2022, 07:34:52 am
content: Quote from: LoveLaika on August 26, 2022, 08:59:07 pmHi,You can actually buy an AC power supply but they will be more expensive than DC types.Using an H bridge, you would basically be building a synthesized sine wave generator. That's an AC converter circuit. it takes in a DC voltage and outputs an AC voltage. Since they regulate voltage it's not too hard to add current limit too. Just a way to sense current instead of voltage, and use that as feedback.Unfortunately it's not that simple to build an AC converter circuit with an H bridge. There are various peculiarities that have to be addressed such as timing and heat dissipation and transistor voltage spike suppression. You should probably find a design already made and tested if you go that route.

author: Ian.M
author_rank: Super Contributor
timestamp: August 27, 2022, 10:40:31 am
content: True - if sinusoidal AC is a requirement. (Probably *YES*, as the user rejected clipped AC) However, square wave AC is *MUCH* simpler, as I described when suggesting an H bridge.If I had to build a low voltage medium current AC source from scratch, I'd use a class D amplifier module as the output stage as it does all the hard work for you. Feed it an AC signal, through a voltage controlled attenuator so you can implement current and voltage control feedback loops, and you are well on the way. Alternatively use a multiplying DAC as the attenuator and use a MCU to implement the control loops digitally.

author: LoveLaika
author_rank: Frequent Contributor
timestamp: August 29, 2022, 03:25:14 pm
content: Thanks for the reply. Thank you for your suggestion, though it really depends on what my colleague really wants (they never seem to adequately explain exactly what they want in situations like this). I don't know how much money they want to spend on this, but buying an AC voltage/current-controlled power supply would be expensive. Using a make-shift current-control circuit leads to clipping, which is what they wish to avoid (though I think I might be able to persuade them on this). I told them about how it would be difficult to implement current controlling for an AC source without some sort of external control like others have suggested, and all they said was that if there is no easy way for the output to be a pure sine wave, then the makeshift control with clipping would work. (Running simple tests for this circuit doesn't seem to be working, as I always seem to blow a fuse every time I plug into my VARIAC, making me feel that this is not safe).Putting aside the aforementioned issues, I thought an H-Bridge converter would be okay, turning the DC output to an AC-square wave (with the MOSFETs being controlled by something running at 60 Hz). Thinking about it, the DC supply's current limit would naturally reduce the voltage as the supply reaches the current limit. It could work, but the problem is it would still rely on external power to power the external switch controller. That's kind of concerning when you don't have many sources available. I was thinking of some sort of very simple crystal oscillator at first, but as I looked into H-Bridge designs, the switch controllers can be quite complex (or simple). I don't know how much my colleague wants to spend on this solution. (also, if I may ask, do you have resulting spikes at the transition points when the square wave changes voltage/current?)Putting the above comments aside, I've never worked withclass D amplifiersbefore (or any amplifiers for that matter). What kind of amplifier modules would you recommend? If I could use it as a voltage/current control for an incoming AC signal, then this might be what I need. Looking at them, they seem to be used for audio supplies. What pre-made modules would you recommend that offer voltage and/or current control of the output?

author: LoveLaika
author_rank: Frequent Contributor
timestamp: August 30, 2022, 03:53:50 am
content: Quote from: Robert763 on August 25, 2022, 07:52:30 amI ran a test on a breadboard using variable resistors as my mock load, and it seems to work more or less. There were some issues using the output with my VARIAC (mainly blowing fuses), but I managed to get it somewhat working through use of a signal transformer. I see the output voltage clipping as more current is being drawn, though I'll need to run more tests. Thanks for all the help. I think I have an idea on how it works, but I wanted to ask about it here in order to make sure.To start off with what I know, from the DC perspective, the LM317 is configured as a precision current-limiter circuit, with the current being controlled via the resistor. The resistor is in between the IC's output and the adjustment terminal. Looking at the internals, there's a 1.25-volt reference Zener diode between ADJ and the internal op-amp's non-inverting pin. The output terminal gets current from the BJT's releasing more current, but it's also connected to the internal op-amp's inverting terminal. So, if there is a difference in voltage between a point in the circuit and the ADJ terminal, the op-amp will cause more current to flow through the IC in order to keep a 1.25 voltage difference between the internal op-amp's two terminals. Is that correct?On that note, relating the DC circuit to the AC one, it's still the same operation. You're essentially placing the current-limiter circuit in between the DC-terminals of the bridge rectifier. If you draw too much current, normally, the DC-output will drop in response to that, so as to keep the current flow at the maximum current output set by the current limiting resistor. The key thing is that the controlled current has a return path to the AC-input, which is in series with the Z-load. So, whatever current limit you set with the precision current limiter, it will also affect the AC-current going through the Z-load. Is that logic right? At first, I thought it was confusing, but after thinking it through, following the return paths of current, I see how it easily make sense.After testing and thinking about it, I realize how it works now. Straightforward enough if you think about it. One thing that bothers me though....at what reference voltage does the AC voltage load draw current from? I mean, breadboarding it, as the load increases drawing more current, the RMS voltage does not stay the same (assuming that the AC load is simply a resistor). So, how do you measure how much current is drawn by hand without a voltage reference?

author: Geoff-AU
author_rank: Frequent Contributor
timestamp: August 30, 2022, 05:27:37 am
content: If clipped AC is OK then it's easy, you've got some good and cheap solutions already.If clipped AC is not OK then it sounds like you're building a variable gain amplifier, and that can be as fancy/expensive as you have the patience for

author: LoveLaika
author_rank: Frequent Contributor
timestamp: August 30, 2022, 08:37:36 pm
content: Even clipped AC comes with its own problems. I get the logic behind how a makeshift current control circuit would work via a linear regulator, but breadboarding it isn't yielding the results I had hoped for. Testing the LM317 circuit recommended by others, it's not capping the circuit at the specified current limit. For example, I use a resistor to cap it at a peak of 10 mA. This was verified with an ammeter and using a DC power supply. But, when using it in the AC configuration, I found I could draw more than 10 mA. (With 10 mA peak, that's 7.07 mA RMS; I was able to draw 8.2 mA RMS with noticeable voltage clipping).......perhaps I'm thinking about this in the wrong way. If the current waveform has flattened at the peak, then the RMS calculations won't hold, so I can't really rely on the multimeter for an accurate reading at that point.

author: tggzzz
author_rank: Super Contributor
timestamp: August 30, 2022, 11:26:37 pm
content: Quote from: LoveLaika on August 30, 2022, 08:37:36 pm"Clipping/limiting" is an instantaneous operation. The alternative is a variable-gain component, where something varies the gain to keep the peak and/or RMS within limits."RMS" implies averaging over a period, which raises the question of what period.

author: LoveLaika
author_rank: Frequent Contributor
timestamp: August 31, 2022, 01:47:52 pm
content: No no, it was just my poor assumptions...I was assuming the RMS value given a regular sine wave, $$V_{RMS} = \frac{A_0}{\sqrt{2}}$$. I naively thought that it would hold consistently, so I could use the readings from the multimeter to calculate the peak current being drawn from my load. However, as the resistance got lower, I saw that I was drawing more current than I should have given my current limiter. This made me think that my circuit was broken at first, but then as I typed out my problem, I realized that maybe it was working correctly after all. With the "clipping/limiting" occurring when the current draw is exceeding its set value, the regular RMS formula for a sine wave no longer holds, so I cannot use the equation above to calculate the peak current amplitude. This is what I get for using a variable resistor; you can see the voltage change, but you don't know the resistance at the time. So, using fixed resistor values, I can observe the voltage waveform and calculate the peak current passing through via Ohm's law. That way shows that the circuit is indeed working.If I may ask.....this configuration turns the voltage source into a current source. Thinking of it like that, how can we calculate the voltage across the load? The regulator will set a limit to the max current that can be drawn from the load. However, if you know very well that the current being drawn by the load is less than the max current limit (but you don't know the current being drawn), can you calculate the expected voltage beforehand?

author: tggzzz
author_rank: Super Contributor
timestamp: August 31, 2022, 07:39:28 pm
content: A suitable starting point is the search term "load line".

author: LoveLaika
author_rank: Frequent Contributor
timestamp: August 31, 2022, 08:12:21 pm
content: Thanks. I feel like I forget the basics as I work on more and more problems. Thanks for reminding me. Some engineer I am.Well, for a bit of good news, they like the idea. I demo'ed it to them, and they said it would work. Thanks everyone for helping me out! Now for the next part: designing. Setting a 10 mA limit worked well, but now they want to have a limit up to 2 amps. I just got to beef it up accordingly....and hope they'll approve the budget for it.

author: LoveLaika
author_rank: Frequent Contributor
timestamp: August 31, 2022, 10:51:15 pm
content: Quote from: MrAl on August 25, 2022, 10:48:16 amMrAl, thanks for your reply. Sorry for not replying sooner. I was testing one of the solutions that was suggested, using the LM317. It effectively functions as a current source, and they said they said that the resulting load voltage doesn't matter to them, only the output current. The solution works, and after demo-ing it using simpler parameters, my colleague liked it and they want me to build something for it outside of a breadboard. You were right about the voltage drop though. I thought that the drop was from the diodes, but I should have accounted for the regulator itself. They didn't seem too concerned about it though.Nevertheless, they're extremely insistant on going ahed with using the circuit with incandescent bulbs. In a way, it's like how you said: if the variac is the controllable voltage source, then the proposed circuit would be to have one or two paralleled incandescent bulbs in series with the load they wish to test. It's shown in the image below. As you said, resistance will increase with current, and one should be prepared for a surge in current when cold, but could you control this surge with a variac, start it off at a low voltage and slowly build it up to get the resulting current you want (as measured by a multimeter)? Honestly, they're so insistant on this that they wanted me to say that this is a good idea, but I'm not confident at all about how to control current at all in this manner, or at least accurately assume what current will be outputted.

author: Vovk_Z
author_rank: Super Contributor
timestamp: September 01, 2022, 08:13:23 am
content: Quote from: LoveLaika on August 25, 2022, 03:30:35 amWhen you want to control current you possibly need a constant current source? Respectively, for your task you need AC current source. (?)

author: LoveLaika
author_rank: Frequent Contributor
timestamp: September 01, 2022, 03:01:57 pm
content: Do you know of any? We don't have any AC power supplies that can control current. At best, we have a AC voltage source (a variac) which can control voltage with no current control at all. That's why they have this idea of using bulbs to act as resistors in such a way to function as a makeshift current limiter.I tried looking for test equipment, but the only one I found was theBK 9801. Their current control sets a limit, but rather than stay at that limit, it shuts off the output when the current limit is exceeded. I don't think this offers the current control that is needed.

author: LoveLaika
author_rank: Frequent Contributor
timestamp: September 01, 2022, 07:06:06 pm
content: Well, aside from looking for a bench power supply for this purpose, I had a look at some other circuits using light bulbs. I was rather surprised that others seem to have the same idea. Thinking about it, I guess this makes sense now after researching the idea. After watching some videos, I came up with the circuit below. This wasn't exactly the way they had things, but I figured that this would be a good way to safely do it.So, my isolated VARIAC will act as an adjustable voltage source, and the light bulbs in series will act as limiters for current flow based on the bulb's wattage (and the incoming voltage). For example, for a 60 watt bulb in series with 120 Vac, the bulb will limit the max current drawn to 0.5 amps. This is basically the principal behind how all this works? Put more light bulbs in parallel with each other, and as you close switches, more current will be released? Gotta say, I feel a bit more understanding about this now after researching the idea. Now the question is: how to do it safely?

author: Jwillis
author_rank: Super Contributor
timestamp: September 02, 2022, 06:00:42 pm
content: Your circuit is sound for current limiting . Given the parameters from the Opening post and reading the other posts , I get a 12Vrms and 2A requirement. A 100W incandescent would pass about 150mA at 12 V .This would mean that your would require 13 to 14 parallel bulbs to get close to the required 2 Amps.But if I may another option. Your will need a power op amp like the LM675Thttps://www.ti.com/lit/ds/symlink/lm675.pdf?HQS=dis-dk-null-digikeymode-dsf-pf-null-wwe&ts=1662139948122&ref_url=https%253A%252F%252Fwww.ti.com%252Fgeneral%252Fdocs%252Fsuppproductinfo.tsp%253FdistId%253D10%2526gotoUrl%253Dhttps%253A%252F%252Fwww.ti.com%252Flit%252Fgpn%252Flm675or similar. Some power op amps have built in adjustable current limiting .which is cool.I did a simulation onhttps://www.falstad.com/circuit/with following circuit and it seems sound. Unfortunately, the simulator is restricted to the LM741 and LM324 so the voltages and current is also restricted . Another problem I found was the simulators scopes seem squirrely and require resetting often to function correctly . Ill include the text to import the circuit into simulator. Give it a try . I'm interested in any input from others as well Including suggestions , improvements and running in a better simulation.Remember that it 's Alternating Current. So the current will be sinusoidal as is the voltage at 60 hertz. So run the sim slowly and you will see that the current peak won't change with change in load but the the voltage will.Ohms law.Best regards