Measuring electrolytic caps with power supply and analog multimeter? - Page 1
Forum section: electronics / beginners

author: RiRaRi
author_rank: Regular Contributor
timestamp: October 05, 2023, 09:40:31 am
content: Hello. I heard that only right way to measure electrolytic cap is by charge it with rated voltage and then measure the discharge with analog multimeter so falling needle can be seen. Here is the video of my test on 470uF 10V. Is this good test, if not why?Did this cap failed test? Needle looks almost like without cap, falling instantly.Edit: when i unplug from power supply rather then turning it off, needle falls much slowly. It starts to fall faster and then more slow. It almost stops at 2V. What to look for? How does needle look on good cap and how on bad one?https://youtube.com/shorts/DK4XlPYQVR4?si=Pica2CsS9_ZNkNA8

author: u666sa
author_rank: Frequent Contributor
timestamp: October 05, 2023, 11:21:58 am
content: I guess you have to have a good one and know how needle falls with a good one, to know how a bad one looks like.Spend $100 for aHantek 1833C.

author: Kleinstein
author_rank: Super Contributor
timestamp: October 05, 2023, 11:52:57 am
content: The test with turning off the lab supply is pretty much useless. Many lab supplies have quite some output capacitance that would be in parallel to the DUT and there is a not so well defined internal load to discharge the capacitors.One should ideally check both the capacitance and the ESR. For the capacitance some DMMs can measure this, at least for the smaller ones. For the ESR it needs a more specialized meter or a test setup with generator and scope.

author: MrAl
author_rank: Super Contributor
timestamp: October 05, 2023, 12:00:38 pm
content: Quote from: RiRaRi on October 05, 2023, 09:40:31 amHi,If you just want to test the value (like 470uf, 10uf, 1000uf, etc.) then you can do it that way, but you have to use a resistor in parallel to discharge it. When you take the resistor value R and multiply it times the capacitor C value in Farads, you get the time constant TC.The TC is the time it takes to discharge to about 37 percent of the maximum voltage, and that time is:TC=R*CThis means if you charge it to 10 volts then disconnect the power supply, the time it takes to discharge to 3.7 volts is the time T, so just then divide by R to get C:C=T/RIf you just charge to 1 volt then the time it takes to discharge to 0.37 volts is the time T and again:C=T/RYou choose R such that you can see the meter needle do down slow enough so you can measure the time when it gets to 0.37 volts approximately.You can also do the charge test. With the capacitor voltage zero, connect R in series with C and then turn on the power supply. The time it takes C to reach 63 percent of max voltage (which could be 10v for example) is the time you divide:C=T/RSo for charging it is 63 percent, and for discharging it is 37 percent.The best test is with an oscilloscope and function generator but not sure if you have those two pieces of test equipment.

author: RiRaRi
author_rank: Regular Contributor
timestamp: October 05, 2023, 12:15:40 pm
content: Thanks. I dont have scope and fun gen but plan to get them and learn how to use them. Can you explain how would i test cap with fun gen and scope?I use capitance od dmm and esr for now.

author: Kleinstein
author_rank: Super Contributor
timestamp: October 05, 2023, 01:36:03 pm
content: With a scope one can measure capacitors by applying a square wave (could be a simple NE555 or HC14 based oscillator) to an RC series circuit and than look at the decay / charge curve. This especially also works for smaller capacitors - really large one would need relatively low resistors (e.g. 50-100 Ohms so that R_out of the gererator may be relevant) and maybe a powerful enough square wave source with relatively low frequency.For electrolytics that were not used for a longer time it may be a good idea to apply some voltage for a an hour or so to reform the oxide layer. Without it the capacitance may read high, but could well drift down on use.

author: Wallace Gasiewicz
author_rank: Super Contributor
timestamp: October 05, 2023, 01:42:51 pm
content: Your test suggests a "Leaky" cap. This is not the Equivalent Series Resistance but a parallel resistance. But I do not think that is the case here because of your voltmeter.If you just disconnect the cap from the PS the voltage should drop in manner described By MrAlYou must remove the voltage supply because either the supply will continue to supply voltage as its internal components discharge, or it will be a drain on the cap being examined.If you just disconnect the voltage supply the dropping of voltage in the cap will be dictated by the resistance of the meter mostly. Or by an internal leak in the cap.I do not know what the input impedance of your meter is but I think it is maybe 1 K ohm per volt or even less and that would cause a rapid discharge of the cap. If you had a 10 Meg ohm input voltmeter the discharge would be over minutes not seconds.Your meter is discharging the cap, in my opinion.

author: wasedadoc
author_rank: Super Contributor
timestamp: October 05, 2023, 02:07:00 pm
content: Compared to a DMM an analogue meter has much lower resistance and it usually changes with the voltage range selected. The meter in the video has 20 kOhm/Volt marked in the lower right corner of the scale. That means when set to 10 Volts (as in the video) the meter is a 200 kOhm load across the points it is measuring.The discharge of a capacitor by a resistor is one of the early things one should have learnt. After the time equal to the product of the capacitance and resistance the voltage wil have decayed to 36.8% of initial value.470uF times 200k is 470E-6 x 200E3 is 47 x 2 which is 94 seconds.So 94 seconds after removing power supply (disconnect wire, not operate switch) the meter should be reading about 3.7 Volts. Bear in mind that even new electroltyics typically have +/- 20% tolerance.(Edited to correct minor typos and use more accurate calculation of time constant.)

author: Wallace Gasiewicz
author_rank: Super Contributor
timestamp: October 05, 2023, 03:04:34 pm
content: Here is something you could do:Charge up the cap using PS without any connection to the voltmeter.Remove the cap from PS entirely, don't touch the leads with your fingers or anything else.Then measure the cap voltage with your voltmeter.I'll bet the voltmeter goes to almost full voltage and declines rapidly.

author: rstofer
author_rank: Super Contributor
timestamp: October 05, 2023, 04:23:31 pm
content: Here is an image of a square wave forcing a 0.1 ufd capacitor to charge/discharge through a 10k resistor.Tau, the time constant, is 1 ms.The charge/discharge current will. be too high to use 1 ms with 100 ufd and will be limited by the nominal 50 Ohm output impedance of the function generator. I would try to get the total driving impedance to 1k (50 Ohm internal and 950 Ohm series) and then realize that Tau = 100 ms. Something like 0.8 Hz would drive 6 Tau for charge and 6 Tau for discharge like the attached waveform.

author: TimFox
author_rank: Super Contributor
timestamp: October 05, 2023, 05:37:01 pm
content: Analog voltmeters (test equipment):Common input resistance values for moving-needle meters:"VTVM" (and solid-state versions): typically 11 megohm total, with 1 megohm in the probe near the tip (for AC isolation)."VOM" (totally passive): good ones are "20,000 ohms/volt", meaning 50 uA movement.Cheap ones might be only "1,000 ohms/volt", meaning 1 mA movement.Expensive ones were "100,000 ohms/volt", meaning 10 uA movement.Modern DMMs (digital readout) are typically 10 megohms, but some have very high resistance on low voltage scales.

author: wasedadoc
author_rank: Super Contributor
timestamp: October 05, 2023, 05:45:33 pm
content: Quote from: TimFox on October 05, 2023, 05:37:01 pmWhich of "high" or "low" or "different" is missing from that sentence?

author: TimFox
author_rank: Super Contributor
timestamp: October 05, 2023, 05:47:28 pm
content: Quote from: wasedadoc on October 05, 2023, 05:45:33 pmOops: I corrected the post.

author: Wallace Gasiewicz
author_rank: Super Contributor
timestamp: October 05, 2023, 08:06:44 pm
content: A question to all;With the fancy digital supply used what would you expect a normal current be thru a good electrolytic cap, after sharging up?? It would be very little but how little?I think this is more important with electrolytics than the ESR.

author: wasedadoc
author_rank: Super Contributor
timestamp: October 05, 2023, 10:50:57 pm
content: Quote from: Wallace Gasiewicz on October 05, 2023, 08:06:44 pmThere is no one answer that covers all uses of electrolytics. Two examples:1. As a reservoir capacitor in a PSU, ESR is typicaly much more important than leakage.2. As a coupling capacitor between stages in a low power amplifier, ESR usually has no impact at all whereas leakage could drastically affect bias voltage.(And what has the supply being "fancy digital" got to do with capacitor leakage current?)

author: Wallace Gasiewicz
author_rank: Super Contributor
timestamp: October 05, 2023, 10:57:27 pm
content: Because you might be able to see the current flow (leakage) on the display.

author: wasedadoc
author_rank: Super Contributor
timestamp: October 05, 2023, 11:15:07 pm
content: Quote from: Wallace Gasiewicz on October 05, 2023, 10:57:27 pmFor a good 470uF cap operated below or at its rated voltage I would expect the current to stabilise at much less than the 1 mA that PSU's current meter can resolve.

author: MrAl
author_rank: Super Contributor
timestamp: October 06, 2023, 12:50:37 pm
content: Quote from: RiRaRi on October 05, 2023, 12:15:40 pmHello again,To get an idea what the capacitance is you could do the same thing but use a scope to measure the voltage, that's all. But before you do that you should check to make sure the cap is still good. With a scope you can get a much better idea if the cap is any good to begin with.The better tests are shown in the attachment.Here we apply an 8v peak to peak square wave to the cap through a 600 Ohm resistor. Those specs are not critical, but you have to have the square wave amplitude higher than the peak you measure with the scope. As you can see in the diagrams, the measured voltages are very low compared to the 8vpp square wave being applied. You could get away with a 4vpp square wave for example but probably dont go any lower than that.As you can see on the right, the measured voltages are very low, much lower than 0.1 volts, even with that 8vpp square wave. That's the first thing to note. The measured voltage has to be low in order to see a fairly straight line with a good cap. It also has something to do with the frequency too though, because if you go too low in frequency you will get a curve, but that curve will go much higher in voltage, like maybe as high as 2 or 3 volts. That would be too high to see the straight line.You can see on the left, the voltage is low (around 0.2 volts peak) but the wave is still curved. That indicates that something is wrong. You can also see, and this is important, the wave curves up and then comes down very very sharply, a straight line down. That indicates the ESR of the capacitor. By measuring that fast drop in voltage and the current at the time, you can calculate the approximate ESR. Since resistance is voltage divided by current, the ESR is:ESR=v/iwhere 'v' here is the voltage only from that fast drop, and 'i' is the current through the 600 Ohm resistor.Since the cap on the left has a voltage for that fast drop of about 0.1 volts and the current is about 4/600=6.6ma, the ESR of that cap is about 15 Ohms. That would be far too high for a computer power supply filter cap.The cap on the right has a voltage for that fast drop of about 0.000600 volts, and since the current is still about 0.0066 amps, the ESR is about 0.000600/0.0066=0.09 Ohms approximately.You can see the difference. The bad cap has ESR about 15 Ohms and the good cap has about 0.1 Ohms, much different. Some filter caps will have an ESR even lower than that.Be aware that most function generators have some sort of internal resistance also, and that has to be considered with the external resistor used for the tests. If the internal resistance is 50 Ohms and the external resistance is 100 Ohms just for example, the total resistance (used to help calculate the current) is not 100 Ohms it is 150 Ohms simply because 50+100 equals 150. For my generator the total was 600 Ohms.To get the peak to peak (or peak) voltage, you have to measure the square wave when the function generator is unloaded (no external connections other than the scope).Once you get the scope and function generator you can come back and we can talk about this in more detail if that isnt enough, and im sure others here can help too.Note i used these techniques when i troubleshooted several devices such as a computer power supply, a TV, and other stuff, and because of these rather simple measurements i was able to fix all those devices because all they needed was new electrolytic capacitors. Once replaced, the devices worked like new again.

author: the_cake_is_a_lie
author_rank: Contributor
timestamp: October 06, 2023, 04:30:32 pm
content: Yeah that is a bad way to do it like MrAl was saying and you're not interested in measuring leakage current. Makes more sense with a function generator and oscilloscope but still not what I would do outside of a curiosity. But this out of curiosity since multimeters with 5% capacitance accuracy are cheap and plentiful? I know sometimes it's nice having a second method as a sanity check. The best method is with an ESR meter. I bought an Atlas ESR Gold meter that displays capacitance and ESR at the same time. General hobbyist doesn't need a pricy ESR meter but if you're repairing electronics, it becomes essential.Anyway, the other method I've done as a sanity check is a series RC circuit. With polarized electrolytic capacitors, they can handle at least 300mVrms of negative voltage at the positive terminal for an extended time without degrading them. I have a 12V AC power supply that is a simple transformer. I recommend a 1kohm resistor that should be measured. Don't assume is exactly 1k. 1/4 watt is fine here, not 1/8 watt. You can put the resistor and capacitor in either order but resistor first might feel safer as a beginner. So then voltage measured at the capacitor Vc is a voltage divider:Code:Vc = Vs * Zc / (Zc + R)where Vs the source voltage you measure before the resistor as more or less 12Vrms and R as more or less 1000 ohmssolve for Zc then plug in:Code:Zc = 1/(2*pi*f*C)where f = 60 Hz for me in the USsolve for C the capacitance that will display something like 0.00010972 for 100uFI use the freePrecise Calculatoryou can download but Windows Calculator can get the job done if you're careful. Some margin of error due to multimeter inaccuracy on Vc, Vs and R readings, resistance of the power supply, capacitor tolerance and the ESR that isn't being accounted for. But that's all well and good to verify the electrolytic capacitor is within its 20% tolerance and that the capacitance setting of your multimeter isn't too far off the mark.Makes more sense using a spreadsheet to automate this if you're measuring multiple capacitors. Non-polarized capacitors could get away with a smaller resistor but beware the i^2*R power draw. I haven't tried below 100uF and less point in 1% ceramic capacitors when the compounded multimeter error is greater than that.

author: Peabody
author_rank: Super Contributor
timestamp: October 06, 2023, 08:03:49 pm
content: I've used an analog meter to match a capacitor of unknown value to another with a known value. But that's done in resistance mode, not voltage mode. Setting the meter to typically 1K resistance, the larger the capacitor, the stronger the deflection toward zero ohms as the capacitor charges from the meter's battery. So you just find the capacitor of known value that behaves most like the mystery one. That method should also work to detect any significant leakage since over time the needle should come back to infinite resistance as the capacitor becomes fully charged.But for polarized capacitors, remember that in resistance mode analog meters typically have polarity reversed - positive voltage is actually on the black lead, and ground is on the red lead.