Measure capacitance without an LCR meter
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- Опубликовано: 7 дек 2022
- In this video I will show how to measure capacitance using a multi meter and a signal generator.
All the math required is packed behind a free tool I developed at
funstacklabs.github.io/capaci...
Depending on your multi meter and your test circuit this method can beat multi meter capacitance measure functions in precision. During my tests, I got results in 22 pF - 22uF range well within component tolerances. Using the ideas here you can build your own LCR meter or component tester and understand the boundaries of what you can with maths with simple voltage measurements in AC domain.
Complementary video on how to measure inductance similarly
• How to measure inducta...
Multimeter used Brymen BM786 True RMS DMM. Signal generator BitScope Micro Oscilloscope & Analyzer. 
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One final observation. I was thinking about the bandwidth limitation of the DMM and speculated that it might effect the calculation at higher frequencies. However looking at the formula for capacitance
C = Vr/(2*pi*f*sqrt(V^2 - Vr^2))
Where Vr is resistor voltage and V is circuit voltage. We can divide numerator and denominator Vr to get
C = 1/(2*pi*f*sqrt((V/Vr)^2 - 1))
We see that the capacitance depends upon voltage only as a ratio of V to Vr meaning that if the DMM has the frequency dependence of a low pass filter (which it does) then it will reduce both V and Vr in the same proportion and thus the V/Vr ratio and calculated capacitance will be unaffected by the DMM frequency defendant attenuation in amplitude. So it may very well be possible to use the DMM at higher frequencies for this application and not have to worry about it’s bandwidth. In my case the DMM I use has a useful bandwidth of about 2 k Hz corresponding to the -3dB point. Another thing I would suggest when dealing with small capacitance values is to measure the resistor voltages by at least 2 significant digits for better accuracy by using higher frequencies. Although you got good results with the 82 and 22 pF caps it makes me a little nervous to see the resistor voltage carried out to only one significant digit when being compared to the circuit voltage.
Excellent video! Thanks!
As a footnote to my last post here's the way to determine the output impedance of your AWG. First set the output to some arbitrary voltage (setting it in RMS value makes the math easier because your DMM measures only RMS). Then measure the open circuit voltage of the AWG. If your AWG has the option to choose Hi-z versus 50 ohm (or some other value) no problem. Just use the open circuit voltage as your reference. Then place a 1000 ohm potentiometer across the output terminals and set to 1000 ohms. While measuring the voltage across the pot decrease the resistance until the measured voltage is one half the open circuit voltage. Remove the pot and measure its resistance. That is the output impedance of your AWG. At this point one half the output voltage is being dropped across the pot and one half is being dropped across the output resistance of the AWG. In my case I have a 50 ohm output impedance so I only need to use a 100 ohm pot. The results I get are within 1 ohm of the nominal output impedance. One caveat is that, if you are using a handheld DMM to measure voltage, consider that they typically have limited bandwidth. Eg the measured voltage from my Brennen 235 falls off quite a bit beyond about 1700 Hz (sine wave) so you need to consider this when doing your measurement so as not to use too high of a frequency. An option if you need to use higher frequencies for lower value caps would be to use your scope as the output which doesn't have this limitation. As an aside it is typical for older AWGs to have output impedances of 600 ohm or so. The newer ones are typically 50 ohms. Based upon your measurement in example #1 in the video I estimated the output impedance of your AWG to be 460 ohm. It would be interesting to see it's measured result.
这表AC毫伏很好
There is something wrong about your measurement at 5:16. You are measuring the RMS value of the total voltage delivered by a sinusoidal source with V peak to peak of 1 V that should give you an RMS voltage of about 1/(2*sqrt(2)) = 354 mV which is what you obtained for this on subsequent measurements. Since current (I) is Vr/R and Xc = 1/(2*pi*f*C) we have for a basic formula Vc = V - Vr. So V - Vr = Xc*I
V - Vr = (Vr/R)/(2*pi*f*C) or rearranging:
C = Vr/(2*pi*f*R*(V - Vr))
Using 111.18 mV in the first capacitor (as measured) would give one a capacitance of 53 microF. Using a more realistic V of 354 mV would give 6.2 microF neither of which is correct.
Thanks for the comment baby boo. You are close but the tool works a bit different or perhaps you are missing a parenthesis somewhere. I will try to type the maths happening at 5:16 behind the tool as clear as I can.
I = V/R
I = 79.61 / 149.92 = 0.534 mA
R = V/I
|Zrc| = 111.18 mV / 0.534 mA = 208.202
|Zrc| = sqrt(R*R + 1/((2*pi*f*C) * (2*pi*f*C)))
43348.0728 = 22476.0064 + (1/(2pifC) * (2pifC) )
20872.0664 = 1 / ((2pifC) * (2pifC))
0.00004791 = (2pifC) * (2pifC)
0.006921 = 2pifC
C = 0.006921 / (2pif) = 0.000022030 F
@@funstacklabs I see where my mistake was. It was in assuming that the voltage across the capacitor was simply the difference between the voltage across the circuit and the voltage across the resistor which it is not. Instead the voltage across the capacitor is equal to the sqrt(V^2 - Vr^2). So the correct formula is
C =Vr/(2*pi*f*R*sqrt(V^2 - Vr^2)).
I still don't understand how you got an RMS voltage across the circuit of 111.18 mV in example #1 (5:16) when inputting a 1 V peak to peak sine wave which should have given you 1/(2*sqrt(2)) = 353 mV.
I think that the reason for the discrepancy in example number 1 is that you were not actually inputting 1 volt peak to peak. Your function generator was actually outputting 111.18*2*sqrt(2) = 313.5 volt peak to peak. That is the reason the calculation came out correctly. With that assumption the measured values make sense. One would then expect 111.18 volts RMS across the circuit (both resistor and capacitor).
Think of the cap as a voltage source in the opposite direction. The effect is there with the other measurements as well though smaller since I was measuring smaller caps. If your measurements are different let me know, there might be a difference in the output impedance of the signal generator.
@@funstacklabs Take a look at the circuit voltages (across both resistor and capacitor) you obtained in all other measurements: at 6:12 you got 342.21. At 7:16 you got 348.77. At 9:21 you got 349.46. At 10:42 you got 349.22. All of these are within the experimental error of the RMS value of a 1 volt peak to peak sine wave which is equal to 0.5/sqrt(2) = 355 mV. The only discrepancy is the first measurement of 111.18 at 5:16. What that tells me is that your actual peak to peak sine wave at 5:16 was 2*sqrt(2)*111.18 = 314.5 mV not 1 volt. That's still OK as you can measure these voltages at any signal level and get reasonably good results. Your DMM will still give you accurate results. All I'm implying is that if your signal generator is claiming 1 volt peak to peak (in the first measurement) it is not actually outputting that voltage but, instead is outputting only 314.5 mV.