4
\$\begingroup\$

How do digital mulitimeters (DMM) measure capacitance through their typical 10M Ohm input/output impedance?

Providing a logic level of 3.3V, attempting to measure 1F would mean a time constant of 10M seconds (R x C) thus the voltage rise in the capacitor being in immeasurable (in the noise floor.) They also do it within a second or so at 3% accuracy. How on earth is this achieved?

\$\endgroup\$
4
  • \$\begingroup\$ I don't think they go as high as 1F \$\endgroup\$
    – Jasen
    May 31 '19 at 20:51
  • 1
    \$\begingroup\$ Only the voltmeter ranges are 10 MΩ. Ampere range input impedance is very low and diode testing will be a couple of mA - see if it lights an LED. \$\endgroup\$
    – Transistor
    May 31 '19 at 20:51
  • \$\begingroup\$ Even at 1mF you're still looking at a constant of 10ks. And it's not low impedance as they use the voltmeter terminals which are high impedance. Seems unlikely to me they would have a second path in parallel with the high impedance due to leakage and requring high voltage withstand components which are expensive. Something like a diode would not be accurate and cheap. \$\endgroup\$ May 31 '19 at 20:55
  • 2
    \$\begingroup\$ "And it's not low impedance as they use the voltmeter terminals which are high impedance." Nope. There isn't a 10 MΩ resistor between the VΩmA (and diode and continuity) input socket and the PCB. There is (effectively) a direct connection to the range select switch. It's 10 MΩ on voltmeter only. Did you try the diode test? (Use a red LED as that has the lowest Vf of the visible LEDs and it might be very faint.) \$\endgroup\$
    – Transistor
    May 31 '19 at 21:40
5
\$\begingroup\$

There are many ways to measure capacitance, If you have a waveform generator you can either use a square wave and measure the rise time. Or a sine wave and measure the current and voltage. If you know current and voltage, you know what your load is. If the load is a capacitor, you'd also need phase information. The links below go into more depth on how this is done. Instead of an waveform generator, the DMMs usually have a simpler circuit (usually only generating one or a few frequencies). Instead of an oscilloscope circuits that measure phase and amplitude to do the calculations.

The cool thing is, if you have an oscilloscope and waveform generator, you can also measure capacitance, sometimes better than a DMM. This also works for inductance to.

enter image description here
Source: https://meettechniek.info/passive/capacitance.html

enter image description here
Source: https://meettechniek.info/passive/capacitance.html

\$\endgroup\$
4
  • \$\begingroup\$ You also find alot of information on this using the keywords "impedance spectroscopy" \$\endgroup\$
    – Jens
    May 31 '19 at 21:23
  • \$\begingroup\$ I've measured inductors/transformers this way, it's kind of fun \$\endgroup\$
    – Voltage Spike
    May 31 '19 at 21:36
  • \$\begingroup\$ Me too, you learn a lot :) \$\endgroup\$
    – Jens
    May 31 '19 at 21:39
  • \$\begingroup\$ For hobbyists- if you can’t afford a nice function generator, you can use a 555 IC configured as astable in the top drawing. Not as nice, but will work for the basement Doc Browns. \$\endgroup\$
    – JoeBob_OH
    Jun 1 '19 at 14:24
2
\$\begingroup\$

Measuring a value of 1 Farad in 1 second with a DMM 1% resolution of 0.1mV and a measured value of 10 mV requires battery current when using pulse measurement techniques.

Even though RLC meters use a more precise constant current sinewave at selected frequencies to measure voltage amplitude and phase shift to compute all values, they still do not go up to 1 Farad.

Ic=CdV/dt= 1F * 10mV/1s = 10 mA which is more current than the DMM normally draws and would reduce battery life. So the Fluke 115 only measures up to 9999 uF.enter image description here

The Keysight portable meters measure only up to 199.99 mF enter image description here

However, if you follow Maxwell's ultracap test procedure, you won't need a $1k+ rack mount RLC meter that does offer 1F readings.

But it does take more than 1 second.

\$\endgroup\$
2
\$\begingroup\$

Edit: The high impedance is only for the voltage measurement setting. The impedance is much much lower when measuring capacitance.

According to Fluke:

A multimeter determines capacitance by charging a capacitor with a known current, measuring the resulting voltage, then calculating the capacitance.

They are not waiting for the RC time constant. It applies a known current for a certain time and looks at the ΔV. They may also do the same thing discharging the cap, in a loop.

Higher ΔV means lower capacitance.

\$\endgroup\$
3
  • \$\begingroup\$ If you are going to down-vote, please provide a comment pointing out where I went wrong with this answer. \$\endgroup\$ May 31 '19 at 21:48
  • 1
    \$\begingroup\$ Not wrong, so I would never down-vote an answer like this, but you didn't answer the 1st question. As is, it is "useful", so it definitely deserves an up-vote. Some people are down-voting answers that "aren't the best", this is contradictory to the guidelines. \$\endgroup\$
    – Mattman944
    May 31 '19 at 22:13
  • \$\begingroup\$ @Mattman944 Thank you for pointing that out, I have edited my answer. \$\endgroup\$ May 31 '19 at 22:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.