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?
Best Answer
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.
Source: https://meettechniek.info/passive/capacitance.html
Source: https://meettechniek.info/passive/capacitance.html