# Capacitance Measurement With Oscilloscope simulate this circuit – Schematic created using CircuitLab

I'm trying to determine the capacitance of a sensor, but the oscilloscope's internal capacitance is loading the measurement too much.

What is an appropriate method to determine a low picofarad capacitance using an o'scope in this situation? (o'scope internal capacitance is 24pF).

• how are you determining the capacitance? (which "detour" are you taking?) – Marcus Müller Jul 18 '18 at 12:55
• Calibration: make sure to get an accurate known capacitor value. – Oldfart Jul 18 '18 at 12:55
• 1) Are you trying to use an oscilloscope to measure a capacitance? That cannot be done directly so explain how you're doing that, draw a schematic of your setup. 2) Are you using a 10:1 probe? If not perhaps you should, that will lower the input capacitance. 3) if a 10:1 probe does not have a low enough capacitance, consider an active probe. – Bimpelrekkie Jul 18 '18 at 12:56
• To be frank, depending on the level of precision and repeatability you want, the best way to use an oscilloscope to measure capacitance might be to turn it off and get an LCR meter or a VNA (depending on the frequency and values you are interested in) – Joren Vaes Jul 18 '18 at 13:06
• What frequency range ? Resistance? DC bias? SRF? is expected. You can use Step response, CC sine voltage sweep, RC Oscillator, LC resonance and a Wheatsone style bridge method with a known fix Cap ref. for differential response. – Tony Stewart Sunnyskyguy EE75 Jul 18 '18 at 13:53

Make a measurement of 'nothing', that is your scope's input capacitance. Then make a measurement of the 'sensor', that is your scope's input capacitance + the sensor's capacitance. Subtract.

• Cheers Neil, the knee of the curve oscillates a little, and the sensor capacitance appears to be quite small, so this is generally not an option. – Himmel Jul 18 '18 at 13:05

Hopefully you know, how a 10:1 passive probe can show the waveform right regardless the input capacitance of the oscilloscope and the probe wire. Make your own same doing compensation circuit. Connect the signal source to the input of the scope through a resistor and your sensor connected in parallel. Find a resistor which placed in parallel with your sensor shows square wave perfectly.

Let that resistor be =Rx. The unknown capacitance of the sensor is Cx. Solve the following balance-equation:

Rx * Cx = R4 * C2 where R4 and C2 are the input resistance and capacitance of the scope.

Be sure there's no long wires making unknown extra capacitance in parallel with Cx or C2. You can add an extra resistor in parallel with the scope input to make a smaller Rx be enough. Otherwise you must have Rx=several megaohms.

Here's what comes to my mind:

• based on your description, you seem to be trying to measure with a cable connected directly to the scope. If you use an actual scope probe, it will reduce the capacitance load on the circuit. A 10x passive probe will cancel out the effects of the scope capacitance by adding a series capacitance (take a look at http://www.ni.com/product-documentation/14825/en/ if you want to learn more about it), but you'll still be left with some capacitance, typically in the order of 15pF. If you use an active probe it can get down to just ~1pF;

• consider using some sort of buffer amplifier (with low parasitic input capacitance, of course) between the capacitor and the scope probe;

• if you have more than one identical scope probe, check whether there is any significant difference in the waveform when only one probe is connected versus when two probes are connected. If there is no significant difference, ignore the probe's input capacitance;

• calculate the total equivalent capacitance when the probe is connected based on the waveform characteristics then subtract the known probe capacitance (this may be tricky if you don't have confidence in the value of the probe capacitance, you could measure with 1 probe and 2 probes to have 2 data points);

• also keep in mind that the time constant for R=1kΩ and C=24pF (assuming the unknown capacitance is not much higher than that) is only 24ns, which means that your signal generator needs to provide rise and fall times shorter than that for you to have any hope to measure anything.

Good luck! :^)