Issues with Designed Capacity Measurement Circuit

Question

I made a circuit to determine the capacity of an unknown capacitor. To test if the circuit is working I wanted to test it with a known capacity first. The circuit and the derivation for the calculation of the capacity are:

I measured the voltages across the resistor and the capacitor through an oscilloscope. For the example, I used: $$v_{cc} = 2.5\text{V} \text{sin}(2\pi 1\text{kHz} \cdot t) \quad ; \quad C = 1\text{nF} \quad ; \quad R = 100\text{k}\Omega$$ According to the equation in the image I should see on the oscilloscope $$\frac{\hat{u}_R}{\hat{u}_C} = 0.62 $$ The measurements deviate strongly from that value, however. All the voltage drops on the capacitor and there is practically 0 voltage on the resistor. I have got deviations for many resistor values and many frequencies.

In the derivation I assumed an ideal capacitor, so as far as I understand I should stay on low frequencies to prevent the parasitic resistance of the capacitor to have an influence. This is why I chose 1kHz and 100kOhm as parameters.

I would appreciate your help.

Solution to the Problem

I didn't realize that the circuit I had was different from the one I drew here. I wired by a mistake R and C in the wrong positions, so in the actual circuit C was above R with V_C>V_R. I connected the probes just like in the schematics, assuming they were differential. However, the GND clip of the probe must be connected to the GND of the circuit. By wiring it the way I did, I pulled the voltage above the resistor to 0 and effectively removed it from the circuit. By following the recommendation from @JRE, the circuit worked. After the measurements I calculated C=0.86nF, which is in line with the 20% tolerance of the capacitor.

Thank you all.

Answer 1

Your oscilloscope input impedance will affect this measurement. At those frequency and measuring those capacitance, you should mostly be concern by the resistance of the probe. Most oscilloscope probe are around 1Mohm of input impedance but it varies on the probe and the multiplicator. Redo your calculation while considering the oscilloscope probe in parallel with your circuit and you should be much closer.

Also, tolerance on capacitors are pretty bad, so even at low frequency, you might have a hard time to measure it to it's nominal value.

Finally, if the two previous aren't the cause of issue, you might want to check your setup for parasitic.

Answer 2

1. Clip the ground of the oscilloscope to the ground of the circuit. This is required.
2. Measure the voltage at the signal generator output (where the generator connects to the resistor.)
3. Measure the voltage at the junction of the resistor and the capacitor.
4. The voltage from (3) is the voltage across the capacitor.
5. The difference of (2) and (3) is the voltage across the resistor.

This method of measuring capacitance will work, but you have to make the measurements properly.

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The method you are using is the basis of a simple capacitance/inductance meter I cobbled together a while back.

You can measure inductors and resistors with the same circuit. Keep in mind that the non-ideal aspects of the components will mess with the measurements.

This is a measurement I made of a 40nF capacitor:

An ideal capacitor would show the same capacitance at all frequencies. As you can see, at low frequencies this part is very far from ideal.

This is a plot of an inductor made the same way:

Again, it is very far off at low frequencies and closer to the expected value at higher frequencies.

Check your connections as described above, and consider switching to maybe 10kHz for your driving signal. It may help to use a 10X probe when making your measurements. A 10X probe has an impedance of 10 Mohm rather than the 1 Mohm of a regular probe. The 10X will put less of a load on the circuit, disturbing your measurements less.