Issues with Designed Capacity Measurement Circuit

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.

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.

• Circuit diagram doesn't show GND. I'm guessing that the 1kHz source might have one terminal at GND (perhaps not). It is certain that one terminal of the oscilloscope is GND. Could GND connection at signal source conflict with oscilloscope GND? Commented Aug 31, 2023 at 18:38
• All the voltage drops on the capacitor and there is practically 0 voltage on the resistor. Draw a schematic diagram showing how exactly was your oscilloscope probe connected to the circuit for both of those measurements. Both the probe tip and GND clip must be shown. If the AC source is an earth-referenced function generator, then the scope's GND can only be connected with the GND of the generator's output - also the GND node of the circuit - and not to any other point. Commented Aug 31, 2023 at 18:53
• @glen_geek Yes, you are correct. I am sorry; I forgot the ground trace. I corrected the schematics. The ground is on the negative side of the voltage source. I connected the probes of the oscilloscope to the legs of the resistors and the capacitor, so I am just measuring the voltage drop on them. Could you please explain how such a conflict can occur?
Commented Aug 31, 2023 at 18:55
• @Kubahasn'tforgottenMonica The probes were connected exactly as in the schematics (probe tip to the upper part of the voltage arrow in the schematics, GND clip to the other side of the component). Both the oscilloscope and the voltage source (a function generator) are connected to the power outlet.
Commented Aug 31, 2023 at 19:20
• @Kubahasn'tforgottenMonica The GND of the oscilloscope is not connected to any part of the circuit, only the probes are connected. Do you think this is the problem? I read I should not play with the GND of the oscilloscope because it can damage the oscilloscope or the components (maybe that's wrong, though). But anyway, since I measure the voltage drop across the components, the GND of the oscilloscope doesn't really matter, does it?
Commented Aug 31, 2023 at 19:21

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.

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.

• Thank you! it works now, I edited the main post