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I would like to correct the values that I am measuring with an LCR meter (specifically the LCR6002 from GW Instek).

I am a bit acquainted with electronics in a very basic level, but here, my knowledge in probing and measurement correction is really lacking.

The measurements that I am doing are in a relatively noisy field, and the capacitances I am interested in are quite low as well, therefore, correcting the measurements is kind of crucial for me to get the small deviations that I am getting when I am changing the environment of the measurement. I know my description is vague here, sorry for that, please ask any question in comments if I can clear anything up.

I am measuring in Cp-D mode, with 1VRMS and 1 kHz.

The values that I am getting when the sample is not connected is as following:

System cap = 19.64 pF System Dval = 0.0029

In Cp-D, mode, the circuit should look something like this:

System Idle

Now, when I connect a sample, I close the two relays shown in the following image, and I get the following reading:

cap1 = 33.89 pF Dval1 = 0.0422

Image with sample

What I know so far is that I need to model the system as a capacitor and a resistor in "idling" mode, i.e., when no sample is connected. Then, I need to imagine that another capacitor and resistor are added to the parallel circuit, equally in parallel, to then calculate the corrected value.

What do I need to do in the last step? Do I consider cap1 and Dval1 as the equivalent of all 4 components (R system, C system, R sample, C sample), or how should I interpret these results?

I would be interested in any insight, and I would really appreciate a technical explanation.

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    \$\begingroup\$ What do you mean by "equally in parallel"? What does the meter manual tell you about this subject? What do you know about placing components in parallel and calculating the net impedance? \$\endgroup\$ Commented Nov 17 at 14:10
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    \$\begingroup\$ I would expect 'correction' to start with measurements of your sample jig either empty or containing a known capacitor of similar size to your unknowns. \$\endgroup\$ Commented Nov 17 at 14:10
  • \$\begingroup\$ @Andyaka hey, thank you so much for your time. By equally in parallel, I mean that the LCR meter treats the new sample as a capacitor and resistor in parallel. From the manual page 12, I can see the same equivalent circuit of one capacitor in parallel with one resistor. About your last question, I do not really know how to do it. Hence my question \$\endgroup\$ Commented Nov 17 at 14:54
  • \$\begingroup\$ @Neil_UK Hey, thank you for your time! That is always possible to do after my measurements stop, for now, the measurement started and I am recording data till January, for project related reasons. \$\endgroup\$ Commented Nov 17 at 14:55

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I will post this answer in case someone with the same problem might find it. I hope that the community will correct me if I am wrong.

So given that the circuit looks like a parallel circuit with one capacitor and one resistor when the system is not connected to any samples. This can be seen like this:

System/ Idle

From this system, I calculated the admitance, by using the following equations:

# system/ idle admittance B1 = w * C1 # C1 is the measured capacitance G1 = D1 * B1 # D1 is the measured DVal

When the system is connected to a probe, we can consider that our measurements are actually the equivalent circuit of the system and the sample, as seen here:

Circuit when a sample is measured

So I then calculated the admittance here:

# measured admittance with sample and system B2 = w * C2 # C2 is the measured capacitance of sample + system G2 = D2 * B2 # D2 is the measured Dval of sample + system

So now we can recalculate the values to the resistance and the reactance of sample, which is after all what I am interested in:

# sample admittance = difference? TBC Gs = G2 - G1 Bs = B2 - B1 # complex sample admittance Ys = Gs + 1j * Bs # actually or - since it is C? # invert admittance --> 1/impedance Zs = 1 / Ys # real and imaginary parts Rs = np.real(Zs) Xs = np.imag(Zs)

I am still not sure if this is really the correct way to do this type of correction, however my data looks much better now, and the low-frequency "waves" that I am seeing over time in the data is no longer there.

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It seems that you did not perform the OPEN/SHORT/SPOT corrections?

If you did, LCR6002 will automatically perform the corrections on the measured samples .

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  • \$\begingroup\$ Wow, thank you so much for the hint. No I didn't... Is there still a way to correct the measured values? I am also measuring the system values together with the samples \$\endgroup\$ Commented Nov 21 at 6:52
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This is a low frequency meter. At low C you need only be concerned about series R and parallel R as parasitics. You also haven't specified measurement frequency, without that we can't convert D to parallel R (or do anything else).

You haven't explained what the "idle system" is. If it's just cables then the D can only come from cable resistance and can't be modeled as a parallel R.

Also you have told us that the DUT can be modeled as parallel RC. Maybe you're right, maybe you're not, without detail we don't know. So overall insufficient detail to reason about.

Finally- this is a 4 terminal meter, if you want to do seriously accurate measurements you need to preserve the 4 terminals right up to the device under test.

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    \$\begingroup\$ Thanks for the answer! That's true, I forgot the frequency. So the voltage level is at 1VRMS and the frequency is 1 kHz. The waveform is a nice sinesoid. The idling system is basically unconnected probes in my measurement environment. The sample is a printed sensor in a capacitance design. The 4 terminals are being used right now with the equipment that was delivered with the LCR meter, I can include info tomorrow \$\endgroup\$ Commented Nov 17 at 19:49
  • \$\begingroup\$ The 4 terminals are connected to this measurement lead, which is connected to the sample as discussed before \$\endgroup\$ Commented Nov 18 at 7:15

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