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Basic Capacitance Measurement Question

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Newbie level 4
May 22, 2011
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I have a circuit containing a capacitor and probably also a resistor, but I am not sure of the configuration.

I have tried to measure the capacitance by passing a triangular waveform potential through the circuit. If it were just a capacitor I would expect to get a square waveform out. However I see the following: -

**broken link removed**
*The inputs in the graphs are potential vs time, the outputs are current vs time.

I think this may be due to a resistor in the circuit.

If it were a simple square waveform output I would simply calculate the capacitance to be C=i/(dV/dt), where the value of 'i' would be the amplitude of the square wave.

So I have the following questions: -

  1. which points on the graph should I be using to measure 'i'?
  2. what is the cause of the slope?
  3. If this is caused by a resistance, would it be in series or in parallel?
  4. what other information can I get from this graph? - i.e. does the gradient of the slope allow calculation of the resistance of the resistor?

Thank you for any help you can give me.

Sorry, may I ask how you are sure that your circuit should be ideal and give the expected output trace?

Hi KerimF

Thank you for replying.
I probably haven't explained this very well. The square waveform is what I would expect if the circuit consisted of just a capacitor, the fact that the output is not this makes me think there is a resistor there too.

I don't expect it to be ideal in the sense that it is a real system and there is noise present in the recordings. The attached image is just a sketch of what I see on screen.

That aside, are you able to help with the questions I have? - Is this the output you would expect to see from a resistor and capacitor in series? or in parallel? - if neither, then what would you expect in these situations?

Is there anyone else out there that has any ideas?

Thank you

I also use to face such situation at work once a while; expecting something and getting something else.
But this happens to me when I work while checking the signals on the scope on real circuits (I design).

I am not sure if I can guess the right answers without knowing anything about the circuit. But you are right to be confused, looking for good answers.

By the way, it seems to me you are using single rail opamp since only the top line has a slope.


Hi again.

Designing circuits must be a nightmare!

I'm not sure what I can tell you about the circuit. It should be just a capacitor and a resistor in parallel. I just have a black box with a connector on either side.

A couple of people that have looked at this graph have both said 'yes, of course! there's a resistance in there causing the slopes, that is exactly what you would expect to see' - I'm not sure why you would expect to see the slopes, this is what really puzzles me.

Thank you for coming back to this thread.
Last edited:

Ok... I got your situation... so I have to think now rather in theory to find out the good answer... I am working on it ;)

Ah yes,

I should have been clearer at the start. Thank you for helping. I will be back tomorrow.


There is some problem at the output you supose be expected.
A square wave current may not be taken at a capacitor.

**broken link removed**

The usual way to measure the capacitance is applying a constant current across device and measuring slope :

C = ( I . T ) / V

Last edited:

Let us suppose there is a shunt resistor Rp on the capacitor.
So the triangular voltage source will have to supply both Rp and C.

I_total = I_r + I_c

I_r = V/Rp
I_c = dV/dt * C

But dV/dt is constant, let us call it k

I_total = V/Rp + k*C

From this simple equation we see that when the slope of V changes its sign, the current will start with a sudden jump equal to +k*C (it was at -k*C) and the ramp V/Rp is added to it which lets the total current increase linearly.

But so far, this explains just the rising part of your trace above...
The falling edge shows as if there is no shunt resistor since it is followed by an horizontal trace as it is expected in the presence of an ideal capacitor (no leakage resistance).

A series resistance Rs affects the rate of change of k*C which follows an exponential function having a time constant Rs*C with k*C being its limit value. From the above trace, it seems Rs is negligible.

Last edited:
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Thank you Kerim It's so obvious now. About the lower half of the sketch, I should have drawn the lower flat lines also as upward slopes - I was very rushed when I drew it. I understand why it would be this way now.

Thank you for all of your help.

---------- Post added at 15:49 ---------- Previous post was at 15:47 ----------


There is some problem at the output you supose be expected.
A square wave current may not be taken at a capacitor.

**broken link removed**

The usual way to measure the capacitance is applying a constant current across device and measuring slope :


Thank you also. I wasn't expecting many replies.

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