Finding the Capacitance from Lissajour plot info?

It's not clear from your schematic or your text -- you have a signal generator that is connected to a resistor and capacitor in series, and the resistor is grounded? Or is the capacitor grounded?

The following answers assume the resistor is grounded.

Method 1:

Get the RMS voltages of both the signal and the resistor voltage. Ignore phase, which means that you assume the capacitor is purely reactive. Then

V_c^2 + V_r^2 = V_s^2,

where V_c is the capacitor voltage, V_r is the resistor voltage and V_s is the signal generator voltage. This implies that V_c = sqrt(V_s^2 - V_r^2).

Now, the current is I = V_r/R, and the capacitive reactance is X_c = V_c/I, so you get X_c = R*sqrt(V_s^2 - V_r^2)/V_r. The capacitance can be found from the capacitive reactance and frequency, C = 1/(2*pi*X_c).

Method 2:

Assume that the capacitor isn't purely reactive. Measure the amplitude and phase of V_r respective to V_s, and solve for the capacitive impedance Z_c (note that you have to use complex arithmetic):

V_r Z_c = -----------. V_s - V_r

Z_c will, in general, be complex, so it will be Z_c = R_c - jX_c, where R_c is the capacitor's equivalent series resistance and X_c is the capacitive reactance. If Z_c is purely imaginary all is well and good. If Z_c has a significant resistive component then it is up to you to decide if this is measurement error or the fault of the capacitor, and whether it is best modeled as a series resistance, a parallel resistance, or something more complicated.

Thanks Tim.

The resistor is in series with the capacitor and the signal gen is across one lead of the capacitor and one lead of the resistor. In other word, if they were both resistors it would be a voltage divider. The resistor or the capacitor can be ground. Forgive me for asking a dumb question but why are V_c^2 + V_r^2 = V_s^2 to the power of 2? If this was a voltage divider then you would have V?=Vs(R/R+Rc)?

In method 2 could you give me an example with the following figures?

Vs=10v p-p Vr=4v p-p Phase = 45deg (with respect Vs)

Cheers

Wayne

Adjust the signal generator frequency (lower) until Vr=0.707*Vs= which makes Vc=Vr and therefore Cx= 1/(2*pi*Freq*Rshunt).

I read in sci.electronics.design that Wayne wrote (in ) about 'Finding the Capacitance from Lissajour plot info?', on Sun, 26 Sep 2004:

Because the two voltages have a phase difference of 90 degrees (or, in practice, very nearly; the capacitor won't be perfectly loss-free).

It's helpful to use a 'phasor diagram' to see what is going on. The current I is in phase with the voltage across the resistor, and the voltage across the capacitor is at 90 degrees to the current. So our diagram is (*use Courier font*):

/| / | / | / | Vs = 10 / | / | Vc =? / | / | / | / 45 | /__________| ----------> Current I Vr = 4 V

Now, I hope you can see that this is not, in theory, possible. The angle at the apex of the triangle must be 45 degrees as well and the triangle must be isosceles. So Vc must be 4 V as well, and then by Pythagoras (which is the same as the squared values in Tim's equation), Vs = sqrt(4^2 + 4^2) = sqrt(32) = 5.66 V.

One practical explanation, if your figures are correct, is that the capacitor is very lossy, thus having a lot of resistance of its own. This add to the horizontal side of the triangle (without altering the 4 V across the resistor). Another 3.07 V brings the figure right. The Vc must also be 7.02 V, of course, because of the 45 degrees. We then get by Pythagoras, Vs = sqrt(7.07^2 + 7.07^2) = sqrt(100) = 10.

It's a very poor capacitor if this is true. There are other possibilities. What frequency are you working at and what are the resistor and capacitor values? Have you actually measured 10 V at the signal generator, or is that just what the output control says? If the signal generator source impedance isn't very much smaller than the impedance of your RC circuit, the Vs value will be way off, indeed it may well be 5.66 V!

I am working from 1Hz to 1Mz. This is a psudo capacitor, in that I mean that I am measuring the capacitance of a cell - such as a batery and trying to model it in the same way as a capacitor. So there will be no XL factors.

Have you actually measured 10 V at the signal generator, or is that just what the output control says? The voltage can be anything from 0.01V to 10V. I have measured it with a scope.

Wayne

If the signal generator source impedance isn't very much smaller than the impedance of your RC circuit, the Vs value will be way off, indeed it may well be 5.66 V!

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- snip -

The voltages are squared because you know they're 90 degrees out of phase, which means that you need to use the pythagorean theorem to add them, with the generator voltage being the "hypotenuse". See any good text on basic electronics that includes "phasors" (the ARRL handbook used to go into this, I sure hope they still do -- or get a 30 year old ARRL handbook).

I was solving dividing by the cap voltage instead of multiplying -- here's the solution with the right formula:

Vs = 10V + j0V Vr = 2.83V + j2.83V (j = sqrt(-1), 2.83 = 4*sin(45 deg)).

Vc 7.17V - j2.83V Zc = ----*R = ----------------*R = (0.76 - j1.76)*R. Vr 2.83V + j2.83V

This indicates a capacitor with quite a large resistive component, or a failure in your phase measurement. If you use method 1 you get a capacitive reactance of 9.17*R, which would give you a phase shift of around 66 degrees.

You may want to double check your measurement or calculations, particularly if you're like me and you occasionally divide when you should be multiplying :).

Ah. In that case, the source impedance of the thing could be quite lossy, and your measurements may not be that far off.

You'll want to characterize it at a number of different frequencies, though -- batteries don't look like simple RC combinations (more like a voltage source in series with an RC which is in series or parallel with another RC, etc.). Add to that that for most batteries the source voltage changes a bit and the apparent resistances change a _lot_ with charge state, and you're in for an interesting modeling problem.

I read in sci.electronics.design that Wayne wrote (in ) about 'Finding the Capacitance from Lissajour plot info?', on Sun, 26 Sep 2004:

What do you mean by 'XL factors'? Inductance?

Your cell may indeed have 3.07 ohms of loss resistance.