Re: Finding the Capacitance from Lissajour plot info?
From: Tim Wescott (tim_at_wescottnospamdesign.com)
Date: 09/26/04
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Date: Sun, 26 Sep 2004 10:40:24 -0700
Wayne wrote:
> If I have the Phase then say 40deg and the magnitude say 10 then how can I
> convert that into capasitance.
> I have the following circuit:
>
> ___SIGGEN_____
> ¦ ¦
> ¦--/\/\/\/\-----¦ ¦---¦
> R_shunt Cx
> ¦ ¦
> ----V?----
>
> I am measuring the voltage drop accross R_shunt and comparing that with the
> siggen so I have V_R_shunt - SIGGEN = Z and on a scope I can measure
> the phase. In other words I have the equiverlant info as you would find
> when creating a Lissajour plot.
> How can I find the capacitance of Cx with the info I have here?
>
> Cheers
>
> Wayne
>
>
>
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.
-- Tim Wescott Wescott Design Services http://www.wescottdesign.com
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