Re: Finding the Capacitance from Lissajour plot info?
From: Fred Bloggs (nospam_at_nospam.com)
Date: 09/26/04
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Date: Sun, 26 Sep 2004 19:55:01 GMT
Wayne wrote:
> 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
>
>
> Coul
> "Tim Wescott" <tim@wescottnospamdesign.com> wrote in message
> news:10ldvo2errd1829@corp.supernews.com...
>
>>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
>
>
>
Adjust the signal generator frequency (lower) until Vr=0.707*Vs= which
makes Vc=Vr and therefore Cx= 1/(2*pi*Freq*Rshunt).
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