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Finding the Capacitance from Lissajour plot info?

Discussion in 'Electronic Basics' started by Wayne, Sep 26, 2004.

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  1. Wayne

    Wayne Guest

    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:

    ¦ ¦
    ¦--/\/\/\/\-----¦ ¦---¦
    R_shunt Cx
    ¦ ¦

    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?


  2. Tim Wescott

    Tim Wescott Guest

    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 -

    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):

    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.
  3. Wayne

    Wayne Guest

    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)



  4. Fred Bloggs

    Fred Bloggs Guest

    Adjust the signal generator frequency (lower) until Vr=0.707*Vs= which
    makes Vc=Vr and therefore Cx= 1/(2*pi*Freq*Rshunt).
  5. I read in that Wayne <>
    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!
  6. Wayne

    Wayne Guest

    What frequency are you working at and what are the resistor and capacitor
    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


    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!
  7. Tim Wescott

    Tim Wescott Guest

    - 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 :).
  8. Tim Wescott

    Tim Wescott Guest

    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.
  9. I read in that Wayne <>
    What do you mean by 'XL factors'? Inductance?

    Your cell may indeed have 3.07 ohms of loss resistance.
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