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Homework Help: Calculate Y-parameters of a capacitor in parallel

  1. Jun 22, 2013 #1
    1. The problem statement, all variables and given/known data
    Calculate the Y parameters of a capacitor in parallel (see the attached figure).

    2. Relevant equations


    [itex]Y_{11} = \frac{I_1}{V_1}[/itex] when [itex]V_2=0[/itex]
    [itex]Y_{12} = \frac{I_1}{V_2}[/itex] when [itex]V_1=0[/itex]
    [itex]Y_{21} = \frac{I_2}{V_1}[/itex] when [itex]V_2=0[/itex]
    [itex]Y_{22} = \frac{I_2}{V_2}[/itex] when [itex]V_1=0[/itex]

    3. The attempt at a solution

    I've tried to use formulas written above, but when [itex]V_2=0[/itex], then [itex]V_1=0[/itex], so [itex]Y_{11}\to\infty[/itex]?

    The same happens when I try to find Y-parameters of a resistor in parallel.

    Thank you.

    Attached Files:

  2. jcsd
  3. Jun 22, 2013 #2


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    Staff: Mentor

    Since there's only the one essential node in the circuit, V1 = V2 always. So finding the Y-Parameters by shorting a port and expecting the other port to have some voltage other than zero is, well, problematical!

    You may have more luck with the Z-Parameters... then consider how the Y-matrix is related to Z-matrix. What conclusions can you draw?
  4. Jun 22, 2013 #3
    [itex][/itex][itex][/itex]OK. Thanks.

    [itex]V_1 = Z_{11}I_1 + Z_{12}I_2[/itex]
    [itex]V_2 = Z_{21}I_1 + Z_{22}I_2[/itex]

    I have no problem finding [itex]Z_{11}[/itex]:
    [itex]Z_{11} = \left.\frac{V_1}{I_1}\right|_{I_2=0} = \frac{1}{j\omega C}[/itex]

    However, when finding [itex]Z_{12}[/itex] I'm not sure if I draw the circuit properly in order to calculate the parameters. I mean [itex]I_1=0[/itex] and [itex]V_1=V_2[/itex], so [itex]Z_{12} = \left.\frac{V_1}{I_2}\right|_{I_1=0} = \left.\frac{V_2}{I_2}\right|_{I_1=0}= \frac{1}{j\omega C}[/itex].

    If that's true, all Z-parameters are equal to [itex]Z_{11} = \frac{1}{j\omega C}[/itex].

    Is this correct?

  5. Jun 22, 2013 #4


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    Staff: Mentor

  6. Jun 22, 2013 #5
    In that case I have the same problem as before.

    I know that [itex]Y_{11}=\frac{Z_{22}}{\Delta_Z}[/itex], where [itex]\Delta_Z = Z_{11}Z_{22}-Z_{12}Z_{21}[/itex].

    Because [itex]Z_{11} = Z_{12} = Z_{21} = Z_{22} = \frac{1}{j\omega C}[/itex], [itex]\Delta_Z = 0[/itex], so [itex]Y_{11}\to\infty[/itex].

    Where's the mistake?

    Thank you.
  7. Jun 22, 2013 #6


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    Staff: Mentor

    No mistake. Since the determinant of the impedance matrix is zero, there's no inverse, hence no finite Y-parameters.

    If you consider the circuit, the two ports are directly wired together. Hence in reality it's a single port device.
  8. Jun 22, 2013 #7
    Thank you.

    I really wanted to find the Y-parameter matrix of a Schottky diode working as mixer, whose model is this: fig309_01.jpg

    Should I conclude that it isn't posible to get it by adding capacitor Cj and resistor Rj Y-matrices, going to Z-parameters, adding the result with Rs Z-matrix, and finally going back to Y-parameters?

  9. Jun 22, 2013 #8

    rude man

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    Homework Helper
    Gold Member

    The Schottky is obviously part of a larger circuit - if you show us that then possibly we can advise you as to exploiting the conveniences of 2-terminal matrices, if any.
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