1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    Y-Parameters:
    [itex]I_1=Y_{11}V_1+Y_{12}V_2[/itex]
    [itex]I_2=Y_{21}V_1+Y_{22}V_2[/itex]

    [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

    gneill

    User Avatar

    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.

    So,
    [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?

    Thanks!
     
  5. Jun 22, 2013 #4

    gneill

    User Avatar

    Staff: Mentor

    Yup.
     
  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

    gneill

    User Avatar

    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?

    Thanks.
     
  9. Jun 22, 2013 #8

    rude man

    User Avatar
    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Calculate Y-parameters of a capacitor in parallel
Loading...