# Calculate Y-parameters of a capacitor in parallel

1. Jun 22, 2013

### Bromio

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:
$I_1=Y_{11}V_1+Y_{12}V_2$
$I_2=Y_{21}V_1+Y_{22}V_2$

$Y_{11} = \frac{I_1}{V_1}$ when $V_2=0$
$Y_{12} = \frac{I_1}{V_2}$ when $V_1=0$
$Y_{21} = \frac{I_2}{V_1}$ when $V_2=0$
$Y_{22} = \frac{I_2}{V_2}$ when $V_1=0$

3. The attempt at a solution

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

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

Thank you.

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2. Jun 22, 2013

### 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?

3. Jun 22, 2013

### Bromio

OK. Thanks.

So,
$V_1 = Z_{11}I_1 + Z_{12}I_2$
$V_2 = Z_{21}I_1 + Z_{22}I_2$

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

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

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

Is this correct?

Thanks!

4. Jun 22, 2013

### Staff: Mentor

Yup.

5. Jun 22, 2013

### Bromio

In that case I have the same problem as before.

I know that $Y_{11}=\frac{Z_{22}}{\Delta_Z}$, where $\Delta_Z = Z_{11}Z_{22}-Z_{12}Z_{21}$.

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

Where's the mistake?

Thank you.

6. Jun 22, 2013

### 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.

7. Jun 22, 2013

### Bromio

Thank you.

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

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.

8. Jun 22, 2013

### rude man

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.