Hi all,

I have a doubt regarding the result of h12 from following example:



which is is taken from here.

The example shown is based on the hybrid parameters matrix (wikipedia link).
Now, I tried to determine the value for h12 starting from this circuit:



The circuit equation I write is:

\(
V_2\left( \frac{ \frac{1}{sC_2} }{\frac{1}{sC_2} + (\frac{1}{sC_1} \parallel R_1)} \right ) - \left( \frac{g_mV_1 R_2}{R_2 + \frac{1}{sC_2} + \left (\frac{1}{sC_1} \parallel R_1 \right )} \right )\left(\frac{1}{sC_1} \parallel R_1 \right ) = V_1
\)

which solved for

\(
h_{12} = \frac{V_1}{V_2}
\)

does not results the same as the h12 parameter obtained from the book linked.
So the question is: what's wrong in my reasoning ? I am a little bit confused



Thank you in advance
thumb2

 

The circuit equation I write is:

\(
V_2\left( \frac{ \frac{1}{sC_2} }{\frac{1}{sC_2} + (\frac{1}{sC_1} \parallel R_1)} \right ) - \left( \frac{g_mV_1 R_2}{R_2 + \frac{1}{sC_2} + \left (\frac{1}{sC_1} \parallel R_1 \right )} \right )\left(\frac{1}{sC_1} \parallel R_1 \right ) = V_1
\)

You must have made a mistake in this equation. In order to help you, show the details of how you got this.

 

Hi The Electrician,

I tried to apply superposition with VCCS in one step.
Given the equation:

\(\boxed{V_2\left( \frac{ \frac{1}{sC_2} }{\frac{1}{sC_2} + (\frac{1}{sC_1} \parallel R_1)} \right ) - \left( \frac{g_mV_1 R_2}{R_2 + \frac{1}{sC_2} + \left (\frac{1}{sC_1} \parallel R_1 \right )} \right )\left(\frac{1}{sC_1} \parallel R_1 \right ) = V_1}\)

the therm

\(
V_2 \left( \frac{ \frac{1}{sC_2} }{\frac{1}{sC_2} + (\frac{1}{sC_1} \parallel R_1)} \right )
\)

comes from applyng the voltage dividirer rule with the impedances given by C2 & R1 || C1:



while with the therm

\(
- \left( \frac{g_mV_1 R_2}{R_2 + \frac{1}{sC_2} + \left (\frac{1}{sC_1} \parallel R_1 \right )} \right )\left(\frac{1}{sC_1} \parallel R_1 \right )
\)

referring to the following circuit:


I would apply the current divider to determine the current \(i_1\) multplied by the impedance given by R1 || C1, which should give the value of V1.

Thank you.

 

R2 and the gm V1 source are shorted out by your V2 test source and they can be eliminated. Then what would you get for the voltage appearing at the input?

 

OK, seems I have to review my circuits concepts

Then what would you get for the voltage appearing at the input?

In this case V1 results from the voltage divider rule, and h12 results to be :

\(
\begin{align}
& R_1 || C_1 = \frac{\frac{R1}{sC_1}}{R_1 + \frac{1}{sC_1}} = \frac{R_1}{sR_1C_1 + 1}\\
& V_2 \left( \frac{ \frac{1}{sC_2} }{\frac{1}{sC_2} + (\frac{1}{sC_1} \parallel R_1)} \right ) = V_2\left(\frac{sR_1C_1 + 1}{sR_1C_1 + sR_1C_2 + 1} \right ) = V_1\\
& h_{12} = \frac{V_1}{V_2} = \left(\frac{sR_1C_1 + 1}{sR_1C_1 + sR_1C_2 + 1} \right )
\end{align}
\)

but again, I don't see the equivalence into the numerator.
To get the same denominator from the book it's obvious I can divide by \(R_1\).

 

OK, seems I have to review my circuits concepts


In this case V1 results from the voltage divider rule, and h12 results to be :

\(
\begin{align}
& R_1 || C_1 = \frac{\frac{R1}{sC_1}}{R_1 + \frac{1}{sC_1}} = \frac{R_1}{sR_1C_1 + 1}\\
& V_2 \left( \frac{ \frac{1}{sC_2} }{\frac{1}{sC_2} + (\frac{1}{sC_1} \parallel R_1)} \right ) = V_2\left(\frac{sR_1C_1 + 1}{sR_1C_1 + sR_1C_2 + 1} \right ) = V_1\\
& h_{12} = \frac{V_1}{V_2} = \left(\frac{sR_1C_1 + 1}{sR_1C_1 + sR_1C_2 + 1} \right )
\end{align}
\)

but again, I don't see the equivalence into the numerator.
To get the same denominator from the book it's obvious I can divide by \(R_1\).

You're not doing the voltage divider rule correctly. The numerator of that expression inside the big parentheses should be \(\frac{1}{sC_1} \parallel R_1\), not \( \frac{1}{sC_2}\)

 

Doh ! What a shame !
Someone call me a doctor.



Solved. Thank you The Electrician.

 

A quick way to get the h matrix of a two-port is to first write the Y matrix which is usually easy to do by inspection. This, of course, probably wouldn't be an acceptable short cut if your goal is to show the actual analysis of the circuit leading to each element of the h matrix.

 

In that book the admittance matrix is also explained, later than the h matrix.
But I need a good review of circuits analysis ... and the best way to do it is by solving exercises and learning from the errors, as always.

 

Doh ! What a shame !
Someone call me a doctor.

You're a doctor!

It's getting late. Signing off.