# How to find Thevenin voltage of two-port network from admittance matrix

The two-port network is powered by 10 mV voltage source and has admittance matrix: $Y = \left| \begin{matrix} y_{11} & y_{12} \\ y_{21} & y_{22} \end{matrix} \right| = \left| \begin{matrix} 2 \cdot 10^{-3} & -3 \cdot 10^{-5} \\ 0.5 & 2 \cdot 10^{-4} \end{matrix} \right|$

From admittance matrix I know, that

$\left[ \begin{matrix} i_1 \\ i_2 \end{matrix} \right] = \left[ \begin{matrix} 2 \cdot 10^{-3} & -3 \cdot 10^{-5} \\ 0.5 & 2 \cdot 10^{-4} \end{matrix} \right] \times \left[ \begin{matrix} u_1 \\ u_2 \end{matrix} \right]$

$i_1 = 2 \cdot 10^{-3} \cdot u_1 - 3\cdot 10^{-5}\cdot u_2 \\ i_2 = 0.5 \cdot u_1 + 2\cdot 10^{-4}\cdot u_2$

The task is to draw Thevenin's circuit for this and to compute its parameters. All I can imagine when I hear "Thevenin" is circuit like this:

I suppose, the parameters are $U$ and $R_i$. I would compute $R_i$ as $\frac{1}{y_{11}} = \frac{10^3}{2} = 500 \space \Omega$. Is that right? If not, how to compute that?

And I have no idea how to compute $U$. I know it is equal to voltage between $2$ and $2'$ nodes in this picture:

Any hint or explanation?

• Can you post the complete circuit? – Martin Petrei Feb 5 '14 at 15:11
• What do you mean? This is all I have. We have no other information. – user50222 Feb 5 '14 at 15:18
• sorry. I thought you had the diagram associated to the matrix. – Martin Petrei Feb 5 '14 at 15:32

To calculate the Thevenin voltage the port 2 has to be opened, thus the current is zero: $i_2=0$.
You get: $0.5⋅u_1+2⋅10^{-4}⋅u_2=0 \to u_2= -2500⋅u_1 \to u_2=10mV*2500=25V$.
To calculate the Thevenin resistance, do the same but with $u_2=0 \to i_2=0.5u_1 \to i_2=5mA$.
Then, since the voltage drop is all on the Thevenin resistance $$R_{th}=\frac{u_2}{i_2} \to \frac{25}{5m}=5k\Omega$$
Notice it is the $(y_{22})^{-1}$.
• Do you mean to put $u_2 = 0$ in the first equation with $i_1$ or in the second, as in your answer? Is $R_i = \frac{u_1}{i_1}$ or $\frac{u_1}{i_2}$? I could understand the first case. If the second is right, why? – user50222 Feb 5 '14 at 15:56