Compute the voltage gain of this bipolar junction transistor circuit

[Peter Alexander]

Homework Statement

Hi!

This is one of the tasks that appeared in our exam preparation script. There is a sample solution path given, but I am unable to understand how to get there.

Idea is that given a bipolar junction transistor (BJT), you need to compute voltage gain $A_u$ from the following parameter values: $g_{11} = 2mS, g_{12} \approx 0S, g_{21} = 25mS, g_{22} = 0.1mS$ and the load has resistence $R_b = 2500\Omega$.

Relevant Equations

To solve this task, you only need to remember that voltage gain is computed as $$A_u = \frac{u_2}{u_1}$$ where $u_2$ is the output and $u_1$ is the input voltage.

This is the given schematic. The result should be $$A_u = \frac{R_b g_{21}}{1 + R_b g_{22}} = -50$$ and the way I wanted to tackle this problem is by stating $$A_u = \frac{-R_b i_2}{u_1} = \frac{-R_b (g_{21} u_1 + g_{22} u_2)}{u_1} = \frac{-R_b (g_{21} u_1 + g_{22} u_2)}{g_{11} u_1 + g_{12} u_2} = \frac{-R_b g_{21} + g_{22} u_2}{g_{11}}$$ but that leaves me with $u_2$. I am kind of stuck here.

I do not expect anyone to solve this task for me, I am just looking for guidance on how to continue, what have I missed... Any sort of help will be greatly appreciated!

 

[LvW]

At first, the denominator is a voltage (u1)...and in the third expression it is suddenly a current (g*u).
This is impossible.

 

[NascentOxygen]

Inside your first set of brackets, that + should be –.
Keep the denominator as u₁ all through, until it finally cancels with the same in the numerator.