I want to find the amplification, $$U_{out}/U_{in}$$ of the pictured circuit. My electronics knowledge is very basic, but I tried to solve this using the ground node method, with the bottom of the circuit being grounded. I noticed that
$$V_y=U_{out}$$
$$V_1=U_{in}-V_x$$
Then I write equations for the node voltages (I switched resistances to conductances for clarity):
$$V_x(G_E +g_{CE}+g_{be})-V_y (g_{CE}) = gV_1 = gU_{in}-gV_x$$
$$V_y (g_{CE}+G_C)-V_x (g_{CE})=-gV_1=gV_x-gU_{in}$$
Adding the two equations I got:
$$V_x (G_E+g_{be})=-V_y (G_C) \implies V_x=-\frac{G_C}{G_E +g_{be}}V_y$$
Now from the first equation, after substituting:
$$V_y \frac{G_C (G_E+g_{CE}+g_{be})}{g_{be}+G_E} =-gU_{in}$$
$$\frac{V_y}{U_{in}}=\frac{U_{out}}{U_{in}}=-g\frac{ g_{be}+G_E }{G_C (G_E+g_{CE}+g_{be})}$$
Is this solution correct?
I also want to find the input resistance, but here I'm completely out of ideas.
simulate this circuit – Schematic created using CircuitLab
Best Answer
This is (for the time being) more a comment than an answer.
The first KCL equation should be:
$$(V_x - U_{in})g_{be} + (V_x - V_y)g_{ce} + V_xG_e = (U_{in} - V_x)g$$
which simplifies to:
$$V_x(g_{be} + g_{ce} + G_e + g) -V_yg_{ce} = U_{in}(g_{be} + g) $$
For input resistance, find the current through \$r_{be}\$ which will be in terms of \$U_{in}\$.
The input resistance is then easily found to be:
$$r_{in} = \dfrac{U_{in}}{i_{r_{be}}}$$