# Common emitter ampl. with emitter resistor degen.

1. Apr 11, 2006

### Corneo

I would like to find the transconductance of such circuit. I am choosing to ignore r_0 for the time being and going with the definition

$$G_m = \frac {i_0}{v_i} \bigg |_{v_i = 0}$$,

What I did first was notice that $$i_0= - g_m v_\pi[/itex] [tex]v_i$$ appears as the top node voltage at $r_\pi$ w.r.t to ground. Then I have a voltage divider and using the impedance reflection rule, I have the relationship $$v_\pi = \frac {r_\pi}{r_\pi + (\beta + 1)R_{EA}}v_i$$, then $$G_m = -\frac {g_m v_\pi r_\pi}{r_\pi + (\beta + 1) R_{EA} } = - \frac {g_m}{1+( \frac{\beta + 1}{r_\pi})R_{EA}}$$
Then I use $$r_\pi = \frac {\beta}{g_m}$$ and $$\beta >> 1$$, so $$G_m = -\frac {g_m}{1 + g_mR_{EA}}$$. Is this correct?

Last edited: Apr 11, 2006