Output Stages - 2 -

The minimum voltage $V_o$ can ever get is $V_o = -I_1R_1$ so the maximum peak voltage of $V_o$ is $V_p = I_1R_1$ . We know that the power disipated in a resistor is equal to $P = \frac{V_{rms}}{R} = \left(\frac{V_p}{\sqrt{2}}\right)^2*\frac{1}{R} = \frac{V_p^2}{2R}$

So the total power dicipated on this circuit is equal to

$P = \frac{I_1^2R_1}{2}$

The small signal gain of this topology can be extracted using the hybrid-pi model and it follows that:

$A_v = \frac{R1}{R1 + \frac{1}{gm}}$

by rearanging it we get that

$gm = \frac{A_v}{R_1(1-A_v)}$

from $gm = \frac{I_C}{U_T}$ we get

$I_C = \frac{U_TA_v}{R_1(1-A_v)} \Rightarrow P = {1 \over 2R_1} \cdot \left[\frac{U_TA_v}{(1-A_v)}\right]^2$

replacing the values with $A_v = 0.7$ , $R_1 = 4\Omega$ and $U_T = 26mV$ we get a power output of $0.46mW$ .