Output Stages - 1 -

By the hybrid pi model we can get that the small signal gain of this circuit is $\frac{V_o}{V_i} = \frac{R1}{R1 + {1\over gm}} = 0.8$ by rearanging this we get $gm = 0.5 S$ . Since the thermal voltage is $U_T = 26mV$ , it follows that $I_C = gm \cdot U_T = 13mA$ , since $\beta >> 1$ , $I_C \approx I_E$ . Now consider the case where $V_o = 0V$ , all of the current in $I_E$ is going thrugh $I1$ , so $I_E = I1 = 13mA$ , since this current don't change, $I_E = 13mA$