In the case of small BJT model with the common emitter, the colle
| In the case of small BJT model with the common emitter, the collector current ic is 1.3 mA, when the collector-emitter voltage is Vce of 2.6 V. The output conductance of the circuit is
A. 2.0 mΩ
B. 2.0 m℧
C. 0.5 mΩ
D. 0.5 m℧
Please scroll down to see the correct answer and solution guide.
Right Answer is: D
SOLUTION
Concept:
The conductance of an amplifier is defined as the ratio of the collector to emitter voltage to the collector current of the BJT, i.e.
\(g_0=\frac{I_c}{V_{ce}}\)
IC = Collector current
Vce = Collector to emitter voltage
Calculation:
With Vce = 2.6 V and Ic = 1.3 mA, the conductance will be:
\(g_0=\frac{1.3m}{2.6}\)
g0 = 0.5 m℧, where ℧ is the inverse of resistance (Ω).