A load resistor of 2 kΩ is connected in the collector branch of an amplifier circuit using a transistor in common-emitter mode. The current gain β = 50. The input resistance of the transistor is 0.50 kΩ. If the input current is changed by 50 μA,

(a) by what amount does the output voltage change,


(b) by what amount does the input voltage change and


(c) what is the power gain?


The current gain β=50

Input current in a common emitter mode is the base current. Change in base current or input current = δIb=50μA


Output voltage, V0 =β×RG=50×20.5=200V


(a) Voltage Gain, VG=V0/V1


Input voltage, V1= δlb+Ri


Input resistance, Ri = 0.5kΩ = 500Ω


VG = 200/(50×10−6+500)


VG=8000V


(b) Change in input voltage, δV1 = δlb×Ri


= 50×10−6×5×102


=25×10−3 V


=25 mV


(c) Resistance Gain, RG= Load Resistance/Input resistance


= RL/Ri


Power gain =β2×RG


=β2×RL/Ri


=2500×2/0.5=2500×20.5
=2500×205=104


1