Semiconducting Devices Question 46
Question: A transistor is used as an amplifier in CB mode with a load resistance of 5 k W the current gain of amplifier is 0.98 and the input resistance is 70 W, the voltage gain and power gain respectively are
[Pb. PET 2003]
Options:
A) 70, 68.6
B) 80, 75.6
C) 60, 66.6
D) 90, 96.6
Show Answer
Answer:
Correct Answer: A
Solution:
In common base mode a = 0.98, R = 5 kW, Rin = 70W
voltage gain $ A _{v}=\alpha \times \frac{R}{R _{in}}=0.98\times \frac{5\times 10^{3}}{70}=70 $
Power gain = Current gain - Voltage gain = 0.98 - 70 = 68.6
