A npn transistor is connected in common emitter configuration in a given amplifier. A load resistance of 800 Ω is connected in the collector circuit and the voltage drop across it is 0.8 V. If the current amplification factor is 0.96 and the input resistance of the circuit is 192 Ω, the voltage gain and the power gain of the amplifier will respectively be:(a) 4, 3.84(b) 3.69, 3.84(c) 4, 4(d) 4, 3.69
Answers
Answer:
Voltage gain =4
power gain =3.84
Explanation:
Given, resistance across load, RL=800Ω
Voltage drop across load, VL=0.8V
Input resistance of circuit,Ri=192Ω
Collector current Ic= VL/RL= 0.8/800
=8/8000= 1m A
Current Amplification = output current / Input current
Ic/Ib=0.96
Ib= Ic/ 0.96= 1mA/0.96
Voltage gain Av= Vl/Vin=Vl/Ib Ri
=0.8 x(0.96)/10⁻³ x 192
=4
Av= 4
Powe gain
Ap=[ I²c RL/ I²B] Ri =( Ic/ Ib)² RL/Ri=(0.96)² x800/192
=(0.96)² x 800/192
Ap=3.84
Answer:
a) 4, 3.84
Explanation:
Resistance across load Rl = 800Ω (Given)
Voltage drop across load Vl - 0.8V (Given)
Input resistance of the circuit = 192Ω (Given)
Collector current -
Ic = Vl/Rl = 0.8/800 = 8/8000 = 1mA
Thus, current amplification - Output current/ Input current
= Ic/Ib = 0.96
= Ib = 1ma/0.96
Voltage gain = 0.8 × 0.96/ 10-³× 192
= 4
Av = 4
Power gain = (0.96)²×800/192
= 3.84