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NEET PHYSICSSEMICONDUCTOR ELECTRONICSMedium

Question

In a common emitter (CE) amplifier having a voltage gain G, the transistor used has transconductance 0.03 mho0.03 \text{ mho} and current gain 25. If the above transistor is replaced with another one with transconductance of 0.02 mho0.02 \text{ mho} and current gain of 20, the voltage gain will be:

A

1.5G1.5G

B

13G\frac{1}{3}G

C

54G\frac{5}{4}G

D

23G\frac{2}{3}G

Step-by-Step Solution

The voltage gain (AvA_v) of a common emitter (CE) amplifier is given by the relation: Av=gmRLA_v = g_m R_L where gmg_m is the transconductance of the transistor and RLR_L is the load resistance. When the transistor is replaced in the same amplifier circuit, the load resistance RLR_L remains constant. Therefore, the voltage gain is directly proportional to the transconductance (Av∝gmA_v \propto g_m).

Given: Initial transconductance, gm1=0.03 mhog_{m1} = 0.03 \text{ mho} Initial voltage gain, Av1=GA_{v1} = G New transconductance, gm2=0.02 mhog_{m2} = 0.02 \text{ mho}

Taking the ratio of the new voltage gain to the initial voltage gain: Av2Av1=gm2gm1\frac{A_{v2}}{A_{v1}} = \frac{g_{m2}}{g_{m1}} Av2G=0.020.03=23\frac{A_{v2}}{G} = \frac{0.02}{0.03} = \frac{2}{3} Av2=23GA_{v2} = \frac{2}{3}G

Note: The current gain (β\beta) values provided in the question are extraneous information and are not needed to find the voltage gain when transconductance is given.

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from SEMICONDUCTOR ELECTRONICS. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSSEMICONDUCTOR ELECTRONICScommonemitteramplifierhavingvoltage

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