GATE Electrical Engineering - Complete Master Library

 

GATE Electrical – Analog Electronics

Practice Problems – Page 5

This section includes numerical problems based on Transistor Biasing and Small Signal Amplifiers.

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Problem 41

Question:

A transistor has β = 120 and base current is 25 μA. Find collector current.

Solution:

Ic = β × Ib

Ic = 120 × 25 μA

Ic = 3000 μA = 3 mA

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Problem 42

Question:

If collector current Ic = 1.5 mA and VT = 25 mV, calculate transconductance (gm).

Solution:

gm = Ic / VT

gm = 1.5 mA / 25 mV

gm = 0.06 S

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Problem 43

Question:

Find small signal emitter resistance if Ic = 2 mA.

Formula:

re = VT / Ic

Solution:

re = 25 mV / 2 mA

re = 12.5 Ω

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Problem 44

Question:

If gm = 0.05 S and load resistance RL = 2 kΩ, find voltage gain of CE amplifier.

Solution:

Av = − gm × RL

Av = −0.05 × 2000

Av = −100

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Problem 45

Question:

If β = 100 and re = 20 Ω, find input resistance of common emitter amplifier.

Solution:

rin = β × re

rin = 100 × 20

rin = 2000 Ω

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Problem 46

Question:

Which biasing method provides best stability for transistor amplifier?

Answer:

Voltage Divider Bias

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Problem 47

Question:

Define Q-point of a transistor amplifier.

Answer:

Q-point is the quiescent operating point where no input signal is applied.

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Problem 48

Question:

Why is biasing necessary in transistor amplifiers?

Answer:

To maintain the transistor in the active region for proper amplification.

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Problem 49

Question:

What happens if Q-point shifts to saturation region?

Answer:

Output waveform becomes distorted due to clipping.

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Problem 50

Question:

What happens if Q-point shifts to cutoff region?

Answer:

Transistor stops conducting and output signal is lost.