GATE Electrical – Analog Electronics

Practice Problems – Page 6

This section contains numerical and conceptual problems based on Small Signal BJT Model and Amplifier Analysis.

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Problem 51

Question:

A transistor amplifier has collector resistance Rc = 4 kΩ and transconductance gm = 0.04 S. Find the voltage gain.

Solution:

Av = − gm × Rc

Av = −0.04 × 4000

Av = −160

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Problem 52

Question:

If Ic = 4 mA, calculate small signal emitter resistance re.

Formula:

re = VT / Ic

Solution:

re = 25 mV / 4 mA

re = 6.25 Ω

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Problem 53

Question:

If β = 150 and re = 10 Ω, find input resistance.

Solution:

rin = β × re

rin = 150 × 10

rin = 1500 Ω

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Problem 54

Question:

A CE amplifier has Rc = 3 kΩ and re = 25 Ω. Find voltage gain.

Formula:

Av ≈ − Rc / re

Solution:

Av = −3000 / 25

Av = −120

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Problem 55

Question:

If collector current increases, what happens to transconductance?

Answer:

Transconductance increases because:

gm = Ic / VT

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Problem 56

Question:

Which amplifier configuration has the highest bandwidth?

Answer:

Common Base Amplifier

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Problem 57

Question:

Which amplifier configuration is used as a buffer?

Answer:

Common Collector (Emitter Follower)

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Problem 58

Question:

What is the typical value of thermal voltage at room temperature?

Answer:

VT ≈ 25 mV

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Problem 59

Question:

Why is bypass capacitor used in CE amplifier?

Answer:

To increase voltage gain by bypassing emitter resistor for AC signals.

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Problem 60

Question:

What is the main advantage of emitter follower amplifier?

Answer:

High input impedance and low output impedance.

 

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.

 

GATE Electrical – Analog Electronics

Practice Problems – Page 4

This section contains problems based on BJT Amplifier Configurations and Gain Calculations.

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Problem 31

Question:

A BJT has current gain β = 100. If base current is 20 μA, find collector current.

Solution:

Ic = β × Ib

Ic = 100 × 20 μA

Ic = 2000 μA = 2 mA

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Problem 32

Question:

If α = 0.98, find β.

Formula:

β = α / (1 − α)

Solution:

β = 0.98 / (1 − 0.98)

β = 49

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Problem 33

Question:

A common emitter amplifier has voltage gain of −50. What does the negative sign indicate?

Answer:

It indicates 180° phase shift between input and output.

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Problem 34

Question:

Which amplifier configuration provides highest current gain?

Answer:

Common Emitter Amplifier

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Problem 35

Question:

Which amplifier configuration has voltage gain approximately equal to 1?

Answer:

Common Collector (Emitter Follower)

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Problem 36

Question:

Which amplifier configuration has the lowest input impedance?

Answer:

Common Base Amplifier

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Problem 37

Question:

Which amplifier configuration has the highest output impedance?

Answer:

Common Emitter amplifier.

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Problem 38

Question:

Define transconductance (gm) of a transistor.

Answer:

gm = Ic / VT

Where VT ≈ 25 mV at room temperature.

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Problem 39

Question:

If collector current Ic = 2 mA, find transconductance.

Solution:

gm = Ic / VT

gm = 2 mA / 25 mV

gm = 0.08 S

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Problem 40

Question:

Why is emitter resistor used in CE amplifier?

Answer:

Emitter resistor provides thermal stability and stabilizes operating point.

 

GATE Electrical – Analog Electronics

Practice Problems – Page 3

This section contains numerical problems based on Rectifiers and Diode Circuits.

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Problem 21

Question:

A half wave rectifier has an input peak voltage of 20 V. Find the average DC output voltage.

Solution:

Vdc = Vm / π

Vdc = 20 / 3.14

Vdc ≈ 6.37 V

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Problem 22

Question:

A full wave rectifier has an input peak voltage of 30 V. Find the average output voltage.

Solution:

Vdc = 2Vm / π

Vdc = (2 × 30) / 3.14

Vdc ≈ 19.1 V

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Problem 23

Question:

Find the peak inverse voltage (PIV) of a half wave rectifier if the peak voltage is 50 V.

Answer:

PIV = Vm = 50 V

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Problem 24

Question:

Find the peak inverse voltage of a center-tapped full wave rectifier if peak voltage is 40 V.

Answer:

PIV = 2Vm

PIV = 2 × 40 = 80 V

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Problem 25

Question:

What is the peak inverse voltage of a bridge rectifier if input peak voltage is 60 V?

Answer:

PIV = Vm = 60 V

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Problem 26

Question:

If ripple factor of a half wave rectifier is 1.21, what does it indicate?

Explanation:

It indicates that the AC component in the output is greater than the DC component.

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Problem 27

Question:

A diode has a forward resistance of 10 Ω and load resistance is 1 kΩ. Find rectification efficiency approximately.

Answer:

Efficiency decreases slightly due to diode resistance but remains close to theoretical value.

≈ 40%

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Problem 28

Question:

What is the transformer utilization factor (TUF) of a bridge rectifier?

Answer:

TUF ≈ 0.812

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Problem 29

Question:

Which rectifier provides the highest DC output voltage?

Answer:

Full wave rectifier.

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Problem 30

Question:

Why is a filter used after a rectifier?

Answer:

To reduce ripple and produce smoother DC output.

 

GATE Electrical – Analog Electronics

Practice Problems – Page 2

This section contains practice problems based on PN Junction Diode, Rectifiers and Amplifier Fundamentals.

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Problem 11

Question:

What is the reverse saturation current of a diode mainly caused by?

Answer: Minority carriers.

Explanation:

In reverse bias condition, current is caused by minority carriers moving across the junction.

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Problem 12

Question:

Which type of diode is commonly used in voltage regulation circuits?

Answer: Zener diode

Explanation:

Zener diodes operate in reverse breakdown region and maintain constant voltage across the load.

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Problem 13

Question:

What is the ripple frequency of a full wave rectifier if the input AC frequency is 50 Hz?

Solution:

Ripple Frequency = 2f

= 2 × 50

= 100 Hz

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Problem 14

Question:

What is the maximum efficiency of a half wave rectifier?

Answer:

40.6%

Explanation:

Half wave rectifiers conduct only during one half of the AC cycle.

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Problem 15

Question:

What is the maximum efficiency of a full wave rectifier?

Answer:

81.2%

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Problem 16

Question:

What is the ripple factor of a full wave rectifier?

Answer:

0.482

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Problem 17

Question:

What is the ripple factor of a half wave rectifier?

Answer:

1.21

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Problem 18

Question:

Which amplifier configuration has the highest input impedance?

Answer: Common Collector (Emitter Follower)

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Problem 19

Question:

Which amplifier configuration has the highest voltage gain?

Answer: Common Emitter amplifier

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Problem 20

Question:

Write the formula for thermal voltage.

Answer:

VT = kT / q

At room temperature:

VT ≈ 25 mV

 

GATE Electrical – Analog Electronics

Practice Problems – Page 1

This section contains conceptual and numerical problems based on PN Junction Diode and Basic Amplifier Concepts. Each question includes a clear explanation.

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Problem 1

Question:

What is the typical cut-in voltage of a silicon diode?

Options:

• A) 0.2 V
• B) 0.3 V
• C) 0.7 V
• D) 1.5 V

Answer: C) 0.7 V

Explanation:

For silicon diodes, the cut-in voltage is approximately 0.7 V, while for germanium diodes it is about 0.3 V.

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Problem 2

Question:

Which region of a diode is responsible for current conduction?

Answer: Depletion region reduces and carriers move across the junction.

Explanation:

When forward bias is applied, the depletion region narrows, allowing charge carriers to cross the junction.

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Problem 3

Question:

What is the thermal voltage at room temperature?

Answer:

VT ≈ 25 mV

Explanation:

Thermal voltage is given by:

VT = kT/q

At room temperature it is approximately 25 mV.

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Problem 4

Question:

A CE amplifier has:

• hfe = 100
• hie = 1 kΩ
• RL = 4 kΩ

Find voltage gain.

Solution:

Av = − (hfe × RL) / hie

Av = − (100 × 4000) / 1000

Av = − 400

Negative sign indicates 180° phase shift.

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Problem 5

Question:

What is the main function of a rectifier?

Answer: Convert AC into DC.

Explanation:

Rectifiers allow current flow in only one direction, thereby converting alternating current into direct current.

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Problem 6

Question:

Which amplifier configuration provides the highest voltage gain?

Answer: Common Emitter (CE)

Explanation:

The CE amplifier provides high voltage gain and moderate input/output impedance, making it the most widely used amplifier configuration.

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Problem 7

Question:

What does CMRR stand for?

Answer: Common Mode Rejection Ratio

Explanation:

CMRR measures the ability of a differential amplifier to reject common mode signals.

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Problem 8

Question:

Write the formula for differential amplifier gain.

Answer:

Ad = RC / (2re)

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Problem 9

Question:

What is the ideal input impedance of an operational amplifier?

Answer: Infinite

Explanation:

An ideal op-amp draws no input current.

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Problem 10

Question:

What is the ideal output impedance of an op-amp?

Answer: Zero

Explanation:

An ideal op-amp can deliver output voltage without internal voltage drop.

 

GATE Electrical Engineering – 1000 Problems Master Plan

Yes, it is absolutely possible to create 1000 problems with solutions and explanations from the topics covered so far in Analog Electronics. This will help students preparing for GATE, PSU and Engineering Exams.

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Topics Covered So Far

• PN Junction Diode
• Diode Characteristics
• Rectifiers
• CE Amplifier
• Frequency Response
• Bode Plot
• Miller Effect
• Differential Amplifier
• CMRR
• Current Mirror
• Widlar Current Source
• Wilson Current Mirror
• Operational Amplifier (Op-Amp)

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Possible Question Distribution

Topic	Number of Questions
PN Junction Diode	80
Rectifiers	120
Amplifiers (CE, CB, CC)	120
Frequency Response	80
Bode Plot	60
Miller Effect	60
Differential Amplifier	120
Current Mirrors	120
Operational Amplifier	160
Mixed Concept Problems	80

Total ≈ 1000 Problems

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Types of Problems Included

• Conceptual MCQ Questions
• Numerical Problems
• Derivation Based Questions
• Circuit Analysis Problems
• GATE Previous Year Questions

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Example Numerical Problem

Question:

A CE amplifier has the following parameters:

• h_fe = 100
• h_ie = 1kΩ
• R_L = 4kΩ

Find the voltage gain.

Solution:

Av = − (hfe × RL) / hie

Av = − (100 × 4000) / 1000

Av = − 400

The negative sign indicates 180° phase shift.

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Recommended Blog Structure

• Page 30 – 50 MCQ Questions (Diode)
• Page 31 – 50 Numerical Problems (Rectifiers)
• Page 32 – CE Amplifier Problems
• Page 33 – Differential Amplifier Problems
• Page 34 – Current Mirror Problems
• Page 35 – Operational Amplifier Problems

If each page contains 50 problems, then

20 Pages × 50 Problems = 1000 Problems

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Final Goal

This structured question bank will create a complete GATE Electrical Digital Library that includes:

• Theory
• Derivations
• Worked Examples
• 1000 Practice Problems
• Mock Tests

This approach will help students build strong conceptual understanding and solve complex exam problems.