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.

 

GATE Electrical – Analog Electronics

Page 23 : Wilson Current Mirror

The Wilson Current Mirror is an improved version of the basic current mirror that provides better current accuracy and higher output resistance.

It uses three transistors instead of two and reduces the error caused by base currents.

                                                 

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Why Wilson Current Mirror?

In a simple current mirror, the output current is slightly smaller than the reference current because of base current losses.

The Wilson current mirror compensates for this error.

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Circuit Structure

• Three matched transistors (Q1, Q2, Q3)
• Reference resistor R
• Feedback connection through Q3

The third transistor provides feedback that stabilizes the output current.

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Working Principle

The circuit works by feeding back the collector current through the third transistor. This feedback increases the effective output resistance and improves current accuracy.

As a result:

Iout ≈ Iref

with much smaller error compared to the simple current mirror.

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Advantages

• Higher output resistance
• Improved current accuracy
• Reduced base current error
• Better current stability

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Applications

• Analog integrated circuits
• Operational amplifiers
• Precision current sources
• Bias circuits in IC design

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Important GATE Points

• Wilson current mirror uses three transistors.
• Provides higher output resistance.
• Improves current mirror accuracy.
• Widely used in analog IC design.

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Next Page → Operational Amplifier (Op-Amp) Introduction

 

GATE Electrical – Analog Electronics

Page 22 : Widlar Current Source

                                                   

The Widlar Current Source is a modification of the basic current mirror that allows the generation of very small output currents without using large resistor values.

It is widely used in integrated circuits and operational amplifiers.

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Basic Idea

In a simple current mirror, the output current is nearly equal to the reference current. However, sometimes we require a much smaller current.

The Widlar current source solves this problem by adding an emitter resistor to the output transistor.

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Circuit Structure

• Two matched transistors (Q1 and Q2)
• Reference resistor R
• Emitter resistor RE in output transistor

The resistor in the emitter of Q2 reduces the output current.

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Working Principle

Since the bases of both transistors are connected together:

VBE1 = VBE2 + IE2 RE

Because of this voltage drop across RE, the collector current of Q2 becomes smaller than the reference current.

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Widlar Current Equation

The relationship between reference current and output current is:

Iref / Iout = e^(Iout RE / VT)

Where:

• VT = thermal voltage ≈ 25 mV

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Advantages

• Generates very small currents
• Suitable for IC design
• Requires smaller resistors

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Applications

• Operational amplifier bias circuits
• Integrated circuit current sources
• Analog IC design

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Important GATE Points

• Widlar source produces smaller current than reference current.
• Emitter resistor reduces output current.
• Widely used in integrated circuits.

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Next Page → Wilson Current Mirror

 

Common Emitter Amplifier Gain using h-Parameters

The h-parameter model is widely used to analyze transistor amplifiers. For a Common Emitter (CE) amplifier, four parameters are used:

                                        

• hie → input resistance
• hre → reverse voltage gain
• hfe → forward current gain
• hoe → output admittance

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Step 1 : Input Voltage Equation

From the h-parameter model:

V1 = hie I1 + hre V2

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Step 2 : Output Current Equation

Output current is given by:

I2 = hfe I1 + hoe V2

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Step 3 : Output Voltage

The output voltage across load resistor RL:

V2 = − I2 RL

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Step 4 : Substitute I2

Substituting the current equation:

V2 = − RL ( hfe I1 + hoe V2 )

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Step 5 : Solve for Voltage Gain

After simplification:

Av = V2 / V1

For practical CE amplifier where hre and hoe are very small:

Av ≈ − hfe RL / hie

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Important Result

Voltage Gain (CE amplifier)

Av ≈ − (hfe × RL) / hie

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Important GATE Points

• CE amplifier gain is negative (phase inversion).
• Higher hfe increases gain.
• Larger load resistance increases gain.
• Small hre and hoe usually neglected.