GATE Electrical – Analog Electronics

Practice Problems – Page 10

This section includes practice problems based on Operational Amplifier fundamentals and ideal characteristics.

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Problem 91

Question:

Define an operational amplifier.

Answer:

An operational amplifier is a high-gain differential amplifier used to perform mathematical operations such as addition, subtraction, integration, and differentiation.

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Problem 92

Question:

What are the ideal characteristics of an op-amp?

Answer:

• Infinite open loop gain
• Infinite input impedance
• Zero output impedance
• Infinite bandwidth
• Infinite CMRR
• Infinite slew rate

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Problem 93

Question:

What is the typical open loop gain of an op-amp?

Answer:

≈ 100,000 (10⁵)

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Problem 94

Question:

Define input offset voltage.

Answer:

Input offset voltage is the small differential voltage required between input terminals to make output zero.

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Problem 95

Question:

What is slew rate?

Answer:

Slew rate is the maximum rate of change of output voltage.

Slew Rate = dV/dt

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Problem 96

Question:

What is the ideal input current of an op-amp?

Answer:

Zero

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Problem 97

Question:

Why is input impedance of op-amp very high?

Answer:

To ensure negligible current flows into the input terminals.

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Problem 98

Question:

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

Answer:

Zero

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Problem 99

Question:

What is the function of feedback in op-amp circuits?

Answer:

Feedback stabilizes gain and improves amplifier performance.

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Problem 100

Question:

Name two basic op-amp configurations.

Answer:

• Inverting amplifier
• Non-inverting amplifier

 

GATE Electrical – Analog Electronics

Practice Problems – Page 9

This section includes numerical problems based on Differential Amplifier Gain, Output Voltage and CMRR.

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Problem 81

Question:

A differential amplifier has differential gain Ad = 150. If V1 = 20 mV and V2 = 5 mV, find output voltage.

Solution:

Vd = V1 − V2

Vd = 20 mV − 5 mV

Vd = 15 mV

Vo = Ad × Vd

Vo = 150 × 15 mV = 2.25 V

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Problem 82

Question:

If common mode gain Ac = 0.2 and common mode voltage is 1 V, find output voltage.

Solution:

Vo = Ac × Vc

Vo = 0.2 × 1 = 0.2 V

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Problem 83

Question:

A differential amplifier has CMRR = 500 and differential gain Ad = 200. Find common mode gain.

Solution:

CMRR = Ad / Ac

Ac = Ad / CMRR

Ac = 200 / 500

Ac = 0.4

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Problem 84

Question:

If differential gain Ad = 300 and input difference voltage = 10 mV, find output voltage.

Solution:

Vo = Ad × Vd

Vo = 300 × 10 mV

Vo = 3 V

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Problem 85

Question:

If differential amplifier output is zero, what can be said about the inputs?

Answer:

Both input voltages are equal.

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Problem 86

Question:

Which type of signal is rejected by differential amplifier?

Answer:

Common Mode Signal

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Problem 87

Question:

What is the purpose of current mirror in differential amplifier?

Answer:

To provide constant bias current.

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Problem 88

Question:

Which amplifier stage in an operational amplifier provides high gain?

Answer:

Differential Amplifier Stage

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Problem 89

Question:

What is the ideal input impedance of differential amplifier?

Answer:

Very high.

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Problem 90

Question:

What is the ideal output impedance of differential amplifier?

Answer:

Very low.

 

GATE Electrical – Analog Electronics

Practice Problems – Page 8

This section contains problems based on Differential Amplifier and Common Mode Analysis.

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Problem 71

Question:

Define differential amplifier.

Answer:

A differential amplifier amplifies the difference between two input signals.

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Problem 72

Question:

Write the formula for differential input voltage.

Answer:

Vd = V1 − V2

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Problem 73

Question:

What is common mode input voltage?

Answer:

Vc = (V1 + V2) / 2

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Problem 74

Question:

Define CMRR.

Answer:

CMRR (Common Mode Rejection Ratio) is the ratio of differential gain to common mode gain.

CMRR = Ad / Ac

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Problem 75

Question:

If differential gain Ad = 200 and common mode gain Ac = 0.5, calculate CMRR.

Solution:

CMRR = Ad / Ac

CMRR = 200 / 0.5

CMRR = 400

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Problem 76

Question:

Express CMRR in decibels if CMRR = 400.

Solution:

CMRR(dB) = 20 log10 (CMRR)

CMRR(dB) = 20 log10 (400)

≈ 52 dB

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Problem 77

Question:

Why is high CMRR desirable in amplifiers?

Answer:

It rejects noise and interference present in common mode signals.

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Problem 78

Question:

Which circuit forms the basic stage of operational amplifier?

Answer:

Differential Amplifier

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Problem 79

Question:

What is the ideal value of common mode gain?

Answer:

Zero

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Problem 80

Question:

What is the ideal value of CMRR?

Answer:

Infinite

 

GATE Electrical – Analog Electronics

Practice Problems – Page 7

This section contains problems based on h-parameter model of transistor amplifiers.

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Problem 61

Question:

Write the four h-parameters of a transistor amplifier.

Answer:

• h_ie – Input impedance
• h_re – Reverse voltage gain
• h_fe – Forward current gain
• h_oe – Output admittance

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Problem 62

Question:

What does h_fe represent in transistor amplifier?

Answer:

Forward current gain in common emitter configuration.

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Problem 63

Question:

If h_fe = 120 and base current = 30 μA, find collector current.

Solution:

Ic = hfe × Ib

Ic = 120 × 30 μA

Ic = 3600 μA = 3.6 mA

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Problem 64

Question:

Which h-parameter represents output conductance?

Answer:

h_oe

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Problem 65

Question:

What is the approximate voltage gain of CE amplifier using h-parameters?

Formula:

Av ≈ − hfe × RL / hie

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Problem 66

Question:

If h_fe = 100, R_L = 2 kΩ and h_ie = 1 kΩ, find voltage gain.

Solution:

Av = − (100 × 2000) / 1000

Av = −200

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Problem 67

Question:

If h_ie increases, what happens to input impedance?

Answer:

Input impedance increases.

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Problem 68

Question:

Why is h_re usually neglected in amplifier analysis?

Answer:

Because its value is extremely small.

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Problem 69

Question:

Which parameter mainly controls current amplification?

Answer:

h_fe

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Problem 70

Question:

What is the main advantage of h-parameter model?

Answer:

It simplifies transistor amplifier analysis using linear equations.

 

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.