Analog Electronics – Page 34

Low Frequency Analysis of Amplifiers

At low frequencies, the gain of an amplifier decreases due to the effect of capacitors present in the circuit.

The capacitors responsible for low frequency behavior are:

                                          

• Input Coupling Capacitor
• Emitter Bypass Capacitor
• Output Coupling Capacitor

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1️⃣ Input Coupling Capacitor

The input capacitor blocks DC and allows AC signals to pass to the amplifier.

At low frequencies:

• Capacitive reactance increases
• Signal attenuation occurs
• Amplifier gain decreases

XC = 1 / (2πfC)

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2️⃣ Emitter Bypass Capacitor

The emitter bypass capacitor provides AC ground for the emitter resistor.

At low frequencies:

• Capacitor reactance becomes large
• Emitter degeneration increases
• Voltage gain reduces

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3️⃣ Output Coupling Capacitor

This capacitor transfers AC output to the load while blocking DC.

At low frequencies:

• Capacitive reactance increases
• Output signal reduces

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Lower Cutoff Frequency

The lower cutoff frequency occurs when gain drops to 0.707 of midband gain.

fL = 1 / (2πRC)

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Example Problem

If coupling capacitor C = 10 µF and resistance R = 1 kΩ.

Find lower cutoff frequency.

fL = 1 / (2πRC)

fL = 1 / (2π × 1000 × 10×10⁻⁶)

fL ≈ 15.9 Hz

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Important GATE Points

• Low frequency roll-off caused by capacitors
• Three capacitors control low-frequency response
• Coupling capacitors block DC components
• Bypass capacitor increases voltage gain

 

Analog Electronics – Page 33

Frequency Response of Amplifiers

The frequency response of an amplifier describes how the gain of the amplifier varies with signal frequency.

It shows the relationship between amplifier gain and input signal frequency.

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Three Frequency Regions

• Low Frequency Region
• Mid Frequency Region
• High Frequency Region

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Low Frequency Region

• Gain decreases at low frequencies
• Due to coupling capacitors and bypass capacitors
• Capacitive reactance becomes large

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Mid Frequency Region

• Gain remains constant
• Amplifier operates normally
• This region is called midband gain

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High Frequency Region

• Gain decreases at high frequencies
• Due to internal transistor capacitances
• Parasitic capacitances affect performance

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Cutoff Frequencies

The frequencies at which the gain drops to 70.7% of maximum gain are called cutoff frequencies.

• Lower cutoff frequency = fL
• Upper cutoff frequency = fH

Gain at cutoff = 0.707 × Maximum Gain

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Bandwidth

The bandwidth of an amplifier is the difference between upper and lower cutoff frequencies.

Bandwidth = fH − fL

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Example Problem

If an amplifier has:

• Lower cutoff frequency = 100 Hz
• Upper cutoff frequency = 100 kHz

Find bandwidth.

BW = fH − fL

BW = 100000 − 100

BW = 99.9 kHz

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Important GATE Points

• Midband gain is constant
• Cutoff occurs at 70.7% gain
• Bandwidth determines amplifier performance
• Higher bandwidth means better amplifier

 

Analog Electronics – Page 32

Multistage Amplifiers

A multistage amplifier consists of two or more amplifier stages connected in cascade to increase overall gain.

Each stage amplifies the signal and passes it to the next stage.

                                               

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Why Multistage Amplifiers are Used

• To obtain very high voltage gain
• To improve power amplification
• To drive heavy loads
• To increase signal strength

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Types of Coupling

• RC Coupling
• Transformer Coupling
• Direct Coupling

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Overall Voltage Gain

The total gain of a multistage amplifier is the product of gains of individual stages.

Av = Av1 × Av2 × Av3 × ... × Avn

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Gain in Decibels

Gain is often expressed in decibels.

Gain(dB) = 20 log10(Av)

For multistage amplifier:

Total Gain(dB) = Gain1 + Gain2 + Gain3

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Example Problem

A two-stage amplifier has gains:

• First stage gain = 50
• Second stage gain = 20

Find total voltage gain.

Av = Av1 × Av2

Av = 50 × 20

Av = 1000

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Gain in dB

Gain(dB) = 20 log10(1000)

Gain = 60 dB

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Key Points for GATE

• Overall gain is multiplication of stage gains
• In decibel scale gains are added
• RC coupling is most common in voltage amplifiers
• Transformer coupling used for power amplification

 

Analog Electronics – Page 31

Comparison of BJT Amplifier Configurations

There are three basic transistor amplifier configurations:

• Common Emitter (CE)
• Common Base (CB)
• Common Collector (CC)

Each configuration has different gain and impedance characteristics.

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1️⃣ Common Emitter Amplifier

• High voltage gain
• High current gain
• Moderate input impedance
• Moderate output impedance
• Phase shift = 180°

Voltage gain: Av ≈ − (hfe RL) / hie

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2️⃣ Common Base Amplifier

• High voltage gain
• Current gain ≈ 1
• Low input impedance
• High output impedance
• No phase shift

Current gain: α ≈ Ic / Ie

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3️⃣ Common Collector Amplifier (Emitter Follower)

• Voltage gain ≈ 1
• High current gain
• Very high input impedance
• Low output impedance
• No phase shift

Voltage gain ≈ 1

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Important Comparison Table

Parameter	CE	CB	CC
Voltage Gain	High	High	≈1
Current Gain	High (β)	≈1 (α)	High
Input Impedance	Medium	Low	High
Output Impedance	Medium	High	Low
Phase Shift	180°	0°	0°

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GATE Important Points

• CE is most widely used amplifier
• CB used for high-frequency applications
• CC used as buffer amplifier

 

Analog Electronics – Page 30

Common Emitter Amplifier Gain using h-Parameters

The hybrid h-parameter model is widely used for analyzing transistor amplifiers. It simplifies the transistor into a linear two-port network.

                                      

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CE h-Parameter Model

The equations for h-parameter model are:

V1 = h11 I1 + h12 V2

I2 = h21 I1 + h22 V2

For Common Emitter configuration:

• h11 = hie (Input resistance)
• h21 = hfe (Current gain)
• h12 = hre (Reverse voltage gain)
• h22 = hoe (Output admittance)

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Voltage Gain Derivation

Output voltage:

Vo = − Ic RL

Collector current:

Ic = hfe Ib

Input voltage:

Vi = Ib hie

Therefore voltage gain:

Av = Vo / Vi

Av = − (hfe RL) / hie

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Current Gain

Current gain of CE amplifier:

Ai = Ic / Ib = hfe

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Power Gain

Power gain is:

Ap = Av × Ai

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Example Problem

Given:

• hfe = 120
• hie = 1.2 kΩ
• RL = 3 kΩ

Find Voltage Gain

Av = − (hfe RL) / hie

Av = − (120 × 3000) / 1200

Av = −300

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Key Observations

• Voltage gain increases with higher hfe
• Voltage gain increases with larger load resistance
• Higher hie reduces voltage gain

 

Analog Electronics – Problems Page 29

h-Parameter Analysis & Gain Calculations

This section focuses on transistor amplifier analysis using hybrid h-parameters such as hie, hfe, hre and hoe. These parameters are widely used in small-signal analysis of BJT amplifiers.

                                       

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Basic h-Parameter Equations

V1 = h11 I1 + h12 V2

I2 = h21 I1 + h22 V2

For common emitter configuration:

h11 = hie (input resistance)

h21 = hfe (current gain)

h12 = hre (reverse voltage gain)

h22 = hoe (output admittance)

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Voltage Gain Formula

The approximate voltage gain of a CE amplifier using h-parameters is:

Av ≈ − (hfe RL) / hie

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Problem 1

A transistor amplifier has hfe = 100, hie = 1 kΩ and load resistance RL = 4 kΩ. Calculate voltage gain.

Solution

Av = − (hfe × RL) / hie

Av = − (100 × 4000) / 1000

Av = −400

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Problem 2

If hie = 2 kΩ and input voltage Vin = 10 mV, find base current.

Solution

Ib = Vin / hie

Ib = 10mV / 2kΩ

Ib = 5 μA

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Problem 3

If hfe = 120 and base current Ib = 20 μA, find collector current.

Solution

Ic = hfe × Ib

Ic = 120 × 20 μA

Ic = 2.4 mA

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Problem 4

A CE amplifier has hie = 1.5 kΩ, hfe = 80 and RL = 3 kΩ. Calculate voltage gain.

Solution

Av = − (hfe × RL) / hie

Av = − (80 × 3000) / 1500

Av = −160

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Problem 5

If reverse voltage gain hre is very small, what is the effect?

Solution

• Reverse feedback becomes negligible
• Amplifier analysis becomes simpler
• Voltage gain mainly depends on hfe and hie

 

Analog Electronics – Problems Page 28

Rapid Fire MCQs for GATE Revision

This section contains quick multiple choice questions from analog electronics useful for rapid revision before examinations.

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MCQ 1

The thermal voltage at room temperature is approximately:

A) 10 mV B) 25 mV C) 50 mV D) 100 mV

Answer: B (25 mV)

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MCQ 2

Which diode is used for voltage regulation?

A) Tunnel diode B) Zener diode C) LED D) Photodiode

Answer: B (Zener diode)

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MCQ 3

In a BJT transistor:

A) IC = β IB B) IC = α IB C) IC = IE D) IC = IB

Answer: A

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MCQ 4

The input impedance of an ideal op-amp is:

A) Zero B) Infinite C) Very small D) 1 kΩ

Answer: B (Infinite)

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MCQ 5

The output impedance of an ideal op-amp is:

A) Infinite B) Zero C) 1 kΩ D) Very large

Answer: B (Zero)

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MCQ 6

The Barkhausen criterion for oscillation is:

A) Aβ = 0 B) Aβ = 1 C) Aβ = 10 D) Aβ = 100

Answer: B

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MCQ 7

The oscillator commonly used for audio frequency generation is:

A) Hartley oscillator B) Colpitts oscillator C) RC phase shift oscillator D) Crystal oscillator

Answer: C

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MCQ 8

The cutoff frequency of an RC filter is:

A) 1/RC B) 1/(2πRC) C) RC D) 2πRC

Answer: B

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MCQ 9

Which oscillator uses capacitive divider?

A) Hartley oscillator B) Colpitts oscillator C) Wien bridge oscillator D) RC oscillator

Answer: B

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MCQ 10

The ripple factor of full wave rectifier is:

A) 0.121 B) 0.482 C) 1.21 D) 2

Answer: B