Analog Electronics – Problems Page 22

Oscillator Conceptual & GATE Level Problems

This section includes conceptual questions and solved problems on oscillator circuits useful for GATE and PSU examinations.

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Problem 1

State the Barkhausen criterion for sustained oscillations.

Solution

The Barkhausen criterion states that for sustained oscillations:

|Aβ| = 1

and the total phase shift around the loop must be:

360° (or 0°)

This ensures that the feedback signal reinforces the input signal.

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Problem 2

Why are LC oscillators used for high frequency applications?

Solution

• LC circuits have very low losses
• They provide high Q factor
• They can operate efficiently at radio frequencies

Hence LC oscillators are used in RF transmitters and communication systems.

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Problem 3

Calculate the oscillation frequency of an LC oscillator if L = 2 mH and C = 50 pF.

Solution

Formula:

f = 1 / (2π√LC)

Substitute values:

f = 1 / (2π √(2×10⁻³ × 50×10⁻¹²))

Result:

f ≈ 503 kHz

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Problem 4

What is the advantage of crystal oscillators compared to RC oscillators?

Solution

Crystal oscillators provide:

• Very high frequency stability
• Very high Q factor
• Low frequency drift

Hence they are used in digital clocks, microprocessors and communication systems.

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Problem 5

Which oscillator is commonly used in audio frequency applications?

Solution

RC Phase Shift Oscillator

Because RC networks are suitable for low frequency ranges.

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Problem 6

What happens if the loop gain Aβ is greater than 1?

Solution

• Oscillations grow in amplitude
• Signal distortion occurs
• The amplifier eventually saturates

Therefore practical circuits stabilize the gain.

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Problem 7

What is the main function of the feedback network in an oscillator?

Solution

The feedback network:

• Provides the required phase shift
• Controls the oscillation frequency
• Maintains the feedback ratio β

 

Analog Electronics – Problems Page 21

Oscillators Numerical Problems (GATE Level)

This page contains solved problems from oscillator circuits including RC phase shift, Wien bridge, Hartley and Colpitts oscillators.

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Problem 1

An RC phase shift oscillator uses R = 10 kΩ and C = 0.01 μF. Find the frequency of oscillation.

Solution

Formula:

f = 1 / (2πRC√6)

Given:

R = 10 × 10³ Ω
C = 0.01 × 10⁻⁶ F

Substitute values:

f = 1 / (2π × 10⁴ × 0.01×10⁻⁶ × √6)

Result:

f ≈ 650 Hz

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Problem 2

In a Wien bridge oscillator R = 5 kΩ and C = 0.02 μF. Calculate the oscillation frequency.

Solution

Formula:

f = 1 / (2πRC)

Substitute values:

f = 1 / (2π × 5000 × 0.02×10⁻⁶)

Result:

f ≈ 1590 Hz

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Problem 3

A Hartley oscillator has L1 = 2 mH, L2 = 8 mH and C = 100 pF. Find the frequency.

Solution

Total inductance:

L = L1 + L2 = 10 mH

Formula:

f = 1 / (2π√LC)

Substitute values:

f = 1 / (2π √(10×10⁻³ × 100×10⁻¹²))

Result:

f ≈ 159 kHz

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Problem 4

A Colpitts oscillator uses C1 = 200 pF, C2 = 300 pF and L = 1 mH. Calculate the frequency.

Solution

Equivalent capacitance:

Ceq = (C1C2)/(C1 + C2)

Substitute values:

Ceq = (200×300)/(500)

Ceq = 120 pF

Frequency:

f = 1/(2π√LC)

Result:

f ≈ 145 kHz

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Problem 5

For a Wien bridge oscillator, what minimum gain is required for oscillations?

Solution

From Barkhausen criterion:

Aβ = 1

For Wien bridge network:

β = 1/3

Therefore

A = 3

Minimum amplifier gain required is:

A = 3

 

Oscillators – Page 3

Crystal Oscillator & Advanced Oscillators

Crystal oscillators are used when very high frequency stability is required.

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1. Crystal Oscillator

A crystal oscillator uses the piezoelectric effect of quartz crystal.

When voltage is applied:

• Crystal vibrates mechanically
• Mechanical vibration generates electrical signal

Thus the crystal acts like a very high Q resonant circuit.

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Equivalent Circuit of Crystal

A quartz crystal is electrically equivalent to:

• Series inductance (L)
• Series capacitance (C)
• Series resistance (R)
• Parallel capacitance Cp

Thus the equivalent circuit contains both series resonance and parallel resonance.

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Series Resonant Frequency

At series resonance:

XL = XC

Thus

ωL = 1 / (ωC)

Therefore

fₛ = 1 / (2π √LC)

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Parallel Resonant Frequency

Due to parallel capacitance Cp:

fₚ ≈ fₛ √(1 + C/Cp)

Parallel resonance frequency is slightly higher than series frequency.

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2. Clapp Oscillator

Clapp oscillator is a modified version of the Colpitts oscillator.

An additional capacitor is added in series with the inductor.

Advantages:

• Better frequency stability
• Reduced transistor parameter effects

Frequency of oscillation:

f = 1 / (2π √(L Ceq))

Where

1/Ceq = 1/C1 + 1/C2 + 1/C3

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3. Oscillator Applications

• Radio transmitters
• Signal generators
• Clock circuits in digital systems
• Microprocessor timing circuits
• Communication systems

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4. Quick Oscillator Formula Revision

Oscillator	Frequency
RC Phase Shift	f = 1 / (2πRC√6)
Wien Bridge	f = 1 / (2πRC)
Hartley	f = 1 / (2π√LC)
Colpitts	f = 1 / (2π√LC)
Crystal	f = 1 / (2π√LC)

 

Oscillators – Page 2

LC Oscillators (Hartley & Colpitts)

LC oscillators generate sinusoidal signals using an inductor (L) and capacitor (C) resonant tank circuit.

Energy continuously transfers between:

• Magnetic field of inductor
• Electric field of capacitor

The oscillation frequency depends on the resonance of the LC circuit.

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1. Resonant Frequency of LC Tank Circuit

At resonance:

XL = XC

Inductive reactance:

XL = ωL

Capacitive reactance:

XC = 1 / (ωC)

At resonance:

ωL = 1 / (ωC)

Multiply both sides:

ω² = 1 / LC

Angular frequency:

ω = 1 / √LC

Frequency:

f = 1 / (2π√LC)

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2. Hartley Oscillator Derivation

Hartley oscillator uses:

• Two inductors (L1, L2)
• One capacitor (C)

Total inductance:

L = L1 + L2 + 2M

Where M is mutual inductance.

Frequency of oscillation:

f = 1 / (2π √(LC))

Substitute L:

f = 1 / (2π √((L1 + L2 + 2M) C))

Condition for oscillation:

A ≥ L2 / L1

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3. Colpitts Oscillator Derivation

Colpitts oscillator uses:

• One inductor (L)
• Two capacitors (C1, C2)

Equivalent capacitance:

1/C = 1/C1 + 1/C2

Therefore

C = (C1 C2) / (C1 + C2)

Frequency of oscillation:

f = 1 / (2π √(LC))

Substitute equivalent capacitance:

f = 1 / (2π √( L (C1C2/(C1+C2)) ))

Condition for oscillation:

A ≥ C2 / C1

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Quick Comparison

Oscillator	Feedback Element	Frequency
RC Phase Shift	Resistor + Capacitor	1/(2πRC√6)
Wien Bridge	RC Bridge	1/(2πRC)
Hartley	Inductive divider	1/(2π√LC)
Colpitts	Capacitive divider	1/(2π√LC)

 

Oscillator Derivations – Analog Electronics

Oscillators are circuits that generate periodic signals without external input. They use positive feedback to sustain oscillations.

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1. Barkhausen Criterion Derivation

Consider an amplifier with gain A and feedback factor β.

Output voltage = A × input

Feedback signal:

Vf = β Vo

For sustained oscillations:

Input = Feedback signal

Therefore

Vi = βVo

But

Vo = A Vi

Substitute Vi

Vo = A (βVo)

Simplify

Aβ = 1

Thus the two conditions are:

• Loop gain |Aβ| = 1
• Total phase shift = 360°

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2. RC Phase Shift Oscillator Derivation

The RC phase shift oscillator uses three RC sections to produce 180° phase shift.

Each RC section produces:

60° phase shift

Total phase shift:

3 × 60° = 180°

Amplifier (CE stage) provides:

180°

Total loop phase shift:

360°

Solving the RC network gives frequency:

f = 1 / (2πRC√6)

Minimum amplifier gain required:

A ≥ 29

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3. Wien Bridge Oscillator Derivation

The Wien bridge oscillator uses a bridge network of resistors and capacitors.

At resonance condition:

R1 = R2 = R C1 = C2 = C

The frequency of oscillation becomes:

f = 1 / (2πRC)

For sustained oscillations:

Loop gain Aβ = 1

For Wien bridge:

Amplifier gain A = 3

Thus oscillations occur when amplifier gain equals 3.

 

GATE Electrical Engineering

Analog Electronics – Complete Formula Sheet

This page provides important formulas for quick revision from the entire Analog Electronics syllabus.

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1. Diode Equations

Diode Current Equation

I = Is (e^(V/ηVT) − 1)

Thermal Voltage

VT = kT/q ≈ 25 mV

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2. Rectifier Formulas

Half Wave Rectifier

Vdc = Vm / π

Efficiency = 40.6%

Ripple Factor = 1.21

Full Wave Rectifier

Vdc = 2Vm / π

Efficiency = 81.2%

Ripple Factor = 0.482

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3. BJT Transistor Formulas

Ic = β Ib

β = α / (1 − α)

α = β / (1 + β)

gm = Ic / VT

re = VT / Ic

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4. CE Amplifier Gain

Av = −Rc / re

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5. h-Parameter Gain

Av = − (hfe RL / hie)

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6. Differential Amplifier

Vd = V1 − V2

Vc = (V1 + V2)/2

CMRR = Ad / Ac

CMRR(dB) = 20 log (Ad/Ac)

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7. Op-Amp Gains

Inverting Amplifier

Av = −Rf / Rin

Non-Inverting Amplifier

Av = 1 + (Rf / R1)

Voltage Follower

Av = 1

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8. Integrator

Vo = − (1/RC) ∫ Vin dt

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9. Differentiator

Vo = − RC (dVin/dt)

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10. Filters

fc = 1 / (2πRC)

First Order Slope = −20 dB/decade

Second Order Slope = −40 dB/decade

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11. Oscillators

RC Phase Shift Oscillator

f = 1 / (2πRC√6)

Wien Bridge Oscillator

f = 1 / (2πRC)

Barkhausen Criterion

Loop Gain = 1 Phase Shift = 360°

 

GATE Electrical – Analog Electronics

Practice Problems – Page 19

This section contains mixed problems from Diodes, Transistors, Operational Amplifiers, Filters and Oscillators.

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Problem 181

Question:

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

Answer:

≈ 0.7 V

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Problem 182

Question:

What is the cut-in voltage of a germanium diode?

Answer:

≈ 0.3 V

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Problem 183

Question:

If β = 100 and base current is 40 μA, find collector current.

Solution:

Ic = β × Ib

Ic = 100 × 40 μA = 4 mA

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Problem 184

Question:

What is the ripple factor of a full wave rectifier?

Answer:

0.482

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Problem 185

Question:

What is the efficiency of a full wave rectifier?

Answer:

81.2%

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Problem 186

Question:

If op-amp gain = 20 and input voltage = 0.2 V, find output voltage.

Solution:

Vo = Av × Vin

Vo = 20 × 0.2 = 4 V

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Problem 187

Question:

What is the slope of first order filter?

Answer:

−20 dB/decade

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Problem 188

Question:

What is the slope of second order filter?

Answer:

−40 dB/decade

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Problem 189

Question:

Which oscillator uses positive feedback?

Answer:

All oscillator circuits use positive feedback.

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Problem 190

Question:

What is the ideal CMRR of an op-amp?

Answer:

Infinite