Analog Electronics – Problems Page 25

Very Hard Numerical Problems (GATE Level)

This section contains advanced numerical problems combining diode circuits, BJT amplifiers, operational amplifiers and oscillators.

---

Problem 1

A silicon diode has reverse saturation current Is = 10⁻¹² A. Calculate diode current when forward voltage is 0.7 V at room temperature.

Solution

Diode equation: I = Is (e^(V/VT) − 1)

At room temperature:

VT ≈ 25 mV

Substitute values:

I = 10⁻¹² (e^(0.7/0.025) − 1)

I ≈ 1.2 mA

---

Problem 2

A BJT amplifier has Rc = 4 kΩ and re = 25 Ω. Find the voltage gain of the CE amplifier.

Solution

Av = −Rc / re

Av = −4000 / 25 Av = −160

---

Problem 3

An inverting op-amp has Rin = 5 kΩ and Rf = 50 kΩ. If input voltage is 0.1 V, calculate output voltage.

Solution

Av = −Rf / Rin

Av = −50k / 5k = −10

Vo = Av × Vin Vo = −10 × 0.1 = −1 V

---

Problem 4

An LC oscillator uses L = 2 mH and C = 200 pF. Find oscillation frequency.

Solution

f = 1 / (2π√LC)

f ≈ 251 kHz

---

Problem 5

An op-amp differential amplifier has Ad = 2000 and Ac = 0.5. Calculate CMRR in dB.

Solution

CMRR = Ad / Ac

CMRR = 2000 / 0.5 = 4000

CMRR(dB) = 20 log(4000)

CMRR ≈ 72 dB

---

Problem 6

A Wien bridge oscillator uses R = 10 kΩ and C = 0.01 μF. Find oscillation frequency.

Solution

f = 1 / (2πRC)

f ≈ 1591 Hz

 

Analog Electronics – Problems Page 24

Mixed Concept & Numerical Problems

This section includes mixed questions from diode circuits, BJT amplifiers, operational amplifiers and oscillators for GATE preparation.

---

Problem 1

Find the thermal voltage at room temperature (300 K).

Solution

VT = kT / q

VT ≈ 25 mV

---

Problem 2

A BJT has β = 120 and base current 20 μA. Find collector current.

Solution

Ic = β × Ib

Ic = 120 × 20 μA Ic = 2.4 mA

---

Problem 3

Calculate the voltage gain of an inverting op-amp with Rf = 100 kΩ and Rin = 10 kΩ.

Solution

Av = −Rf / Rin

Av = −100k / 10k Av = −10

---

Problem 4

An LC oscillator has L = 5 mH and C = 100 pF. Find oscillation frequency.

Solution

f = 1 / (2π√LC)

f ≈ 225 kHz

---

Problem 5

What is the ripple factor of a full wave rectifier?

Ripple Factor = 0.482

---

Problem 6

What is the ideal input impedance of an operational amplifier?

Infinite

---

Problem 7

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

Zero

---

Problem 8

For oscillations to occur, what should be the loop gain?

Aβ = 1

This is the Barkhausen criterion.

 

Analog Electronics – Problems Page 23

Oscillator Advanced Numerical Problems

This section contains advanced numerical problems on oscillator circuits useful for GATE and PSU examinations.

                                     

---

Problem 1

An RC phase shift oscillator uses R = 5 kΩ and C = 0.02 μF. Find the oscillation frequency.

Solution

Formula:

f = 1 / (2πRC√6)

Substitute values:

f = 1 / (2π × 5000 × 0.02×10⁻⁶ × √6)

Result:

f ≈ 650 Hz

---

Problem 2

A Hartley oscillator has inductors L1 = 4 mH and L2 = 6 mH with capacitor C = 200 pF. Calculate the oscillation frequency.

Solution

Total inductance:

L = L1 + L2 = 10 mH

Formula:

f = 1 / (2π√LC)

Substitute:

f = 1 / (2π √(10×10⁻³ × 200×10⁻¹²))

Result:

f ≈ 112 kHz

---

Problem 3

A Colpitts oscillator has capacitors C1 = 100 pF, C2 = 400 pF and inductor L = 2 mH. Find the oscillation frequency.

Solution

Equivalent capacitance:

Ceq = (C1C2)/(C1 + C2)

Substitute:

Ceq = (100×400)/(500)

Ceq = 80 pF

Frequency:

f = 1/(2π√LC)

Result:

f ≈ 125 kHz

---

Problem 4

A crystal oscillator has inductance L = 0.5 H and capacitance C = 0.02 pF. Calculate the series resonant frequency.

Solution

Formula:

f = 1/(2π√LC)

Substitute values:

f = 1/(2π √(0.5 × 0.02×10⁻¹²))

Result:

f ≈ 1.59 MHz

---

Problem 5

An oscillator has amplifier gain A = 10 and feedback factor β = 0.1. Check whether oscillations will occur.

Solution

Loop gain:

Aβ = 10 × 0.1

Aβ = 1

Since loop gain equals 1,

Oscillations will occur.

 

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.

---

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.

---

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.

---

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

---

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.

---

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.

---

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.

---

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.

---

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

---

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

---

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

---

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

---

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.

---

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.

---

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.

---

Series Resonant Frequency

At series resonance:

XL = XC

Thus

ωL = 1 / (ωC)

Therefore

fₛ = 1 / (2π √LC)

---

Parallel Resonant Frequency

Due to parallel capacitance Cp:

fₚ ≈ fₛ √(1 + C/Cp)

Parallel resonance frequency is slightly higher than series frequency.

---

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

---

3. Oscillator Applications

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

---

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.

---

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)

---

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

---

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

---

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)