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CBSE Class 12 › Chapter 14 Notes

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1 Semiconductor Electronics: Question Bank

1.1 Part 1: Multiple Choice Questions (1 Mark)

1.2 Part 2: Assertion-Reason Questions (1 Mark)

1.3 Part 3: Short & Long Answer Questions

1.4 Part 4: Numericals

1.5 Part 5: Case Studies

Semiconductor Electronics: Question Bank

Comprehensive Practice Set: MCQs, Assertion-Reason, Numericals & Case Studies.

45 Questions Strictly NCERT Board Pattern

Instructions: Use this page for self-assessment. Click the “Show Answer” button to verify your solution.

Part 1: Multiple Choice Questions (1 Mark)

1. When an intrinsic semiconductor is doped with a small amount of pentavalent impurity, it becomes:

(a) p-type semiconductor
(b) n-type semiconductor
(c) Insulator
(d) Superconductor

Answer: (b)
Pentavalent impurities (Group 15) donate extra electrons, creating n-type (negative type) semiconductors.

2. In a p-n junction diode, the direction of the electric field in the depletion region is:

(a) From p-side to n-side
(b) From n-side to p-side
(c) Random
(d) Perpendicular to junction

Answer: (b)
The n-side has positive donor ions, and the p-side has negative acceptor ions. Electric field points from positive to negative (n to p).

3. The energy gap between the valence band and conduction band in a semiconductor at 0 K is:

(a) Zero
(b) Infinite
(c) Very small (< 3 eV)
(d) Very large (> 3 eV)

Answer: (c)
Semiconductors have a small forbidden gap ( $E_g < 3$ eV). At 0K, they act as insulators because no electrons can jump this gap.

4. Which of the following statements is true for a p-type semiconductor?

(a) Electrons are majority carriers; trivalent atoms are dopants.
(b) Electrons are minority carriers; pentavalent atoms are dopants.
(c) Holes are minority carriers; pentavalent atoms are dopants.
(d) Holes are majority carriers; trivalent atoms are dopants.

Answer: (d)
p-type is formed by Trivalent (Group 13) doping, creating holes as majority carriers.

5. In a half-wave rectifier, if the input frequency is 50 Hz, the output ripple frequency is:

(a) 25 Hz
(b) 50 Hz
(c) 100 Hz
(d) 0 Hz

Answer: (b)
Half-wave rectifiers produce one pulse per cycle, so output frequency = input frequency = 50 Hz.

6. A photodiode is used to detect optical signals. It is preferably operated in:

(a) Forward bias
(b) Reverse bias
(c) Unbiased condition
(d) Breakdown region

Answer: (b)
Sensitivity to light intensity changes is highest in reverse bias (dark current region).

7. The depletion layer in a p-n junction diode is caused by:

(a) Drift of holes
(b) Diffusion of charge carriers
(c) Migration of impurity ions
(d) Drift of electrons

Answer: (b)
Diffusion of majority carriers across the junction leaves behind immobile ions, forming the depletion layer.

8. At absolute zero temperature, a pure Germanium crystal behaves as:

(a) A perfect conductor
(b) A semiconductor
(c) A perfect insulator
(d) A superconductor

Answer: (c)
With no thermal energy, the conduction band is empty.

9. Two p-n junction diodes are connected in series with a battery of 5 V and a resistor of 10 $\Omega$ . If one diode is forward biased and the other is reverse biased (ideal), current is:

(a) 0.5 A
(b) 5 A
(c) Zero
(d) Infinity

Answer: (c)
The reverse-biased diode acts as an open switch ( $R = \infty$ ), blocking current flow.

10. The potential barrier of a silicon p-n junction is approximately:

(a) 0.3 V
(b) 0.7 V
(c) 1.1 V
(d) 1.5 V

Answer: (b)
For Silicon, $V_b \approx 0.7$ V. For Germanium, $V_b \approx 0.3$ V.

Part 2: Assertion-Reason Questions (1 Mark)

Directions:
(A) Both A and R are true and R is correct explanation of A.
(B) Both A and R are true but R is NOT correct explanation of A.
(C) A is true but R is false.
(D) A is false but R is true.

1. Assertion (A): A p-n junction diode conducts current only when forward biased.
Reason (R): The depletion layer width decreases in forward bias.

Answer: (B)
Both are true. Forward bias reduces barrier width, allowing diffusion. However, the *conduction* happens because the barrier potential is lowered, not *just* because width decreases.

2. Assertion (A): The conductivity of an intrinsic semiconductor increases with temperature.
Reason (R): With increase in temperature, more covalent bonds break, creating electron-hole pairs.

Answer: (A)
Correct explanation. Semiconductors have a negative temperature coefficient of resistance.

3. Assertion (A): A photodiode is operated in reverse bias.
Reason (R): In reverse bias, the fractional change in minority carrier current due to light is easily measurable.

Answer: (A)
Correct explanation.

4. Assertion (A): Carbon is an insulator while Silicon is a semiconductor.
Reason (R): The energy gap of Carbon (5.4 eV) is much larger than that of Silicon (1.1 eV).

Answer: (A)
Correct explanation.

5. Assertion (A): In a full-wave rectifier, the output frequency is double the input frequency.
Reason (R): A full-wave rectifier conducts during both positive and negative half-cycles.

Answer: (A)
Correct explanation.

Part 3: Short & Long Answer Questions

1. Why can’t we measure the potential barrier of a p-n junction by simply connecting a voltmeter across its terminals?

Ideally, there are no free charge carriers in the depletion region. When a voltmeter is connected, metal-semiconductor contacts are formed at the terminals. The contact potentials at these junctions exactly cancel out the barrier potential of the p-n junction, resulting in a zero reading.

2. Draw the energy band diagram of an n-type semiconductor at $T > 0$ K.

[Image of n-type energy band diagram]

Energy band diagram of n-type semiconductor showing donor level.

Show the Conduction Band, Valence Band, and the **Donor Energy Level ( $E_D$ )** located just below the Conduction Band ( $E_c - E_D \approx 0.01$ eV).

3. Explain the term ‘dynamic resistance’ of a diode.

It is the resistance offered by the diode to a small AC signal. It is defined as the ratio of a small change in voltage ( $\Delta V$ ) to the corresponding small change in current ( $\Delta I$ ) at a particular operating point: $r_d = \frac{\Delta V}{\Delta I}$ .

4. Draw the circuit diagram of a Full Wave Rectifier and explain the role of the center-tapped transformer.

Circuit diagram and waveforms of a full-wave rectifier.

The center-tapped transformer provides two out-of-phase AC voltages relative to the center tap. This allows diode $D_1$ to conduct during the positive half-cycle and diode $D_2$ to conduct during the negative half-cycle, ensuring unidirectional current through the load at all times.

5. What is the effect of forward biasing on the width of the depletion layer? Explain with a diagram.

Diagram showing how depletion layer width changes with biasing.

In forward bias, the positive terminal is connected to the p-side and negative to the n-side. The external field opposes the internal barrier field. This pushes holes and electrons towards the junction, neutralizing the uncovered ions. Consequently, the **depletion width decreases**.

Part 4: Numericals

1. A Zener diode having breakdown voltage 10 V is used as a voltage regulator in a circuit. The current through the diode is 10 mA. What is the power dissipated in the diode?

Given $V_Z = 10$ V, $I_Z = 10 \text{ mA} = 0.01$ A.
Power Dissipated $P = V_Z \times I_Z = 10 \times 0.01 = 0.1 \text{ W}$ .

2. Find the current through the circuit if a silicon diode (barrier potential 0.7 V) is connected in series with a resistor of 1 k $\Omega$ across a 5 V battery.

$V_{source} = 5$ V, $V_{barrier} = 0.7$ V.
Net Voltage $V_{net} = 5 - 0.7 = 4.3$ V.
Current $I = \frac{V_{net}}{R} = \frac{4.3}{1000} = 4.3 \times 10^{-3}$ A $= 4.3 \text{ mA}$ .

3. The number density of donor atoms in an n-type silicon semiconductor is $4 \times 10^{20} m^{-3}$ . If the intrinsic carrier concentration is $1.5 \times 10^{16} m^{-3}$ , find the number density of holes.

Given $N_D \approx n_e = 4 \times 10^{20} m^{-3}$ and $n_i = 1.5 \times 10^{16} m^{-3}$ .
Using Mass Action Law: $n_e n_h = n_i^2$ .
$n_h = \frac{n_i^2}{n_e} = \frac{(1.5 \times 10^{16})^2}{4 \times 10^{20}} = \frac{2.25 \times 10^{32}}{4 \times 10^{20}} = 5.6 \times 10^{11} m^{-3}$ .

4. A p-n photodiode is fabricated from a semiconductor with a band gap of 2.8 eV. Can it detect a wavelength of 6000 nm?

Energy of photon $E = \frac{hc}{\lambda} = \frac{1240 \text{ eV nm}}{6000 \text{ nm}}$ .
$E = 0.206 \text{ eV}$ .
Since $E (0.206 \text{ eV}) < E_g (2.8 \text{ eV})$ , the photon cannot excite an electron. It **cannot** detect this wavelength.

5. Determine the value of current $I$ in the circuit if the diode is ideal. (Diagram: 10V battery, Diode forward biased, 2 resistors 5 $\Omega$ and 5 $\Omega$ in series).

Since diode is ideal, resistance $R_d = 0$ .
Total Resistance $R_{eq} = 5 + 5 = 10 \Omega$ .
Current $I = \frac{V}{R_{eq}} = \frac{10}{10} = 1 \text{ A}$ .

Part 5: Case Studies

Case Study 1: Zener Diode
Zener diodes are specialized diodes designed to work in the breakdown region. They are heavily doped to have a narrow depletion width.

What is the main mechanism of breakdown in low voltage Zener diodes?
How is a Zener diode connected in a voltage regulator circuit?
What happens to the series resistor current if the input voltage increases?

1. Zener Breakdown (Quantum Tunneling).
2. In parallel with the load, in reverse bias.
3. It increases, but the Zener current absorbs the excess, keeping load voltage constant.

Case Study 2: Rectification
Rectification is the process of converting AC to DC. A full-wave rectifier uses two diodes and a center-tapped transformer.

What is the efficiency of a full-wave rectifier?
Why is a capacitor used at the output?
Can we use a Zener diode for rectification?

1. 81.2%.
2. To filter/smooth the pulsating DC into steady DC.
3. No, Zener diodes are used for regulation, not rectification.

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Semiconductor Electronics: Important Questions & Numericals

Semiconductor Electronics: Question Bank

Part 1: Multiple Choice Questions (1 Mark)

Part 2: Assertion-Reason Questions (1 Mark)

Part 3: Short & Long Answer Questions

Part 4: Numericals

Part 5: Case Studies

Editorial Team

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Part 1: Multiple Choice Questions (1 Mark)

Part 2: Assertion-Reason Questions (1 Mark)

Part 3: Short & Long Answer Questions

Part 4: Numericals

Part 5: Case Studies

Editorial Team

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