Chapter 10 — Exercise Solutions
Electronics is highly scoring because many answers are diagram, table, and truth-table based.
How to Use This Page
- Start with the Key Concepts table to quickly recall definitions before attempting questions.
- Work through MCQs and Fill-in-the-Blanks first — they test recall and are the fastest to score.
- Draw diagrams and truth tables on paper while studying the detail answers; visual practice locks in memory.
- Re-solve the Worked Examples without looking at the solution, then compare your steps.
- Review Common Mistakes before every exam to avoid losing marks on known traps.
Key Concepts (Quick Recap)
| Term | Definition |
|---|
| Semiconductor | Material with conductivity between conductor and insulator (e.g., Si, Ge); conductivity increases with temperature |
| Intrinsic semiconductor | Pure semiconductor with equal number of electrons and holes at room temperature |
| Extrinsic semiconductor | Semiconductor doped with impurity to increase conductivity; classified as p-type or n-type |
| p-type semiconductor | Doped with trivalent impurity (B, In, Ga); majority carriers are holes |
| n-type semiconductor | Doped with pentavalent impurity (P, As, Sb); majority carriers are electrons |
| p-n junction | Interface between p-type and n-type regions; forms a depletion region with potential barrier |
| Forward bias | External voltage that reduces the depletion barrier; diode conducts |
| Reverse bias | External voltage that widens the depletion region; only small leakage current flows |
| Half-wave rectifier | Uses one diode; allows only one half-cycle of AC to pass; efficiency ≈ 40.6% |
| Full-wave rectifier | Uses two diodes (centre-tap) or four diodes (bridge); both half-cycles are rectified; efficiency ≈ 81.2% |
| Transistor | Three-terminal semiconductor device (emitter, base, collector) used for amplification and switching |
| Transistor configurations | Common base (CB), common emitter (CE), common collector (CC); CE is most widely used |
| Current gain alpha (α) | Ratio of collector current to emitter current in CB configuration; α = Ic / Ie; always less than 1 |
| Current gain beta (β) | Ratio of collector current to base current in CE configuration; β = Ic / Ib; typically 50–300 |
| Logic gates | Digital circuits performing Boolean operations — AND, OR, NOT, NAND, NOR, XOR, XNOR |
| Modulation | Process of varying a carrier signal's amplitude, frequency, or phase to transmit information |
| Demodulation | Recovering the original signal from the modulated carrier at the receiver |
| Bandwidth | Range of frequencies within which a communication system operates effectively |
Logic Gate Pattern
1. Draw the symbol if asked. 2. Write input combinations. 3. Fill output column. 4. State the Boolean expression.
Short Answer Points
- p-type semiconductor has holes as majority carriers.
- n-type semiconductor has electrons as majority carriers.
- A diode conducts mainly in forward bias.
- Modulation is needed for long-distance communication.
I. Choose the Best Answer (MCQs)
1. In an n-type semiconductor, the majority carriers are (a) holes (b) electrons (c) protons (d) neutrons
Answer(b) electrons — n-type is doped with a pentavalent impurity, which donates free electrons.
2. The depletion region in a p-n junction is formed due to (a) recombination of electrons and holes near the junction (b) applied voltage (c) temperature rise (d) light exposure
Answer(a) recombination of electrons and holes near the junction — Diffusion of carriers across the junction causes recombination, leaving behind immobile ions.
3. A full-wave rectifier uses (a) one diode (b) two diodes (c) three diodes (d) no diode
Answer(b) two diodes — A centre-tap full-wave rectifier uses two diodes; a bridge rectifier uses four.
4. In a common emitter configuration, the current gain β equals (a) Ic / Ie (b) Ic / Ib (c) Ib / Ic (d) Ie / Ib
Answer(b) Ic / Ib — β is defined as the ratio of collector current to base current in CE configuration.
5. The output of an AND gate is 1 only when (a) any one input is 1 (b) all inputs are 1 (c) any input is 0 (d) all inputs are 0
Answer(b) all inputs are 1 — The AND gate outputs HIGH only when every input is HIGH.
6. NAND gate is the combination of (a) AND gate followed by OR gate (b) AND gate followed by NOT gate (c) OR gate followed by NOT gate (d) NOT gate followed by AND gate
Answer(b) AND gate followed by NOT gate — NAND = NOT(AND); the output of AND is inverted.
7. Modulation is required to (a) increase the power of the signal (b) reduce noise (c) transmit signals over long distances (d) convert AC to DC
Answer(c) transmit signals over long distances — Low-frequency signals cannot travel far; modulation shifts them to higher carrier frequencies.
8. The relation between α and β of a transistor is (a) β = α / (1 − α) (b) α = β / (1 + β) (c) β = α (1 + β) (d) both (a) and (b)
Answer(d) both (a) and (b) — Both formulas are equivalent derivations; β = α/(1−α) and α = β/(1+β).
9. In amplitude modulation, the quantity that varies is (a) frequency (b) phase (c) amplitude (d) wavelength
Answer(c) amplitude — In AM, the amplitude of the carrier wave varies in proportion to the modulating signal.
10. The truth table output of a NOR gate for inputs A = 1, B = 0 is (a) 1 (b) 0 (c) undefined (d) 0.5
Answer(b) 0 — NOR is NOT(OR); OR of 1 and 0 is 1, so NOR output is 0.
II. Fill in the Blanks
1. In a p-type semiconductor, the majority carriers are \_\_\_\_\_\_\_\_\_.
Answerholes 2. The potential barrier for Si p-n junction is approximately \_\_\_\_\_\_\_\_\_ V. Answer: 0.7 3. The process of converting AC to DC is called \_\_\_\_\_\_\_\_\_. Answer: rectification 4. The transistor configuration with the highest current gain is \_\_\_\_\_\_\_\_\_. Answer: common emitter (CE) 5. The Boolean expression for OR gate is \_\_\_\_\_\_\_\_\_. Answer: Y = A + B 6. The range of frequencies that a communication channel can carry is called its \_\_\_\_\_\_\_\_\_. Answer: bandwidth 7. In FM, the \_\_\_\_\_\_\_\_\_ of the carrier wave varies according to the modulating signal. Answer: frequency 8. A bridge rectifier uses \_\_\_\_\_\_\_\_\_ diodes. Answer: four
III. True or False (with Corrections)
1. In an intrinsic semiconductor, the number of electrons equals the number of holes. True. In a pure (intrinsic) semiconductor, thermally generated electron-hole pairs are equal.
2. A diode allows current in both directions. False. A p-n junction diode conducts predominantly in forward bias; in reverse bias only a negligible leakage current flows.
3. The efficiency of a full-wave rectifier is higher than that of a half-wave rectifier. True. Full-wave rectifier efficiency is about 81.2%, compared to 40.6% for half-wave.
4. The current gain α of a transistor in CB configuration is always greater than 1. False. α is always less than 1 because Ic < Ie (some current is lost to the base). α = β / (1 + β).
5. A NAND gate is a universal gate. True. NAND (along with NOR) can be used to realise any Boolean function, hence it is called a universal gate.
6. Modulation increases the frequency of the baseband signal. False. Modulation does not increase the baseband frequency; it impresses the baseband signal onto a higher-frequency carrier.
7. In a transistor, the base region is very thin and lightly doped. True. The thin, lightly doped base ensures that most carriers from the emitter pass through to the collector, giving high current gain.
IV. Match the Following
| Column A | Column B |
|---|
| 1. p-type semiconductor | a. Converts AC to DC using four diodes |
| 2. Forward bias | b. Majority carriers are holes |
| 3. Bridge rectifier | c. Boolean expression: Y = (A·B)' |
| 4. Common emitter | d. Depletion region narrows, current flows |
| 5. NAND gate | e. β = Ic / Ib |
| 6. Frequency modulation | f. Carrier frequency varies with signal |
Answer1–b, 2–d, 3–a, 4–e, 5–c, 6–f
V. Answer Briefly (2-mark)
1. What is a p-n junction? How is it formed?
Answer
A p-n junction is the boundary formed when a p-type semiconductor is joined to an n-type semiconductor. At the junction, electrons from the n-side diffuse into the p-side and recombine with holes, creating a depletion region devoid of free carriers. The immobile ions left behind set up a potential barrier (approximately 0.7 V for Si) that prevents further diffusion, establishing equilibrium.
2. Distinguish between half-wave and full-wave rectifiers.
Answer
A half-wave rectifier uses a single diode and allows only one half-cycle of the AC input to pass; the output has a low ripple frequency equal to the input frequency and efficiency is about 40.6%. A full-wave rectifier uses two diodes (centre-tap) or four diodes (bridge) to rectify both half-cycles; its ripple frequency is double the input frequency and efficiency is about 81.2%.
3. Define current gain α and β. State the relation between them.
Answer
Current gain α (alpha) is the ratio of collector current to emitter current in the common base configuration: α = Ic / Ie. Current gain β (beta) is the ratio of collector current to base current in the common emitter configuration: β = Ic / Ib. The two are related by the formulas:
α = β / (1 + β)
β = α / (1 − α)
4. What is modulation? Why is it needed?
Answer
Modulation is the process of varying one or more properties of a high-frequency carrier wave (amplitude, frequency, or phase) in accordance with the baseband information signal. It is needed because low-frequency signals cannot propagate efficiently over long distances; modulation shifts the signal to a higher carrier frequency, reduces antenna size requirements, allows frequency-division multiplexing, and improves signal-to-noise ratio.
5. State De Morgan's theorems.
Answer
De Morgan's theorems provide equivalence between NAND/NOR combinations and their dual expressions:
- First theorem: (A · B)' = A' + B' — the NAND of two inputs equals the OR of their complements.
- Second theorem: (A + B)' = A' · B' — the NOR of two inputs equals the AND of their complements.
These theorems are useful for simplifying Boolean expressions and converting between gate implementations.
Worked Examples
Example 1 — Transistor current gain
A transistor has a collector current of 4.95 mA and a base current of 50 μA. Find β.
β = Ic / Ib = (4.95×10⁻³) / (50×10⁻⁶) = 4.95×10⁻³ / 5×10⁻⁵ = 99
Example 2 — Relation between α and β
If β = 99, find the common-base current gain α.
α = β / (1 + β) = 99 / 100 = 0.99
Example 3 — Logic gate (NAND truth table)
A | B | A·B | NAND = (A·B)'
0 | 0 | 0 | 1
0 | 1 | 0 | 1
1 | 0 | 0 | 1
1 | 1 | 1 | 0
NAND output is 0 only when both inputs are 1.
VI. Answer in Detail (5-mark)
1. Explain the formation of a p-n junction diode and describe its V-I characteristics.
Answer
When a p-type semiconductor (doped with trivalent impurity) and an n-type semiconductor (doped with pentavalent impurity) are brought into contact, electrons from the n-side diffuse across the junction into the p-side and recombine with holes. This creates a depletion region near the junction that is free of mobile carriers but contains immobile positive ions on the n-side and immobile negative ions on the p-side. These ions produce an internal electric field that opposes further diffusion, creating a potential barrier (0.7 V for Si, 0.3 V for Ge).
Under forward bias, the external voltage opposes the barrier, narrowing the depletion region. Once the applied voltage exceeds the barrier potential, majority carriers cross the junction and a large forward current flows. Under reverse bias, the external voltage adds to the barrier, widening the depletion region; only a very small reverse saturation current (due to minority carriers) flows. The V-I graph shows a sharp rise in current after the knee voltage in forward bias, and negligible current in reverse bias until breakdown occurs.
2. Explain half-wave and full-wave rectifier circuits with diagrams.
Answer
A half-wave rectifier consists of a single diode connected in series with the load resistor across an AC source. During the positive half-cycle, the diode is forward biased and conducts; current flows through the load. During the negative half-cycle, the diode is reverse biased and blocks current. The output is a pulsating DC with only positive half-cycles. Its ripple frequency equals the input frequency, and its rectification efficiency is about 40.6%.
A full-wave rectifier can be centre-tap or bridge type. In the centre-tap version, two diodes work on alternate half-cycles through a centre-tapped transformer. During the positive half-cycle, diode D1 conducts; during the negative half-cycle, diode D2 conducts, so both half-cycles contribute to the output. In the bridge rectifier, four diodes are arranged in a bridge configuration; during each half-cycle, two diagonally opposite diodes conduct. The output has a ripple frequency twice the input frequency, and efficiency is about 81.2%.
Half-wave rectification efficiency: η = 40.6%
Full-wave rectification efficiency: η = 81.2%
3. Explain transistor action and compare CB, CE, and CC configurations.
Answer
A bipolar junction transistor (BJT) has three regions — emitter (heavily doped), base (thin and lightly doped), and collector (moderately doped, larger area). In normal active mode, the emitter-base junction is forward biased and the collector-base junction is reverse biased. Electrons (in an npn transistor) are injected from the emitter into the base; because the base is very thin, about 98–99% of these electrons cross into the collector, producing a large collector current controlled by a small base current — this is transistor action (amplification).
The three configurations differ in which terminal is common to input and output:
- Common Base (CB): Input at emitter, output at collector, base common. Current gain α < 1, voltage gain high, no phase inversion.
- Common Emitter (CE): Input at base, output at collector, emitter common. Current gain β is high (50–300), moderate voltage gain, 180° phase shift. Most widely used in amplifiers.
- Common Collector (CC): Input at base, output at emitter, collector common. Voltage gain ≈ 1, high current gain, no phase inversion. Used as an impedance-matching buffer (emitter follower).
Relation: α = β / (1 + β)
Current gain: α (CB) < 1 ; β (CE) >> 1
4. Explain AND, OR, NOT, NAND, and NOR logic gates with truth tables.
Answer
AND gate: Output is 1 only when all inputs are 1. Boolean expression: Y = A · B. Truth table shows output 1 only for input combination (1, 1).
OR gate: Output is 1 when at least one input is 1. Boolean expression: Y = A + B. Truth table shows output 0 only for input combination (0, 0).
NOT gate (inverter): Output is the complement of the input. Boolean expression: Y = A'. Single-input gate; output is 1 when input is 0 and vice versa.
NAND gate: Universal gate; output is complement of AND. Boolean expression: Y = (A · B)'. Output is 0 only when all inputs are 1. Any Boolean function can be implemented using only NAND gates.
NOR gate: Universal gate; output is complement of OR. Boolean expression: Y = (A + B)'. Output is 1 only when all inputs are 0. Like NAND, NOR alone can realise any logic function.
AND: Y = A·B NAND: Y = (A·B)'
OR: Y = A+B NOR: Y = (A+B)'
NOT: Y = A'
Common Mistakes to Avoid
1. ❌ Writing that p-type semiconductors have electrons as majority carriers. ✅ p-type has holes as majority carriers; n-type has electrons as majority carriers.
2. ❌ Drawing the diode symbol with the arrow pointing from cathode to anode. ✅ The arrow in a diode symbol points from the p-side (anode) to the n-side (cathode), showing the direction of conventional current flow in forward bias.
3. ❌ Forgetting that α is always less than 1 and confusing it with β. ✅ α = Ic / Ie is less than 1 (CB config); β = Ic / Ib is much greater than 1 (CE config). Use α = β / (1 + β) to convert.
4. ❌ Writing that a half-wave rectifier uses two diodes. ✅ Half-wave rectifier uses one diode; full-wave centre-tap uses two; bridge rectifier uses four.
5. ❌ Confusing AM and FM — saying AM varies frequency. ✅ AM varies the amplitude of the carrier; FM varies the frequency of the carrier.
6. ❌ Omitting rows in truth tables for logic gates. ✅ A 2-input truth table always has 4 rows (00, 01, 10, 11); a 3-input table has 8 rows. Never skip input combinations.
Exam Checklist
- Draw neat circuit symbols for diode, npn/pnp transistors, and all logic gates — label terminals clearly.
- Write complete truth tables with all input combinations; never skip rows.
- State the correct Boolean expression alongside every truth table.
- Remember the potential barrier values: 0.7 V for Si, 0.3 V for Ge.
- Memorise the α–β relation: α = β/(1+β) and β = α/(1−α).
- Know the difference between half-wave (1 diode, 40.6%) and full-wave (2/4 diodes, 81.2%) rectifiers.
- Distinguish AM (amplitude varies), FM (frequency varies), and PM (phase varies).
- Identify the three transistor configurations and their key properties (current gain, voltage gain, phase).
- Practise numerical problems on β, α, Ic, Ib, and Ie using Ie = Ic + Ib.
- Review De Morgan's theorems and universal gate concepts — they are frequently tested.
