Chapter 14: Semiconductor Electronics
Q1. Distinguish between intrinsic and extrinsic semiconductors.
Solution
Intrinsic: Pure semiconductor with equal number of electrons and holes.
Extrinsic: Doped semiconductor with majority carriers (electrons or holes).
Extrinsic types:
n-type: Doped with pentavalent (P, As), majority carriers = electrons
p-type: Doped with trivalent (B, Ga), majority carriers = holes
Q2. Draw the V-I characteristic of a p-n junction diode in forward and reverse bias.
Solution
In forward bias, current increases exponentially after knee voltage (~0.7V for Si, ~0.3V for Ge). In reverse bias, a small saturation current flows until breakdown voltage is reached.
Q3. What is the logic gate equivalent of the Boolean expression Y = A · B + C?
Solution
A · B: AND gate with inputs A, B
(A · B) + C: OR gate with inputs (A · B) and C
Circuit: A, B → AND gate → output with C → OR gate → Y
Q4. Explain the working of an n-p-n transistor as an amplifier in common emitter configuration.
Solution
In CE configuration, the emitter is common, input is applied to base-emitter junction (forward biased), and output is taken from collector-emitter junction (reverse biased).
When a small AC signal is applied at the base:
• Base current changes by ΔIB
• Collector current changes by ΔIC = β ΔIB
• Voltage gain: Av = βRC/ri
The transistor amplifies the weak input signal at the output.
Q5. Draw the truth table for NAND gate and NOR gate. Show that NAND gate is a universal gate.
Solution
NAND Gate: Y = (A · B)'
A B | Y
0 0 | 1
0 1 | 1
1 0 | 1
1 1 | 0
NOR Gate: Y = (A + B)'
A B | Y
0 0 | 1
0 1 | 0
1 0 | 0
1 1 | 0
NAND is universal: NOT = NAND(A,A), AND = NAND(NOT(A),NOT(B)), OR = NAND(NAND(A,A),NAND(B,B))
Q6. Draw the circuit diagram of a half-wave rectifier using a p-n junction diode. Explain its working.
Solution
Circuit: AC supply → transformer → p-n diode (in series) → load resistor.
Working:
• During positive half-cycle: Diode is forward biased, current flows through load. Output ≈ input.
• During negative half-cycle: Diode is reverse biased, no current flows. Output = 0.
Output is pulsating DC with frequency equal to input frequency.
Efficiency: η = 40.6% (maximum)
Q7. What is a Zener diode? How does it work as a voltage regulator? Draw the circuit diagram.
Solution
A Zener diode is a specially designed p-n junction diode that operates in the reverse breakdown (Zener) region.
As voltage regulator:
• Connected in reverse bias across the load
• When Vin increases, current through Zener increases but VZ remains constant
• Series resistor absorbs excess voltage
• Load voltage remains constant at VZ
Used in power supplies to maintain constant output voltage despite input variations.
Q8. The output of an OR gate is connected to the input of a NOT gate. Draw the logic circuit and truth table of the combination.
Solution
This combination is called a NOR gate: Y = (A + B)'
Truth table:
A B | A+B | Y = (A+B)'
0 0 | 0 | 1
0 1 | 1 | 0
1 0 | 1 | 0
1 1 | 1 | 0
Circuit: OR gate → NOT gate
Q9. An n-p-n transistor has a current amplification factor (β) of 100. If the base current is 20 μA, find (a) collector current (b) emitter current.
Solution
(a) IC = β IB = 100 × 20 μA = 2 mA
(b) IE = IB + IC = 20 μA + 2 mA
= 0.02 + 2 = 2.02 mA
Also: IE = (1 + β)IB = 101 × 20 μA = 2.02 mA ✓
Q10. Draw the V-I characteristics of a Zener diode and explain how it is used as a voltage regulator. What is the significance of the knee voltage?
Solution
V-I characteristics:
• Forward bias: Similar to ordinary diode (exponential rise after ~0.7V)
• Reverse bias: Very small leakage current until knee (breakdown) voltage, then sharp increase in current while voltage remains nearly constant.
Knee voltage significance:
• It is the Zener breakdown voltage VZ
• Beyond this, voltage is constant regardless of current
• Used as reference voltage in regulator circuits
• Typical values: 3V to 200V
Q11. Implement the Boolean expression Y = A + (B·C) using basic logic gates. Write the truth table.
Solution
Circuit: B, C → AND gate → output with A → OR gate → Y
Truth table:
A B C | B·C | Y = A+(B·C)
0 0 0 | 0 | 0
0 0 1 | 0 | 0
0 1 0 | 0 | 0
0 1 1 | 1 | 1
1 0 0 | 0 | 1
1 0 1 | 0 | 1
1 1 0 | 0 | 1
1 1 1 | 1 | 1
Q12. Explain the working of a full-wave rectifier with a neat diagram. Compare it with a half-wave rectifier.
Solution
Full-wave rectifier: Uses two diodes and a center-tapped transformer.
Working:
• Positive half-cycle: D₁ conducts, D₂ off → current through load in one direction
• Negative half-cycle: D₂ conducts, D₁ off → current through load in same direction
Comparison:
• Ripple frequency: HW = f, FW = 2f
• Efficiency: HW = 40.6%, FW = 81.2%
• Transformer utilization: FW is better
• PIV: HW = V₀, FW = 2V₀
Q13. Define the terms (i) current gain (ii) voltage gain (iii) power gain of a transistor amplifier. How are they related?
Solution
(i) Current gain: β = ΔIC/ΔIB (CE configuration)
(ii) Voltage gain: Av = ΔVC/ΔVB = βRC/ri
(iii) Power gain: Ap = β × Av
Relation: Ap = β × Av = β²RC/ri
A transistor amplifier provides power gain, making it useful for signal amplification.
Q14. Draw the transfer characteristic of a NOT gate and an AND gate. What do you understand by the terms 'logic family' and 'fan-out'?
Solution
NOT gate transfer characteristic: Vout vs Vin — high output when input is low, low output when input is high (inverted).
AND gate transfer characteristic: Output is high only when all inputs are high.
Logic family: A group of logic circuits that share similar technology, voltage levels, and characteristics (e.g., TTL, CMOS, ECL).
Fan-out: The maximum number of logic gate inputs that the output of a single gate can drive without degrading its output voltage levels. Typical CMOS fan-out = 50.
Q15. An LED emits light of wavelength 5800 Å. Find (i) the energy gap in eV (ii) the material used (iii) its forward bias voltage.
Solution
(i) E = hc/λ = 12400/5800 = 2.14 eV
(ii) For λ = 5800 Å (yellow-orange light), the material is likely GaAsP (Gallium Arsenide Phosphide) or GaP doped with ZnO.
(iii) Forward bias voltage is approximately equal to the energy gap:
Vf ≈ 2.0 − 2.2 V