JEE Physics Semiconductor Electronics MCQs Set 04

Answer: (a) 1, 3

Explanation:
1. In the active region the emitter injects electrons into the base
2. Most of them cross the thin base and are swept into the collector, so statements (1) and (3) are true
3. Holes do not move to the collector, and only a very small fraction of the electrons leave through the base, so (2) and (4) are not true. Answer: (a)

Teacher's Note:
1. In an npn transistor the majority carriers are electrons from the emitter.
2. The base is thin and lightly doped, so almost all electrons reach the collector.

Answer: (d) the collector current increases slightly because the neutral base width reduces

Explanation:
1. A larger reverse collector-base voltage widens the collector depletion region
2. This reduces the effective neutral base width (the Early effect)
3. A thinner base means fewer recombinations and a slightly larger collector current. Answer: (d)

Teacher's Note:
1. The Early effect explains the slight slope of the output characteristic in the active region.
2. A narrower base reduces recombination in the base.

Answer: (d) \( I_C = \beta I_B \)

Explanation:
1. Current gain in common-emitter configuration is defined as \( \beta = \frac{I_C}{I_B} \)
2. So \( I_C = \beta I_B \). Answer: (d)

Teacher's Note:
1. Beta is the ratio of collector current to base current.
2. Alpha is the ratio of collector current to emitter current.

Answer: (b) 4.9 V

Explanation:
1. \( I_E = I_B + I_C \) and \( I_C = \alpha I_E \), so \( I_C = \frac{\alpha}{1-\alpha}I_B = 49\times0.02 = 0.98 \) mA
2. \( V = I_C R_L = 0.98\times10^{-3}\times5000 = 4.9 \) V. Answer: (b)

Teacher's Note:
1. Express collector current in terms of base current using alpha.
2. Output voltage across the load is collector current times load resistance.

Answer: (d) Base 1 V, collector 2 V, emitter 0 V

Explanation:
1. Active region: emitter-base junction forward biased and collector-base junction reverse biased
2. Case (a): V_B = 2 V is above V_C = 1 V, so the collector junction is forward biased (saturation)
3. Case (b): both junctions are forward biased (saturation)
4. Case (c): the emitter is above the base, so the emitter junction is reverse biased (cut-off)
5. Case (d): the base is 1 V above the emitter (forward) and the collector at 2 V is above the base (reverse), which is the active region. Answer: (d)

Teacher's Note:
1. For an npn transistor the collector must be at a higher potential than the base for the active region.
2. The base must be a little above the emitter.

Answer: (c) AND (A = 1, B = 1, Y = 1)

Explanation:
1. The first gate gives \( \overline{A\cdot B} \)
2. A NAND gate with both inputs joined acts as a NOT gate
3. So \( Y = \overline{\overline{A\cdot B}} = A\cdot B \), which is an AND gate
4. For A = 1, B = 1 the output is Y = 1. Answer: (c)

Teacher's Note:
1. NAND followed by a NOT gives AND.
2. Check with A = 1 and B = 1 to confirm Y = 1.

Answer: (b) D2 is forward biased and D1 is reverse biased, hence no current flows from B to A and vice versa

Explanation:
1. B (0 V) is at a higher potential than A (-10 V), so the driving current would flow from B to A
2. D2 conducts for B to A current, so it is forward biased
3. D1 conducts only for A to B current, so it is reverse biased and blocks the flow
4. With one diode blocking, no current flows in either direction. Answer: (b)

Teacher's Note:
1. Diodes in series must all be forward biased for current to flow.
2. One reverse-biased ideal diode is enough to stop the current.

Answer: (b) \( 200\,\Omega \)

Explanation:
1. Voltage across the resistor \( = 1.5 - 0.3 = 1.2 \) V
2. \( R = \frac{1.2}{6\times10^{-3}} = 200\,\Omega \). Answer: (b)

Teacher's Note:
1. A germanium diode drops about 0.3 V when conducting.
2. Use Ohm's law on the series resistor.

Answer: (d) \( 220\sqrt2\ \mathrm{V} \)

Explanation:
1. The diode charges the capacitor to the peak value of the a.c. voltage
2. 220 V is the rms value, so the peak is \( 220\sqrt2 \approx 311 \) V
3. With no load, the capacitor holds this peak value. Answer: (d)

Teacher's Note:
1. Quoted a.c. voltages are rms values.
2. The capacitor holds the peak value because the diode stops it from discharging back.

Answer: (c) 40 mW

Explanation:
1. The voltage across the parallel combination is fixed at 4 V
2. Current through the series resistor \( I_1 = \frac{12-4}{400} = 20 \) mA
3. Current through the other resistor \( I_3 = \frac{4}{400} = 10 \) mA
4. Current through the Zener \( I_2 = 20 - 10 = 10 \) mA
5. Power \( = V_z I_2 = 4\times10\times10^{-3} = 40 \) mW. Answer: (c)

Teacher's Note:
1. The Zener keeps the voltage across it at 4 V.
2. Use Kirchhoff's current law to find the Zener current.

Answer: (d) the reverse breakdown region, where the voltage stays nearly constant for a large change in current

Explanation:
1. A voltage regulator needs a nearly constant voltage across the load while the current changes
2. In the reverse breakdown region of a Zener diode the voltage remains almost constant although the current varies widely
3. So this region is used for regulation. Answer: (d)

Teacher's Note:
1. Zener diodes are operated in reverse bias at breakdown.
2. The steep, almost vertical part of the curve gives the constant voltage.

Answer: (b) \( \frac{200}{11}\,\Omega \) and \( \frac{200}{7}\,\Omega \)

Explanation:
1. (i) A higher: D₁ is forward (20) and D₂ is reverse (200) in parallel: \( R = \frac{20\times200}{220} = \frac{200}{11}\,\Omega \)
2. (ii) B higher: D₁ is reverse (600) and D₂ is forward (30) in parallel: \( R = \frac{600\times30}{630} = \frac{200}{7}\,\Omega \). Answer: (b)

Teacher's Note:
1. Decide which diode conducts for each polarity, then combine the two resistances in parallel.
2. A forward diode has the small resistance.

Answer: (d) 12.5 mA, 7.5 mA, 5 mA

Explanation:
1. The Zener holds the load at 10 V
2. \( I = \frac{60-10}{4\,\mathrm{k\Omega}} = 12.5 \) mA
3. \( I_L = \frac{10}{2\,\mathrm{k\Omega}} = 5 \) mA
4. \( I_Z = I - I_L = 7.5 \) mA. Answer: (d)

Teacher's Note:
1. The voltage across the load equals the Zener voltage.
2. The current splits between the Zener and the load.