JEE Physics Semiconductor Electronics MCQs Set 13

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. In reverse bias the dark current is tiny, so the extra current produced by light is a large fractional change and is easily measured: A is true.
2. R is also true (forward current is much larger), but it does not explain why photodiodes are used in reverse bias.

Teacher's Note:
1. Photodiodes work best when the dark current is very small.
2. A forward-biased diode carries a large current irrespective of illumination.

Answer: (c) A is false but R is true

Explanation:
1. Photodiodes are operated in reverse bias to measure light intensity, so A is false.
2. R is true: beyond the threshold voltage (and below breakdown) forward current is far greater than reverse current.

Teacher's Note:
1. The fractional change in reverse current under illumination is large.
2. Forward current is dominated by diffusion and hides the photocurrent.

Answer: (d) Statement I is correct but Statement II is incorrect

Explanation:
1. Statement I is correct: boron (trivalent) gives p-type, arsenic (pentavalent) gives n-type.
2. Statement II is incorrect: at equilibrium the diffusion and drift currents cancel, so no net current flows and none is detected by an ammeter.

Teacher's Note:
1. In equilibrium the junction has a built-in potential but zero net current.
2. A real current needs an external source.

Answer: (c) C

Explanation:
1. A, B and D are true statements about an LED.
2. C is false: an LED emits light only in forward bias.

Teacher's Note:
1. Light is emitted when electrons and holes recombine across the junction, which requires forward bias.
2. Photodiodes (not LEDs) are used in reverse bias.

Answer: (b) an XOR gate

Explanation:
1. \( X=A\overline B+\overline AB \).
2. This is 1 only when the inputs differ, which is the XOR function \( A\oplus B \).

Teacher's Note:
1. Make the truth table: (0,0)=0, (0,1)=1, (1,0)=1, (1,1)=0.
2. The XOR gate is the 'exclusive OR'.

Answer: (d) 00:0, 01:1, 10:1, 11:0

Explanation:
1. \( P=\overline{AB} \). Gate 2 gives \( \overline{AP} \) and gate 3 gives \( \overline{PB} \).
2. \( Y=\overline{\overline{AP}\cdot\overline{PB}}=AP+PB=P(A+B)=\overline{AB}(A+B) \).
3. This equals 1 when exactly one input is 1: (0,0) gives 0, (0,1) gives 1, (1,0) gives 1, (1,1) gives 0 (XOR).

Teacher's Note:
1. Work out the intermediate outputs for each input pair.
2. Four NANDs connected this way are the standard XOR construction.

Answer: (c) n_e increases and resistance decreases

Explanation:
1. Heating breaks more covalent bonds, so more electrons reach the conduction band.
2. More charge carriers mean lower resistance, so a semiconductor has a negative temperature coefficient of resistance.

Teacher's Note:
1. Metals show the opposite trend because collisions increase.
2. Carrier density rises exponentially with temperature in a semiconductor.

Answer: (c) Statement I is incorrect but Statement II is correct

Explanation:
1. Statement I is incorrect: the emitter is heavily doped, the base lightly doped and the collector moderately doped.
2. Statement II is correct: the base is very thin and the collector is the largest region (it must dissipate heat).

Teacher's Note:
1. A thin, lightly doped base keeps recombination low.
2. The emitter must supply a large number of carriers.

Answer: (b) A-II, B-I, C-III, D-IV

Explanation:
1. Intrinsic: Fermi level in the middle of the gap (II).
2. n-type: electrons are donated, so the Fermi level moves near the conduction band (I).
3. p-type: near the valence band (III).
4. Metals: bands overlap and the Fermi level lies inside the conduction band (IV).

Teacher's Note:
1. The Fermi level moves toward the band that has more carriers.
2. For metals the conduction band is partly filled.

Answer: (a) It works as a voltage regulator in reverse bias and behaves like a simple p-n junction diode in forward bias.

Explanation:
1. A Zener diode is operated in reverse bias at its breakdown voltage, where the voltage stays almost constant over a wide range of current: this is voltage regulation.
2. In forward bias it conducts like an ordinary p-n junction diode.

Teacher's Note:
1. The breakdown (Zener) voltage is fixed by the doping.
2. A series resistor limits the current.