JEE Physics Semiconductor Electronics MCQs Set 15

Answer: (d) the resistance of copper decreases while that of germanium increases

Explanation:
1. In a metal, lower temperature means fewer lattice collisions, so the resistance of copper decreases.
2. In a semiconductor, fewer electrons reach the conduction band at low temperature, so the resistance of germanium increases.

Teacher's Note:
1. Metals have positive, semiconductors negative temperature coefficients of resistance.

Answer: (c) B and C

Explanation:
1. Trivalent impurities (Al, B) create holes and give a p-type semiconductor.
2. Pentavalent impurities (As, P) give n-type material.

Teacher's Note:
1. Group III dopants are acceptors, group V dopants are donors.

Answer: (a) current is mainly carried by holes

Explanation:
1. Trivalent doping creates many holes, which are the majority carriers.
2. The crystal remains electrically neutral.

Teacher's Note:
1. Majority carriers in p-type are holes.

Answer: (c) there are no free electrons at 0 K

Explanation:
1. At 0 K the valence band is full and the conduction band is empty, so there are no free electrons.
2. At higher temperatures some electrons are thermally excited across the gap.

Teacher's Note:
1. Intrinsic semiconductors act as insulators at 0 K.

Answer: (a) \( \mu_e>\mu_h \)

Explanation:
1. Electrons move freely in the conduction band, whereas holes move through the valence band by successive electron jumps, which is slower.
2. Hence \( \mu_e>\mu_h \).

Teacher's Note:
1. Mobility is always positive.

Answer: (b) increases the potential barrier

Explanation:
1. The reverse voltage adds to the built-in barrier voltage, widening the depletion region.
2. So the potential barrier increases.

Teacher's Note:
1. Forward bias lowers the barrier; reverse bias raises it.

Answer: (a) ions

Explanation:
1. The depletion layer contains immobile ionised donor and acceptor atoms, which set up the barrier field.
2. Free carriers have been swept out of this region.

Teacher's Note:
1. The barrier arises from fixed space charge.

Answer: (c) there is an electric field at the junction directed from the n-type side to the p-type side

Explanation:
1. The p-side has negative acceptor ions and the n-side has positive donor ions near the junction.
2. The field points from positive ions to negative ions, i.e. from the n side to the p side, with the n-side at higher potential.

Teacher's Note:
1. Field direction is from positive to negative charge.

Answer: (c) The majority carriers in an n-type semiconductor are holes

Explanation:
1. In n-type material the majority carriers are electrons, not holes, so statement (c) is false.
2. The other three statements are correct.

Teacher's Note:
1. N-type has electron majority; p-type has hole majority.

Answer: (b) diffusion in forward bias, drift in reverse bias

Explanation:
1. With forward bias, majority carriers diffuse across the lowered barrier.
2. In reverse bias only minority carriers are swept across by the field, which is drift current.

Teacher's Note:
1. Forward current is large diffusion; reverse current is tiny drift.

Answer: (a) decreases exponentially with increasing band gap

Explanation:
1. The number of thermally excited electrons varies as \( e^{-E_g/2kT} \).
2. So it falls exponentially with larger gap and rises with temperature.

Teacher's Note:
1. A wider gap means fewer carriers.

Answer: (c) 100 Hz

Explanation:
1. A full-wave rectifier produces one output pulse per half-cycle of the input.
2. So the ripple frequency is twice the input frequency: 100 Hz.

Teacher's Note:
1. A half-wave rectifier ripple has the input frequency.

Answer: (a) electrons move from base to collector

Explanation:
1. In an NPN transistor the emitter injects electrons into the thin base; most of them are swept across the reverse-biased base-collector junction.
2. From the options given, the movement of electrons from base to collector is the correct one.

Teacher's Note:
1. Majority carriers in the NPN emitter are electrons.

Answer: (b) after electrons are lost in the base, only the remaining ones reach the collector

Explanation:
1. A small fraction of the electrons from the emitter recombine with holes in the base and form the base current.
2. Only the rest reach the collector, so \( I_C<I_E \).

Teacher's Note:
1. I_E = I_B + I_C.

Answer: (a) 0

Explanation:
1. In common base mode, increasing the emitter-base forward bias raises the collector current, and the voltage drop across the load rises in step.
2. Input and output are in phase (the common emitter circuit has a phase difference of \( \pi \)).

Teacher's Note:
1. CE inverts the signal; CB and CC do not.

Answer: (b) \( \beta=\frac{\alpha}{1-\alpha} \)

Explanation:
1. Using \( I_E=I_B+I_C \): \( \frac1\alpha=\frac{I_B+I_C}{I_C}=\frac1\beta+1 \).
2. Hence \( \beta=\frac{\alpha}{1-\alpha} \).

Teacher's Note:
1. Divide I_E = I_B + I_C by I_C.

Answer: (b) NOR gate

Explanation:
1. The output is 1 only when both inputs are 0.
2. That is the NOR function \( y=\overline{A+B} \).

Teacher's Note:
1. NOR is OR followed by NOT.

Answer: (c) a NOT gate

Explanation:
1. The output is the complement of the input.
2. This is a NOT gate.

Teacher's Note:
1. NOT has one input.

Answer: (d) NAND

Explanation:
1. The output is 0 only when both inputs are 1, the opposite of AND.
2. This is a NAND gate.

Teacher's Note:
1. NAND = AND followed by NOT.

Answer: (a) 0, 0

Explanation:
1. For the product to be 1, both factors must be 1.
2. \( \overline{A+B}=1 \) requires A = B = 0, and then \( \overline{A\cdot B}=1 \) also holds.

Teacher's Note:
1. NOR is 1 only when both inputs are 0.

Answer: (c) 1.0 V

Explanation:
1. Voltage gain \( =\beta\frac{R_L}{R_{in}}=100\times\frac{10}{1}=1000 \).
2. Output = \( 1\ \text{mV}\times1000=1.0 \) V.

Teacher's Note:
1. Voltage gain = beta x (R_L / R_in).

Answer: (b) an XOR gate

Explanation:
1. Y is 1 only when exactly one input is 1.
2. That is the exclusive-OR function.

Teacher's Note:
1. XOR = A xor B.