JEE Physics Semiconductor Electronics MCQs Set 21

Answer: (c) 11.5 mA

Explanation:
1. \( I=\frac{3-0.7}{200}=11.5\ \text{mA} \)

Teacher's Note:
1. A silicon diode drops 0.7 V when conducting.
2. Check units and orders of magnitude before choosing an option.

Answer: (b) \( 6\times10^{5}\ \text{V/m} \)

Explanation:
1. \( E=\frac{V}{d}=\frac{0.6}{10^{-6}}=6\times10^{5}\ \text{V/m} \)

Teacher's Note:
1. Convert micrometres to metres: 1 µm = 10⁻⁶ m.
2. Compare with the corresponding statement for the other kind of semiconductor or junction to avoid mixing them up.

Answer: (a) 81.2%

Explanation:
1. For a full-wave rectifier \( \eta_{max}=\frac{8}{\pi^2}=0.812 \), i.e. 81.2% (a half-wave rectifier gives 40.6%).

Teacher's Note:
1. Not all the input power becomes d.c. power.
2. Questions of this kind are asked directly in JEE Main, so keep the key fact at your fingertips.

Answer: (d) 49

Explanation:
1. \( \beta=\frac{\alpha}{1-\alpha}=\frac{0.98}{0.02}=49 \); the emitter current is not needed.

Teacher's Note:
1. Do not be misled by the extra data.
2. Link the answer to the physical picture of carriers and junctions rather than memorising it as a bare fact.

Answer: (c) \( 1:10^{4} \)

Explanation:
1. Forward resistance is small (tens of ohms) while reverse resistance is of the order of mega-ohms, so the ratio is very small, about 1 : 10⁴ or less.

Teacher's Note:
1. This large ratio is what makes rectification possible.
2. A quick sketch of the band diagram or the circuit makes this type of question easy.

Answer: (b) as a voltage regulator

Explanation:
1. In its reverse breakdown region the voltage across a zener stays nearly constant while the current varies, so it holds a steady output voltage.

Teacher's Note:
1. It is always operated in reverse bias.
2. Check units and orders of magnitude before choosing an option.

Answer: (a) the peak voltage of the a.c. source

Explanation:
1. The capacitor charges through the diode up to the peak value of the source. When it reaches that value no further current flows, so the ammeter reads zero.

Teacher's Note:
1. The diode prevents it from discharging back through the source.
2. Questions of this kind are asked directly in JEE Main, so keep the key fact at your fingertips.

Answer: (d) 100 Hz

Explanation:
1. A bridge rectifier is full wave, so the ripple frequency is twice the input frequency.

Teacher's Note:
1. The fundamental ripple frequency is 100 Hz.
2. Eliminate the options that contradict a basic property of the device, then check the remaining one.

Answer: (c) a light emitting diode (LED)

Explanation:
1. An LED converts electrical energy into light; it is a light source, not a light detector.

Teacher's Note:
1. The other three change their electrical behaviour with light intensity.
2. Link the answer to the physical picture of carriers and junctions rather than memorising it as a bare fact.

Answer: (b) circuits 2 and 3

Explanation:
1. In circuits 2 and 3 the two junctions face the same way, so both are forward biased or both reverse biased and carry the same current; identical diodes then have equal drops.

Teacher's Note:
1. In circuit 1 the junctions face opposite ways: one is forward and the other reverse, so the drops differ greatly.
2. Check units and orders of magnitude before choosing an option.

Answer: (a) there is an electric field at the junction directed from the N-type side to the P-type side

Explanation:
1. Uncovered positive donor ions on the N side and negative acceptor ions on the P side produce a field from N to P. The N side is therefore at the higher potential.

Teacher's Note:
1. This is the built-in barrier field.
2. Compare with the corresponding statement for the other kind of semiconductor or junction to avoid mixing them up.

Answer: (d) reverse

Explanation:
1. In reverse bias the only current is the drift of minority carriers: holes of the N region move to the P region (and electrons the other way).

Teacher's Note:
1. In forward bias holes flow from P to N.
2. Questions of this kind are asked directly in JEE Main, so keep the key fact at your fingertips.

Answer: (c) increases and decreases respectively

Explanation:
1. In a conductor more lattice vibration increases scattering, so the resistivity rises; in a semiconductor the carrier density rises exponentially, so the resistivity falls.

Teacher's Note:
1. Metals: positive temperature coefficient; semiconductors: negative.
2. Link the answer to the physical picture of carriers and junctions rather than memorising it as a bare fact.

Answer: (b) insulators

Explanation:
1. Insulators have the widest forbidden gap, semiconductors a narrow one and metals none.

Teacher's Note:
1. Wide gap means no thermal excitation at room temperature.
2. A quick sketch of the band diagram or the circuit makes this type of question easy.

Answer: (a) emitter

Explanation:
1. The emitter supplies the majority carriers that are injected into the base, so it is heavily doped.

Teacher's Note:
1. The base is lightly doped and the collector moderately doped.
2. Check units and orders of magnitude before choosing an option.

Answer: (d) the variation of the number of charge carriers with temperature

Explanation:
1. In metals the carrier number is almost constant and scattering rises with temperature. In a semiconductor the carrier number rises exponentially and this dominates.

Teacher's Note:
1. The scattering effect exists in both but is overwhelmed in semiconductors.
2. Questions of this kind are asked directly in JEE Main, so keep the key fact at your fingertips.

Answer: (c) both the depletion region and the barrier height are reduced

Explanation:
1. The forward voltage opposes the built-in potential, which lowers the barrier and narrows the depletion layer.

Teacher's Note:
1. Both change together; they cannot move in opposite directions.
2. Eliminate the options that contradict a basic property of the device, then check the remaining one.

Answer: (b) 0.5 eV

Explanation:
1. \( E_g=\frac{1240\ \text{eV nm}}{2480\ \text{nm}}=0.5\ \text{eV} \)

Teacher's Note:
1. The threshold wavelength corresponds to the gap energy.
2. Link the answer to the physical picture of carriers and junctions rather than memorising it as a bare fact.

Answer: (a) 100 Hz

Explanation:
1. Both half cycles are used, so the pulsating output repeats twice per mains cycle: 100 Hz.

Teacher's Note:
1. Half-wave would give 50 Hz.
2. Check units and orders of magnitude before choosing an option.

Answer: (d) 5/4

Explanation:
1. \( I\propto nv\Rightarrow\frac{v_e}{v_h}=\frac{I_e/I_h}{n_e/n_h}=\frac{7/4}{7/5}=\frac54 \)

Teacher's Note:
1. Current is proportional to concentration times drift velocity.
2. Compare with the corresponding statement for the other kind of semiconductor or junction to avoid mixing them up.

Answer: (c) 49

Explanation:
1. \( \alpha=\frac{5.488}{5.60}=0.98 \)
2. \( \beta=\frac{\alpha}{1-\alpha}=\frac{0.98}{0.02}=49 \)

Teacher's Note:
1. Compute alpha first.
2. Questions of this kind are asked directly in JEE Main, so keep the key fact at your fingertips.

Answer: (b) an OR gate

Explanation:
1. If either input is high, its diode conducts and raises the output to the high level. The output is low only when both inputs are low.

Teacher's Note:
1. This is the OR function.
2. Link the answer to the physical picture of carriers and junctions rather than memorising it as a bare fact.