JEE Physics Semiconductor Electronics MCQs Set 19

Answer: (c) negative on the P side and positive on the N side

Explanation:
1. The P side loses holes and gains negative acceptor ions; the N side loses electrons and gains positive donor ions.

Teacher's Note:
1. The field points from N to P and opposes further diffusion.
2. Compare with the corresponding statement for the other kind of semiconductor or junction to avoid mixing them up.

Answer: (b) reverse bias

Explanation:
1. Reverse bias adds to the built-in potential, widening the depletion region and raising the barrier.

Teacher's Note:
1. Forward bias lowers the barrier.
2. Questions of this kind are asked directly in JEE Main, so keep the key fact at your fingertips.

Answer: (a) from the N-side to the P-side

Explanation:
1. The junction field points from N to P and drives minority carriers, so the drift (conventional) current flows from n to p.

Teacher's Note:
1. It is opposite to the diffusion current.
2. Link the answer to the physical picture of carriers and junctions rather than memorising it as a bare fact.

Answer: (d) the P region is negative (connected to the negative terminal) and the current is due to both electrons and holes (minority carriers)

Explanation:
1. Reverse bias connects the p side to the negative terminal. The tiny current is carried by minority electrons in p and minority holes in n.

Teacher's Note:
1. Reverse current is of the order of microamperes.
2. A quick sketch of the band diagram or the circuit makes this type of question easy.

Answer: (c) 0.3 V, opposing the forward current

Explanation:
1. Ge has a barrier of about 0.3 V and Si about 0.7 V. The barrier opposes the flow of majority carriers, i.e. the forward current.

Teacher's Note:
1. Remember Ge: 0.3 V, Si: 0.7 V.
2. Check units and orders of magnitude before choosing an option.

Answer: (b) P-N junction

Explanation:
1. Conduction in only one direction is the defining property of a junction diode (forward and reverse bias).

Teacher's Note:
1. A transistor has three terminals and a resistor conducts both ways.
2. Questions of this kind are asked directly in JEE Main, so keep the key fact at your fingertips.

Answer: (a) the majority carriers

Explanation:
1. The barrier prevents majority holes and electrons from crossing; minority carriers are helped across by the same field.

Teacher's Note:
1. It opposes both electrons in n and holes in p, i.e. the majority carriers.
2. Eliminate the options that contradict a basic property of the device, then check the remaining one.

Answer: (d) in reverse bias

Explanation:
1. In reverse bias the depletion region widens and only a tiny leakage current flows, so the diode behaves like an open circuit.

Teacher's Note:
1. In forward bias it conducts after the knee voltage.
2. Link the answer to the physical picture of carriers and junctions rather than memorising it as a bare fact.

Answer: (c) \( 10^{6}\ \Omega \)

Explanation:
1. The reverse current is only microamperes, so the resistance is of the order of mega-ohms.

Teacher's Note:
1. Forward resistance is only a few ohms to hundreds of ohms.
2. Check units and orders of magnitude before choosing an option.

Answer: (b) the cumulative effect of collisions of conduction-band electrons (carrier multiplication)

Explanation:
1. At high reverse voltage the carriers gain enough energy to knock out more electrons from bonds, which in turn do the same, and the current multiplies like an avalanche.

Teacher's Note:
1. Zener breakdown occurs in heavily doped junctions with a strong field.
2. Compare with the corresponding statement for the other kind of semiconductor or junction to avoid mixing them up.

Answer: (a) there will be no steady current in the circuit

Explanation:
1. No energy source is present; the contact potentials at the metal junctions cancel the junction potential, so the net emf around the loop is zero.

Teacher's Note:
1. A steady current without an energy source would violate energy conservation.
2. Questions of this kind are asked directly in JEE Main, so keep the key fact at your fingertips.

Answer: (d) 11.7 V

Explanation:
1. Ge starts conducting first at 0.3 V and holds the voltage drop across the parallel pair at 0.3 V, so Si never conducts.
2. \( V_Y=12-0.3=11.7\ \text{V} \).

Teacher's Note:
1. Two diodes in parallel: the one with the lower knee voltage clamps the voltage.
2. Link the answer to the physical picture of carriers and junctions rather than memorising it as a bare fact.

Answer: (c) 2.0 eV

Explanation:
1. \( E_g=\frac{hc}{\lambda}=\frac{1240\ \text{eV nm}}{620\ \text{nm}}=2.0\ \text{eV} \)

Teacher's Note:
1. The threshold wavelength corresponds to a photon energy equal to the gap.
2. A quick sketch of the band diagram or the circuit makes this type of question easy.

Answer: (b) 1 kΩ

Explanation:
1. \( r=\frac{\Delta V}{\Delta I}=\frac{1.5}{1.5\times10^{-3}}=10^{3}\ \Omega \)

Teacher's Note:
1. Dynamic resistance is the ratio of the change in voltage to the change in current.
2. Check units and orders of magnitude before choosing an option.

Answer: (a) 2n

Explanation:
1. A full-wave rectifier turns both half cycles into pulses of the same polarity, so there are two output pulses per input cycle.

Teacher's Note:
1. A half-wave rectifier would have ripple frequency n.
2. Questions of this kind are asked directly in JEE Main, so keep the key fact at your fingertips.

Answer: (d) 5 mA

Explanation:
1. Voltage across the 500 Ω resistor is 10 − 5 = 5 V, so the total current is \( I=5/500=10\ \text{mA} \).
2. Load current is \( 5/1000=5\ \text{mA} \), so the zener current is \( 10-5=5\ \text{mA} \).

Teacher's Note:
1. The zener holds 5 V across the load, so the load current is fixed and the rest flows through the zener.
2. Eliminate the options that contradict a basic property of the device, then check the remaining one.

Answer: (c) convert a.c. into d.c.

Explanation:
1. A rectifier uses the one-way conduction of diodes to change alternating current into unidirectional current.

Teacher's Note:
1. An inverter does the opposite.
2. Link the answer to the physical picture of carriers and junctions rather than memorising it as a bare fact.

Answer: (b) Only statements I and III

Explanation:
1. Statement I is true: negative voltage means reverse bias.
2. Statement II is false: the knee of a silicon diode is at about 0.7 V.
3. Statement III is true: reverse current is in microamperes while forward current is in milliamperes, so the two parts are drawn to different scales.

Teacher's Note:
1. Silicon knee voltage is 0.7 V and germanium 0.3 V.
2. Check units and orders of magnitude before choosing an option.

Answer: (a) the base

Explanation:
1. The emitter is heavily doped to supply carriers, the collector is moderately doped and the base is lightly doped and thin.

Teacher's Note:
1. This keeps recombination in the base small.
2. Compare with the corresponding statement for the other kind of semiconductor or junction to avoid mixing them up.

Answer: (d) the base

Explanation:
1. A thin base lets most carriers cross to the collector without recombining.

Teacher's Note:
1. Typical base width is a few micrometres.
2. Questions of this kind are asked directly in JEE Main, so keep the key fact at your fingertips.

Answer: (c) the base-emitter junction is forward biased and the base-collector junction is reverse biased

Explanation:
1. Forward bias at the emitter-base junction injects carriers into the base; reverse bias at the collector-base junction sweeps them into the collector.

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

Answer: (b) the emitter-base depletion layer is thin (small) but the base-collector depletion layer is wide (large)

Explanation:
1. Forward bias narrows the emitter-base depletion layer and reverse bias widens the collector-base one.

Teacher's Note:
1. Wide collector-base layer supports the large reverse voltage.
2. A quick sketch of the band diagram or the circuit makes this type of question easy.