JEE Physics Semiconductor Electronics MCQs Set 22

Answer: (c) 4 A, 2 A

Explanation:
1. Circuit A: two 4 \( \ \Omega \) branches conduct in parallel, resistance 2 \( \ \Omega \), so \( I=8/2=4 \) A.
2. Circuit B: the reverse biased diode blocks its branch, leaving only 4 \( \ \Omega \): \( I=8/4=2 \) A.

Teacher's Note:
1. An ideal forward biased diode is a wire and a reverse biased one is an open switch. Then it is a simple resistor problem.

Answer: (b) 1/7 A

Explanation:
1. The diode conducts, so total resistance = 25 + 10 = 35 \( \ \Omega \).
2. \( I=5/35=1/7\ \text{A} \).

Teacher's Note:
1. Decide first whether the diode is forward or reverse biased. Then use the resistance of that state.

Answer: (a) 5 mA

Explanation:
1. The zener holds the voltage at 5 V. Current through the series resistor \( =\dfrac{25-5}{2000}=10\ \text{mA} \).
2. Load current = \( 5/1000=5\ \text{mA} \).
3. Zener current = 10 - 5 = 5 mA.

Teacher's Note:
1. The voltage across the parallel combination is fixed at the zener voltage. The zener carries what the load does not.

Answer: (c) positive half cycles only, with peak 5 V

Explanation:
1. Only the positive half cycles pass through the diode, so the output has positive half cycles only.
2. The 2 k\( \Omega \) series and 2 k\( \Omega \) load divide the voltage equally, so the peak is 10/2 = 5 V.

Teacher's Note:
1. This is a half wave rectifier with a potential divider. The output is half of the input peak.

Answer: (c) 20 mA

Explanation:
1. The anode is at a higher potential (-1 V) than the cathode side (-3 V), so the diode is forward biased.
2. \( I=\dfrac{-1-(-3)}{100}=\dfrac{2}{100}=20\ \text{mA} \).

Teacher's Note:
1. Potentials below zero are still compared by magnitude and sign. The p-side is at the higher potential, so the diode conducts.

Answer: (b) \( 2\times10^6\ \Omega \)

Explanation:
1. \( R=\dfrac{\Delta V}{\Delta I}=\dfrac{1}{0.5\times10^{-6}}=2\times10^6\ \Omega \).

Teacher's Note:
1. The dynamic resistance is the change in voltage over the change in current. Reverse resistance is very large.

Answer: (a) \( 660\ \Omega \)

Explanation:
1. Voltage across the resistor = 4 - 0.7 = 3.3 V.
2. \( R=\dfrac{3.3}{5\times10^{-3}}=660\ \Omega \).

Teacher's Note:
1. Subtract the diode drop before applying Ohm's law. This is the standard current limiting resistor calculation.

Answer: (a) majority carriers in both regions

Explanation:
1. The barrier potential repels majority carriers (electrons in n, holes in p) from crossing the junction.
2. It assists the flow of minority carriers across the junction.

Teacher's Note:
1. The barrier stops majority carrier diffusion and helps minority carrier drift. Forward bias reduces the barrier.

Answer: (d) \( 400\ \text{k}\Omega \)

Explanation:
1. Voltage across R = 6 - 2 = 4 V.
2. \( R=\dfrac{4}{10\times10^{-6}}=4\times10^5\ \Omega=400\ \text{k}\Omega \).

Teacher's Note:
1. Subtract the LED drop from the battery voltage. Then divide by the current in amperes.

Answer: (b) \( 1\times10^{9}\ \text{m}^{-3} \)

Explanation:
1. Mass action law: \( n_en_h=n_i^2 \) with \( n_i=7.07\times10^{15} \).
2. \( n_e=\dfrac{(7.07\times10^{15})^2}{5\times10^{22}}=\dfrac{5\times10^{31}}{5\times10^{22}}=1\times10^{9}\ \text{m}^{-3} \).

Teacher's Note:
1. The product of electron and hole densities stays constant at a given temperature. More holes mean fewer electrons.

Answer: (a) thermal excitations only

Explanation:
1. Reverse current is carried by minority carriers.
2. These are produced by thermal generation of electron-hole pairs, so it rises with temperature.

Teacher's Note:
1. The small reverse saturation current comes from thermally generated minority carriers. It is nearly independent of the reverse voltage.

Answer: (a) conduct

Explanation:
1. The p-side is at 0 V, higher than the n-side at -2 V, so the diode is forward biased and conducts.

Teacher's Note:
1. Forward bias means the p-side is at the higher potential. Compare the two potentials with their signs.

Answer: (a) \( 3\ \Omega\ \text{and}\ 12\ \Omega \)

Explanation:
1. (i) The diode conducts, so 12 and 4 \( \ \Omega \) are in parallel: \( \dfrac{12\times4}{16}=3\ \Omega \).
2. (ii) The diode blocks the lower branch, leaving only 12 \( \ \Omega \).

Teacher's Note:
1. Reverse biased diode removes its branch from the circuit. Forward biased diode behaves as a plain wire.

Answer: (d) X is undoped while Y and Z have been doped with a fifth group and a third group impurity respectively

Explanation:
1. A donor level just below the conduction band belongs to an n-type semiconductor, made with a fifth group impurity (P, As, Sb).
2. An acceptor level just above the valence band belongs to a p-type semiconductor, made with a third group impurity (B, Al, In).
3. So Y is fifth group doped and Z is third group doped.

Teacher's Note:
1. Donor levels sit near the conduction band, acceptor levels near the valence band. Fifth group atoms are donors.

Answer: (d) (a), (d)

Explanation:
1. Avalanche breakdown occurs in lightly doped junctions: a wide depletion region lets minority carriers gain a large velocity and ionise atoms by collision, which is (a).
2. Zener breakdown occurs in heavily doped junctions: a very thin depletion region has a strong electric field that pulls electrons out of bonds, which is (d).

Teacher's Note:
1. Remember the two breakdown mechanisms. Light doping gives avalanche and heavy doping gives zener.