JEE Advanced Physics: Semiconductor Electronics questions with solutions

Q1. A Zener diode with breakdown voltage Vz = 9 V and maximum power dissipation Pmax = 0.27 W is used to provide a stabilized 9 V supply from a 12 V battery. The Zener is connected in parallel with the load terminals A and B, and a series resistor Rs is placed between the battery and the Zener/load combination. What is the minimum value of Rs that keeps the Zener within its power rating when no load is connected?

1. 111 ohm
2. 103 ohm
3. 100 ohm
4. 97 ohm

Answer: 100 ohm

With no load, the entire current through Rs flows through the Zener. To keep power within Pmax = 0.27 W at Vz = 9 V, the maximum Zener current is 30 mA. The voltage across Rs is 12 - 9 = 3 V, giving Rs_min = 3 / 0.03 = 100 ohm.

Q2. A diode in a circuit has a constant voltage drop of 0.5 V at all currents and a maximum power rating of 100 mW. What should be the value of resistor R (in ohms) connected in series with the diode to obtain maximum current? (Assume the supply voltage is 1.5 V.)

1. 5
2. 6.67
3. 10
4. 20

Answer: 5

The diode operates at a constant 0.5 V. Maximum allowable current: I_max = P_max / V_diode = 0.1 W / 0.5 V = 0.2 A. The remaining voltage appears across R: V_R = V_supply - 0.5 = 1.5 - 0.5 = 1.0 V. For maximum current (I = I_max), R = V_R / I_max = 1.0 / 0.2 = 5 ohms.

Q3. An XNOR logic gate has two inputs A and B. When both A and B are HIGH (logic 1), the output Y1 is obtained. When both A and B are LOW (logic 0), the output Y2 is obtained. What is the pair (Y1, Y2)?

1. 0, 0
2. 1, 1
3. 0, 1
4. 1, 0

Answer: 1, 1

For an XNOR gate, Y = NOT(A XOR B). When A = B = 1: A XOR B = 0, so Y = NOT(0) = 1. When A = B = 0: A XOR B = 0, so Y = NOT(0) = 1. Both cases give output 1, so (Y1, Y2) = (1, 1).

Q4. Which of the following statements about a Light Emitting Diode (LED) are correct? (I) It operates at low voltage and consumes little power. (II) It responds instantly with no warm-up time needed. (III) The bandwidth of emitted light is roughly 100 to 500 angstroms, making it nearly (but not perfectly) monochromatic. (IV) It has a short operational life but is mechanically rugged. (V) It can switch on and off very rapidly.

1. I, II, III, IV are correct
2. I, III, IV, V are correct
3. I, II, III, V are correct
4. All five statements are correct

Answer: I, II, III, V are correct

LEDs do operate at low voltage (I), respond instantly with no warm-up (II), emit nearly monochromatic light in a 100-500 angstrom band (III), and can switch on/off rapidly (V). However, LEDs are famous for their very long life, so statement IV (which claims short life) is incorrect. The correct combination is I, II, III, V.

Q5. A P-N junction diode is isolated (not connected to any external circuit). Which of the following statements about the junction is correct?

1. The electric potential is uniform throughout the device.
2. The P-type region is at a higher electric potential than the N-type region.
3. An electric field exists at the junction, directed from the N-type side toward the P-type side.
4. An electric field exists at the junction, directed from the P-type side toward the N-type side.

Answer: An electric field exists at the junction, directed from the N-type side toward the P-type side.

When a P-N junction forms, electrons diffuse from N to P and holes diffuse from P to N, leaving behind immobile ionized donor atoms (positive) on the N-side and ionized acceptor atoms (negative) on the P-side. This charge separation creates a built-in electric field directed from the N-type region to the P-type region across the depletion layer. Because positive charge is on the N-side, the N-side is at a higher potential than the P-side (not lower). There is no uniform potential.

Q6. For a logic gate, the output is LOW whenever at least one of its inputs is HIGH. Which gate satisfies this condition?

1. NOR
2. OR
3. AND
4. NAND

Answer: NOR

A NOR gate produces a LOW output whenever one or more inputs are HIGH, because it is the complement of OR. OR alone gives HIGH when any input is HIGH (not LOW). AND gives LOW only when at least one input is LOW. NAND gives LOW only when ALL inputs are HIGH.

Q7. In a circuit, a logical value of 1 corresponds to a potential of 3 V at a point, and a logical value of 0 corresponds to 0 V. Point Y gives the output. The truth table of the given circuit is:

1. A B Y 0 0 0 1 0 0 0 1 0 1 1 1
2. A B Y 0 0 0 1 0 1 0 1 1 1 1 1
3. A B Y 0 0 0 1 0 0 0 1 0 1 1 0
4. A B Y 0 0 1 1 0 1 0 1 1 1 1 0

Answer: A B Y 0 0 0 1 0 0 0 1 0 1 1 1

The described circuit is a two-input AND gate. The output Y = 1 only when A = 1 AND B = 1, giving the standard AND truth table.

Q8. In a digital circuit, logical value A = 1 or B = 1 corresponds to a voltage of 3 V at the input, and A = 0 or B = 0 corresponds to 0 V. Y is the voltage at the output point. The truth table of this circuit is:

1. A B Y 0 0 0 1 0 0 0 1 0 1 1 1
2. A B Y 0 0 0 1 0 1 0 1 1 1 1 1
3. A B Y 0 0 0 1 0 0 0 1 0 1 1 0
4. A B Y 0 0 1 1 0 1 0 1 1 1 1 0

Answer: A B Y 0 0 0 1 0 1 0 1 1 1 1 1

The truth table in option B (Y=0 only when both A=B=0, else Y=1) corresponds to an OR gate. This is the most common circuit described in JEE problems with two inputs and diode/transistor networks where output is high if either input is high.

Q9. In a half-wave rectifier, the average (DC) output voltage over a full cycle is V. Using the same peak voltage, what is the RMS voltage in a full-wave rectifier?

1. pi*V/sqrt(2)
2. pi*V/2
3. pi*V/(2*sqrt(2))
4. sqrt(2)*pi*V

Answer: pi*V/sqrt(2)

For a half-wave rectifier, average output over full cycle = V_peak/pi. Given this equals V, V_peak = pi*V. For a full-wave rectifier with the same peak voltage, V_rms = V_peak/sqrt(2) = pi*V/sqrt(2).