Unit 19: Electronic Devices — Online MCQ Test
PHYSICS · NEET (All) · NEET
Practice Unit 19: Electronic Devices with a free chapter-wise online MCQ test.
This chapter covers: Semiconductors; semiconductor diode: I-V characteristics in forward and reverse bias; diode as a rectifier; I-V characteristics of LED. the photodiode solar cell and Zener diode; Z....
AI-generated questions from basic to board-exam level, with instant results and explanations.
Unit 19: Electronic Devices — Important Questions & Answers
The solids which have the negative temperature coefficient of resistance are:
- A. insulators only
- B. semiconductors only
- C. insulators and semiconductors
- D. metals
Answer: C. insulators and semiconductors
Both insulators and semiconductors exhibit a decrease in resistance (or an increase in conductivity) with an increase in temperature, resulting in a negative temperature coefficient of resistance.
Both insulators and semiconductors exhibit a decrease in resistance (or an increase in conductivity) with an increase in temperature, resulting in a negative temperature coefficient of resistance.
For the logic circuit shown, the truth table is:
- A. A 0 0 1 1; B 0 1 0 1; Y 1 1 1 0
- B. A 0 0 1 1; B 0 1 0 1; Y 0 1 1 1
- C. A 0 0 1 1; B 0 1 0 1; Y 1 0 0 0
- D. A 0 0 1 1; B 0 1 0 1; Y 0 0 0 1
Answer: A. A 0 0 1 1; B 0 1 0 1; Y 1 1 1 0
The circuit consists of two NOT gates feeding into a NOR gate. Y = NOT(NOT A OR NOT B) = A AND B is not correct. The gates are NOT gates followed by a NOR gate. Y = NOT(NOT A OR NOT B) which is equivalent to A AND B? No, Y = NOT(NOT(A) OR NOT(B)) = A AND B. Wait, looking at symbol, it is a NOR gate (OR with circle). Output Y = NOT(NOT A OR NOT B) = A AND B? No, De Morgan's: NOT(NOT A OR NOT B) = A AND B. Actually, the NOR gate with inverted inputs is an AND gate.
The circuit consists of two NOT gates feeding into a NOR gate. Y = NOT(NOT A OR NOT B) = A AND B is not correct. The gates are NOT gates followed by a NOR gate. Y = NOT(NOT A OR NOT B) which is equivalent to A AND B? No, Y = NOT(NOT(A) OR NOT(B)) = A AND B. Wait, looking at symbol, it is a NOR gate (OR with circle). Output Y = NOT(NOT A OR NOT B) = A AND B? No, De Morgan's: NOT(NOT A OR NOT B) = A AND B. Actually, the NOR gate with inverted inputs is an AND gate.
The output (Y) of the given logic implementation is similar to the output of an/a _______gate.
- A. AND
- B. NAND
- C. OR
- D. NOR
Answer: D. NOR
Determined by logic gate truth table reduction.
Determined by logic gate truth table reduction.
A full wave rectifier circuit with diodes (D1) and (D2) is shown in the figure. If input supply voltage Vin = 220sin(100πt) volt, then at t = 15 m sec
- A. D1 is forward biased, D2 is reverse biased
- B. D1 is reverse biased, D2 is forward biased
- C. D1 and D2 both are forward biased
- D. D1 and D2 both are reverse biased
Answer: B. D1 is reverse biased, D2 is forward biased
At t = 15ms, phase = 100π * 0.015 = 1.5π radians = 270 degrees. sin(270) = -1. For the standard configuration, this negative peak biases D2 forward.
At t = 15ms, phase = 100π * 0.015 = 1.5π radians = 270 degrees. sin(270) = -1. For the standard configuration, this negative peak biases D2 forward.
Consider the junction diode as ideal. The value of current flowing through AB is:
- A. 10⁻² A
- B. 10⁻¹ A
- C. 10⁻³ A
- D. 0 A
Answer: A. 10⁻² A
The diode is forward biased (+4V at anode, -6V at cathode). Potential difference across the 1kΩ resistor is 10V. I = V/R = 10/1000 = 10⁻² A.
The diode is forward biased (+4V at anode, -6V at cathode). Potential difference across the 1kΩ resistor is 10V. I = V/R = 10/1000 = 10⁻² A.
A npn transistor is connected to common emitter configuration in a given amplifier. A load resistance of 800 Ω is connected in the collector circuit and the voltage drop across it is 0.8 V. If the current amplification factor is 0.96 and the input resistance of the circuit is 192 Ω, the voltage gain and the power gain of the amplifier will respectively be:
- A. 3.69, 3.84
- B. 4, 4
- C. 4, 3.69
- D. 4, 3.84
Answer: D. 4, 3.84
Voltage gain Av = β * (RL/Ri) = 0.96 * (800/192) = 4. Power gain Ap = β * Av = 0.96 * 4 = 3.84.
Voltage gain Av = β * (RL/Ri) = 0.96 * (800/192) = 4. Power gain Ap = β * Av = 0.96 * 4 = 3.84.
Consider the following statements A and B and identify the correct answer: A. For a solar cell, the I-V characteristics lies in the IV quadrant of the given graph. B. In a reverse biased pn junction diode, the current measured in (µA), is due to majority charge carriers.
- A. Both A and B are correct.
- B. Both A and B are incorrect.
- C. A is correct but B is incorrect.
- D. A is incorrect but B is correct.
Answer: C. A is correct but B is incorrect.
A solar cell delivers power, so its I-V curve lies in the fourth quadrant. In reverse bias, the small current is due to minority carriers, not majority carriers.
A solar cell delivers power, so its I-V curve lies in the fourth quadrant. In reverse bias, the small current is due to minority carriers, not majority carriers.
In a common emitter transistor amplifier the audio signal voltage across the collector is 3 V. The resistance of collector is 3 kΩ. If current gain is 100 and the base resistance is 2 kΩ, the voltage and power gain of the amplifier is
- A. 15 and 200
- B. 150 and 15000
- C. 20 and 2000
- D. 200 and 1000
Answer: B. 150 and 15000
Voltage gain A_v = β * (R_c/R_b) = 100 * (3000/2000) = 150. Power gain = A_v * β = 150 * 100 = 15000.
Voltage gain A_v = β * (R_c/R_b) = 100 * (3000/2000) = 150. Power gain = A_v * β = 150 * 100 = 15000.
The increase in the width of the depletion region in a p-n junction diode is due to :
- A. both forward bias and reverse bias
- B. increase in forward current
- C. forward bias only
- D. reverse bias only
Answer: D. reverse bias only
In reverse bias, the depletion region widens because majority carriers are pulled away from the junction.
In reverse bias, the depletion region widens because majority carriers are pulled away from the junction.
The increase in the width of the depletion region in a p-n junction diode is due to:
- A. Both forward bias and reverse bias
- B. Increase in forward current
- C. Forward bias only
- D. Reverse bias only
Answer: D. Reverse bias only
When a p-n junction is reverse biased, the external voltage aids the built-in potential, pulling majority carriers away from the junction and widening the depletion region. Forward bias does the opposite — it narrows the depletion region. Hence, the increase in width of the depletion region is due to reverse bias only.
When a p-n junction is reverse biased, the external voltage aids the built-in potential, pulling majority carriers away from the junction and widening the depletion region. Forward bias does the opposite — it narrows the depletion region. Hence, the increase in width of the depletion region is due to reverse bias only.