JEE Physics Semiconductor Electronics MCQs Set 20

Answer: (a) slightly more than the collector current

Explanation:
1. \( I_E=I_B+I_C \) and the base current is small, so \( I_E \) is slightly larger than \( I_C \).

Teacher's Note:
1. The ratio \( I_C/I_E=\alpha \) is slightly less than 1.
2. Check units and orders of magnitude before choosing an option.

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

Explanation:
1. Proper bias puts the transistor in the active region where small input changes control the large collector current.

Teacher's Note:
1. Both forward gives saturation; both reverse gives cut-off.
2. Questions of this kind are asked directly in JEE Main, so keep the key fact at your fingertips.

Answer: (c) electrons move from the emitter to the base

Explanation:
1. The forward biased emitter-base junction injects electrons from the n-type emitter into the p-type base; most are then swept to the collector.

Teacher's Note:
1. Holes in the base are few and carry only a small base current.
2. Eliminate the options that contradict a basic property of the device, then check the remaining one.

Answer: (b) 250 mA

Explanation:
1. \( I_B=\frac{5}{1000}=5\ \text{mA} \)
2. \( I_C=\beta I_B=50\times5=250\ \text{mA} \)

Teacher's Note:
1. Find the base current from the input first, then multiply by the current gain.
2. Link the answer to the physical picture of carriers and junctions rather than memorising it as a bare fact.

Answer: (a) The common-emitter configuration is the most commonly used because it gives the largest current gain

Explanation:
1. The common-emitter current gain \( \beta=\alpha/(1-\alpha) \) is large (tens to hundreds) whereas the common-base gain is below 1.

Teacher's Note:
1. It also gives both voltage and current gain.
2. Check units and orders of magnitude before choosing an option.

Answer: (d) \( \beta=\frac{\alpha}{1-\alpha} \)

Explanation:
1. \( \alpha=\frac{I_C}{I_E},\ \beta=\frac{I_C}{I_B},\ I_E=I_B+I_C \)
2. \( \beta=\frac{I_C}{I_E-I_C}=\frac{\alpha}{1-\alpha} \).

Teacher's Note:
1. Also \( \alpha=\frac{\beta}{1+\beta} \).
2. Compare with the corresponding statement for the other kind of semiconductor or junction to avoid mixing them up.

Answer: (c) 9

Explanation:
1. \( \beta=\frac{\alpha}{1-\alpha}=\frac{0.9}{0.1}=9 \)

Teacher's Note:
1. Check that \( \alpha\lt 1 \) and \( \beta\gt 1 \).
2. Questions of this kind are asked directly in JEE Main, so keep the key fact at your fingertips.

Answer: (b) 0.98 and 49

Explanation:
1. Emitter current \( I_E=\frac{10^{10}\times1.6\times10^{-19}}{10^{-6}}=1.6\ \text{mA} \).
2. Base current is 2% of \( I_E \), so \( I_C=0.98I_E \) and \( \alpha=0.98 \).
3. \( \beta=\frac{\alpha}{1-\alpha}=\frac{0.98}{0.02}=49 \)

Teacher's Note:
1. The actual emitter current is not needed; only the fraction lost in the base matters.
2. Link the answer to the physical picture of carriers and junctions rather than memorising it as a bare fact.

Answer: (a) 1.0 V

Explanation:
1. \( A_V=\beta\frac{R_L}{R_{in}}=100\times\frac{10}{1}=1000 \)
2. \( V_{out}=1000\times1\ \text{mV}=1\ \text{V} \)

Teacher's Note:
1. Voltage gain is current gain times the ratio of load to input resistance.
2. A quick sketch of the band diagram or the circuit makes this type of question easy.

Answer: (d) 7.2 V, 4 × 10⁻⁵ A, voltage gain 100, power gain 2500

Explanation:
1. \( V_{CE}=8-0.8=7.2\ \text{V} \); \( I_C=\frac{0.8}{800}=1\ \text{mA} \).
2. \( \beta=\frac{\alpha}{1-\alpha}=25 \), so \( I_B=\frac{1\ \text{mA}}{25}=4\times10^{-5}\ \text{A} \).
3. \( A_V=\beta\frac{R_L}{R_{in}}=25\times\frac{800}{200}=100 \) and \( A_P=\beta A_V=2500 \).

Teacher's Note:
1. Convert alpha to beta first.
2. Check units and orders of magnitude before choosing an option.

Answer: (c) 50 kΩ

Explanation:
1. \( r_o=\frac{\Delta V_{CE}}{\Delta I_C}=\frac{0.2}{4\times10^{-6}}=5\times10^{4}\ \Omega \)

Teacher's Note:
1. Convert mA to A: \( 4\times10^{-3}\ \text{mA}=4\times10^{-6}\ \text{A} \).
2. Questions of this kind are asked directly in JEE Main, so keep the key fact at your fingertips.

Answer: (b) There is no voltage gain in an oscillator

Explanation:
1. An oscillator is an amplifier with positive feedback with loop gain equal to 1; it needs voltage gain to sustain oscillations, so the statement that there is no voltage gain is false.

Teacher's Note:
1. Without an amplifying element the oscillations would die out.
2. Eliminate the options that contradict a basic property of the device, then check the remaining one.

Answer: (a) (iv), (i), (iii)

Explanation:
1. A circle on the output means inversion. AND with a circle is NAND, so NAND is (iv).
2. OR with a circle is NOR, which is (i).
3. The plain OR symbol is (iii).

Teacher's Note:
1. A bubble always means NOT.
2. Link the answer to the physical picture of carriers and junctions rather than memorising it as a bare fact.

Answer: (d) AND

Explanation:
1. The output is 1 only when both inputs are 1, which is the AND operation.

Teacher's Note:
1. NAND would give the complement.
2. Check units and orders of magnitude before choosing an option.

Answer: (c) a universal gate

Explanation:
1. Any other gate (NOT, AND, OR) can be built from NAND gates alone, so it is called universal. (NOR is also universal.)

Teacher's Note:
1. The three basic gates are AND, OR and NOT.
2. Compare with the corresponding statement for the other kind of semiconductor or junction to avoid mixing them up.

Answer: (b) NAND

Explanation:
1. The output is 0 only when both inputs are 1; this is the NAND gate.

Teacher's Note:
1. XOR gives 0 also for (0,0).
2. Questions of this kind are asked directly in JEE Main, so keep the key fact at your fingertips.

Answer: (a) \( C=\overline{A+B} \)

Explanation:
1. NOR is OR followed by NOT.

Teacher's Note:
1. NAND is \( \overline{A\cdot B} \).
2. Link the answer to the physical picture of carriers and junctions rather than memorising it as a bare fact.

Answer: (d) A = 0, B = 0

Explanation:
1. For the product to be 1 both factors must be 1. \( \overline{A+B}=1 \) needs A = B = 0, and then \( \overline{AB}=1 \) holds as well.

Teacher's Note:
1. Only the input (0, 0) satisfies both.
2. A quick sketch of the band diagram or the circuit makes this type of question easy.

Answer: (c) NOR gate

Explanation:
1. A NAND gate with its inputs tied together is a NOT gate, so the combination gives \( Y=\overline{A+B} \).

Teacher's Note:
1. That is the NOR function.
2. Check units and orders of magnitude before choosing an option.

Answer: (b) \( 10^{-2}\ \text{A} \)

Explanation:
1. The P side is at higher potential, so the diode is forward biased and has zero drop.
2. \( I=\frac{4-1}{300}=10^{-2}\ \text{A} \).

Teacher's Note:
1. An ideal forward diode behaves like a wire.
2. Questions of this kind are asked directly in JEE Main, so keep the key fact at your fingertips.

Answer: (a) zero

Explanation:
1. The N side is at a higher potential (2 V) than the P side (1 V), so the diode is reverse biased.

Teacher's Note:
1. An ideal reverse-biased diode carries no current.
2. Eliminate the options that contradict a basic property of the device, then check the remaining one.

Answer: (d) 10 Ω

Explanation:
1. By symmetry C and D are at the same potential, so no current flows through the diode, and it can be removed.
2. The circuit is two branches of 20 Ω in parallel: \( \frac{20\times20}{40}=10\ \Omega \).

Teacher's Note:
1. A balanced bridge needs no current in the middle branch.
2. Link the answer to the physical picture of carriers and junctions rather than memorising it as a bare fact.