JEE Physics Semiconductor Electronics MCQs Set 06

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Chapter-wise Objective Questions: Semiconductor Electronics

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Multiple Choice Questions

Question 1: Which is the correct statement about a diode?
(a) in a full wave rectifier both diodes work alternately
(b) in a full wave rectifier both diodes work simultaneously
(c) the efficiency of a full wave rectifier and a half wave rectifier is the same
(d) a full wave rectifier is bidirectional
Show Answer & Explanation

Answer: (a) in a full wave rectifier both diodes work alternately

Explanation:
1. In a centre-tap full wave rectifier, one diode conducts in each half cycle.
2. So the two diodes conduct alternately.

Teacher's Note:
1. Output current in the load is in the same direction for both half cycles.
2. Efficiency of full wave is about twice that of half wave.

Question 2: When a junction diode is reverse biased, the flow of current across the junction is mainly due to:
(a) diffusion of charges
(b) the nature of the material
(c) drift of charges
(d) both drift and diffusion of charges
Show Answer & Explanation

Answer: (c) drift of charges

Explanation:
1. The barrier is too high for diffusion.
2. Minority carriers drift across under the junction field.

Teacher's Note:
1. The reverse current is a drift current.
2. It is nearly independent of the reverse voltage.

Question 3: The width of the depletion region in a p-n junction diode:
(a) increases when reverse bias is applied
(b) increases when forward bias is applied
(c) decreases when reverse bias is applied
(d) remains the same irrespective of bias voltage
Show Answer & Explanation

Answer: (a) increases when reverse bias is applied

Explanation:
1. Reverse bias pulls carriers away from the junction.
2. More ions are uncovered, so the layer widens.

Teacher's Note:
1. Forward bias narrows it.
2. Width also depends on doping.

Question 4: A 15 V battery is connected to a 500 \( \Omega \) resistor in series with a parallel combination of a 1500 \( \Omega \) resistor and a Zener diode of \( V_Z = 10 \) V. The current through the Zener diode is:
(a) 10 mA
(b) 6.67 mA
(c) 5 mA
(d) 3.33 mA
Show Answer & Explanation

Answer: (d) 3.33 mA

Explanation:
1. Without the diode, the voltage across 1500 \( \Omega \) would be 11.25 V, above 10 V, so the Zener conducts and holds 10 V.
2. Current through 500 \( \Omega \) = 5/500 = 10 mA; through 1500 \( \Omega \) = 10/1500 = 6.67 mA.
3. Zener current = 10 − 6.67 = 3.33 mA.

Teacher's Note:
1. Check first whether the Zener is in breakdown.
2. Then use Kirchhoff's current law at the junction.

Question 5: If a full wave rectifier operates from 50 Hz mains, the fundamental frequency in the ripple will be:
(a) 25 Hz
(b) 50 Hz
(c) 70.7 Hz
(d) 100 Hz
Show Answer & Explanation

Answer: (d) 100 Hz

Explanation:
1. A full wave rectifier produces two output pulses per input cycle.
2. So the ripple frequency is 2 × 50 = 100 Hz.

Teacher's Note:
1. Half wave: ripple frequency equals input frequency.
2. Full wave: twice the input frequency.

Question 6: The electrical circuit used to get smooth dc output from a rectifier circuit is called:
(a) filter
(b) oscillator
(c) logic gate
(d) amplifier
Show Answer & Explanation

Answer: (a) filter

Explanation:
1. A capacitor or LC filter removes the ripple.
2. The output then becomes nearly steady dc.

Teacher's Note:
1. Rectifier gives pulsating dc.
2. Filter gives smooth dc.

Question 7: In a p-n junction photocell, the e.m.f. due to monochromatic light is proportional to:
(a) the p-n potential barrier
(b) the intensity of light
(c) the frequency of light
(d) the p-n applied voltage
Show Answer & Explanation

Answer: (b) the intensity of light

Explanation:
1. More photons create more electron-hole pairs.
2. The generated photocurrent and e.m.f. grow with the intensity of the light.

Teacher's Note:
1. Frequency must exceed a threshold set by the band gap.
2. Above the threshold, intensity decides the output.

Question 8: The efficiency of a half wave rectifier is nearly:
(a) 80 %
(b) 60 %
(c) 40 %
(d) 20 %
Show Answer & Explanation

Answer: (c) 40 %

Explanation:
1. The maximum rectification efficiency of a half wave rectifier is \( 4/\pi^2 \approx 40.6\% \).
2. A full wave rectifier is about 81%.

Teacher's Note:
1. Remember 40.6% and 81.2%.

Question 9: A Zener diode is used for:
(a) rectification
(b) stabilization
(c) amplification
(d) producing oscillations
Show Answer & Explanation

Answer: (b) stabilization

Explanation:
1. In reverse breakdown, the voltage across a Zener diode stays almost constant.
2. So it is used as a voltage stabiliser.

Teacher's Note:
1. Always connect it in reverse bias with a series resistor.

Question 10: In a transistor:
(a) the emitter has the least concentration of impurity
(b) the collector has the least concentration of impurity
(c) the base has the least concentration of impurity
(d) all three regions have equal concentration of impurity
Show Answer & Explanation

Answer: (c) the base has the least concentration of impurity

Explanation:
1. The emitter is heavily doped, the collector moderately doped and the base lightly doped.
2. A lightly doped, thin base lets most carriers pass to the collector.

Teacher's Note:
1. Emitter > collector > base in doping.
2. Collector is the largest in size.

Question 11: In transistor symbols, the arrow shows the direction of:
(a) current in the emitter
(b) electron current in the emitter
(c) hole current in the emitter
(d) electron current in the base
Show Answer & Explanation

Answer: (a) current in the emitter

Explanation:
1. The arrow on the emitter points in the direction of conventional current.
2. It points outwards for npn and inwards for pnp.

Teacher's Note:
1. The arrow shows conventional current, not electron flow.

Question 12: The region of a transistor in which extra impurity is doped to obtain a large number of majority carriers is called:
(a) emitter
(b) base
(c) collector
(d) any one of these depending on the transistor
Show Answer & Explanation

Answer: (a) emitter

Explanation:
1. The emitter is heavily doped so that it can supply many majority carriers.
2. The base and collector are doped less.

Teacher's Note:
1. Emitter emits the carriers.
2. Collector collects them.

Question 13: The input resistance of a common emitter transistor compared with its output resistance is:
(a) less
(b) more
(c) less and more
(d) none of these
Show Answer & Explanation

Answer: (a) less

Explanation:
1. The forward biased base-emitter junction has low resistance.
2. The reverse biased collector junction has high resistance.

Teacher's Note:
1. This difference gives the amplifying property.
2. Typical values: input about 1 k\( \Omega \), output about 50 k\( \Omega \).

Question 14: The current gain \( \beta \) of a transistor is 50. The input resistance in common emitter configuration is 1 k\( \Omega \). The peak value of the collector a.c. current for a peak input voltage of 0.01 V is:
(a) 100 \( \mu \)A
(b) 250 \( \mu \)A
(c) 500 \( \mu \)A
(d) 800 \( \mu \)A
Show Answer & Explanation

Answer: (c) 500 \( \mu \)A

Explanation:
1. Base current = 0.01/1000 = 10 \( \mu \)A.
2. Collector current = 50 × 10 = 500 \( \mu \)A.

Teacher's Note:
1. Find \( i_b \) from the input side.
2. Multiply by \( \beta \).

Question 15: A transistor in CE configuration at \( V_{CC} = 2 \) V: a change in base current from 100 \( \mu \)A to 200 \( \mu \)A produces a change in collector current from 9 mA to 16.5 mA. The current gain \( \beta \) is:
(a) 45
(b) 50
(c) 60
(d) 75
Show Answer & Explanation

Answer: (d) 75

Explanation:
1. \( \beta = \Delta I_C / \Delta I_B = 7.5 \text{ mA} / 100 \ \mu\text{A} = 75 \).

Teacher's Note:
1. Use the changes, not the absolute values.
2. Convert units carefully.

Question 16: For a transistor amplifier, the power gain is 7.5 and the voltage gain is 2.5. The current gain will be:
(a) 0.33
(b) 0.66
(c) 0.99
(d) 3
Show Answer & Explanation

Answer: (d) 3

Explanation:
1. Power gain = voltage gain × current gain.
2. Current gain = 7.5 / 2.5 = 3.

Teacher's Note:
1. Use \( A_P = A_V \times A_I \).

Question 17: In a common emitter npn transistor circuit, \( \beta = 100 \), \( V_{CC} = 15 \) V, the collector resistance is 2 k\( \Omega \), \( V_{CE} = 7 \) V and \( V_{BE} \) is negligible. The base current \( I_B \) is:
(a) 0.01 mA
(b) 0.04 mA
(c) 0.02 mA
(d) 0.03 mA
Show Answer & Explanation

Answer: (b) 0.04 mA

Explanation:
1. \( I_C = (15 - 7)/2\text{ k}\Omega = 4 \) mA.
2. \( I_B = I_C/\beta = 4/100 = 0.04 \) mA.

Teacher's Note:
1. Use Kirchhoff's voltage law in the output loop first.

Question 18: When a transistor is used in a circuit:
(a) both junctions are forward biased
(b) the emitter-base junction is forward biased and the base-collector junction is reverse biased
(c) the emitter-base junction is reverse biased and the base-collector junction is forward biased
(d) both junctions are reverse biased
Show Answer & Explanation

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

Explanation:
1. This is the active region used for amplification.
2. Carriers enter the base from the emitter and are collected by the collector.

Teacher's Note:
1. Both forward: saturation. Both reverse: cut-off.

Question 19: In an npn transistor circuit the collector current is 10 mA. If 90% of the electrons emitted reach the collector, the emitter current \( I_E \) and base current \( I_B \) are about:
(a) \( I_E = 1 \) mA, \( I_B = 11 \) mA
(b) \( I_E = 11 \) mA, \( I_B = 1 \) mA
(c) \( I_E = -1 \) mA, \( I_B = 9 \) mA
(d) \( I_E = 9 \) mA, \( I_B = -1 \) mA
Show Answer & Explanation

Answer: (b) \( I_E = 11 \) mA, \( I_B = 1 \) mA

Explanation:
1. \( I_C = 0.9 I_E \), so \( I_E = 10/0.9 \approx 11 \) mA.
2. \( I_B = I_E - I_C \approx 1 \) mA.

Teacher's Note:
1. Emitter current equals collector plus base current.

Question 20: In a transistor, the base is made very thin and lightly doped:
(a) to save the transistor from heating
(b) to enable the emitter to emit a small number of electrons and holes
(c) to enable the collector to collect most of the carriers coming from the emitter side
(d) none of the above
Show Answer & Explanation

Answer: (c) to enable the collector to collect most of the carriers coming from the emitter side

Explanation:
1. A thin, lightly doped base has little recombination.
2. So almost all carriers from the emitter reach the collector.

Teacher's Note:
1. This keeps \( \alpha \) close to 1.

Question 21: In CB configuration the a.c. current gain is \( \Delta i_C/\Delta i_E = 0.98 \). The current gain in CE configuration is:
(a) 49
(b) 98
(c) 4.9
(d) 24.5
Show Answer & Explanation

Answer: (a) 49

Explanation:
1. \( \beta = \alpha/(1-\alpha) = 0.98/0.02 = 49 \).

Teacher's Note:
1. Learn the relation between alpha and beta.

Question 22: In an npn transistor:
(a) holes move from emitter to base
(b) electrons move from emitter to base
(c) holes move from base to collector
(d) electrons move from collector to base
Show Answer & Explanation

Answer: (b) electrons move from emitter to base

Explanation:
1. The emitter injects electrons into the thin p-type base.
2. Most of them are swept into the collector.

Teacher's Note:
1. In pnp the holes play this part.

Question 23: In the study of a transistor as an amplifier, if \( \alpha = I_C/I_E \) and \( \beta = I_C/I_B \), then:
(a) \( \beta = \dfrac{\alpha}{1+\alpha} \)
(b) \( \beta = \dfrac{\alpha}{1-\alpha} \)
(c) \( \beta = \dfrac{1+\alpha}{\alpha} \)
(d) \( \beta = \dfrac{1-\alpha}{\alpha} \)
Show Answer & Explanation

Answer: (b) \( \beta = \dfrac{\alpha}{1-\alpha} \)

Explanation:
1. \( I_E = I_C + I_B \), so \( 1/\alpha = 1 + 1/\beta \).
2. This gives \( \beta = \alpha/(1-\alpha) \).

Teacher's Note:
1. Derive it from \( I_E = I_B + I_C \).

Question 24: In CB mode of a transistor, when the collector voltage is changed by 0.5 V, the collector current changes by 0.05 mA. The output resistance is:
(a) 10 k\( \Omega \)
(b) 20 k\( \Omega \)
(c) 5 k\( \Omega \)
(d) 2.5 k\( \Omega \)
Show Answer & Explanation

Answer: (a) 10 k\( \Omega \)

Explanation:
1. \( r_o = \Delta V_C/\Delta I_C = 0.5/(0.05 \times 10^{-3}) = 10 \) k\( \Omega \).

Teacher's Note:
1. Output resistance is the ratio of voltage change to current change.

Question 25: An npn transistor conducts when:
(a) both collector and emitter are positive with respect to the base
(b) the collector is positive and the emitter is negative with respect to the base
(c) the collector is positive and the emitter is at the same potential as the base
(d) both collector and emitter are negative with respect to the base
Show Answer & Explanation

Answer: (b) the collector is positive and the emitter is negative with respect to the base

Explanation:
1. The emitter-base junction must be forward biased: emitter negative relative to base.
2. The collector-base junction must be reverse biased: collector positive.

Teacher's Note:
1. For npn, base is the positive reference for the emitter.

Question 26: For the constants \( \alpha \) and \( \beta \) of a transistor:
(a) \( \alpha = \beta \)
(b) \( \beta < 1, \alpha > 1 \)
(c) \( \alpha\beta = 1 \)
(d) \( \beta > 1, \alpha < 1 \)
Show Answer & Explanation

Answer: (d) \( \beta > 1, \alpha < 1 \)

Explanation:
1. \( \alpha = I_C/I_E \) is slightly below 1.
2. \( \beta = \alpha/(1-\alpha) \) is much greater than 1.

Teacher's Note:
1. Typical values: alpha 0.98, beta 49.

Question 27: For a transistor in a common emitter arrangement, the alternating current gain \( \beta \) is given by:
(a) \( \left[\dfrac{\Delta I_C}{\Delta I_E}\right]_{V_{CE}} \)
(b) \( \left[\dfrac{\Delta I_B}{\Delta I_C}\right]_{V_{CE}} \)
(c) \( \left[\dfrac{\Delta I_C}{\Delta I_B}\right]_{V_{CE}} \)
(d) \( \left[\dfrac{\Delta I_E}{\Delta I_C}\right]_{V_{CE}} \)
Show Answer & Explanation

Answer: (c) \( \left[\dfrac{\Delta I_C}{\Delta I_B}\right]_{V_{CE}} \)

Explanation:
1. In CE mode, the input current is \( I_B \) and the output current is \( I_C \).
2. So \( \beta_{ac} = (\Delta I_C/\Delta I_B) \) at constant \( V_{CE} \).

Teacher's Note:
1. Option (a) is alpha for the CB mode.

Question 28:
Assertion (A): In an unbiased p-n junction, holes diffuse from the p-region to the n-region.
Reason (R): The hole concentration in the p-region is more compared to the n-region.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Diffusion occurs from high to low concentration.
2. The p region has many more holes, so they diffuse to the n region. R explains A.

Teacher's Note:
1. Diffusion is driven by a concentration gradient.

Question 29:
Assertion (A): We cannot measure the potential barrier of a p-n junction by putting a sensitive voltmeter across its terminals.
Reason (R): In the depletion layer there are no free electrons or holes and, in the absence of forward bias, it offers infinite resistance.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. The barrier is balanced by contact potentials at the metal junctions, so no emf appears.
2. The depletion layer has very high resistance.
3. Both are true and R explains A.

Teacher's Note:
1. A voltmeter draws a current.
2. The junction cannot supply it.

Question 30:
Assertion (A): For forward biasing, the positive terminal of the external battery is connected to the p-type and the negative terminal to the n-type semiconductor.
Reason (R): This supports the tendency of majority charge carriers to cross the junction.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. The external field opposes the barrier field.
2. This lowers the barrier and favours the majority carriers crossing.

Teacher's Note:
1. Forward bias reduces the barrier.

Question 31:
Assertion (A): Two p-n junction diodes placed back to back will work as an n-p-n transistor.
Reason (R): The p-region of two p-n junction diodes placed back to back forms the base of the n-p-n transistor.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation

Answer: (d) A is false, but R is true.

Explanation:
1. A thin base is needed for transistor action, which two separate diodes cannot provide, because the carriers recombine.
2. Both statements are therefore taken as false.

Teacher's Note:
1. Transistor action needs a thin, lightly doped base shared by two junctions.

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