Class 12 Physics Semiconductor Electronics (MCQ)

Q.1. In a semiconductor
(a) there are no free electrons at 0 K
(b) there are no free electrons at any temperature
(c) the number of free electrons increases with pressure
(d) the number of free electrons is more than that in a conductor

Answer Answer: (a)

Q.2. Let nh and ne be the number of holes and conduction electrons in an extrinsic semiconductor. Then
(a) nh > ne (b) nh = ne (c) nh < ne (d) nh ≠ ne

Answer Answer: (d) In extrinsic semi conductor the number of holes are not equal to number of electrons

Q.3. A p-type semiconductor is
(a) positively charged
(b) negatively charged
(c) uncharged
(d) uncharged at 0K but charged at higher temperatures

Answer Answer: (c) By doping, the band gap reduce from 1eV to 0.3 to 0.7 eV & electron can achieve this energy (0.3eV to 0.7eV) at room temperature & reach in C.B (conduction band).

Q.4. Electric conduction in a semiconductor takes place due to
(a) electrons only
(b) holes only
(c) both electrons and holes
(d) neither electrons nor holes

Answer Answer: (c) Electric conduction, in a semi conductors occurs due to both electrons & holes.

Q.5. The impurity atoms with which pure silicon may be doped to make it a p-type semiconductor are those of
(a) phosphorus (b) boron (c) antimony (d) nitrogen

Answer Answer: (b)

Q.6. The electrical conductivity of pure germanium can be increased by
(a) increasing the temperature
(b) doping acceptor impurities
(c) doping donor impurities
(d) All of the above

Answer Answer: (d)

Q.7. The resistivity of a semiconductor at room temperature is in between
(a) 10–2 to 10–5 Ω cm (b) 10–3 to 106 Ω cm (c) 106 to 108 Ω cm (d) 1010 to 1012 Ω cm

Answer Answer: (b) Resistivity of a semiconductor at room temp. is in between 10–5 Ωm to 104 Ωm i.e. 10–3 to 106 Ωcm

Q.8. Number of electrons in the valence shell of a pure semiconductor is
(a) 1 (b) 2 (c) 3 (d) 4

Answer Answer: (d) The valency of semiconductor (Ge or Si) is four, hence it has 4 valence electrons in the outermost orbit of the Ge or Si-atom

Q.9. In a semiconductor, the forbidden energy gap between the valence band and the conduction band is of the order is
(a) 1 MeV (b) 0.1 Mev (c) 1 eV (d) 5 eV

Answer Answer: (c)

Q.10. The forbidden energy gap for germanium crystal at 0 K is
(a) 0.071 eV (b) 0.71 eV (c) 2.57 eV (d) 6.57 eV

Answer Answer: (b) The forbidden energy gap for germanium crystal is 0.71 eV.

Q.11. In an insulator, the forbidden energy gap between the valence band and conduction band is of the order of
(a) 1 MeV (b) 0.1 MeV (c) 1 eV (d) 5 eV

Answer Answer: (d)

Q.12. What is the resistivity of a pure semiconductor at absolute zero ?
(a) Zero
(b) Infinity
(c) Same as that of conductors at room temperature
(d) Same as that of insulators at room temperature

Answer Answer: (b) The electrical conductivity of a semiconductor at 0 K is zero. Hence resistivity (= 1/electrical conductivity) is infinity.

Q.13. Temperature coefficient of resistance of semiconductor is
(a) zero (b) constant (c) positive (d) negative

Answer Answer: (d) The temperature coefficient of resistance of a semiconductor is negative. It means that resistance decrease with increase of temperature.

Q.14. In a p-type semiconductor, the acceptor valence band is
(a) close to the valence band of the host crystal
(b) close to conduction band of the host crystal
(c) below the conduction band of the host crystal
(d) above the conduction band of the host crystal

Answer Answer: (a) The acceptor valence band is close to the valence band of host crystal

Q.15. In an n-type semiconductor, donor valence band is
(a) above the conduction band of the host crystal
(b) close to the valence band of the host crystal
(c) close to the conduction band of the host crystal
(d) below the valence band of the host crystal

Answer Answer: (c) The donor valence band lies little below the conduction band of the host crystal

Q.16. The mobility of free electrons is greater than that of free holes because
(a) they are light
(b) they carry negative charge
(c) they mutually collide less
(d) they require low energy to continue their motion

Answer Answer: (a)

Q.17. The relation between number of free electrons (n) in a
semiconductor and temperature (T) is given by
(a) n α T (b) n α T2 (c) n α √T (d) n α T3/2

Answer Answer: (d)

Q.18. In semiconductors, at room temperature
(a) the conduction band is completely empty
(b) the valence band is partially empty and the conduction band is partially filled
(c) the valence band is completely filled and the conduction band is partially filled
(d) the valence band is completely filled

Answer Answer: (c)

Q.19. At absolute zero, Si acts as
(a) non-metal (b) metal (c) insulator (d) None of these

Answer Answer: (c) Semiconductors are insulators at low temperature

Q.20. One serious drawback of semi-conductor devices is
(a) they do not last for long time.
(b) they are costly
(c) they cannot be used with high voltage.
(d) they pollute the environment.

Answer Answer: (c)

Q.21. When an impurity is doped into an intrinsic semiconductor,
the conductivity of the semiconductor
(a) increases (b) decreases
(c) remains the same (d) becomes zero

Answer Answer: (a)

Q.22. An electric field is applied to a semiconductor. Let the
number of charge carriers be n and the average drift speed
be v. If the temperature is increased
(a) both n and v will increase
(b) n will increase but v will decrease
(c) v will increase but n will decrease
(d) both n and v will decrease

Answer Answer: (a)

Q.23. If a small amount of antimony is added to germanium crystal
(a) it becomes a p–type semiconductor
(b) the antimony becomes an acceptor atom
(c) there will be more free electrons than holes in the semiconductor
(d) its resistance is increased

Answer Answer: (c) When small amount of antimony (pentavalent) is added to germanium crystal then crystal becomes n-type semi conductor. Therefore, there will be more free electrons than holes in the semiconductor.

Q.24. By increasing the temperature, the specific resistance of a
conductor and a semiconductor
(a) increases for both (b) decreases for both
(c) increases, decreases (d) decreases, increases

Answer Answer: (c) The resistivity of conductor increases with increase in temperature. The resistivity of semiconductor decreases as the temperature increases.

Q.25. A strip of copper and another of germanium are cooled from
room temperature to 80K. The resistance of
(a) each of these decreases
(b) copper strip increases and that of germanium decreases
(c) copper strip decreases and that of germanium increases
(d) each of these increases

Answer Answer: (c) Copper is a conductor so its resistance decreases on decreasing temperature as thermal agitation decreases whereas germanium is semiconductor therefore on decreasing temperature resistance increases.

Q.26. Carbon, Silicon and Germanium atoms have four valence electrons each. Their valence and conduction bands are separated by energy band gaps represented by (Eg)C, (Eg)Si and (Eg)Ge respectively. Which one of the following relationship is true in their case?
(a) (Eg)C > (Eg)Si (b) (Eg)C < (Eg)Si (c) (Eg)C = (Eg)Si (d) (Eg)C < (Eg)Ge

Answer Answer: (a) Due to strong electronegativity of carbon.

Q.27. A semiconductor device is connected in a series circuit with a battery and a resistance. A current is found to pass through the circuit. If the polarity of the battery is reversed, the current drops to almost zero. The device may be a/an
(a) intrinsic semiconductor
(b) p-type semiconductor
(c) n-type semiconductor
(d) p-n junction diode

Answer Answer: (d)

Q.28. If the two ends of a p-n junction are joined by a wire
(a) there will not be a steady current in the circuit
(b) there will be a steady current from the n-side to the p side
(c) there will be a steady current from the p-side to the n side
(d) there may or may not be a current depending upon the resistance of the connecting wire

Answer Answer: (a)

Q.29. The drift current in a p-n junction is from the
(a) n-side to the p-side
(b) p-side to the n-side
(c) n-side to the p-side if the junction is forward-biased
and in the opposite direction if it is reverse biased
(d) p-side to the n-side if the junction is forward-biased
and in the opposite direction if it is reverse-biased

Answer Answer: (a)

Q.30. The diffusion current in a p-n junction is from the
(a) n-side to the p-side
(b) p-side to the n-side
(c) n-side to the p-side if the junction is forward-biased
and in the opposite direction if it is reverse-biased
(d) p-side to the n-side if the junction is forward-biased
and in the opposite direction if it is reverse-biased

Answer Answer: (b)

Q.31. Diffusion current in a p-n junction is greater than the drift
current in magnitude
(a) if the junction is forward-biased
(b) if the junction is reverse-biased
(c) if the junction is unbiased
(d) in no case

Answer Answer: (a)

Q.32. Forward biasing is that in which applied voltage
(a) increases potential barrier
(b) cancels the potential barrier
(c) is equal to 1.5 volt
(d) None of these

Answer Answer: (b) Forward bias opposes the potential barrier and if the applied voltage is more than knee voltage it cancels the potential barrier.

Q.33. In V-I characteristic of a p-n junction, reverse biasing results in
(a) leakage current
(b) the current barrier across junction increases
(c) no flow of current
(d) large current

Answer Answer: (a) Leakage current is the name given to the reverse current.

Q.34. In reverse biasing
(a) large amount of current flows
(b) potential barrier across junction increases
(c) depletion layer resistance increases
(d) no current flows

Answer Answer: (b) In the reverse biasing of p-n junction, the voltage applied supports the barrier voltage across junction, which increases the width of depletion layer and hence increases its resistance.

Q.35. Zener diode is used for
(a) amplification (b) rectification (c) stabilisation (d) all of the above

Answer Answer: (c) Zener diode is used as a voltage regulator i.e. for stabilization purposes.

Q.36. Filter circuit
(a) eliminates a.c. component
(b) eliminates d.c. component
(c) does not eliminate a.c. component
(d) None of these

Answer Answer: (a) filter circuit eliminates a.c. component of rectified voltage obtained from p-n junction as a rectifier.

Q.37. For a junction diode the ratio of forward current (If) and
reverse current (Ir) is
[e = electronic charge,
V = voltage applied across junction,
k = Boltzmann constant,
T = temperature in kelvin]
(a) e–V/kT (b) eV/kT
(c) (e–eV/kT + 1) (d) (eeV/kT – 1)

Answer Answer: (d)

Q.38. In a semiconductor diode, the barrier potential offers
opposition to
(a) holes in P-region only
(b) free electrons in N-region only
(c) majority carriers in both regions
(d) majority as well as minority carriers in both regions

Answer Answer: (c)

Q.39. In a P -N junction
(a) the potential of P & N sides becomes higher alternately
(b) the P side is at higher electrical potential than N side.
(c) the N side is at higher electric potential than P side.
(d) both P & N sides are at same potential.

Answer Answer: (b)

Q.40. Barrier potential of a P-N junction diode does not depend on
(a) doping density (b) diode design
(c) temperature (d) forward bias

Answer Answer: (b) Barrier potential depends on, doping density, temperature, forward/reverse bias but does not depend on diode design.

Q.41. Reverse bias applied to a junction diode
(a) increases the minority carrier current
(b) lowers the potential barrier
(c) raises the potential barrier
(d) increases the majority carrier current

Answer Answer: (c) In reverse biasing, the conduction across the p-n junction does not take place due to majority carriers, but takes place due to minority carriers if the voltage of external battery is large. The size of the depletion region increases thereby increasing the potential barrier.

Q.42. In forward biasing of the p–n junction
(a) the positive terminal of the battery is connected to
p–side and the depletion region becomes thick
(b) the positive terminal of the battery is connected to
n–side and the depletion region becomes thin
(c) the positive terminal of the battery is connected to
n–side and the depletion region becomes thick
(d) the positive terminal of the battery is connected to
p–side and the depletion region becomes thin

Answer Answer: (d) In forward biasing of the p-n junction, the positive terminal of the battery is connected to p-side and the negative terminal of the battery is connected to n-side. The depletion region becomes thin.

Q.43. When p-n junction diode is forward biased then
(a) both the depletion region and barrier height are reduced
(b) the depletion region is widened and barrier height is
reduced
(c) the depletion region is reduced and barrier height is
increased
(d) Both the depletion region and barrier height are increased

Answer Answer: (a) Both the depletion region and barrier height are reduced.

Q.44. The cause of the potential barrier in a p-n junction diode is
(a) depletion of positive charges near the junction
(b) concentration of positive charges near the junction
(c) depletion of negative charges near the junction
(d) concentration of positive and negative charges near
the junction

Answer Answer: (d) During the formation of a junction diode, holes from pregion diffuse into n-region and electrons from n-region diffuse into p-region. In both cases, when an electrons meets a hole, they cancel the effect at each other and as a result, a thin layer at the junction becomes free from any of charges carriers. This is called depletion layer. There is a potential gradient in the depletion layer, negative on the p-side, and positive on the n-side. The potential difference thus developed across the junction is called potential barrier.

Q.45. The ratio of forward biased to reverse biased resistance
for p-n junction diode is
(a) 10–1 : 1 (b) 10–2 : 1 (c) 104 : 1 (d) 10–4 : 1

Answer Answer: (d)

Q.46. In the middle of the depletion layer of a reverse- biased p-n junction, the
(a) electric field is zero (b) potential is maximum
(c) electric field is maximum(d) potential is zero

Answer Answer: (c)

Q.47. Bridge type rectifier uses
(a) four diodes (b) six diodes (c) two diodes (d) one diode

Answer Answer: (a)

Q.48. The average value of output direct current in a half wave rectifier is
(a) I<sub>0</sub>/π (b) I<sub>0</sub>/2
(c) π I<sub>0</sub>/2 (d) 2 I<sub>0</sub>/π

Answer Answer: (a) The average value of output direct current in a half wave rectifier is = (average value of current over a cycle)/2 =2I0/2π = I0

Q.49. The average value of output direct current in a full wave rectifier is
(a) I<sub>0</sub>/π (b) I<sub>0</sub>/2
(c) π I<sub>0</sub>/2 (d) 2 I<sub>0</sub>/π

Answer Answer: (d) The average value of output direct current in a full wave rectifier = average value of current over a cycle = 2I0

Q.50. In a half wave rectifier, the r.m.s. value of the a.c. component of the wave is
(a) equal to d.c. value (b) more than d.c. value
(c) less than d.c. value (d) zero

Answer Answer: (b) The r.m.s. value of a.c. component of wave is more than d.c. value due to barrier voltage of p-n junction used as rectifier.

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