Semiconductor &

Electronic devices
XII ( PHYSICS )
( Important Ques)

#4008, SEC 32 A , OPPOSITE GURU AMARDAS HOSPITAL, CHANDIGARH ROAD,LUDHIANA

1. In the given diagram, is the diode D forward or reverse biased? 1

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Ans : The diode D is reverse biased.

2. What happens to the width of depletion layer of a p–n junction when it is (i) forward biased, (ii) 1

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reverse biased?

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Ans : (i) The width of depletion layer decreases.
(ii) The width of depletion layer increases.
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3. Name the type of biasing of a p-n junction diode so that the junction offers very high resistance. 1

Ans : Reverse biasing.

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4. Name one impurity each, which when added to pure Si, produces (i) n-type, and (ii) p-type 1
semiconductor.

Ans : (i) for n-type, arsenic.

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(ii) for p-type, Indium.

5. Why is the conductivity of n-type semi-conductor greater than that of the p-type semi-conductor 1
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even when both of these have same level of doping?

Ans : In n-type semiconductor charge carriers are electrons and mobility of electrons is more
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than that of holes.

6. Name two factors on which electrical conductivity of a pure semiconductor at a given 1

temperature depends.
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Ans : (i) Band gap (ii) Biasing

7. What is a hole? What is its physical significance? 1

Ans : Hole is the vacancy of electron in valence band. The vacancy with the hole behaves as
an apparent free particle with effective positive charge.

8. Draw energy band diagram for an intrinsic semiconductor at T = 0 K. 1

Ans :

9. Draw energy band diagram for an intrinsic semiconductor at T > 0 K. 1

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Ans :

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10. Why are the elemental dopants mainly taken from 13th and 15th group, for doping Silicon or 1
Germanium?

Ans : The dopant has to be such that it does not distort the original pure semiconductor lattice.
So that the sizes of the dopant and the semiconductor atoms should be nearly the same.
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11. What is an ideal diode? 1

Ans : It is a p-n junction diode which offer zero resistance in forward biasing and infinite
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resistance in reverse biasing, i.e. current flows through it in one direction only.

12. What do you understand by a dynamic resistance of p-n junction diode. 1

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Ans : Dynamic resistance is the ratio of a small change in voltage ∆V to a small change in
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current

13. What happens when a forward bias is applied to a p–n junction? 1

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Ans : It conducts current.

14. Which one of the two diodes D1 and D2 in the given figures (i) forward biased, (ii) reverse biased 1
?
Ans : D1 is reverse biased and D2 is forward biased.

15. Distinguish between ‘intrinsic’ and ‘extrinsic’ semiconductors. 2

Ans : Intrinsic semiconductors

(i) There are no impurity atoms.
(ii) The number density of electrons is equal to the number density of holes, i.e. ne = nh.
Extrinsic semiconductors
(i) They are doped with impurity atoms (trivalent/pentavalent).
(ii) ne ≠ nh.

16. Carbon and silicon both have four valence electrons each. How then are they distinguished? 2

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Ans : Although both carbon and silicon have same lattice structure, the four bonding electrons

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of carbon and silicon are respectively in the second and third orbits. Thus, ionisation
energy is less for silicon than carbon. Hence, the number of free electrons in silicon is
more than of carbon. Thus, they can be distinguished on the bases of their conductivity.

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17. Draw energy band diagrams of an n-type and a p-type semiconductor at temperature T > 0 K. 2
Mark the donor and acceptor energy levels with their energies.

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Ans :
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C
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18. Distinguish between an intrinsic semiconductor and p-type semiconductor. Give reason, why a p- 2
type semiconductor crystal is electrically neutral, although nh >> ne?
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Ans :
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In the formation of p-type semiconductor, the number of negatively charged electrons is

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equal to positively charged holes. So the p-type semiconductor thus formed is electrically
neutral.

19. Draw the circuit diagrams showing how a p–n junction diode is 2
(i) forward biased and
(ii) reverse biased. How is the width of depletion layer affected in the two cases?
Ans : The following figures show the required circuit diagrams.

When the p–n junction is forward biased as in Figure (i), the width of the depletion layer
decreases. When an external voltage is applied, the barrier potential is reduced thereby,
decreasing the width of depletion layer. When the p–n junction is reverse biased as in
Figure (ii), the barrier potential increases, thereby, increasing the thickness of the

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depletion layer.

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20. Draw the circuit diagram of an illuminated photodiode in reverse bias. How is photodiode used to 2
measure light intensity ?

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Ans :

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21. Draw a circuit diagram showing the biasing of an LED. State the factor which controls (i) 2
wavelength of light, and (ii) intensity of light emitted by the diode.

Ans : The following circuit shows the biasing of an LED.

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(i) The wavelength of emitted light depends upon the nature of the material of diode/band
gap.
(ii) Biasing of LED.
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22. C, Si and Ge have same lattice structure. Why is C insulator, while Si and Ge are intrinsic 2
semiconductors?
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Ans : The 4 bonding electrons of C, Si or Ge lie, respectively, in the second, third and fourth
orbit. Hence, energy required to take out an electron from these atoms (i.e. ionisation
energy Eg) will be least for Ge, followed by Si and highest for C. Hence, the number of
free electrons for conduction in Ge and Si are significant but negligibly small for C.
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23. What is doping? Why is it needed? 2

Ans : (a) The deliberate addition of a desirable impurity is called doping.

(b) At room temperature the conductivity of intrinsic semiconductor is very low. Hence,
there is a necessity of improving their conductivity. This can be done by doping.

24. The current in the forward bias is known to be more (mA) than the current in the reverse bias 2
(~µA). What is the reason, then, to operate the photodiode in reverse bias ?
Ans : The fractional change in minority charge carriers is more than the fractional change in
majority charge carriers. This means change in reverse bias current is more prominent
than the change in forward bias current for a change in intensity of incident light.

25. Name the device D which is used as a voltage regulator in the given circuit and give its symbol. 2

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Ans : The device D in the circuit is a zener diode being used as voltage regulator. The symbol
of zener diode is shown below.

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26. Mention the important considerations required while fabricating a p–n junction diode to be used 2
as a light emitting diode (LED). What should be the order of band gap of an LED, if it is required to
emit light in the visible range?

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Ans : Diode is encapsulated with a transparent cover so that emitted light came out. We must
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have heavily doped p–n junction under forward biased. If an LED is required to emit light
in the visible range, semiconductor must have band gap of 1.8 eV.

27. (a) Explain with the help of a diagram, how depletion region and potential barrier are formed in a 3
junction diode.
(b) If a small voltage is applied to a p–n junction diode how will the barrier potential be affected
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when it is (i) forward biased, and (ii) reverse biased?

Ans : (a) As soon as a p-type semiconductar is joined with an n-type semiconductor, the
diffusion of free charges across the junction starts. The free charges move from the
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higher to lower concentration side.

Formation of potential barrier: Electrons are the majority charge carriers in n-type
semiconductor. They move towards p-type semiconductor
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leaving behind the positive charged ions.

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Similarly, holes being in majority in the p-type semiconductor, move towards the n-type
semiconductor. They leave behind the negatively charged ions. This way the
accumulation of charges takes place near the junction. This stops further diffusion of the
charges and the potential drop across the junction due to these fixed charges is called
potential barrier.
(b) (i) In forward bias, the barrier potential decreases.
(ii) In reverse bias, the barrier potential increases.

28. With what considerations in view, a photodiode is fabricated? State its working with the help of a 3
suitable diagram.
Even though the current in the forward bias is known to be more than in the reverse bias, yet the
photodiode works in reverse bias. What is the reason?
Ans : A photodiode is fabricated with a transparent window so that the light falling on it is able
to generate the electron-hole pairs.

When the photodiode is illuminated with light (photons) of energy (hv > Eg) greater than

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the energy gap of the semiconductor, it generates electronhole pairs in the junction
region. The junction electric field separates these charges before they recombine. The
electrons move towards n-side and the holes move towards p-side.

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The accumulation of charges develops an emf when joined in external circuit and the
current begins to flow.
The fractional change in the minority charge carriers is more than the fractional change
in the majority charge carriers, that is why, the change in current in reverse bias is more

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observable than the change in current in the forward bias.

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29. Draw V–I characteristics of a p–n junction diode. Answer the following questions, giving reasons: 3
(i) Why is the current under the reverse bias almost independent of the applied potential up to a
critical voltage?
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(ii) Why does the reverse current show a sudden increase at the critical voltage?
Name any semiconductor device which operates under the reverse bias in the breakdown region.

Ans :
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(i) The reverse current is not limited by the magnitude of applied voltage, but it is limited
due to the concentration of minority carriers on either side of the junction.
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(ii) At the critical voltage, the electric field across the junction is high enough to pull
valence electrons from the host atoms. As a result, a large number of electrons are
available for the conduction which shows a sudden increase of the reverse current at
breakdown voltage.

30. (a) Describe the working of light emitting diodes (LEDs). 3

(b) Which semiconductors are preferred to make LEDs and why?
(c) Give two advantages of using LEDs over conventional incandescent low power lamps.
Ans : (a) An LED is a forward biased p–n junction semiconducting material which emits a
visible light when forward biased, the electrons from the n-region cross the junction and
recombine with the holes in the p-region. During the recombination, their energy
difference is given out in the form of heat and light. The energy of radiation emitted is
less than or equal to Eg.
(b) Materials like gallium phosphide, GaAs, etc. are used. They emit the maximum
amount of energy in the form of light.
(c) The following are advantages of LEDs over conventional incandescent low power
lamps:
(i) Low operating voltage.
(ii) Low power consumption and quick action.

31. (a) Why is a photodiode operated in the reverse bias mode? 3

(b) For what purpose is a photodiode used?

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(c) Draw its I-V characteristics for different intensities of illumination.

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Ans : (a) The fractional change in the reverse bias current is more than that of the forward bias
current.
So, the change in reverse bias current with intensity of light is more prominent.

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(b) (i) As a detector: It is a fast photon detector having a switching time of the order of
nanoseconds.
(ii) In demodulators and in logic encoder.
(c) The I–V characteristics of a photodiode are as shown:

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32. Draw a block diagram of a full-wave rectifier with capacitor filter. Draw input and output (filtered) 3
voltage of rectifier.
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Ans :
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33. What is a filter? Explain the role of capacitor in filtering. 3

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Ans : To get steady dc output from the pulsating voltage normally a capacitor is connected
across the output terminals (parallel to the load RL). One can also use an inductor in
series with RL for the same purpose. Since these additional circuits appear to filter out
the ac ripple and give a pure dc voltage, so they are called filters.
When the voltage across the capacitor is rising, it gets charged. If there is no external
load, it remains charged to the peak voltage of the rectified output. When there is a load,
it gets discharged through the load and the voltage across it begins to fall. In the next
half-cycle of rectified output it again gets charged to the peak value. The rate of fall of the
voltage across the capacitor depends upon the inverse product of capacitor C and the
effective resistance RL used in the circuit and is called the time constant. To make the
time constant large value of C should be large. So capacitor input filters use large
capacitors. The output voltage obtained by using capacitor input filter is nearer to the
peak voltage of the rectified voltage. This type of filter is most widely used in power
supplies.

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34. With the help of energy band diagrams, distinguish between conductors, semiconductors and 3
insulators.

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Ans : Refer to Point no. 1 (b) (i), (ii) and (iii) [Important Terms, Definitions and Formulae].

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35. Write briefly the important processes that occur during the formation of p–n junction. With the 3
help of necessary diagrams, explain the term ‘barrier potential’.

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Ans : Two important processes involved during the formation of p-n junction are:
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(i) diffusion and (ii) drift.
As soon as p-type semiconductor comes in contact with n-type semiconductor due to the
different concentration gradient of charge carriers, the electrons start moving towards p-
side and the holes start moving from p-side to n-side. This process is called diffusion.
Due to diffusion, the positive space charge region is created on the n-side of the junction
and the negative space charge region is created on the p-side of the junction. This
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charge develops an electric field (junction field) from n-side to p-side. This field forces
the free charges to move. This process is called drift.
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The loss of electrons from n-side and gain of electrons by p-side cause a difference of
potential across the junction, which is called barrier potential. Its polarity is such that it
opposes the flow of majority charges through the junction.
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36. (a) How is a photodiode fabricated? 3

(b) Briefly explain its working. Draw its V – I characteristics for two different intensities of
illumination.
Ans : (a) A photodiode is a special purpose p–n junction diode fabricated with a transparent
window to allow light to reach the junction. Semiconductors used in the fabrication of
photodiode must have energy gap (Eg) lesser than the energy of photons (hn) used.
(b) Working: Three processes are involved in the working of a photodiode.

(i) Generation of electron-hole pairs

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(ii) Separation of electrons and holes
(iii) Collection of electrons and holes
Photons generate electron-hole pairs in the junction region of the diode. The electric field

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in the junction region separates these electrons and holes. The n–side collects electrons
and p-side collects holes giving rise to an emf. When a load is connected, the reverse
current begins to flow. For a slight variation in intensity of light, there is a significant

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variation in reverse current. So, it is used to detect optical signals.

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37. (a) State briefly the processes involved in the formation of p–n junction explaining clearly how 5
the depletion region is formed.
(b) Using the necessary circuit diagrams, show how the V-I characteristics of a p–n junction are
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obtained in
(i) Forward biasing (ii) Reverse biasing
How are these characteristics made use of in rectification?
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Ans :

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38. (a) Distinguish between an intrinsic semiconductor and a p-type semiconductor. Give 5
reason why a p-type semiconductor is electrically neutral, although nh >> ne.
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(b) Explain, how the heavy doping of both p- and n-sides of a p–n junction diode results in the
electric field of the junction being extremely high even with a reverse bias voltage of a few volts.
Explain, with the help of a circuit diagram, how this property is used in voltage regulator.
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Ans : (a) In case of an intrinsic semiconductor the number density of free electrons is equal to
number density of holes. There are no external impurity atoms in the intrinsic
semiconductors and the conductivity is low in comparision to that of doped (extrinsic
semiconductors). When the intrinsic semiconductor is doped with acceptor impurity
atoms, a p-type semiconductor is formed. In case of p-type semiconductor, nh >> ne.
Therefore, the conductivity of p-type semiconductor is higher than that of the intrinsic
semiconductor.
The number of negatively charged immobile ions is equal to the number of oppositively
charged holes. Therefore, p type semiconductor is electrically neutral.
(b) The heavy doping of p and n sides of a p–n junction results into a reduced width of
depletion region. As, for the same value of barrier potential, the width decreases and the
electric field across the junction increases.

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As, the electric field of the junction acquires very high value for a large variation of
current, the potential difference in breakdown region remains almost same.

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This property of zener diode is used in voltage regulation.
Zener diode as a voltage regulator—Refer to Ans. 57.

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39. The circuit shown in the figure has two oppositely connected ideal diodes connected in parallel. 4
Find the current flowing through each diode in the circuit.

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Ans : Since D1 is in reversed biased and offers infinite resistances so no current will flow
through the diode D1. Only diode D2 will conduct as it is in forward biased which offers
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zero resistance.
So, current through D1, i1 = 0 and current through D2,
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40. You are given three semiconductors: A, B and C with respective bandgaps of 3 eV, 2 eV and 1 4
eV for use in photodetector to detect λ = 1200 nm. Select the suitable semiconductor. Give
reasons.
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Ans :
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41. A photodetector is made of a semiconductor— In 0.53Ga0.47; as with Eg = 0.73 eV. What is the 4
maximum wavelength, which it can detect?
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42. The circuit shown in the figure contains two diodes each with a forward resistance of 50 Ω and 4
infinite backward resistance. Find the current through the 100 Ω resistance.

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Ans : In the given circuit, diode D1 is forward biased with forward resistance of 50 Ω and D2 is
reversed biased, which offers infinite backward resistance.

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Therefore, current through 100 Ω resistance = 0.02 A

43. The conductivity of a semiconductor increases with increase in temperature, because 1

(a) number density of free current carriers increases.
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(b) relaxation time increases.

(c) both number density of carriers and relaxation time increase.
(d) number density of carriers increases, relaxation time decreases but effect of decrease in
relaxation time is much less than increase in number density.
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Ans : (d)
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44. In the given figure V0 is the potential barrier across a p-n junction, when no battery is connected 1
across the junction.
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(a) 1 and 3 both correspond to forward bias of junction.

(b) 3 corresponds to forward bias of junction and 1 corresponds to reverse bias of junction.
(c) 1 corresponds to forward bias and 3 corresponds to reverse bias of junction.
(d) 3 and 1 both correspond to reverse bias of junction.

Ans : (b) Height of potential barrier is decreases when p-n junction is forward bias.

45. In figure given, assuming the diodes to be ideal 1

 (a) D1 is forward biased and D2 is reverse biased and hence current flows from A to B.
(b) D2 is forward biased and D1 is reverse biased and hence no current flows from B to A and vice
versa.
(c) D1 and D2 are both forward biased and hence current flows from A to B.
(d) D1 and D2 are both reverse biased and hence no current flows from A to B and vice versa.

Ans : (b) Explanation: 10 V is the lower voltage in the circuit. Now p side of the p-n junction

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diode D1 is connected to lower voltage and n-side of D1 to higher voltage, thus D1 is
reverse biased. In D2 p-side of p-n junction diode is at higher potential and n-side is at
lower potential, therefore D2 is forward biased. Hence current flows through junction from

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B to A.

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46. Silicon is a semiconductor. If a small amount of As is added to it, then its electrical 1
conductivity_______.

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Ans : increases
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47. When the electrical conductivity of a semiconductor is due to the breaking of its covalent bonds, 1
then the semiconductor is said to be ____________.

Ans : intrinsic
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48. Electrical conductivity of a semiconductor 1

(a) decreases with the rise in its temperature.
(b) increases with the rise in its temperature.
(c) does not change with the rise in its temperature.
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(d) first increases and then decreases with the rise in its temperature.

Ans : (b) With temperature rise, the conductivity of semiconductor increases.

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49. The forbidden energy band gap in conductors, semiconductors and insulators are EG1,EG2 and 1
EG3 respectively. The relation among them is
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(a) EG1 = EG2 = EG3

(b) EG1 < EG2 < EG3
(c) EG1 > EG2 > EG3
(d) EG1 < EG2 > EG3
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Ans : (b) In insulators, the forbidden energy gap is very large, in case of semiconductor it is
moderate and in conductors the energy gap is zero.

50. In an insulator, the forbidden energy gap between the valence band and the conduction band is 1
of the order of ____________.

Ans : 5 eV

51. A n-type semiconductor is 1

(a) negatively charged. (b) positively charged.
(c) neutral. (d) none of these
Ans : (c) n-type semiconductors are neutral because neutral atoms are added during doping.

52. In the half-wave rectifier circuit shown. Which one of the following waveforms is true for VCD the 1
output across C and D?

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Ans : (b) Half wave rectifier rectifies only the half cycle of the input ac signal and it blocks the
other half.
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53. A full-wave rectifier circuit along with the input and output voltages is shown in the figure The 1
contribution to output voltage from diode 2 is
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(a) A, C (b) B, D
(c) B, C (d) A, D

Ans : (b) In the positive half cycle of input ac signal diode D1 is forward biased and D2 is
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reverse biased so in the output voltage signal, A and C are due to D1. In negative half
cycle of input ac signal, D2 conducts, hence output signals B and D are due to D2

54. A 220 V AC supply is connected between points A and B (figure). What will be the potential 1
difference V across the capacitor?
 (a) 200 V (b) 110 V
(c) 0 V (d) 220 V

Ans : (d) As p-n junction diode will conduct during positive half cycle only and during negative
half cycle diode is reverse biased. During this diode will not give any output. So potential
difference across the capacitor C = peak voltage of the given AC voltage.

55. In the circuit shown in figure below, if the diode forward voltage drop is 0.3 V, the voltage 1
difference between A and B is

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(a) 1.3 V (b) 2.3 V
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(c) 0 (d) 0.5 V

Ans : (b) Suppose the potential difference between A and B is VAB.

Then, VAB – 0.3 = [(r1 + r2)103] × (0.2 × 10–3) [∵ VAB = ir]
= [(5 + 5)103] × (0.2 × 10–3) = 10 × 103 × 0.2 × 10–3 = 2
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⇒ VAB = 2 + 0.3 = 2.3 V

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56. When an electric field is applied across a semiconductor [NCERT Exemplar] 1

(a) holes move from lower energy level to higher energy level in the conduction band.
(b) electrons move from higher energy level to lower energy level in the conduction band.
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(c) holes in the valence band move from higher energy level to lower energy level.
(d) holes in the valence band move from lower energy level to higher energy level.
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Ans : (c) When electric field is applied across a semiconductor, the electrons in the conduction
band move from lower energy level to higher energy level. While the holes in valence
band move from higher energy level to lower energy level, where they will be having
more energy.
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57. In the depletion region of a diode 1

(a) there are mobile charges.
(b) equal number of holes and elections exist, making the region neutral.
(c) recombination of holes and electrons has taken place.
(d) immobile charged ions do not exist.

Ans : (b)

58. The breakdown in a reverse biased p-n junction is more likely to occur due to 1
(a) large velocity of the majority charge carriers if the doping concentration is small.
(b) large velocity of the minority charge carriers if the doping concentration is large.
(c) strong electric field in a depletion region if the doping concentration is small.
(d) strong electric field in the depletion region if the doping concentration is large.
Ans : (d)

59. A p-type semiconductor can be obtained by adding 1

(a) arsenic to pure silicon.
(b) gallium to pure silicon.
(c) antimony to pure germanium.
(d) phosphorus to pure germanium.

Ans : (b) Ga has a valency of 3.

60. In a p-type semiconductor, there is 1

(a) excess of one electron.

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(b) absence of one electron.
(c) a missing atom.
(d) a donar level.

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Ans : (b) Absence of one electron, creates the positive charge of magnitude equal to that of
electronic charge.

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61. A 2V battery is connected across the points A and B as shown in the figure. Assuming that the 1

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resistance of each diode is zero in forward bias and infinity in reverse bias, the current supplied by
the battery when its positive terminal is connected to A is
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(a) 0.2 A (b) 0.4 A

(c) Zero (d) 0.1 A
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Ans : (a) Since diode in upper branch is forward biased and in lower branch is reverse biased.
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So current through circuit i = here rd is diode resistance in forward biasing = 0

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62. Current in the circuit will be 1

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(a) 5/40 A (b) 5/50 A

(c) 5/10 A (d) 5/20 A
Ans : (b) The diode in lower branch is forward biased and diode in upper branch is reverse
biased

63. The ________ biasing in a p-n junction diode increases the potential barrier. 1

Ans : reverse

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