WORKSHEET _ Semiconductor

1- MCQs
1. A p- type semiconductor can be obtained by adding:
(a) Arsenic to pure silicon (b) gallium to pure Si (c) Antimony to pure Ge (d)
Phosphorus to pure Ge.

2. The depletion layer opposes the flow of:

(a) majority charge careers (b) minority charge carriers
(c) both minority and majority charge carriers (d) neither minority nor
majority charge careers

3. The cause of the potential barrier in a p and 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 charge and negative charges near the junction

4. When voltage is applied across a semiconductor, holes will:

(a) flow towards negative terminal (b) flow in external circuit
(c) flow away from negative terminal (d) not flow

5. Doping of silicon with arsenic leads to:

(a) A conductor (b) an insulator (c) A p-type semiconductor
(d) an n-type semiconductor

6. The cause of the barrier layer in PN junction is :

(a) doping (b) recombination (c) barrier (d) ions

7. The formation of depletion region in a PN junction diode is due to :

(a) movement of dopant atoms (b) diffusion of both electrons and holes
(c) drift of electrons only (d) drift of holes only
8. When a voltmeter is connected across a forward biased diode, it will read a
voltage approximately equal to:
(a) bias battery voltage (b) output voltage
(c) diode barrier potential (d) none of these

9. In an extrinsic semiconductor, the number density of holes is4×10²⁰ m-³. If

the number density of intrinsic cariers is 1.2× 10¹⁵ m-³, the number density of
electrons in it is :
(a) 1.8×10⁹ m-³ (b) 2.4×10¹⁰ m-³ (c) 3.6×10⁹ m-³ (d) 3.2×10¹⁰ m-³

10. The current in the external wire for P type semiconductor is due to:
(a) free electrons (b) holes (c) positive ions (d) negative ions

11. The Forbidden energy gaping conductors, semiconductors and insulators

are EG1,EG2 and EG3 respectively. The relation among them is:
(a) EG1=EG2=EG3 (b) EG1<EG2<EG3 (c) EG1>EG2>EG3 (d) EG1<EG2>EG3

12. The conductivity of semiconductor increases with temperature, because:

(a) number density of free current cariers increases (b) relaxation time
increases
(c) both number density of cariers and relaxation time increases
(d) number density of carriers increases, relaxation time decreases but effect
of decrease is much less than the increase in number density.

13. When an electric field is applied across a semiconductor :

(a) holes move from lower energy level to higher energy level in conduction
band
(b) electrons move from higher energy level to lower energy level in
conduction band (c) holes in valance band more from higher to lower energy
level
(d) holes in valance band move from lower to higher energy level

14. Which of the following is responsible for electric conductivity in a

semiconductor:
(a) only electrons(b) holes only (c)both holes and electrons (d) none of these

15. In the depletion reason of a diode:

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

2. ASSERTION & REASON

For the following questions two statements are given- one labelled Assertion
(A) and the other labelled Reason (R). Select the correct answer to these
questions from the codes (a), (b), ( c) and (d) as given below.
(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 and R is also false.
1. Assertion: conductivity of a semiconductor increases on doping with
pentavalent atoms.
Reason: pentavalent atoms can easily donate electrons due to their less
ionization energy.
2. Assertion: conductivity of an n-type semiconductor is greater than that of
p-type semiconductor.
Reason: electrons have greater mobility than holes.
3. Assertion: in a half wave rectifier if diode is short circuited the output from
the rectifier will be identical to the wave of the input primary voltage.
Reason: since the diode is shorted it act as a piece of wire.
4. Assertion: full wave rectifier preferred to half wave rectifier.
Reason: the output of a full wave rectifier is twice as great as the half wave
rectifier.
5. Assertion: the depletion region of a p-n junction is devoid of free electrons
and holes.
Reason: we can take a slab of p-type semiconductor and join it physically to
n- type and form a p-n junction.
6. Assertion: n-type material is negatively charged.
Reason: n-type material has excess of electrons which are negatively
charged.
7. Assertion: the depletion layer in p-n junction under forward bias decreases.
Reason: the electric field due to external voltage supports the electric field
due to potential barrier.
8. Assertion: for the same degree of doping the conductivity of n-type
semiconductor is more than that of p type semiconductor.
Reason: a hole represents a missing valence electron in a covalent bond,
therefore a hole can be thought as a positive charge.
9. Assertion: a semiconductor is virtually an insulator at room temperature.
Reason: at room temperature almost all the valence electrons are engaged in
the formation of covalent bond and there are practically very few free
electrons.
10. Assertion: in a semiconductor, the conduction of electrons have a higher
mobility than holes.
Reason: the electrons experience fewer collisions.

5. CASE STUDY BASED - 5

1. A Silicon diode has forward voltage drop 1.0 V for a DC current of 50 mA. It
has a reverse current of 1 microampere for a reverse voltage 5V. A bilateral
element generally offers same value of resistance in either direction of
current whereas unilateral elements do not offer same resistance.
(i) what is the resistance ofered by the diode in the forward direction?
(a) 10 ohm (b) 20 ohm (c) 30 ohm (d) 40 ohm
(ii) calculate the resistance offered by the diode in reverse direction.
(a) 5G ohm (b) 5 microohm (c) 5M ohm (d) 5m ohm
(iii) Is diode unilateral or bilateral?
(a) it is unilateral element as it does not offer different resistance for either
direction of current.
(b) it is bilateral element as it offers different resistance for either direction of
current.
(c) it is unilateral element as it does not offer same resistance for either
direction of current.
(d) it is bilateral element as it does not offer same resistance for either
direction of current.
(iv) observe the values calculated in part (i) and (ii) which one is higher and
why?
(a) reverse resistance as the reverse current is very low
(b) forward resistance as the reverse current is very low
(c) reverse resistance as the reverse current is very high
(d) forward resistance as the reverse current is very high.
Or
(iv) the forward biased diode current is:
(a) more of drift current than the diffusion current
(b) predominantly drift current (c) predominantly diffusion current
(d) an equal combination of drift and diffusion current.

2. An alternative voltage V= 200 sin Omega t is applied to a divide that offers

an ohmic resistance of 20 ohm to the flow of current in one direction, while
blocked the flow of current in opposite direction.
(i) which device is used here?
(a) AC generator (b) Transformer (c) p-n junction diode (d) none of these
(ii) which type of circuit is this?
(a) inductive circuit (b) capacitive (c) half wave rectifier (d) full wave rectifier
(iii) what is the rms value of current in the above circuit for one cycle?
(a) 5 A (b) 6 A (c) 7A (d) 8 A
Or
What is the average value of current for one cycle?
(a) 1.18 A (b) 2.18 A (c) 3.18 A (iv) 4.18 A
(iv) how many diodes are used by a half wave rectifier circuit for the
transformation?
(a) no diode used (b) one (c) two (d) four

3. There are different techniques of fabrication of p-n junction. In one such

technique, called fused junction techniques, an aluminium film is kept on the
wafer of n-type semiconductor and the combination is then heated to a high
temperature (about 600 °C). As a result, aluminium fused into silicon and
produces p-type
semiconductor and in this way p-n junction is formed.
(i) when a PN junction is reversed by asked then how does the height of
potential barrier change?
(a) no current flows (b) the depletion region is reduced (c) height of potential
barrier is decreased
(d)height of potential barrier is increased
(ii) the cause of potential barrier in PN junction is:
(a) depletion of positive charges near the junction
(b) concentration of negative charges near the junction
(c) concentration of positive and negative charges near the junction
(d) depletion of negative charges near the junction
(iii) the circuit has two oppositely connected ideal diodes in parallel. What is
the current flowing in the circuit?

(a) 1.17 A (b) 2.0 A (c) 2.13 A (d) 1.33 A

(iv) Carbon, Germanium and silicon are 14th group elements:
(a) C and Ge are semiconductors (b) C and Si are semiconductor
(c) all C, Ge and Si are semiconductors (d) Si and Ge are semiconductors
Or
When a PN junction is forward biased then:
(a) only diffusion current flows
(b) both diffusion current and drift current flow but diffusion current is more
than drift current
(c) only drift current flows
(d) both diffusion and drift current flow but drift current exceeds the diffusion
current.

4. In forward bias arrangement the P side of a p-n junction is connected to

positive terminal of battery and and n side to negative terminal of battery,
the current flow increases very slowly tell the certain threshold voltage is
reach. Beyond this value that diode current increases exponentially even for
a very small increment in diode bias voltage. In Reverse bias the current
suddenly increases at very high reverse bias this is called breakdown voltage.
(i) the characteristic curve for p-n junction in forward biased is:

(ii) what is the approximate value of threshold voltage for a silicon diode?
(a) 0.7 V (b) 0.14 V (c) 0.7 eV (d) 0.14 eV
(iii) which of the following is a semiconductor device that emits visible light
when it is forward biased?
(a) transistor (b) LED (c) p-n junction (d) none of these
(iv) which diagram represents the reverse bias of a p-njunction diode

Or
How does current under reverse bias depend on applied voltage?
(a) it varies directly with potential (b)it varies inversely with potential
(c) it is almost independent of applied potential up to critical voltage
(d) it remains unchanged after critical voltage is reached.
Answers: MCQs
1. Ans. (b) Gallium to pure silicon
2. Ans. (a) majority charge carriers.
3. Ans. (d) concentration of positive charge and negative charges near the
junction.
4. Ans. (a) flow towards the negative terminal.
5. Ans. (d)an n-type semiconductor.
6. Ans. (b) recombination.
7. Ans. (b) diffusion of both electrons and holes.
8. Ans. (c)diode barrier potential.
9. Ans. (c)3.6×10⁹ m-³
10. Ans. (a) free electrons.
11. Ans. (b)EG1<EG2<EG3
12. Ans. (d)number density of carriers increases, relaxation time decreases
but effect of decrease is much less than the increase in number density.
13. Ans. (c) when electric field is applied across a semiconductor, electrons in
the conduction band move from lower to higher energy level. While the holes
in valance band move from higher to lower energy level,
where they will be having more energy.
14. Ans. (c) both holes and electrons
15. Ans.(b)equal number of holes and electrons exist, making the region
neutral
16. Ans. (d)strong electric field in a depletion region if doping concentration is
large.

2. ASSERTION & REASON - 10

1. (a) 2. (a) 3. (a) 4. (a) 5. (c) 6. (d) 7. (c) 8. (b) 9. (a) 10.(c)
5. CASE STUDY BASED - 5
1. (i) Ans.(b) Forward resistance = forward voltage drop/ forward DC current
= 1.0/ 50×10-³= 20 ohm
(ii) Ans. (c) reverse resistance = reverse voltage/ reverse current = 5/1×10-⁴
= 5×10⁶ ohm = 5M ohm
(iii) Ans.(c) Diode doesn't offer same resistance for either direction current.
Hence, it is categorised under unilateral element.
(iv) Ans. (a)reverse resistance as the reverse current is very low
Or Ans. (c)predominantly diffusion current.

2. (i) Ans. (c) p-n junction diode

(ii) Ans. (c) half wave rectifier
(iii) Ans. (a) 5 A ,I(rms) = V (rms)/ R, V(rms) = 100V, I(rms) = 100/20 = 5 A
Or Ans.(c) 3.18 A
I(av) = V(av)/R, V(av) = 200/pi = 63.67, I(av) = 3.18 A
(iv) Ans.(b) one
3. (i) Ans. (d) Reason: in reverse biased p-n junction l, potential difference
across a junction becomes [V+V(b)]
(ii) Ans. (c)concentration of positive and negative charges near the junction
(iii) Ans. (b)
Reason: D2 is reverse biased and D1 is forward biased. So, 2phm and 4 ohm
are in series and connected to 12
V. Hence, I= 12/ (2+4) = 2A
(iv) Ans. (d) Si and Ge are semiconductors
Or Ans. (c) current almost remains unchanged till critical voltage is reached.
4. (i) Ans. (b)
(ii) Ans. (a) 0.7 V
(iii) Ans. (c) LED
(iv) Ans. (d) Or Ans. (c) it is almost independent of applied potential up to
critical voltage

LONG ANSWER QUESTIONS

1. Distinguish between a conductor, an insulator and a semiconductor on the basis of
energy band diagrams.
[Ans. Distinction between Conductors (metals), insulators and semiconductors on the
basis of Energy bands
1. Conductors (Metals) :
In conductors either conduction and valence band partly overlap each
other or the conduction
band is partially filled. Forbidden energy gap does not exists ( . This
makes a large number of free electrons
available for electrical conduction. So the metals have high conductivity.
2. Semiconductors :
In semiconductors, conduction band is empty and valance band is totally
filled. is quite small ( 3 eV).
At , electrons are not able to cross this energy gap and semiconductor
behaves as an insulator. But at room
temperature, some electrons are able to jump to conduction band and
semiconductor acquires small conductivity
3. Insulators
In insulators, conduction band is empty and valance band is totally filled.
is very large ( 6 eV). It is not possible to give such large amount of energy
to electrons by any means. Hence conduction band remains total empty
and the crystal remains as insulator
2. What is p-n junction ? Explain briefly, with the help of suitable diagram, how a p-n
junction is formed. Define the term Potential barrier and depletion region.

[Ans. p-n junction : When a semiconductor crystal is so prepared that, it’s one half is p-type and
other is n-type, then the contact surface dividing the two halves, is called p-n junction

Formation of p-n junction : potential barrier & depletion region

Diffusion and drift are the two important processes involved during the
formation of a p-n junction
Due to different concentration gradient of the charge carriers on two
sides of the junction, electrons from n-side starts moving towards p-side
and holes start moving from p-side to n-side . This process is called
diffusion. Due to diffusion, positive space charge region is created on the
n-side of the junction and negative space charge region is created on the
p-side of the junction. Hence an electric field called Junction field is set up
from n-side to p-side which forces the minority charge carriers to cross the
junction. This process is called Drift. The potential difference developed
across the p-n junction due to diffusion of majority charge carriers, which
prevents the further movement of majority charge carriers through it, is
called potential barrier. For Si, VB = 0.7 V and for Ge, VB =0.3 V The small
space charge region on either side of the p-n junction, which becomes
depleted from mobile charge carriers is known as depletion region (10 -6
m)
3. Draw the circuit diagram for studying the V-I characteristics of a p-n
junction diode in (i) forward bias and (ii) reverse bias. Draw the typical V-I
characteristics of a silicon diode

od

e.
[ Ans. V-I characteristics : A graph showing the variation of current
through a p-n junction with the voltage applied across it, is called the
voltage – current (V-I) characteristics of that p-n junction.
For different values of voltages, the value of the current is noted. A graph
between V and I is obtained as in fig. This V-I graph shows that -
(i) At a certain forward bias voltage, current increases rapidly showing the
linear variation. This voltage is known as knee voltage or threshold
voltage or cut-in voltage.
(ii) The ratio of change in forward voltage to the change in forward current
is called dynamic resistance (rd) i,e, rd = Ω
(iii) Under reverse bias, the current is very small (~μA) and remains
almost constant. However, when reverse bias voltage reaches a high
value, reverse current suddenly increases. This voltage is called Zener
breakdown voltage

4. Explain with the help of a circuit diagram, the working of p-n junction
diode as half wave rectifier.
[Ans. Half wave rectifier
During the positive half cycle of ac input signal, the diode is forward
biased and it conducts. Hence, there

Is
current in the load resistance and we get an output voltage. During the
negative half cycle of ac input signal, diode is reverse-biased and it does
not conduct. Hence, there is no current in the load resistance and there is
no output. Thus, we get the output only for half cycle of a.c. input signal.

5. Draw a labelled circuit diagram of a junction diode as a full wave

rectifier. Explain its underlying principle and working.

[Ans. Full wave rectifier

During the positive half cycle of a.c. input signal, diode gets forward
biased and conducts while being reverse biased does not conducts.
Hence, there is a current in due to diode and we get an output voltage.
During the negative half cycle of ac input signal, diode gets reverse
biased and does not conduct while being forward biased conducts. Hence,
now there is a current in due to diode and again we get an output voltage.
Thus, we get output voltage for complete cycle of a.c. input signal in the
same direction