Long Answer Questions

Q. 1. (a) State briefly the processes involved in the formation of p-n junction
explaining clearly how the depletion region is formed.
(b) Using the necessary circuit diagrams, show how the V–I characteristics of a p-
n junction are obtained in
(i) Forward biasing (ii) Reverse biasing How are these characteristics made use of
in rectification? [CBSE Delhi 2014]
OR
Draw the circuit arrangement for studying the V–I characteristics of a p-n junction
diode (i) in forward bias and (ii) in reverse bias. Draw the typical V–I
characteristics of a silicon diode.
Describe briefly the following terms:
(i) “Minority carrier injection” in forward bias
(ii) “Breakdown voltage” in reverse bias. [CBSE Chennai 2015]
Ans. (a)

Two processes occur during the formation of a p-n junction are diffusion and drift. Due
to the concentration gradient across p and n-sides of the junction, holes diffuse from p-
side to n-side (p → n) and electrons diffuse from n-side to p-side (n → p). This
movement of charge carriers leaves behind ionised acceptors (negative charge φ-
immobile) on the p-side and donors (positive charge immobile) on the n-side of the
junction. This space charge region on either side of the junction together is known as
depletion region.
(b) The circuit arrangement for studying the V–I characteristics of a diode are shown in
Fig. (a) and (b). For different values of voltages the value of current is noted. A graph
between V and I is obtained as in Figure (c).
From the V–I characteristic of a junction diode it is clear that it allows current to pass
only when it is forward biased. So if an alternating voltage is applied across a diode the
current flows only in that part of the cycle when the diode is forward biased. This
property is used to rectify alternating voltages.

(i) Minority Carrier Injection: Due to the applied voltage, electrons from n-side cross
the depletion region and reach p-side (where they are minority carriers). Similarly, holes
from p-side cross this junction and reach the n-side (where they are minority carriers).
This process under forward bias is known as minority carrier injection.
(ii) Breakdown Voltage: It is a critical reverse bias voltage at which current is
independent of applied voltage.
Q. 2. Explain, with the help of a circuit diagram, the working of a p-n junction
diode as a half-wave rectifier. [CBSE (AI) 2014]
Ans.

Working
(i) During positive half cycle of input alternating voltage, the diode is forward biased and
a current flows through the load resistor RL and we get an output voltage.
(ii) During other negative half cycle of the input alternating voltage, the diode is reverse
biased and it does not conduct (under break down region).
Hence, AC voltage can be rectified in the pulsating and unidirectional voltage.
Q. 3. State the principle of working of p-n diode as a rectifier. Explain with the
help of a circuit diagram, the use of p-n diode as a full wave rectifier. Draw a
sketch of the input and output waveforms. [CBSE Delhi 2012]
OR
Draw a circuit diagram of a full wave rectifier. Explain the working principle. Draw
the input/output waveforms indicating clearly the functions of the two diodes
used.
[CBSE (AI) 2011]
OR
With the help of a circuit diagram, explain the working of a junction diode as a full
wave rectifier. Draw its input and output waveforms. Which characteristic
property makes the junction diode suitable for rectification?
[CBSE Ajmer 2015, North 2016]
Ans. Rectification: Rectification means conversion of ac into dc. A p-n diode acts as a
rectifier because an ac changes polarity periodically and a p-n diode allows the current
to pass only when it is forward biased. This makes the diode suitable for rectification.
Working: The ac input voltage across secondary s1 and s2 changes polarity after each
half cycle. Suppose during the first half cycle of input ac signal, the terminal s1 is
positive relative to centre tap O and s2 is negative relative to O. Then diode D1 is
forward biased and diode D2 is reverse biased. Therefore, diode D1conducts while
diode D2 does not. The direction of current (i1) due to diode D1 in load resistance RL is
directed from A to B In next half cycle, the terminal s1 is negative and s2 is positive
relative to centre tap O. The diode D1 is reverse biased and diode D2 is forward biased.
Therefore, diode D2 conducts while D1 does not. The direction of current (i2) due to
diode D2 in load resistance RL is still from A to B. Thus, the current in load
resistance RL is in the same direction for both half cycles of input ac voltage. Thus for
input ac signal the output current is a continuous series of unidirectional pulses.

In a full wave rectifier, if input frequency is f hertz, then output frequency will be 2f hertz
because for each cycle of input, two positive half cycles of output are obtained.
Q. 4. Answer the following questions.

(i) 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. [CBSE (F) 2013]

Ans. (i) If p-type and n-type semiconductor are heavily doped. Then due to diffusion of
electrons from n-region to p-region, and of holes from p-region to n-region, a depletion
region formed of size of order less than 1 µm. The electric field directing from n-region
to p-region produces a reverse bias voltage of about 5V and electric field becomes very
large.

Q. 5. Why is a Zener diode considered as a special purpose semiconductor

diode?
Draw the I–V characteristic of a zener diode and explain briefly how reverse
current suddenly increases at the breakdown voltage.
Describe briefly with the help of a circuit diagram how a Zener diode works to
obtain a constant dc voltage from the unregulated dc output of a rectifier. [CBSE
(F) 2012]
OR
How is Zener diode fabricated? What causes the setting up of high electric field
even for small reverse bias voltage across the diode?
Describe with the help of a circuit diagram, the working of Zener diode as a
voltage regulator. [CBSE Panchkula 2015]
Ans. A Zener diode is considered as a special purpose semiconductor diode because it
is designed to operate under reverse bias in the breakdown region.
Zener diode is fabricated by heavy doping of its p and n sections. Since doping is high,
depletion layer becomes very thin.
𝑣
Hence, electric field (= ) becomes high even for a small reverse bias.
𝑙
We know that reverse current is due to the flow of electrons (minority carriers)
from p → n and holes from n → p. As the reverse bias voltage is increased, the electric
field at the junction becomes significant. When the reverse bias voltage V = VZ, then the
electric field strength is high enough to pull valence electrons from the host atoms on
the p-side which are accelerated to n-side. These electrons causes high current at
breakdown.

Working:

The unregulated dc voltage output of a rectifier is connected to the zener diode through
a series resistance Rs such that the Zener diode is reverse biased. Now, any
increase/decrease in the input voltage results in increase/decrease of the voltage drop
across Rs without any change in voltage across the Zener diode. Thus, the Zener diode
acts as a voltage regulator.

Explanation of voltage regulator.

If reverse bias voltage V reaches the breakdown voltage VZ of zener diode, there is a
large change in the current. After that (just above VZ there is a large change in the
current by almost insignificant change in reverse bias voltage. This means diode voltage
remains constant.

For example: If unregulated voltage is supplied at terminals A and B, and input voltage
increases, the current through resistor RZ and diode also increases. This current
increases the voltage across RZ without any change in the voltage across diode.
Thus, we have a regulated voltage across load resistor RL.

Q. 6. In the figure given alongside, is (i) the emitter and (ii) the collector forward
or reverse biased? With the help of a circuit diagram explain the action of npn
transistor.
[CBSE (AI) 2012]

Ans. The given transistor is p-n-p transistor. The emitter is reverse biased and the
collector is forward biased.

Action of n-p-n transistor

An npn transistor is equivalent to two p-n junction diodes placed back to back with their
very thin p-regions connected together. The circuit diagram for the operation
of npn transistor is shown in fig. The two batteries VEE and VCC represent emitter supply
and collector supply respectively.

Transistor Action: Transistor works only when its emitter-base junction is forward
biased and collector-emitter junction is reversed biased. Due to this the majority charge
carriers from the emitter, accelerate to collector side and create IE, IB and IC such
that IE = IB + IC.

Base Current and Collector Current: Under forward bias of emitter-base junction, the
electrons in emitter and holes in base are compelled to move towards the junction, thus
the depletion layer of emitter-base junction is eliminated. As the base region is very thin,
most electrons (about 98%) starting from emitter region cross the base region and
reach the collector while only a few of them (about 2%) combine with an equal number
of holes of base-region and get neutralised. As soon as a hole (in -region) combines
with an electron, a covalent bond of crystal atom of base region breaks releasing an
electron-hole pair. The electron released is attracted by positive terminal of emitter
battery VEE, giving rise to a feeble base current (IB). Its direction in external circuit is
from emitter to base. The hole released in the base region compensates the loss of hole
neutralised by electrons.

The electrons crossing the base and entering the collector, due to reverse biasing of
collector-base junction, are attracted towards the positive terminal of collector
battery VCC. In the process an equal number of electrons leave the negative terminal of
battery VCC and enter the positive terminal of battery VEE. This causes a current in
collector circuit, called the collector current. In addition to this the collector current is
also due to flow of minority charge carriers under reverse bias of base-collector junction.
This current is called the leakage current.
Thus, collector current is formed of two components:

(i) Current (Inc) due to flow of electrons (majority charge carriers) moving from emitter to
collector.

(ii) Leakage current (Ileakage) due to minority charge carriers, i.e., Ic=Inc +Ileakage.

Emitter Current: When electrons enter the emitter battery VEE from the base causing
base current or electrons enter the collector battery VCC from the collector causing
collector current, an equal number of electrons enter from emitter battery VEE to emitter,
causing the emitter current. The process continues.

Relation between Emitter, Base and Collector Currents:

Applying Kirchhoff’s I law at terminal O, we get

IE = IB + IC

That is, the emitter current IE is the sum of base current IB and the collector current IC.
This is the fundamental relation between currents in the bipolar transistor circuit.

Q. 7. Draw the circuit diagram to study the characteristics of npn transistor in

common emitter configuration. Sketch typical (i) input characteristics (ii) output
characteristics for such a configuration. Explain how the current gain of
transistor is calculated from output characteristics. [CBSE Delhi 2009]

OR

Draw the circuit arrangement for studying the input and output characteristics of
an n-p-n transistor in CE configuration. With the help of these characteristics
define (i) input resistance, (ii) current amplification factor.
[CBSE (AI) 2010; (F) 2013, Delhi 2015]

Ans. Characteristic Curves: The circuit diagram for determining the static
characteristic curves of an n-p-n transistor in common-emitter configuration is shown in
figure.
Common Emitter Characteristics:

(i) Input characteristics are obtained by recording the values of base current IB for
different values of base-emitter voltage VBE at constant collector emitter voltage VCE.

(ii) Output characteristics are obtained by recording the values of collector

current IC for different values of collector emitter voltage VCE at constant base current IB.

The characteristic curves show:

When collector-emitter voltage VCE is increased from zero, the collector

current IC increases as VCE increases from 0 to 1 V only and then the collector current
becomes almost constant and independent of VCE. The value of VCE upto which
collector current IC changes is called the knee voltage Vknee

Definition of Input resistance:

Refer to Point 11(a) of Basic Concepts.

Definition of Amplification Factor: Refer to Point 12 of Basic Concepts.

Determination of Current Gain

We take the active region of output characteristics i.e, the region where collector current
(IC) is almost independent of VCE.

Now we choose any two characteristic curves for given values of IB and find the two
corresponding values of IC.
Q. 8. Draw the transfer characteristics of a base biased transistor in common
emitter configuration. Explain briefly the meaning of the term active region in
these characteristics. For what practical use, do we use the transistor in this
active region?

Explain clearly how the active region of the V0 versus Vi curve in a transistor is
used as an amplifier. [CBSE Delhi 2011]

Ans. Transfer characteristics of a base biased transistor in common-emitter

configuration:

These are the curves representing the variation of output voltage (VBE) with input
voltage (VCE)
Circuit Diagram: It is shown in fig. (a).

For plotting the curve the input voltage (Vi) is changed in small steps and the
corresponding output voltage is measured. The curve obtained is shown in fig. Curve is
nearly linear.

In active region the transistor is used as an amplifier.

The switching circuits are designed in such a way that the transistor does not remain in
active state.

In the active region, a small increase of Vi results in a large (almost linear) increase
in IC. This results in an increase in the voltage drop across RC.

Q. 9. Draw a labelled circuit diagram of a common emitter amplifier using a p-n-p

transistor. Define the term voltage gain and write an expression for it. Explain
how the input and output voltages are out of phase by 180° for a common-emitter
transistor amplifiers.

Ans. Common-Emitter Transistor Amplifier: Given below is the circuit for a p-n-p
transistor. In this circuit, the emitter is common to both the input (emitter-base) and
output (collector-emitter) circuits and is grounded. The emitter-base circuit is forward
biased and the base-collector circuit is reverse biased.

In a common-emitter circuit, the collector-current is controlled by the base-current rather

than the emitter-current. Since in a transistor, when input signal is applied to base, a
very small change in base-current provides a much larger change in collector-current
and thus extremely large current gains are possible.
When positive half cycle is fed to the input circuit, it opposes the forward bias of the
circuit which causes the collector current to decrease. It decreases the voltage drop
across load RL and thus makes collector voltage more negative. Thus, when input cycle
varies through a positive half cycle, the output voltage developed at the collector varies
through a negative half cycle and vice versa. Thus, the output voltage in common-
emitter amplifier is in antiphase with the input signal or the output and input voltages are
180° out of phase.

Current Gain. The ratio of change in collector current (∆IC) to the change in base
current (∆IB) is called the alternating current gain denoted by β Thus,

β has positive values and is generally greater than 20.

Voltage Gain. The voltage gain of common- emitter transistor amplifier is given by

Q. 10. (a) Differentiate between three segments of a transistor on the basis of

their size and level of doping.

(b) How is a transistor biased to be in active state? [CBSE Delhi 2014]

(c) With the help of necessary circuit diagram describe briefly how npn transistor
in CE configuration amplifies a small sinusoidal input voltage. Write the
expression for ac current gain.

OR
Explain with the help of a circuit diagram the working of npn transistor as a
common emitter amplifier. [CBSE Delhi 2009, South 2016]

OR

Draw the circuit diagram of a common-emitter amplifier using an npn transistor.

What is the phase difference between the input signal and output voltage? Draw
the input and output waveforms of the signal. Write the expression for its voltage
gain. State two reasons why a common emitter amplifier is preferred to a
common base amplifier. [CBSE (AI) 2009, Allahabad 2015]

Ans. (a) Emitter: It is of moderate size and heavily doped.

Base: It is very thin and lightly doped.

Collector: The collector side is moderately doped and larger in size as compared to the
emitter.

(b) Transistor is said to be in active state when its emitter-base junction is suitably
forward biased and base-collector junction is suitably reverse biased.

(c) The circuit of common emitter amplifier using n-p-n transistor is shown below:

Working: If a small sinusoidal voltage, with amplitude Vs, is superposed on dc basic

bias (by connecting the sinusoidal voltage in series with base supply VBB), the base
current will have sinusoidal variations superposed on the base current IB. As a
consequence the collector current is also sinusoidal variations superimposed on the
value of collector current IC, this will produce corresponding amplified changes in the
value of output voltage V0. The ac variations across input and output terminals may be
measured by blocking the dc voltage by large capacitors.

The phase difference between input signal and output voltage is 180°.

The input and output waveforms are shown in fig.

Reasons for using a common emitter amplifier:

(i) Voltage gain is quite high.

(ii) Voltage gain is uniform over a wide frequency range or power gain is high.

Q. 11. Answer the following questions.

(i) Show the output waveforms (Y) for the following inputs A and B of

(i) OR gate (ii) NAND gate [CBSE Delhi 2012]

Ans. (i) Output waveforms for the following inputs A and B of OR gate and NAND gate.
Q. 12. Draw a simple circuit of a CE transistor amplifier. Explain its working. Show
𝑩𝒂𝒄 𝑹𝑳
that the voltage gain AV, of the amplifier is given by AV =– , where βac is the
𝒓𝒊
current gain, RL is the load resistance and ri is the input resistance of the
transistor. What is the significance of the negative sign in the expression for the
voltage gain?
[CBSE Delhi 2012]

Ans. Circuit diagram of CE transistor Amplifier.

When an ac input signal Vi (to be amplified) is superimposed on the bias VBB, the
output, which is measured between collector and ground, increases.

We first assume that Vi = 0. Then, applying Kirchhoff's law to the output loop.
VCC = VCE + IC RL

Similarly, the input loop gives

VBB = VBE + IB RB

When Vi is not zero, we have

VBE + Vi = VBE + IB RB + ∆IB (RB+Ri)

⇒Vi = ∆IB (RB + Ri) ⇒ Vi = r∆IB

Change in IB causes a change in IC

Negative sign in the expression shows that output voltage and input voltage have phase
difference of π.

Q. 13. In fig., the circuit symbol of a logic gate and two input waveforms A and B
are shown:
(i) Name the logic gate.
(ii) Write its truth table.
(iii) Give the output waveform.

Ans. (i) The logic gate shown is OR gate.

(ii) Truth table of OR gate is

(iii) The input waveforms A and B are discrete square waves. The components of
waveforms A and B are shown by vertical dotted lines.
Between a and b, A = 1, B = 1 → Y =1

Between b and c, A = 0, B = 0 → Y = 0

Between c and d, A = 0, B = 1 → Y = 1

Between d and e, A =1, B = 0 → Y = 1

Between e and f, A = 1, B = 1 → Y = 1

Between f and g, A = 0, B = 0 → Y = 0

Between g and h, A = 0, B = 1 → Y = 1

Accordingly, the waveform Y is shown as above.

Q. 14. Draw the output signals C1 and C2 in the given combination of gates (Fig.)
[HOTS][NCERT Exemplar]
Ans. The output signals C1 and C2 are as shown.

Q. 15. Input signals A and B are applied to the input terminals of the ‘dotted box’
set-up shown here. Let Y be the final output signal from the box.

Draw the wave forms of the signals labelled as C1 and C2 within the box, giving (in
brief) the reasons for getting these wave forms. Hence draw the wave form of the
final output signal Y. Give reasons for your choice.
What can we state (in words) as the relation between the final output signal Y and
the input signals A and B? [HOTS]

Ans.

A B

From 0 to 1 1 0 0
From 1 to 2 1 1 0
From 2 to 3 0 1 1
From 3 to 4 1 0 0
From 4 onwards 1 0 0
 A B

From 0 to 1 1 0 1
From 1 to 2 1 1 0
From 2 to 3 0 1 0
From 3 to 4 1 0 1
From 4 onwards 1 0 1

A B

From 0 to 1 0 1 0
From 1 to 2 0 0 1
From 2 to 3 1 0 0
From 3 to 4 0 1 0
From 4 onwards 0 1 0
The gate shown in circuit is NOT XOR gate. According to definition the output Y is
obtained only if either both signals are 0 or 1.

Q. 16. Identify which logic gate OR, AND and NOT is represented by the circuits in
the dotted line boxes 1, 2 and 3. Give the truth table for the entire circuit for all
possible values of A and B. [HOTS]

Ans. The dotted line box 1 represents NOT gate.

The dotted line box 2 represents OR gate.

The dotted line box 3 represents AND gate.

The output of box 1 is 𝐴̅

The inputs of box 2 are A and 𝐵̅

As box 2 is OR gate, therefore, output of box 2 is E = (𝐴̅ + 𝐵).

The inputs of box 3 are E and B Box 3 represents AND gate; therefore, output of box 3
is
Truth table of the entire circuit is