Basic Electronics Sample Paper and Answers

Karmaveer Bhaurao Patil Polytechnic, Satara
Mr. Gade M.R. Page 1
Scheme- I
Sample Question paper Model Answer
Program Name- Diploma in Electronics Program Group.
Semester- second
Course title- Basic Electronics
Marks:- 70 Time:- 3 Hrs.
______________________________________________________________________________
Q.1) Attempt any five of the Following 10 Marks
a) Name of the components of following symbols:
1]
Ans:- Photodiode
2]
Ans:- LED
b) State any two applications of FET.
Ans:- 1) as a amplifier
2) Buffer amplifier
3) Electronic Switch
4) Phase shift Oscillator
5) Constant current source.
6) Voltage variable resistor or voltage dependent resistor.
22216

c) State type of transistor configuration for obtaining highest current gain?
Ans: - Types of transistor configuration is as follow:
1) Common Base
2) Common Emitter
3) Common Collector
Common emitter configuration is used for obtaining highest current gain.
d) Sketch the symbol of P-channel and N-channel enhancement type MOSFET?
Ans:-
e) State any two limitations of Zener diode regulator?
Ans:- 1)Maximum load current provided to load is limited to Iz(max)-Iz(min).
2) Minimum internal impedance of circuit is limited due to inherent characteristics of
Zener diode.
f) Define load regulation and line regulation?
Ans:- 1) Load Regulation : It is defined as the change in output D.C. voltage for change in load
current and it is expressed in mV or percentage of output voltage.
L.R. = VNL-VF

g) Identify the type of diode for the given V-I characteristics shown in the figure 1
Ans:- Zener diode
Q.2) Attempt any three of the following. 12 Marks
a) Sketch the block diagram of regulated DC power supply, explain working of each block
with input and output waveform.
Ans:-
Transformer: - The AC input signal is applied to the transformer input and the output of
transformer is step-down.
Rectifier: - The step downed output is applied to the rectifier input and the rectifier used may be
half-wave, center tap full wave or bridge type rectifier. The input AC voltage is applied the
rectifier and this AC voltage is converted into pulsating DC voltage. Thus the function of a
rectifier is to rectify the input AC voltage to provide the pulsating DC voltage.
Filter: - The pulsating DC voltage contains large ripple. This voltage is applied to the LC filter
circuit and removes the ripple. The function of a filter circuit is to remove the ripples to provide
DC voltage at its output. The voltage so obtained is the unregulated DC voltage.

Voltage Regulator: - The unregulated DC voltage is applied to voltage regulator. It makes this
DC voltage steady and independent of variation in load and mains AC voltage. This improves the
load and line regulation and provides the regulated DC voltage across the load.
Load: - The constant DC voltage is obtained across load.
b) Sketch fixed bias and self-bias B JT biasing circuit
a) Fixed bias circuit b) Self bias circuit
c) Differentiate Zener breakdown and Avalanche breakdown on basis of:
1. Definition
2. Break down characteristics
Zener breakdown Avalanche breakdown
Definition:-A large number of minority carriers
are generated and a large current flows through
the P-N junction. This mechanism of
breakdown is called Zener breakdown
Definition: - when minority carriers collide
with semiconductor atoms of P-N junction they
impart energy to break covalent bond and
generate additional carriers. This cumulative
process of carrier generation is known as
avalanche breakdown.
Breakdown characteristics: Breakdown characteristics:
Fig. Avalanche breakdown

d) Explain the thermal runaway phenomenon for BJT
1. The increase in the collector current increases the power dissipated at the collector junction.
This in turn further increases the temperature of the junction and hence increases in the collector
current. The process is cumulative and it is referred to as self-heating.
2. The excess heat produced at the collector base junction may even burn and destroy the
transistor. This situation is called 'Thermal runaway' of the transistor.
Fig. Thermal Runway
Q.3) Attempt any THREE of the following. 12 Marks
a) Sketch input and output characteristics of CE configuration. Label various regions
on characteristics.
Fig. Input characteristics of CE configuration.
O A

Fig. Output characteristics of CE configuration.
b) Explain the working of negative clipper with circuit diagram.
Working:- 1) The diode is kept in series with the load.
2) During the negative half cycle of the input waveform, the diode ‘D’ is
connected in reverse polarity, which maintains the output
voltage at 0 Volts.
3) Thus causes the negative half cycle to be clipped off.
4) During the positive half cycle of the input, the diode is reversing biased
and so the positive half cycle appears across the output.

c) A JFET has a drain current of 5mA .If IDSS = 10mA and VGS(off) = -6V .Find the
value of i) VGS ii) Vp
Ans:-
Given :-
Id= 5 ma
IDSS=10 ma
Vgs (off)=-6V
We know that
ID= IDSS [1- ]2
Calculate VGS
VGS = VGS (off) [1-√ ]
= - 6[1-√ ]
VGS = -1.75V
Calculate Vp
ID= IDSS [1- ]2
5=10 [ 1-(-1.75/Vp] 2
Vp =
√
=
Vp= -5.97V

d) Explain working of Zener as a voltage regulator with circuit diagram.
Ans:-
1) When we apply a reverse voltage to a Zener diode, a negligible amount of current flows
through the circuit.
2) When a voltage higher than Zener breakdown voltage is applied, Zener breakdown
occurs.
3) Zener breakdown is a phenomenon where a significant amount of current flows through
the diode with a negligible drop in voltage.
4) When we increase the reverse voltage further, the voltage across the diode remains at the
same value of Zener breakdown voltage whereas the current through it keeps on rising
as seen in the graph above.
5) Here in the graph Vz refers to the Zener breakdown voltage. Zener breakdown voltage
typically can range from 1.2 V to 200 V depending on its application.

Q.4) Attempt any THREE of the following. 12 Marks
a) Define the following parameters of rectifier:-
1. Peak Inverse Voltage (PIV)
2. Ripple factor
3. Efficiency
4. Transformer Utilization Factor.
Ans:-
1. Peak Inverse Voltage (PIV):- Peak Inverse Voltage (PIV) is defined as the maximum
negative voltage which appears across non conducting reverse biased diode.
2. Ripple factor:- Ripple Factor is defined as the ratio of RMS value of the AC
component of output to the DC or average value of the output.
R=
3. Efficiency:- The rectification efficiency of a rectifier is a ratio of the output DC power
delivered to load to input AC power supplied to the rectifier circuit.
η=
4. Transformer Utilization Factor:- Transformer Utilization Factor (TUF) is defined as
the ratio of DC output power to the AC power ratings of the transformer.
Mathematically it is expressed as,
TUF=
b) Describe operation of voltage divider biasing with circuit diagram.
Ans:-

I) The name voltage divider is derived from the fact that resistor R1 and R2 form a potential
divider across the VCC supply.
II) The voltage drop across resistor R2 forward biases the base – emitter junction of a transistor.
III) The emitter resistor (RE) provides the D.C. stability.
IV) It is evident from that the voltage at the transistor base (due to the voltage divider network of
resistors R1 and R2).
VB=Vcc*R2/(R2+R1)
Neglecting VBE
Therefore value of emitter current,
IE = VE/RE
And the value of collector current,
IC = IE
The voltage drop across the collector resistor,
V RC = IC* RC
And the voltage at the collector (measured with respect to the ground)
VC = VCC- VRC = VCC – IC * RC
The voltage from collector – to – emitter.
VCE = VC – VE =VCC – IC * RE
VCE = VCC – IE(RC + RE) ………………(IC = IE)
c) Compare CB and CC configuration of transistor with respect to
1. Voltage Gain
2. Input – output terminals
3. Input Impedance
4. Output Impedance
Ans:-
Parameters CB CC
Voltage Gain High Above 150 Less than 1
Input-Output terminals Input- Emitter
Output- Collector
Input- Base
Output- Emitter
Input Impedance Low (100 ohm) Very high(750kohm)
Output Impedance Very high (450 K ohm) Low (50 Ohm)
d) Calculate input impedance of JFET if reverse gate source voltage of 15V and gate
Current is 10–3 uA
Ans:- Given
VGS = 15V
IGSS = 10 -3
uA= 10-9
A
Input Impedance of JFET
Zi=
Zi=
Zi= 15000Mohm

e) Sketch the block diagram of Regulated DC power supply, explain working of each
block with output waveforms.
Ans:-
1. A step down transformer
2. A rectifier
3. A DC filter
4. A regulator
Operation of Regulated Power Supply:-
Step Down Transformer:-
A step down transformer will step down the voltage from the ac mains to the required voltage
level. The turn’s ratio of the transformer is so adjusted such as to obtain the required voltage
value. The output of the transformer is given as an input to the rectifier circuit.
Rectification:-
Rectifier is an electronic circuit consisting of diodes which carries out the rectification process.
Rectification is the process of converting an alternating voltage or current into corresponding
direct (dc) quantity. The input to a rectifier is ac whereas its output is unidirectional pulsating dc.
Usually a full wave rectifier or a bridge rectifier is used to rectify both the half cycles of the ac
supply (full wave rectification)
DC Filter :-
The rectified voltage from the rectifier is a pulsating dc voltage having very high ripple content.
But this is not we want, we want a pure ripple free dc waveform. Hence a filter is used. Different
types of filters are used such as capacitor filter, LC filter, Choke input filter, π type filter.
Regulator:
This is the last block in a regulated DC power supply. The output voltage or current will change
or fluctuate when there is change in the input from ac mains or due to change in load current at
the output of the regulated power supply or due to other factors like temperature changes. This
problem can be eliminated by using a regulator. A regulator will maintain the output constant
even when changes at the input or any other changes occur. Transistor series regulator, Fixed and

variable IC regulators or a zener diode operated in the zener region can be used depending on
their applications. IC’s like 78XX and 79XX are used to obtained fixed values of voltages at the
output. With IC’s like LM 317 and 723 etc we can adjust the output voltage to a required
constant value.
Q.5) Attempt any TWO of the following. 12 Marks
a) Justify ‘for FET amplification factor depends on its trans conductance.
Ans:- Amplification Factor:- It is the ratio of change in drain source voltage (VDS) to the change
in gate source voltage (VGS) at constant drain current.
µ=
Transconductance:- It is the ratio of change in drain current (ID) to the change in gate source
voltage (VGS) at constant drain source voltage.
gm=
ie Amplification factor rearranged as follows
µ = x
µ= rd x gm
This amplification factor µ is equal to the product of drain resistance rd and transconductance gm
b) Explain the working of bridge rectifier connected with capacitor filter, sketch circuit
diagram and output waveforms with respect to ac signal input.
Ans:-

1) When input AC signal is applied, during the positive half cycle both diodes D1 and D3 are
forward biased. At the same time, diodes D2 and D4 are reverse biased.
2) On the other hand, during the negative half cycle, diodes D2 and D4 are forward biased.
At the same time, diodes D1 and D3 are reverse biased.
3) Thus, the bridge rectifier allows both positive and negative half cycles of the input AC
signal.
4) The DC output produced by the bridge rectifier is not a pure DC but a pulsating DC. This
pulsating DC contains both AC and DC components.
5) The AC components fluctuate with respect to time while the DC components remain
constant with respect to time. So the AC components present in the pulsating DC is an
unwanted signal.
6) The capacitor filter present at the output removes the unwanted AC components. Thus, a
pure DC is obtained at the load resistor RL.
c) Compare LED and photo diode on basis of:
1. Function
2. Symbol
3. Construction
Parameters LED Photodiode
Function LED emits photons due to electron-hole
recombination
Photodiode provides
energy to electron and
holes by exposing itself
towards light radiation.
Symbols

Construction
Q.6) Attempt any TWO of the following. 12 Marks
a) Compare P-N Junction diode and Zener diode on following parameters:
1. Doping Level
2. Breakdown voltage
3. Applications
Ans:-
Parameters PN junction diode Zener diode
Doping level Low High
Breakdown voltage Occurs in higher voltage Occurs in lower Voltage
Applications In rectifiers, Wave shaping
circuits etc.
As a voltage regulator, In
wave shaping circuits etc.
b) Draw the circuit diagram of CE amplifier; explain its working with input and output
Characteristics.
Ans:-

1) There are different types of electronic components in the common emitter amplifier which are
R1 resistor is used for the forward bias, the R2 resistor is used for the development of bias, the
RL resistor is used at the output it is called as the load resistance.
2) The RE resistor is used for the thermal stability. The C1 capacitor is used to separate the AC
signals from the DC biasing voltage and the capacitor is known as the coupling capacitor.
3) The figure shows that the bias vs gain common emitter amplifier transistor characteristics, if
the R2 resistor increases then there are an increase in the forward bias and R1 & bias are
inversely proportional to each other.
4) The alternating current is applied to the base of the transistor of the common emitter amplifier
circuit then there is a flow of small base current.
5) Hence there is a large amount of current flow through the collector with the help of the RC
resistance.
6) The voltage near the resistance RC will change because the value is very high and the values
are from the 4 to 10kohm.
7) Hence there is a huge amount of current present in the collector circuit which amplified from
the weak signal, therefore common emitter transistor work as an amplifier circuit.
Input output Waveform:-

c) Identify the given circuits in figure 2 and draw input and output waveforms for
following circuits :
Ans:- i) Series Negative clipper

ii) Combinational Clipper

Any Query Please Contact
Mr. Gade M.R.
(M.E. Signal Processing)
Lecturer
Electronics and Telecommunication Department
Contact No- 9766906075
email- gademangesh07@gmail.com

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