Analog Electronics Test - 3

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Test-03Solutions

ANALOG ELECTRONICS
1. Consider the following statements about a good power supply:

1. The AC ripple should be high.
2. SV, (Voltage stability factor) should be low.
3. ST, (Temperature stability factor) should be low.
Which of the above statements are correct?
a) 1, 2 and 3
b) 3 only
c) 2 only
d) 2and3only
Answer: (d)
Solution:
In a good power supply, the AC ripple should be as low as
possible.
2. Assertion (A): In a Darlington connection, two transistors are
connected in cascade in common emitter configuration.
Reason (R): The Darlington connection aims at making the
current gain very high, almost equal to the product of beta of
individual transistors.
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 but R is true
Answer: (d)

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Solution:
In a Darlington connection, two transistors are connected in
cascade in common collector configuration.
3. If the circuit shown has to function as a clamping circuit, then
which one of the following conditions should be satisfied for the
sinusoidal signal of period T?

a) RC > T
Answer: (d)
Solution:

This Circuit is a Negative Clamper but when a Load Resistor is
connected in shunt, the situation changes. When the diode is

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OFF, the Capacitor discharges through the resistor ‘R’. The time

constant RC.

To avoid that RC >> T.

4. Of the four biasing circuits shown in Fig. For a BJT, indicate the
one which can have maximum bias stability:

3

Answer: (a)
Solution:
In BJT maximum bias stability is given by self-bias (or) potential
Divider circuit.
5. Which of the following will be true for a CE transistor amplifier
if the emitter resistor value is made equal to zero?
1. Its gain will increase.
2. Its stability will increase.
3. Its gain will decrease.
4. Its stability will decrease.

4

Select the correct answer from the codes given below:
a) 1 and 2
b) 2 and 3
c) 3 and 4
d) 1 and 4
Answer: (d)
Solution:
The emitter resistor in CE transistor amplifier gives negative
feedback due to which gain decreases and Stability increases.
Therefore, making emitter resistor equal to zero, gain will
increase and Stability will decrease.
6. In the bistable circuit shown, the ideal op amp has saturation
levels of + 5 V. The value of R1(in kΩ) that gives a hysteresis
width of 500 mV is _______

Answer: 1

Solution:

KCL at non-inverting terminal,

0 − Vin + 0 − V0 = 0
R1 20K

V0 = − 20k Vin
R1

5

Case (i): Assume Vout = Vsat = 5V

⇒ 5 = − 20k Vin ⇒ Vin = − R1
R1 4K

If Vin < − 4RK1, then V0 Changes from Vsat to – Vsat

Case (ii): Assume Vout = −Vsat = – 5V

⇒ – 5 = − 20k Vin ⇒Vin = R1
R1 4K

If Vin < 4RK1, then V0 Changes from Vsat to –Vsat

Transfer Characteristics:

By transfer characteristics,

Hysteresis width = 2R1 = 500 mV
4k

R1 = 500m × 4k = 1kΩ
2

7. Which of the following statements are correct?

1. ICO for germanium is much greater than for silicon.

2. The steady-state temperature rise at the collector junction is

proportional to the power dissipated at the junction.

3. To avoid thermal runaway the required condition is that the

rate at which heat is released at the collector junction must

exceed the rate at which the heat can be dissipated under

steady-state conditions.

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a) 1, 2 and 3
b) 2 and 3 only
c) 1 and 3 only
d) 1 and 2 only
Answer: (d)
Solution:
For the condition to avoid thermal runaway the heat released at
collector junction must not exceed to the rate at which the heat
can be dissipated under steady state condition.
8. In a transconductance, the device output
a) voltage depends upon the input voltage.
b) voltage depends upon the input current.
c) current depends upon the input voltage.
d) current depends upon the input current.
Answer: (c)
Solution:
Transconductance is the ratio of the output current to input
voltage i.e. current depends on the input voltage.
9. A BJT is said to be operating in the saturation region if
a) Both the Junctions are reverse biased.
b) Base-Emitter Junction is reverse biased and Base-Collector

Junction is forward biased.
c) Base - Emitter Junction is forward biased and Base-Collector

Junction is reverse - biased.
d) Both the Junctions are forward biased.

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Answer: (d)
Solution:
Whenever Base-Emitter Junctions and Base-Collector Junction
are forward biased, then BJT operates in Saturation Region.
10. If the input to the circuit of figure is a sine wave the output will
be.

a) A half-wave rectified sine wave.
b) A full-wave rectified sine wave.
c) A triangular wave.
d) A square wave.
Answer: (d)
Solution:
The given OP-Amp is Open Loop Configuration. So it is acting
as a Comparator.

If Sinusoidal input is applied to the high gain comparator. The
comparator generates the square wave output.

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11. fT is the frequency at which the short circuit
a) Common collector current gain has a magnitude of unity
b) Common base current gain has a magnitude of unity
c) Common emitter current gain has a magnitude of unity
d) Common emitter current gain has a magnitude of ½
Answer: (c)

12. The ideal OP-amp has the following characteristics.
a) Ri = ∞, A = ∞, R0 = 0
b) Ri = 0, A = ∞, R0 = 0
c) Ri = ∞, A = ∞, R0 = ∞
d) Ri = 0, A = ∞, R0 = ∞
Answer: (a)
Solution:
The ideal Op-Amp input Resistance is infinite.
The ideal Op-Amp Output Resistance is zero.
The ideal Op-Amp gain is infinite.

13. A source of (power/energy) feeds the input port of an amplifier
and the output port is connected to a ‘load’. The input
impedance of the ideal amplifier should ideally be
a) zero
b) ∞
c) low
d) high
Answer: (b)

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Solution:
The input impedance of the ideal amplifier should ideally be
infinite to avoid loading of the source.
14. The Op-Amp circuit shown above represents

a) high pass filter

b) low pass filter

c) band pass filter

d) band reject filter

Answer: (b)

Solution:

Apply nodal Analysis

0−Vi + 0−V0
R1+L R2∥C

⇒ −Vi + V0
R1 + jωL 1
R2 . jωc

R2 + 1
jωc

⇒ −Vi = V0 × (R2jωC + 1)
R1 + jωL R2

⇒ −Vi �R2jωRC2 + 1� = V0[R1 + jωL]

⇒ V0 = − (R2jωC + R2 + jωL)
Vi 1)(R1

Which equivalent to standard form of transfer function of low

pass filter.

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15. An amplifier has an open loop gain of 1000 ±10. Negative

feedback is provided such that the gain variation remains within

0.1%. What is the amount of feedback βF?

a) 1
10

b) 1
9

c) 9
100

d) 9
1000

Answer: (d)

Solution:

dAf = dA �1+1Aβ�
Af A

⇒ 0.1 = 10 �1 +1Aβ�
100 1000

⇒1 + 1000β = 10

⇒ β = 9
1000

16. In the silicon BJT circuit shown below, assume that the emitter

area of transistor Q1 is half that of transistor Q2. The value of

current I0 is approximately

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a) 0.5mA
b) 2mA
c) 9.3mA
d) 15mA
Answer: (b)
Solution:
Total Current Ii,

9.3 ICi + 0.7 = 0 – (– 10)
9.3 ICi + 0.7 = 10
9.3 ICi = 9.3
ICi = 1mA

Since the Emitter area of transistor Q1 is half that of transistor

Q2 so Ii = I0
2

I0 = 2mA (since IC1 = Ii = 1mA)

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17. Consider the following statements regarding a common emitter
amplifier. It can be converted into an oscillator by:
1. providing adequate positive feedback
2. phase shifting the output by 180oand feeding this phase-
shifted output to the input
3. using only a series tuned circuit as a load on the amplifier
4. using a negative resistance device as a load on the amplifier
Which of the above statements are correct?
a) 1,2,3 and 4
b) 1 and 2
c) 1, 3 and 4
d) 3 and 4
Answer: (b)
Solution:
a) In Oscillators, Positive feedback is used.
b) In Oscillators, total phase shift in a loop is multiple of 3600.
In CE amplifier, phase shift between input and output is 1800.
So, phase shift in the feedback network should be1800.

18. Consider the following statements regarding a common emitter
amplifier. It can be converted into an oscillator by
1. Providing adequate positive feedback.
2. Phase shifting the output by 180o and feeding this phase-
shifted output to the input.
3. Using only a series tuned circuit as a load on the amplifier.

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4. Using the negative resistance device as a load on the
amplifier.

Which of the above statements are correct?
a) 1, 2, 3 and 4
b) 1 and 2 only
c) 2 and 3 only
d) 3 and 4 only
Answer: (b)
Solution:
According to Barkhausen criteria an amplifier will give an
oscillatory response if it is given a positive feedback.
A Common emitter amplifier produces a phase shift of 1800,
thus to make it an oscillator we need to provide it with another
phase shift of 1800.
19. An op-amp circuit is shown in the figure given below.
Different inputs and output are given under List-I and List-II.
Match List-I (inputs) with List-II (Outputs) and select the
correct answer using the codes given below the lists:

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List – I List – II

Codes:
ABCD

a) 2 4 1 3
b) 1 3 2 4
c) 2 3 1 4
d) 1 4 2 3
Answer: (a)
Solution:
The given op-amp circuit is a differentiator circuit. So if input
signal is step signal, then output will be impulse.
If input is ramp signal, then output will be the step signal.

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20. A 2-terminal network consists of one of the R-L-C elements.

The element is connected to an AC supply. The current through

the element is I. When a capacitor is inserted in series between

the source and the element, the current through the element

becomes 2I. The element

a) Is a resistor

b) Is an inductor

c) Is a capacitor

d) Cannot be a single element

Answer: (b)

Solution:

When current increases it implies that total impedance of the

circuit has decreased. If by addition of a capacitor total

impedance decreases, it means other element is inductor.

XL = jωL

XC = 1
jωC

X = XL + XC
= j�ωL − ω1C� < XL

21. Resistor R1 in the circuit below has been adjusted so that I1 = 1

mA. The bipolar transistors Q1 and Q2 are perfectly matched and

have very high current gain, so their base currents are negligible.

The supply voltage VCC is 6 V. The thermal voltage kT/q is 26

mV

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The value of R2 (in Ω) for which I2 = 100 µA is _________.

Answer: 598.67

Solution:

Vbe

IC = Ise Vt
∴ Vbe1 = Vt ln �IICS� = Vt ln �II1S�

Vbe2 = Vt ln �II2S�

From the circuit,

Vbe1 = Vbe2 + I2R2 and Vt = KT = 26 mV
q

∴ R2 = Vbe1 − Vbe2 = Vt ln�II12� = 26mV ln�110m0μAA� = 598.67Ω
I2 I2 100μA

22. Consider the following statements:

Dominant-pole frequency compensation in an Op-Amp

1. increases the slew-rate of the Op-Amp

2. increases the stability of the Op-Amp

3. reduces the bandwidth of the Op-Amp

4. reduces the CMRR of the Op-Amp

Which of the statements given above are correct?

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a) 1 and 3 only
b) 1, 2 and 4
c) 1 and 2 only
d) 2 and 3 only
Answer: (c)
23. Statement (I): An ideal Op-Amp when used to make an
inverting amplifier, has both input terminals at the same
Potential.
Statement (II): CMRR of the Op-Amp is low.
a) Both Statement (I) and Statement (II) are individually true

and Statement (II) is the correct explanation of Statement (I).
b) Both Statement (I) and Statement (II) are individually true

but Statement (II) is not the correct explanation of Statement
(I).
c) Statement (I) is true but Statement (II) is false.
d) Statement (I) is false but Statement (II) is true.
Answer: (c)
Solution:
An ideal op-amp, due to concept of virtual short, both the
inverting and non-inverting terminals of the op-amp should be at
the same voltage i.e. both the input terminals are at the same
potential.

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CMRR = Ad = Differential gain
Acm Common mode gain

Common-mode gain is the gain when some inputs are applied at

both the input terminals of OP-Amps

So VO = Acm(V1 – V2)
∴ Acm = 0

∴ CMRR = ∞ (ideally) or very high practically

24. A three terminal monolithic IC regulator can be used as

a) an adjustable output voltage regulator alone

b) an adjustable output voltage regulator and a current regulator

c) a current regulator and a power switch

d) a current regulator alone

Answer: (a)

25. Match List-I (Name of the circuit) with List-II (Property of

the circuit) and select the correct answer using the code given

below the lists:

List-I

A. Pre-amplifier

B. Power amplifier

C. Rectifier circuit

D. Purely resistive circuit

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List-II

1. Non-linear circuit

2. Lumped, linear, passive, bilateral, finite circuit

3. Large signal amplifier

4. Small signal amplifier

Codes:

A BC D

a) 4 2 1 3

b) 1 3 4 2

c) 4 3 1 2

d) 1 2 4 3

Answer: (c)

26. An n-channel JFET has IDSS = 1 mA and VP = –5V. Its

maximum transconductance is ________ ms.

Solution: 0.4

The maximum transconductance is

gm max = 2IDSS
|VP|

= 2 × 1 × 10−3
|5|

= 0.4 × 10−3

= 0.4 ms

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27. Statement (I): In a transistor designed to be used for power
amplification, the collector size is largest relative to the emitter
and base.
Statement (II): The collector is connected to the body of the
transistor and hence to a heat sink for heat dissipation to be
effective.
a) Both Statement (I) and Statement (II) are individually true
and Statement (II) is the correct explanation of Statement (I).
b) Both Statement (I) and Statement (II) are individually true
but Statement (II) is not the correct explanation of Statement
(I).
c) Statement (I) is true but Statement (II) is false.
d) Statement (I) is false but Statement (II) is true.
Answer: (b)
Solution:

• Collector is provided with the largest area in comparison to
emitter and base because all the electronics emitted from the
emitter are properly collected by the collector in the forward
active mode.

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• The Collector is also made largest to withstand heat
dissipation.

28. Consider the following circuit:

What is the type of circuit given above?
a) Monostable
b) Ramp generator
c) VCO
d) Bistable multivibrator
Answer: (b)
29. The current stability of a common collector amplifier can be
improved by
a) decreasing emitter and base resistance
b) increasing emitter and base resistance
c) decreasing emitter resistance and increasing base resistances
d) increasing resistance and decreasing base resistance
Answer: (d)
30. Monolithic IC design is based on
a) extensive use of transistor and diodes
b) extensive use of RC coupling

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c) using high value resistors and capacitors
d) making the area of circuit elements as large as possible
Answer: (a)

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