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The correct option is (C) The overall output RESISTANCE is less than the lowest output resistance in all amplifiers used

Easy explanation: In cascading, the output of one amplifier is connected to the input of another amplifier. It is used to increase gain while obtaining DESIRED VALUES of input and output resistances. Overall input resistance is the same as input resistance of the first amplifier and net output resistance is the same as output resistance of the last (n^th) amplifier in the network. When amplifiers are connected in cascade, then loading effect does OCCUR.

The CORRECT choice is (a) True

For explanation I would say: Considering each amplifier has lower cut-off frequency fL1 then net lower cut-off frequency for a network of N cascaded amplifiers is fL = \(\frac{f_{L1}}{\sqrt{2^{1/N}}-1}\)

For N>=2, \(\sqrt{2^{1/N}}-1\) < 1, thus fL > fL1.

The correct option is (B) 83 HZ

Easiest explanation: fL1 = 25Hz

FL = \(\FRAC{f_{L1}}{\sqrt{2^{1/N}}-1} = \frac{25}{\sqrt{2^{1/8}} -1} = \frac{25}{\sqrt{0.0905}}\)

fL = 83 Hz.

Right choice is (a) 0.26

Best explanation: To calculate NET bandwidth B2 = B1 x \(\SQRT{2^{1/N}}-1\)

B2/B1 = \(\sqrt{2^{1/N}}-1\)

B2/B1 = 0.26.

Right answer is (d) It is a current buffer

The explanation is: A Darlington AMPLIFIER has a very HIGH input resistance, low OUTPUT resistance, UNITY voltage gain and a high current gain. It is a voltage buffer, not a current buffer.

The CORRECT OPTION is (B) 298 mA

For explanation: IB = 20 μA

IC = β.IB = 15000 x 20μ = 300 mA

IC1 = β1.IB = 100.20μ = 2mA

IC2 = 300 – 2 = 298mA.

Right option is (a) True

Best EXPLANATION: Darlington pair is an emitter follower circuit, in which a DARLING pair is used in PLACE of a single

transistor. It ALSO provides a large β as PER requirements.

Right answer is (b) To prevent an increase in the INPUT resistance due to the biasing NETWORK

To explain I would say: A BOOTSTRAP biasing network is a special biasing circuit USED in the DARLINGTON amplifier to prevent the decrease in input resistance due to the biasing network being used. Capacitors and resistors are added to the circuit to prevent it from happening.

The CORRECT OPTION is (c) A cascade of CE and CB AMPLIFIERS

Explanation: A cascade amplifier is a cascade network of CE and CB amplifiers, or CS and CG amplifiers.

It is used as a wide-band amplifier.

Correct answer is (c) RI=956 Ω, RO=2 kΩ

For EXPLANATION: RO = RC = 2kΩ

Input RESISTANCE = hie||50k||40k = 0.956 kΩ.

Correct option is (C) 10V

The explanation is: RL = 10kΩ

IF = βIL

IL = IF/β = 10/10 = 1mA

VL = ILRL = 10V.

Correct option is (a) True

For EXPLANATION I WOULD say: In any circuit, the feedback depends on the configuration of resistor network and presence of capacitances. Consider a collector to base bias circuit, in which base resistance causes voltage shunt feedback. HOWEVER, presence of an emitter resistance provides a second feedback of current series TYPE.

Correct OPTION is (c) Series-series feedback

To explain I would say: In series-series feedback, the output is current sampled, that is it is in series with the load. Also, INPUT is a voltage MIXER, which is in series with signal source. So feedback factor

Β = VF/IL in OHMS.

Right option is (c) Series-Series feedback

Easiest explanation: Given that input is 14V, feedback is 6V and source is 20 V, we can see

VI = VS – VF, which is voltage mixing. ALSO, β is in ohms that is voltage/CURRENT. Since OUTPUT of feedback is voltage and input is current, the output has current sampling. Thus, CONFIGURATION is a series-series feedback/current – series feedback.

The correct choice is (c) – 46.88 dB

Easiest EXPLANATION: Return ratio is Aβ, where β=FEEDBACK FACTOR, and A=open loop gain.

Amount of feedback is AF/A = 1 + Aβ

In DECIBELS, amount of feedback = -20log10(1+Aβ) = -20log10(221) = -46.88 decibels.

Correct choice is (C) 5000

To elaborate: Feedback FACTOR, β=10.

AF = A/(1+Aβ) = 50

A = 20*50 = 1000

Output is thus 5000.

Right ANSWER is (d) 0.01

To explain: AF = A/(1+Aβ)

Aβ = 24

A = 8

Relative change in GAIN = dAF/AF = dA/A(1+Aβ)

dAF/AF = 2/8*25 = 0.01.

Correct option is (B) Output RESPONSE of Decreased frequency DISTORTION

To explain I would SAY: Output should have increased Bandwidth and decreased frequency distortion. If bandwidth increases, the distortion cannot increase since it’s a case of negative FEEDBACK. Also, the distortion cannot remain the same.

Correct option is (c) Circuit R

To explain: Greater desensitivity INDICATES better stability in gain. More desensitivity means gain becomes SMALLER, but stable.

For circuit Q, desensitivity = 1/S = 10

Circuit R has higher desensitivity, hence most stable.

Right option is (B) 10

To explain I would say: We can SIMPLY USE the RATIO of 0.1 to find the answer.

Loop gain Aβ = 9

1+Aβ = 10

Desensitivity = 1/S = 1+Aβ = 10.

Correct CHOICE is (a) 0.5

To explain I WOULD say: DHF = DH/1+Aβ

1 + Aβ = DH/DHF = 10/5 = 2

Sensitivity = \(\frac{1}{2}\) = 0.5.

Correct OPTION is (c) 0.0041

Easy EXPLANATION: AF = A/1+Aβ

A = A1 x A2 = 1200

Thus 1 + Aβ = 1200/200 = 6

Aβ = 5

β = 5/A = 0.0041.