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Test: Differential Amplifiers - Electronics and Communication Engineering (ECE) MCQ


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10 Questions MCQ Test GATE ECE (Electronics) Mock Test Series 2025 - Test: Differential Amplifiers

Test: Differential Amplifiers for Electronics and Communication Engineering (ECE) 2024 is part of GATE ECE (Electronics) Mock Test Series 2025 preparation. The Test: Differential Amplifiers questions and answers have been prepared according to the Electronics and Communication Engineering (ECE) exam syllabus.The Test: Differential Amplifiers MCQs are made for Electronics and Communication Engineering (ECE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Differential Amplifiers below.
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Test: Differential Amplifiers - Question 1

For an amplifier, which parameter is calculated with Vs = 0?

Detailed Solution for Test: Differential Amplifiers - Question 1

Concept:

VT & IT method:
Deactivate all the independent sources and do not disturb the dependent sources.
Rth = VT / IT

Deactivate source:
voltage source →  short circuit  i.e V = 0
current  source → open  circuit  i.e I = 0

Analysis:
While calculating Ro in the amplifier we apply VT  & IT method at the output.
Apply VT  & IT method to calculate R0
R0 = VT / IT While deactivating sources 
R0 is calculated when V0 is short circuit i.e  Vs = 0. 

Test: Differential Amplifiers - Question 2

When voltage gain Av is greater than 1, the voltage gain in dB is ______.

Detailed Solution for Test: Differential Amplifiers - Question 2

Concept:
Voltage gain (Av):
Voltage gain (Av) is given as 

Av in dB is given as
AvdB = 20 log(Av)
Analysis:
AV dB = 20 log(Av
We know that for x >1, log(x) is always positive 
Therefore Av >1, log(Av) is always positive 
For  Av > 1
AvdB is positive

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Test: Differential Amplifiers - Question 3

A signal that is applied with equal strength to both inputs of a differential amplifier or an operational amplifier

Detailed Solution for Test: Differential Amplifiers - Question 3

Concept:
Common mode signal:

  • A common-mode signal is one that drives both inputs of a differential amplifier equally
  • The common-mode signal is interference, static and other kinds of undesirable pickup etc

Application:

  • The transistors in the differential amplifier are chosen to be closely matched
  • If both the transistors were matched in all respects then the balanced output would be theoretically zero
  • This is the important characteristic of a differential amplifier
  • It discriminates against common-mode input signals i.e. it refuses to amplify the common-mode signals
Test: Differential Amplifiers - Question 4

The negative feedback _______ input impedance of an amplifier.

Detailed Solution for Test: Differential Amplifiers - Question 4
  • Negative feedback reduces the gain of the system by a factor of (1 + Aβ).
  • Thus, because of feedback configuration, voltage gain will be decreased by a factor (1 + Aβ).
  • Voltage series type feedback tells us that the input of the system is series connected with the feedback loop and series connection corresponds to increased impedance.
  • Therefore the input impedance will be increased by a factor (1 + Aβ).
Test: Differential Amplifiers - Question 5

Effect of cascading several amplifier stages is to

Detailed Solution for Test: Differential Amplifiers - Question 5

Multistage amplifier:

  • In practical applications, the output of a single-stage amplifier is usually insufficient, though it is a voltage or power amplifier .hence they are replaced by multi-stage amplifiers.
  • In multi-stage amplifiers, the output of the first stage is coupled to the input of the next stage using a coupling device.
  • These coupling devices usually be a capacitor or a transformer, this process of joining two amplifier stages using a coupling device is called cascading
  • Now the overall gain is equal to the product of individual gains and hence the gain increases.
  • The frequency and bandwidth gets reduced.
Test: Differential Amplifiers - Question 6

An amplifier has a gain of 20 without feedback. If 10% of the output voltage is fed back by means of a resistance negative feedback circuit, the overall gain would be:

Detailed Solution for Test: Differential Amplifiers - Question 6

Concept:
For the negative feedback amplifier, the closed-loop gain is given by:

For the positive feedback amplifier, the closed-loop gain is given by:

Where AOL is the open-loop gain
β is the gain of the feedback
Calculation:
The gain of feedback (β) = 10/100 = 0.1
Gain without feedback = 20
Gain with negative feedback:


⇒ 6.67
Gain with feedback = 6.67

Test: Differential Amplifiers - Question 7

In a common emitter amplifier the unbypassed emitter resistance provides:

Detailed Solution for Test: Differential Amplifiers - Question 7

CE amplifier with unbypassed emitter resistance:

Performance parameters:

 

  • Lower voltage gain which provides better bias stabilization.
  • It works as a current series feedback network.
  • Input impedance increases.
  • Provide linear (undistorted) operation for larger input signals.
Test: Differential Amplifiers - Question 8

Which of the following characteristics is NOT desirable for the DC amplifier?

Detailed Solution for Test: Differential Amplifiers - Question 8

An op-amp has the following characteristics:

  • Input impedance (Differential or Common-mode) = very high (ideally infinity)
  • Output impedance (open loop) = very low (Ideally zero)
  • Voltage gain = very high (ideally infinity)
  • Common-mode voltage gain = very low (ideally zero), i.e. Vout = 0 (ideally), when both the inputs are at the same voltage, i.e. (zero "offset voltage")
  • Output can change instantaneously (Infinite Slew Rate)
  • The purpose of bias current is to achieve the ideal behavior in op-amp which is high CMRR, high differential gain, and high input impedance
Test: Differential Amplifiers - Question 9

In the figure shown below, what is the expected power level at the input to the receiver?

Detailed Solution for Test: Differential Amplifiers - Question 9

Concept:
Expected power level = Input power – Losses + Total Gain
Note: In dB, the gains are simply added.
To convert dBm to dBW and vice-versa:
P(dBm) = P(dBW) + 30
Calculation:
Given:
Input Power = -10 dBm
- 10 = P + 30
Input power = -40 dBW
Net losses = – 2 – 3 – 5 = –10 dBW
Net Gain = 20 + 30 = 50 dBW
Expected power level  will be:
= –40 – (2) + 30 – (3) + 20 – 5 = 0 dBW
P(dBm) = 0 + 30 
∴ 30 dBm will be the correct answer.

Test: Differential Amplifiers - Question 10

In an ideal CC amplifier circuit, output voltage is_______ its input voltage.

Detailed Solution for Test: Differential Amplifiers - Question 10

Concept:
Ideal common collector (CC) amplifier: cc amplifier is also known as 
(1) unity gain amplifier (AV≈ 1)
(2) voltage follower
(3) voltage-Buffer
(4) Impedance matching
Analysis:
In an ideal cc amplifier voltage gain is 1

Vo = Vi

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