# Differential Amplifiers with Multiple Op-Amp MCQ [Free PDF] – Objective Question Answer for Differential Amplifiers with Multiple Op-Amp Quiz

1. Why differential amplifiers are preferred for instrumentation and industrial applications?

A. Input resistance is low
B. Produce amplified output
C. Amplify individual input voltage
D. Reject common-mode voltage

Differential amplifiers are preferred in these applications because they are better able to reject common-mode voltage than single input circuits and present balanced input impedance.

2. Which of the following is a combination of inverting and non-inverting amplifiers?

A. Differential amplifier with one op-amp
B. Differential amplifier with two op-amps
C. Differential amplifier with three op-amps
D. Differential amplifier with four op-amps

In a differential amplifier with one op-amp both the inputs are connected to separate voltage sources. So, if any one of the sources is reduced to zero, a differential amplifier acts as an inverting or non-inverting amplifier.

3. What will be the output voltage when Vx =0v?
(Where Vx –> inverting input terminal of differential amplifier with one op-amp)

A. Vo = -(1+R F/R1)*V1
B. Vo = -(1- R F/ R1)*V1
C. Vo = (1+ R F/ R1)*V1
D. Vo = (R F/ R1)*V1

When Vx =0v, the configuration is a non-inverting amplifier.

Vo = (1+ R F/ R1)*V1

4. Compute the output voltage from the following circuit diagram?

A. -17v
B. -27v
C. -39v
D. -15v

Since VB=0, the configuration becomes an inverting amplifier. Hence, the output due to VA is

Vo = -(RF/R1)*VA

= -(15kΩ/1.5kΩ)*2.7v

= -10*2.7 = -27v.

5. Compute the output voltage if the input voltage is reduced to zero in a differential amplifier with one op-amp?

A. Inverted Voltage
B. Same as the input voltage
C. Amplified inverted voltage
D. Cannot be determined

It is not mentioned clearly whether inverting input or non-inverting input is reduced to zero. Therefore, the output cannot be determined.

6. The difference between the input and output voltage is -1v and 17v. Calculate the closed-loop voltage gain of differential amplifier with one op-amp?

A. -51
B. 34
C. -17
D. 14

Voltage gain of differential amplifier with one op-amp

AD=Output voltage / Difference of input voltage

=> AD = 17v/-1v = -17v.

7. For the differential amplifier given below, determine the Vx and RF value. Assume that the circuit is initially nulled.

A. Vx = -8v, RF = 9.9kΩ
B. Vx = 8v, RF = 9.9kΩ
C. Vx = -8v, RF = -9.9kΩ
D. Vx = 8v, RF = -9.9kΩ

The closed loop voltage gain, AD = -(RF/R1)

=> RF = -3*3.3kΩ = -9.9kΩ

The net output is given is

VO=-(RF /R1)*(Vx-Vy)

=> Vx= Vy– Vo (-R1 /RF)

=> Vx = 6+6(3.3kΩ/9.9kΩ) = 6+2 = 8v.

8. The gain of a differential amplifier with one op-amp is the same as that of

A. The inverting amplifier
B. The non-inverting amplifier
C. Both inverting and non-inverting amplifier
D. None of the mentioned

The gain of the differential amplifier is given as AD= -(RF /R1), which is equivalent to the output voltage obtained from the inverting amplifier.

9. Find the value of input resistance for differential amplifier with one op-amp. If R1 = R2=100Ω and RF = R3 =5kΩ.

A. RIFx = 110Ω; RIFy = 6.7kΩ
B. RIFx = 100Ω; RIFy = 5.1kΩ
C. RIFx = 150Ω; RIFy = 7.2kΩ
D. RIFx = 190Ω; RIFy = 9.0kΩ

The input resistance of inverting amplifier is RIFx = (R1) and the input resistance of non-inverting amplifier is RIFy = (R2+ R3)

=> ∴ RIFx = 100Ω and

=> RIFy =100+5kΩ =5.1kΩ.

10. What is the net output voltage for a differential amplifier with one op-amp

A. Vo = -(RF /R1)*Vx
B. Vo = -(RF /R1)*(Vx -Vy)
C. Vo = (1+RF /R1)*(Vx -Vy)
D. None of the mentioned

The net output voltage for a differential amplifier with one op-amp is given as Vo= -(RF /R1)*(Vx-Vy).

11. The characteristics of a non-inverting amplifier are identical to

A. Differential Amplifier with one op-amp
B. Differential Amplifier with two op-amp
C. Differential Amplifier with three op-amp
D. None of the mentioned

The characteristics of a non-inverting amplifier are identical to the differential amplifier with two op-amps as the gain of the amplifier are same. i.e. A = 1+(RF /R1).

12. How is it possible to vary the voltage from the closed-loop gain to the open-loop gain?

A. By using a differential amplifier with a larger gain
B. By using a differential amplifier with a small gain
C. By using a differential amplifier with variable gain
D. By using a differential amplifier with differential gain

The variable gain in the differential amplifier can be obtained by using a potentiometer. So, depending on the position of the wiper in the potentiometer, voltage gain can be varied from the closed-loop gain to the open-loop gain.

13. A differential amplifier with two op-amps has the following specifications:
R1=R3=1.5kΩ ; RF=R2=5.7kΩ. Compute the input resistance for the following circuit.

A. RIFx = 83GΩ, RIFy = 317GΩ
B. RIFx = 90GΩ, RIFy = 400GΩ
C. RIFx = 59GΩ, RIFy = 269GΩ
D. RIFx = 36GΩ, RIFy = 156GΩ

The input resistance of the first stage is given as RIFy,

=> RIFy = Ri*[1+(A *R2)/( R2+ R3)]

= RIFy = 2MΩ*[(1+200000*5.7kΩ)/(5.7kΩ+1.5kΩ)]

= 3.166*1011 = 317GΩ.

Similarly the input resistance of the second stage amplifier is given as RIFx,

=> RIFx = Ri*[1+(A *R1)/( R1+ RF)]

= RIFx = 2MΩ*[(1+200000*1.5kΩ)/(5.7kΩ+1.5kΩ)] = 83GΩ.

14. Determine the output voltage from the diagram. Where V1=-4.3Vpp and V2 = -5.1Vpp sinewave at 1000hz.

A. 12 Vpp sinewave at 1000hz
B. 13.3 Vpp sinewave at 1000hz
C. 14 Vpp sinewave at 1000hz
D. 11 Vpp sinewave at 1000hz

In differential amplifier with two op-amps, R1=R3 and RF=R2.

The closed loop voltage gain, AD = 1+( RF/ R1) = 1+(7.5kΩ/480Ω) = 16.62.

Output voltage , Vo = AD(V1– V2) = 16.62(-4.3Vpp+5.1Vpp) = 13.29 =13.3 Vpp sinewave at 1000hz.

15. Determine the output resistance of the differential amplifier with three op-amps. The op-amp used is 741c, with A=200000 and Ro. The output and difference of input voltages are 44 and 11.

A. 5.5mΩ
B. 3.5mΩ
C. 2.4mΩ
D. 1.5mΩ

Gain AD = Vo/Vid = 44/11 =4.

The output resistance of differential amplifier is given as

= 75/[1+(200000/4)] = 1.5mΩ.

16. Find the input resistance for the given circuit

A. RIF = Ri*[1+(A*R4+R5)/(2* R4+R5)].
B. RIF = Ri*[1+(A*R4)/(2* R4+R5)].
C. RIF = Ri*[1+(A*R4+R5)/(R4+R5)].
D. None of the mentioned

The input resistance of the op-amp with feedback is given as RIF= Ri*(1+AB.

From the diagram, B=Vf/Vo = [R4+R5)/(2* R4+R5)].

Therefore, RIF = Ri*[1+(A*R4+R5)/(2* R4+R5)].

17. The bandwidth of the differential amplifier increases, if the value of

A. Open loop voltage gain decreases
B. Closed-loop voltage gain decreases
C. Differential voltage gain decreases
D. All of the mentioned

The bandwidth of the differential amplifier is inversely proportional to the closed-loop voltage gain.

18. Calculate the bandwidth for the given circuit. (Where the 741 op-amp has a gain of 200000).

A. 560hz
B. 390khz
C. 25.6khz
D. 1.5Mhz

Gain of the amplifier

= -{[1+(2*5kΩ/2.5kΩ)]*(3.9kΩ/500Ω)} (since, RF = R3 and R2 = R1).

The bandwidth for the circuit is calculated as

= (200000*5hz)/39

(For 741 op-amp 5hz is the break frequency)

=> fF = 25.6khz.

19. Calculate the output voltage. If Vx=3.9Vpp and Vy = 5.5 Vpp sinewave at 1khz for the following circuit.

A. 11 Vpp sinewave at 1Khz
B. -8Vpp Vpp sinewave at 1Khz
C. -10 Vpp sinewave at 1Khz
D. 6 Vpp sinewave at 1Khz

= -15kΩ/3kΩ = -5kΩ.

The output voltage, Vo = – AD*(V x– V y)

= 5*(3.9-5.5) = -8 Vpp sinewvae at 1Khz.

20. Miniaturization of components with superior performance can be obtained in

A. Integrated circuits
B. Discrete circuits
C. Both integrated and discrete circuits
D. None of the mentioned