Linear Integrated Circuit Multiple Choice Questions on “Differential Amplifiers with Multiple Op-Amp – 1”.
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
Answer: D
Clarification: 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 amplifier?
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
Answer: A
Clarification: In differential amplifier with one op-amp both the inputs are connected to separate voltage source. So, if any one of the source is reduced to zero, 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
Answer: C
Clarification: When Vx =0v, the configuration is a non-inverting amplifier.
4. Compute the output voltage from the following circuit diagram?
A. -17v
B. -27v
C. -39v
D. -15v
Answer: B
Clarification: Since VB=0, the configuration becomes as 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 differential amplifier with one op-amp?
A. Inverted Voltage
B. Same as the input voltage
C. Amplified inverted voltage
D. Cannot be determined
Answer: D
Clarification: 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 are -1v and 17v. Calculate the closed loop voltage gain of differential amplifier with one op-amp?
A. -51
B. 34
C. -17
D. 14
Answer: C
Clarification: 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
Answer: D
Clarification: 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 differential amplifier with one op-amp is 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
Answer: A
Clarification: The gain of 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
Answer: B
Clarification: 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 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
Answer: B
Clarification: The net output voltage for differential amplifier with one op-amp is given as Vo= -(RF /R1)*(Vx-Vy).