# Mechatronics Questions and Answers – Analog Signal Conditioning – Operational Amplifier

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This set of Mechatronics Multiple Choice Questions & Answers (MCQs) focuses on “Analog Signal Conditioning – Operational Amplifier”.

1. What is the slew rate of an ideal operational amplifier?
a) 0
b) 1
c) 100
d) Infinite

Explanation: The slew rate of an ideal operational amplifier is infinite. Slew rate of an operational amplifier is defined as the rate of change of output voltage with respect to time (i.e) dv/dt. An ideal operational amplifier takes no time to change the output, therefore dt=0. So dv/dt becomes infinite and hence an ideal operational amplifier has infinite slew rate.

2. What is the open loop gain of an ideal operational amplifier?
a) 0
b) 1
c) Infinite
d) -1

Explanation: The open loop gain of an ideal operational amplifier is infinite. Open loop gain of an operational amplifier is defined as output voltage divide by input voltage (Vout/Vin). In ideal case the operational amplifier produces infinite output voltage when very low input voltage is applied.

3. What is the phase difference between input signal and output signal when input is provided to the inverting terminal of the operational amplifier?
a) 0 degree
b) 90 degree
c) 180 degree
d) 45 degree

Explanation: There 180 degree phase difference between input signal and output signal when input is provided to the inverting terminal of the operational amplifier. When input is provided to the negative terminal it adds 180 degree phase shift to the input signal and inverts the input signal, that is why the terminal is also called as inverting terminal.
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4. What is the slew rate of an operational amplifier whose output voltage increases by 9 volt in 18 micro second?
a) 0.5 Volt/microsecond
b) 5 Volt/microsecond
c) 50 Volt/second
d) 2 Volt/microsecond

Explanation: Given:
Change in voltage(dv)=9 Volt
Change in time(dt)= 18 micro second
Slew rate=dv/dt=9V/18 microsecond=0.5 volt/microsecond

5. A practical operational amplifier has infinite bandwidth.
a) True
b) False

Explanation: A practical operational amplifier does not have infinite bandwidth. It is so because at higher frequency signals the efficiency of amplification starts to decrease. The practical operational amplifier cannot amplify signals with very high frequency. Ideal operational amplifiers have infinite bandwidth.

6.An ideal Operational amplifier should have Infinite output resistance and zero input resistance.
a) True
b) False

Explanation: An ideal Operational amplifier should have zero output resistance and infinite input resistance. Infinite input resistance allows any input signal to drive the operational amplifier and zero output resistance facilitates the operational amplifier to drive infinite number of loads.

7. CMRR stands for ____
a) Common mode rejection ratio
b) Common mains rejection ratio
c) Common mode reluctance ratio
d) Common mode rejection rate

Explanation: CMRR stands for Common mode rejection ratio. It is the ability of the Operational Amplifier to reject the common mode signals in the inverting and non-inverting terminals. It is the ratio of differential voltage gain to common mode voltage gain.

8. Which of the integrated circuit mentioned below is the name of an operational amplifier?
a) BC541
b) LM741H
c) TIP122
d) TIP135

Explanation: LM741H is most commonly used operational amplifier. It comes in 8-pin TO-99 package with an operating voltage of +15 volt to -15 volt. It provides high voltage gain with a bandwidth of 1.5MHz.

9. PSRR stands for ________
a) Power Supply Relaxation Ratio
b) Power Supply Rejection Rate
c) Power Supply Rejection Ratio
d) Power Supply Rejecting Ratio

Explanation: PSRR stands for Power Supply Rejection Ratio. It is defined as the ratio of change in the input supply voltage to the equivalent output voltage it generates. An ideal operational amplifier has infinite power supply rejection ratio.

10. What is the common mode rejection ratio of an operational amplifier which has common mode gain=0.5 and differential gain=1200?
a) 600
b) 1200
c) 2400
d) 1200 db

Explanation: Given:
Common mode gain=0.5
Differential gain=1200
Common mode rejection ratio(CMRR)=differential gain(Ad) / common mode gain(Ac)

11. What is the slew rate of an operational amplifier whose output voltage increases by 6 volt in 12 micro second?
a) 0.5 Volt/microsecond
b) 5 Volt/microsecond
c) 50 Volt/second
d) 2 Volt/microsecond

Explanation: Given:
Change in voltage(dv)=6 Volt
Change in time(dt)=12 micro second
Slew rate=dv/dt=6V/12 microsecond=0.5 volt/microsecond

12. What is the slew rate of an operational amplifier whose output voltage increases by 5 volt in 15 micro second?
a) 0.33 Volt/microsecond
b) 3 Volt/microsecond
c) 50 Volt/second
d) 2 Volt/microsecond

Explanation: Given:
Change in voltage(dv)=5 Volt
Change in time(dt)=15 micro second
Slew rate=dv/dt=5V/15 microsecond=0.33 volt/microsecond

13. What is the common mode rejection ratio of an operational amplifier which has common mode gain=0.2 and differential gain=1300?
a) 650
b) 6500
c) 2400
d) 1200 db

Explanation: Given:
Common mode gain=0.5
Differential gain=1200
Common mode rejection ratio(CMRR)=differential gain(Ad)/common mode gain(Ac)

14. What is the common mode rejection ratio of an operational amplifier which has common mode gain=0.6 and differential gain=1000?
a) 650
b) 6500
c) 1666.66
d) 1200 db

Explanation: Given:
Common mode gain=0.6
Differential gain=1000
Common mode rejection ratio(CMRR)=differential gain(Ad) / common mode gain(Ac)

15. Common mode rejection ratio is defined as (common mode gain)/(differential mode gain).
a) True
b) False

Explanation: Common mode rejection ratio is defined as (differential mode gain)/ (common mode gain). It is the ability of the Operational Amplifier to reject the common mode signals in the inverting and non-inverting terminals.

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