# MCQs on Analog Signal Conditioning – Operational Amplifier

1 - Question

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 - Question

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 - Question

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. advertisement

4 - Question

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 - Question

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 - Question

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 - Question

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 - Question

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 - Question

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 - Question

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) Ad/Ac=2400.

11 - Question

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 - Question

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 - Question

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) Ad/Ac=6500.

14 - Question

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