11. Define the polarity of the output offset voltage in a practical op-amp?
A. Positive polarity
B. Negative polarity
C. Positive or negative polarity
D. None of the mentioned
Answer: C
The output offset voltage is a DC voltage, it may be positive or negative in polarity depending on whether the potential difference between two input terminals is positive or negative.
12. The input offset voltage of 741 op-amps has an absolute maximum value of 6mv, which means
A. Minimum difference between input terminals in 741 op-amps can be large as 6mv DC
B. Minimum difference between input terminals in 741 op-amps can be large as 6mv AC
C. Maximum difference between input terminals in 741 op-amps can be large as 6mv DC
D. Maximum difference between input terminals in 741 op-amps can be large as 6mv AC
Answer: C
Given that, the absolute maximum value for a 741 is Vio= 6mv. Therefore, the voltage at the non-inverting input terminal may differ from that at the inverting input terminal by as much as 6mv dc. Also, the output offset voltage is a DC voltage and it cannot be an AC voltage.
13. If three different 741 op-amps are taken and the corresponding output offset voltage for each of them is measured. The output voltage in these three op-amps have
A. Same amplitude and polarity
B. Different amplitude and polarity
C. Same amplitude and different polarity
D. Different amplitude and same polarity
Answer: B
Even though the op-amps are of the same type, the output voltage in these three op-amps is not of the same amplitude and polarity, because of mass production.
14. To reduce the output offset voltage VooT to zero
A. Input offset voltage compensating network is added at the inverting input terminal
B. Input offset voltage compensating network is added at the non-inverting input terminal
C. Input offset voltage compensating network is added at the output terminal
D. None of the mentioned
Answer: D
To reduce the VooT to zero, the external circuit is added at the input terminal of the op-amp which will give the flexibility of obtaining input offset voltage of proper amplitude and polarity. The input terminal can be inverting or non-inverting.
15. Which of the following op-amp does not need a compensating network?
A. 777
B. 741
C. 748
D. All of the mentioned
Answer: D
The compensating network is not needed for these op-amps because they have offset null pins.
16. What will the condition of the op-amp before applying any external input
A. Compensated
B. Biased
C. Balanced
D. Zero
Answer: C
Before applying external input to the op-amp, the output offset voltage should be reduced to zero with the help of an offset voltage compensating network. At this condition, the op-amp is said to be balanced or nulled.
17. Find Thevenin’s equivalent for resistance and voltage?
A. 1-iii, 2-ii, 3-1
B. 1-ii, 2-I, 3-iii
C. 1-I, 2-ii, 3-iii
D. 1-ii, 2-iii, 3-i
Answer: B
The maximum Thevenin equivalent resistance Rmax occurs when the wiper is at the center of the potentiometer and the maximum Thevenin equivalent voltage Vmax is equal to +Vcc or –Vee when the wiper is uppermost or lowest in the potentiometer.
18. What is done to compensate for the voltage, when V1 > V2?
A. Move the wiper towards +Vcc
B. Move the wiper towards –Vee
C. Keep the wiper at the center of the potentiometer
D. None of the mentioned
Answer: A
V1 > V2 implies that output offset voltage is positive. This means that V2 should be increased until it is equal to V1. The wiper can be moved towards +Vcc until the output offset voltage is reduced to zero.
19. Calculate the maximum Thevenin equivalent resistance, if a 10kΩ potentiometer is used?
A. 0.4kΩ
B. 5 kΩ
C. 2.5kΩ
D. 4kΩ
Answer: C
Rmax= Ra/2 || Ra/2 = Ra/4.
Given potentiometer, Ra = 10kΩ
=> Therefore, Rmax = 10kΩ/4 =2.5kΩ.
20. Find the value of Ra and Rb from the circuit shown?
A. Ra =4.6kΩ ; Rb= 9kΩ
B. Ra =7.3kΩ ; Rb= 3.4kΩ
C. Ra =2.5kΩ ; Rb= 5.1kΩ
D. Ra =4kΩ ; Rb= 10kΩ
Answer: D
We know that input offset voltage, Vio =(Rc*Vmax)/ Rb
=> Rb = Vmax*(Rc / Rb )
= (10v/10mv)*10Ω (∵ Vio specified on the datasheet is 10mv for LM307 op-amp).
=> Rb =10000 = 10kΩ.
Since Rb > Rmax let us choose Rb = 10*Rmax. (Where Rmax = Ra/4).
∴ Rb = (10*Rb)/4 and Ra = Rb/2.5
= 10kΩ/2.5=4kΩ.
21. Why does an op-amp without feedback is not used in linear circuit application?
A. Due to high current gain
B. Due to high voltage gain
C. Due to high output signal
D. All of the mentioned
Answer: B
In an op-amp without feedback, the voltage gain is extremely high (ideally infinite). Because of the high risk of distortion and clipping of the output signal, an op-amp in open loop configuration is not used in linear circuit applications.
