11. What happens if 741 op-amps are configured as a closed-loop inverting amplifier?
A. Gain increases
B. Gain roll-off at a rate of 20dB/decade
C. No gain roll-off takes place
D. Gain decreases
Answer: B
Whether the op-amp is inverting / non-inverting the gain will always roll off at a rate of 20dB/decade, using only resistive components regardless of the value of its closed-loop gain.
12. Op-amp requiring external compensating components is called as
A. Tailored frequency response op-amp
B. Compensating op-amp
C. Transient op-amp
D. High-frequency op-amp
Answer: A
Op-amps using external components like resistor and capacitor to form the compensating network are sometimes called tailored frequency response op-amps because the user has to provide the compensation if it is needed to tailor the response.
13. In the first generation op-amp 709c, the open-loop bandwidth of gain versus frequency curve
A. Increases from the innermost compensated curve to the outermost
B. Decrease from the innermost compensated curve to the outermost
C. Increases from the outermost compensated curve to the innermost
D. Decreases from the outermost compensated curve to the innermost
Answer: D
The gain versus frequency curve of 709c decreases from the outermost compensated curve to the innermost.
For example, if C1 =10pF, R1 = 0Ω and C2 = 3pF, the bandwidth ≅ 5kHz.
While if C1 =5000pF, R1 =1.5Ω and C2= 200pF, the bandwidth will be 100Hz.
14. Which type of op-amp offers relatively broader open-loop bandwidth?
A. Compensated op-amp
B. Uncompensated op-amp
C. Tailored frequency response op-amp
D. Non-compensated op-amp
Answer: B
The uncompensated op-amps offer broader open-loop bandwidth whereas; the internally compensated op-amps have very small open-loop bandwidth.
15. How the performance of an op-amp circuit can be improved?
A. By using a non-compensating network
B. By using frequency network
C. By using compensating network
D. None of the mentioned
Answer: C
The compensating networks are used to improve /modify the performance of an op-amp circuit over the desired frequency range by controlling its gain and phase shift.
16. Which op-amp requires an external compensating network?
A. Op-amp 771
B. Op-amp 351
C. Op-amp 709
D. Op-amp 741
Answer: C
Op-amp 709 is the first-generation op-amp. Generally, a first-generation op-amp is required for an external compensating network.
17. IC 741c op-amp belongs to
A. Compensated op-amp
B. Uncompensated op-amp
C. Non-compensated op-amp
D. None of the mentioned
Answer: A
741c belongs to later generation op-amp and it has an internal compensating network. In an internally compensated op-amp, the compensating network is designed into the circuit to control the gain and phase shift of the op-amp and they are called compensating op-amp.
18. Determine the output voltage for an op-amp with single break frequency.
A. VO ={ jXC /[Ro+(iXC)]} × AVid
B. VO = = AVid / [1+ j2πfRoC].
C. VO = AVid /(Ro+j2πfC.
D. VO = Vid / [Ro-(j2πfRoC.].
Answer: B
The output voltage for an op-amp with single break frequency,
VO = {(-jXC) / [(Ro)-(jXC)} ×AVid
∵ -j=1/j & XC =1/2πfC
=> VO = {(1/j2ΠfC./[Ro + (1/ j2πfC.] } × AVid
= AVid / [1+ j2πfRoC].
19. Compute the break frequency of an op-amp, if the output resistance=10kΩ and the capacitor connected to the output =0.1µF.
A. 159.2Hz
B. 6.28Hz
C. 318.4Hz
D. 1000Hz
Answer: A
Break frequency of the op-amp is given as
fo = 1/(2πRoC.= 1/ (2π×10kΩ×0.1µF)
= 1/ (6.28×10-3) = 159.2Hz.
20. The open-loop voltage gain as a function of frequency is defined as
A. AOL(f) = VO/Vin
B. AOL(f) = VO/Vid
C. AOL(f) = VO/Vf
D. All of the mentioned
Answer: B
The open-loop voltage gain as a function of frequency is defined as the ratio of output voltage to the difference of input voltages.
21. Which of the following factor remain fixed for an op-amp?
A. Open loop voltage gain
B. Gain of the op-amp
C. Operating frequency
D. Break frequency of the op-amp
Answer: D
Break frequency fo depends on the value of capacitors and on output resistance. Therefore, fo is fixed for an op-amp.
22. Find the gain magnitude and phase angle of the op-amp using the specifications:
f= 50Hz; fo=5Hz ; A=140000.
A. AOL(f)= 22.92dB , Φ(f) = – 89.99o
B. AOL(f)= 66dB , Φ(f) = – 90o
C. AOL(f)= 26dB, Φ(f) = – 89.99o
D. AOL(f)= 20dB , Φ(f) = – 84.29o
23. Consider an op-amp where the inverting input voltage =3.7mv, non-inverting input voltage=6.25mv and open-loop voltage gain =142dB. Find the output voltage.
A. 0.21v
B. 0.45v
C. 0.78v
D. 0.36v
Answer: D
Open loop voltage gain, AoL(f) = Vo/Vid
VO = AOL(f) × (Vin1-Vin2)
= 142 dB×(6.25-3.7)
= 142×2.55 = 0.36v.
24. Express the open-loop gain of the op-amp in complex form?
A. A/√ [1+(f/fo)2
B. 20log{A/√[1+(f/fo)2}
C. A/[1+j(f/fo)].
D. None of the mentioned
Answer: C
The open-loop gain of the op-amp AOL(f) is a complex quantity and is expressed as
AOL(f) = A/[1+ j(f/fo)] .
The remaining equations are expressed in polar form.
25. Determine the difference between two AOL(f) at 50Hz and 500Hz frequencies? (Consider the op-amp to be 741C.
A. 40dB
B. 30dB
C. 20dB
D. 10dB
Answer: C
AOL(f) dB= 20log[√ [1+ (f/fo)2]
At f= 50 Hz,
AOL(f) dB = 20log(200000)- 20log(√(1+(50/5)2) = 106.02-20.04 ≅ 86dB
At f= 500Hz
AOL(f) dB =20log(200000)-20log(√(1+(500/5)2) = 106.02-40 ≅ 66dB
Therefore, the difference between AOL(f)dB = 86-66 = 20dB.
26. At what frequency, the phase shift between input &output voltage will be zero?
A. -40Hz
B. 0Hz
C. -22Hz
D. 20Hz
Answer: B
At 0Hz the phase shift between input and output voltage is zero.
At f=0Hz
φ(f) = – tan-1 (f/fo)
= -tan-1(0/5) = 0o
27. At what frequency AOL(f)=A?
A. 50Hz
B. 10Hz
C. 5Hz
D. 0Hz
Answer: D
For any frequency less than break frequency (fo =5Hz) the gain is approximately constant and is equal to A.
