6. 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, A_{oL}(f) = V_{o}/V_{id}

V_{O} = A_{OL}(f) × (V_{in1}-V_{in2})

= 142 dB×(6.25-3.7)

= 142×2.55 = 0.36v.

7. Express the open-loop gain of the op-amp in complex form?

A. A/√ [1+(f/f_{o})^{2}
B. 20log{A/√[1+(f/f_{o})^{2}}
C. A/[1+j(f/f_{o})].
D. None of the mentioned

Answer: C

The open-loop gain of the op-amp A_{OL}(f) is a complex quantity and is expressed as

A_{OL}(f) = A/[1+ j(f/f_{o})] .

The remaining equations are expressed in polar form.

8. Determine the difference between two A_{OL}(f) at 50Hz and 500Hz frequencies? (Consider the op-amp to be 741C.

A. 40dB
B. 30dB
C. 20dB
D. 10dB

Answer: C

A_{OL}(f) dB= 20log[√ [1+ (f/f_{o})^{2}]

At f= 50 Hz,

A_{OL}(f) dB = 20log(200000)- 20log(√(1+(50/5)^{2}) = 106.02-20.04 ≅ 86dB

At f= 500Hz

A_{OL}(f) dB =20log(200000)-20log(√(1+(500/5)^{2}) = 106.02-40 ≅ 66dB

Therefore, the difference between A_{OL}(f)dB = 86-66 = 20dB.

9. 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/f_{o})

= -tan^{-1}(0/5) = 0^{o}

10. At what frequency A_{OL}(f)=A?

A. 50Hz
B. 10Hz
C. 5Hz
D. 0Hz

Answer: D

For any frequency less than break frequency (f_{o} =5Hz) the gain is approximately constant and is equal to A.

For example, f_{o} =0Hz,

Then A_{OL}(f) dB= 20log(200000-20log[√1+(0/5)^{2})]