# Frequency Compensating Network MCQ [Free PDF] – Objective Question Answer for Frequency Compensating Network Quiz

1. Variation in the operating frequency of op-amp causes

A. Variation in gain amplifier
B. Variation in gain phase angle
C. Variation in gain amplitude and its phase angle
D. None of the mentioned

The gain of the op-amp is a function of frequency. It will have a specific magnitude as well as a phase angle.

2. A graph of the magnitude of the gain versus frequency is called

A. Break frequency
B. Frequency response plot
C. Frequency stability plot
D. Transient response plot

A frequency response plot is obtained by plotting the gain of the op-amp responding to different frequencies.

3. In the frequency response plot, the frequency is expressed in

A. Anti-logarithmic scale
B. Logarithmic scale
C. Linear scale
D. Exponential scale

To accommodate large frequency ranges the frequency is assigned to a logarithmic scale.

4. Why the gain magnitude in the frequency response plot is expressed in decibels (dB.

A. To obtain gain > 105
B. To obtain gain < 105
C. To obtain gain = 0
D. To obtain gain = ∞

In the frequency response plot, gain magnitude is assigned a linear scale and is expressed in decibels to accommodate very high gain ( ≅ of the order 105 or higher).

5. Which technique is used to determine the stability of op-amp?

A. Frequency response plot
B. Transient response plot
C. Bode plot
D. All of the mentioned

Although frequency response and bode plots indicate the effect of frequency variation on gain, the Bode plot is generally used for stability determination and network design.

6. How many types of plots can be obtained in the AC analysis of the network using the Bode plot?

A. Five
B. Four
C. Three

Two types of plots can be obtained using the Bode plot. They are magnitude versus frequency and phase angle versus frequency plots.

7. What happens when the operating frequency of op-amp increases?

A. Gain of the amplifier decrease
B. Phase shift between output and input signal decrease
C. Gain and phase shift of amplifier decreases
D. None of the mentioned

When the operating frequency has increased the gain of the amplifier decrease. As it is linearly related to frequency, the phase shift is logarithmically related to frequency.

8. Which of the following causes change in gain and phase shift?

A. Internally integrated Resistor
B. Internally integrated inductors
C. Internally integrated Capacitor
D. All of the mentioned

The change in function of frequency is attributed to the internally integrated capacitor as well as a stray capacitor. These capacitors are due to the physical characteristic of the semiconductor device.

9. Which plot is not provided by the manufacturers?

A. Magnitude plot
B. Phase angle plot
C. Frequency response plot
D. None of the mentioned

Phase angle plots are not generally provided because phase shifts of later generation op-amp are less than 90o even at cross-over frequency.

1. Open-loop bandwidth of an op-amp extends its bandwidth from

A. 0 Hz to fo
B. 20dB to fo
C. 3dB to fo
D. 0.704dB to fo

The gain of the op-amp remains essentially constant from 0 to the break frequency fo and therefore rolls off at a constant rate of 20dB per decade. Thus, the open-loop bandwidth is the frequency band extending from 0Hz to fo.

2. 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

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.

3. 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

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.

4. 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

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.

5. 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

The uncompensated op-amps offer broader open-loop bandwidth whereas; the internally compensated op-amps have very small open-loop bandwidth.

6. 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

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.

7. Which op-amp requires an external compensating network?

A. Op-amp 771
B. Op-amp 351
C. Op-amp 709
D. Op-amp 741

Op-amp 709 is the first-generation op-amp. Generally, a first-generation op-amp is required for an external compensating network.

8. IC 741c op-amp belongs to

A. Compensated op-amp
B. Uncompensated op-amp
C. Non-compensated op-amp
D. None of the mentioned

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.

1. 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.].

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].

2. 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

Break frequency of the op-amp is given as

fo = 1/(2πRoC.= 1/ (2π×10kΩ×0.1µF)

= 1/ (6.28×10-3) = 159.2Hz.

3. 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

The open-loop voltage gain as a function of frequency is defined as the ratio of output voltage to the difference of input voltages.

4. 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

Break frequency fo depends on the value of capacitors and on output resistance. Therefore, fo is fixed for an op-amp.

5. 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

The open loop gain magnitude

|AOL(f)|= 20log[A/√[1+ f/fo)2]

= 20logA-20 log[A/√ [1+(f/fo)2]

= 20log(140000)- 20log[√(1+(50,000/5)2)]

AOL(f) dB= 102.922-80 = 22.92dB.

Phase angle, φ(f) = -tan-1(f/fo) = -tan-1(50000/5) = -89.99o.

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

Open loop voltage gain, AoL(f) = Vo/Vid

VO = AOL(f) × (Vin1-Vin2)

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

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.

8. 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

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.

9. At what frequency, the phase shift between input &output voltage will be zero?

A. -40Hz
B. 0Hz
C. -22Hz
D. 20Hz

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

10. At what frequency AOL(f)=A?

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