# Linear Integrated Circuit MCQ [Free PDF] – Objective Question Answer for Linear Integrated Circuit Quiz

1. An inverting amplifier that amplifies dc input level is called

A. DC and AC amplifier
B. AC amplifier
C. DC amplifier
D. None of the mentioned

In a DC amplifier, the output signal changes in response to changes in its dc input level. A DC amplifier can be inverting, non-inverting or differential.

2. Why DC amplifier uses offset null circuitry?

A. To reduce the distortion in the output
B. To improve the accuracy of the amplifier
C. To get a large output gain
D. All of the mentioned

A DC amplifier uses offset null circuitry to reduce the output offset voltage to zero, that is, to improve the accuracy of the DC amplifier.

3. Why are coupling capacitors used in audio receiver systems?

A. All of the mentioned
B. Thermal drift
C. Component tolerance
D. Variation in signal

An audio receiver system consists of a number of stages because of thermal drift, components tolerance, and variation, which introduces a dc level. To prevent the amplification of such dc level, the coupling capacitors are used.

4. Determine the bandwidth of the AC inverting amplifier for a high cut-off frequency of 15 Hz?

A. 4.321Hz
B. 8.356Hz
C. 7.056Hz
D. 2.334Hz

Input resistance RIF=R1=100Ω (for ideal inverting amplifier).
=> The source resistance RIN=RO=5kΩ.
Therefore, Low frequency cut-off,

fL=1/2πCi(RIF+RO) = 1/ [2π×4.7µF×(5kΩ+100Ω)]

= 6.64Hz. and the bandwidth is calculated as,

BW= fH-fL

= 15 Hz –6.64 Hz = 8.356 Hz.

5. When does the offset voltage compensating network must be used in inverting configuration?

A. When the input is AC voltage
B. When the input is DC voltage
C. When the input is either AC or DC voltage
D. None of the mentioned

To reduce the output offset voltage to zero, the offset minimizing resistor is used to minimize the effect of input bias currents on the output offset voltage. However, when the inputs are DC voltages, the offset compensating network must be used.

6. State the application in which summing, scaling, or averaging amplifiers are used?

A. Multiplexers
B. Counters
C. Audio mixers
D. All of the mentioned

Summing, scaling, or averaging amplifiers are commonly used in audio mixers, in which several inputs are added up to produce the desired output.

7. How an AC amplifier can be powered by a single supply voltage and produces voltage swing?

A. By inserting a voltage divider at the inverting input
B. By inserting a voltage divider at the non-inverting input
C. By inserting a voltage divider at the output
D. By inserting a voltage divider at the feedback circuit

A positive dc level is intentionally inserted using a voltage-divider network at the non-inverting input terminal so that output can swing both positively as well as negatively.

8. Find the maximum output voltage swing of an AC inverting amplifier using op-amp 741C?

A. +15Vpp
B. ±15Vpp
C. ±13Vpp
D. +13Vpp

The value of power supply for 741 op-amp=±15v. Therefore, the ideal maximum output voltage swing for an AC amplifier with a single power supply = +Vcc = +15v.

6. Determine the lower cut-off frequency of the circuit.

A. 21.3Hz
B. 12.15Hz
C. 1.35Hz
D. None of the mentioned

The input resistance of the amplifier is RIF =(R2 ||R3) || [ri×(1+AB.] –> equ 1

As [ri×(1+AB.] >> R2

=> Therefore, equ 1 becomes

RIF ≅ R2 || R3 = 100kΩ || 100kΩ

= (100×100)/(100+100) = 50kΩ.

=> Rin = Ro = 150Ω.

∴ Lower cut off frequency

fL= 1/[2πCi×(RIF+Ro)]

= 1/[2π×0.47µF×(50kΩ+150Ω)] = 6.75Hz.

9. In differential op-amp configuration a subtractor is called as

A. Summing amplifier
B. Difference amplifier
C. Scaling amplifier
D. All of the mentioned

In a subtractor, input signals can be scaled to the desired values by selecting appropriate values for the external resistors. Therefore, this circuit is referred to as a scaling amplifier

10. How the peaking response is obtained?

A. Using a series LC network with op-amp
B. Using a series RC network with op-amp
C. Using a parallel LC network with op-amp
D. Using a parallel RC network with op-amp

The peaking response is the frequency response that peaks at a certain frequency. This can be obtained by using a parallel LC network with the op-amp.

11. The expression for the resonant frequency of the op-amp

A. fp = 1/[2π×√(LC.].
B. fp = (2π×√L)/C
C. fp = 2π×√(LC.
D. fp = 2π/√(LC.

The resonant frequency is also called peak frequency, which is determined by the combination of L and C.

fp = 1/(2π√LC).

12. From the circuit given below find the gain of the amplifier

A. 1.432
B. 9.342
C. 5.768
D. 7.407

Frequency,

fp= 1/[2π×√(LC.]

=1/[2π√(0.1µF×8mH)]

=1/1.776×10-4= 5.63kHz.

=> XL = 2πfpL

= 2π×5.63kHz×8mH = 282.85.

The figure of merit of coil

Qcoil= XL/R1= 282.85/100Ω = 2.8285.

∴ Rp = (Qcoil)2 ×R1

= (2.82852)×100Ω= 800Ω.

The gain of the amplifier at resonance is maximum and given by

AF =-(RF||Rp)/R1

= -(10kΩ||800)/100Ω

=-740.740/100 = -7.407.

13. The parallel resistance of the tank circuit and for the circuit is given below. Find the gain of the amplifier?

A. -778
B. -7.78
C. -72.8
D. None of the mentioned

The gain of the amplifier at resonance is the maximum and given by,

AF =-(RF||Rp)/ R1

=-[(Rp×RF)/ (RF+Rp)] /R1

= -[ (10kΩ × 35kΩ)/ (10kΩ+35kΩ)] /1kΩ

=> AF =- 7.78kΩ/1kΩ= -7.78.

14. The band width of the peaking amplifier is expressed as

A. BW = (fp× XL)/ (RF+Rp)
B. BW =[ fp×(RF+Rp)× XL ] / (RF×Rp)
C. BW =[ fp×(RF+Rp)] / (RF×Rp)
D. BW = [fp×(RF+Rp) ]/ XL

The bandwidth of the peaking amplifier,

BW = fp/Qp,

where Qp – figure of merit of the parallel resonant circuit

= (Rf||Rp)/Xl = (Rf×Rp)/[(Rf+Rp)× Xl]

=> BW = [fp×(Rf+Rp)× Xl] / (Rf×Rp).

15. Design a peaking amplifier circuit to provide a gain of 10 at a peak frequency of 32khz given L=10mH having 30Ω resistance.

Given L=10mH and the internal resistance of the inductor R=30Ω. Assume R1=100Ω. The gain times peak frequency=

10 × 32kHz = 320kHz

fp= 1/2π√LC

=> C = 1/[(2π)2× (fp)2×L]

= 1/ [(2π)2×(320)2×10mH]

= 1/252405.76 = 3.96µF ≅4µF.

Qcoil = xL/R

=(2πfpL)/R

=(2π×320kHz×10mH)/30

= 20096/30 =669.87

=> Rp= (Qcoil)2×R

= (669.88)2×30 = 13.5MΩ

To find Rf,

AF= (RF×Rp)/[R1×(RF+Rp)]

=>RF = (Af ×Rp ×R1)/ (Rp -AF ×R1)

RF = (-10×13.5×106×100) / (13.5×106-(10×100))=1000Ω

=> RF = 1kΩ.

Thus the component values are R1=100Ω, RF= 1kΩ, L=10mH at R=30Ω and C = 4µF.

16. In which amplifier the output voltage is equal to the negative sum of all the inputs?

A. Averaging amplifier
B. Summing amplifier
C. Scaling amplifier
D. All of the mentioned

In the summing amplifier, the output voltage is equal to the sum of all input. Since the total input is a sum of negative input, the amplifier is an inverting summing amplifier.

17. Determine the expression of output voltage for inverting summing amplifier consisting of four internal resistors? (Assume the value of internal resistors to be equal)

A. Vo = -(Rf/R )×(Va +Vb+Vc+Vd)
B. Vo = (RF/R)×(Va +Vb+Vc+Vd)
C. Vo = (R/ RF)×(Va +Vb+Vc+Vd)
D. None of the mentioned

If the internal resistors of the circuit are the same i.e

Ra=Rb=Rc=Rd=R (since there are four internal resistors)

Then, the output voltage for inverting amplifier is given as

Vo= -(Rf/R)×(Va +Vb+Vc+Vd).

18. An inverting amplifier with gain 1 has different input voltage: 1.2v,3.2v, and 4.2v. Find the output voltage?

A. 4.2v
B. 8.6v
C. -4.2v
D. -8.6v

When the gain of the inverting summing amplifier gain is 1 then, the internal resistors and feedback resistors have the same value. So, the output is equal to the negative sum of all input voltages.

VO= -(Va+Vb+Vc)

=-(1.2+3.2+4.2)= -8.6v.

19. In which type of amplifier, the input voltage is amplified by a scaling factor

A. Summing amplifier
B. Averaging amplifier
C. Weighted amplifier
D. Differential amplifier

The weighted amplifier is also called a scaling amplifier. Here each input voltage is amplified by a different factor i.e. Ra, Rb and Rc are different in values ( which are the input resistors at each input voltage).

20. An inverting scaling amplifier has three input voltages Va, Vb and Vc. Find it output voltage?

A. VO= – {[(RF/Ra)×Va] +[(RF/Rb)×Vb]+[(RF/Rc)×Vc]}
B. VO= – [(RF/Ra)+(RF/Rb)+(RF/Rc)]×[( Va +Vb+Vc)].
C. VO = – {[(Ra/RF)×Va] +[(Rb/RF)×Vb]+[(Rc/RF)×Vc]}
D. None of the mentioned