# Operational Amplifier Internal Circuit MCQ [Free PDF] – Objective Question Answer for Operational Amplifier Internal Circuit Quiz

31. For the current repeater shown in the circuit, determine IC4 value, Where β = 75.

A. 0.035mA
B. 0.028mA
C. 0.04mA
D. 0.052mA

The reference current

Iref = VCC-VBE/R1

= (15v-0.7v)/39kΩ = 0.366mA.

⇒ Iref=IC+4×IB

=IC(1+1/β)

∴ IC= Iref×(1+1/β)

=Iref×(1+1/β) = 0.366mA×(1+1/75) =0.347mA

⇒ IC1=IC2=IC3=0.347mA

To determine IC4,

RE=VT/(1+1/β)×IC4×ln(C3/IC4)

⇒ 1.62kΩ = 25mv/(1+1/75)×IC4×ln(0.347mA/IC4)

⇒ IC4=0.035mA (find using trial and error method.

32. The requirements for a good current source is the one in which, (Take Output current – IO and Output resistance – rO )

A. IO is independent upon current gain and should be low
B. rO should be very high
C. IO in the circuit should be low
D. IO independent upon current gain and rO should be very high

The need for high output resistance current source can be seen because the common-mode gain of the differential amplifier can only be reduced by using high resistance current sources.

33. Which current source exhibits a very high output resistance?

A. Simple current mirror
B. Wilson current mirror
C. Widlar current mirror
D. All of the mentioned

The output resistance of Wilson’s current mirror is substantially greater than ≅(β× Output resistance)/2 than Simple or Widlar’s current source.

34. What will be the overall gain in the Darlington circuit, if the individual transistor gain is 200?

A. 10000
B. 40000
C. 8000
D. 1000

Overall current gain, β= β1×β2 (Multiplication of current gain of the individual transistor)
⇒ β=200×200=40000

35. To increase the input resistance in differential amplifier, replace the transistor by

A. Current mirror
B. Current repeater
C. Darlington pair
D. All of the mentioned

A higher value of input resistance can be obtained by using the Darlington pair in place of the transistor.

36. What is the drawback of using Darlington pair in differential amplifier?

A. Large current gain
B. Output current in milliampere
C. Gain is proportional to load resistor
D. High offset voltage

Due to the cascaded stage, the Darlington differential amplifier offers a higher offset voltage which is two times larger than the ordinary two-transistor used in the differential amplifier.

37. Determine the amount of shift that happens in the level shifter?

A. Vcc + 0.7v
B. Vcc – 0.7v
C. -0.7v
D. +0.7v

A level shifter is basically a simple type emitter follower. Hence, the level shifter also acts as a buffer to isolate high gain stages from the output stage. Therefore, the amount of shift obtained is

VO – Vi = -VBE = -0.7v.

38. To increase the input resistance, the differential amplifier replaces the transistor by

A. Current mirror
B. Current repeater
C. Darlington pair
D. All of the mentioned

A higher value of input resistance can be obtained by using the Darlington pair in place of the transistor.

39. In the Darlington pair differential amplifier the current gain is given as 100. Where IB1=5µA and IC1=0.35mA. Determine IC2

A. 0.5mA
B. 1.5mA
C. 2mA
D. 0.15mA

The current gain in the Darlington pair differential amplifier is given as β=( IC1+IC2)/IB1.

Substituting the values in the equation, we get

IC2=(β×IB1)-IC1

=(100×5µA.-0.35mA =0.15mA.

40. In the circuit shown, find the overall current gain?

A. 456218
B. 444878
C. 444210
D. 455734

From the circuit given

IB= IB1 = 5.6µA.

IE1= IB1+ IC1 = 1.43mA + 5.6µA = 1.435mA.

IE1= IB2 = 1.435mA.

The individual current gain values,

β1=IC1/ IB1

=> β1 = 1.43mA/5.6µA= 255.36.

Similarly β2=IC2/ IB2

=> β2 = 2.5A / 1.435mA =1742.16

Therefore, the overall current gain

β = β1 × β2 = 255.36 × 1742.16 = 444878.

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