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

21. Define total current (IQ) equation in differential amplifier with constant current bias current

A. IQ=1/R3 × (VEE/R1+R2)
B. IQ =(VEE × R2)/(R1+R2)
C. IQ=1/R3 × (VEE×R2/R1+R2)
D. IQ)=R3 × (VEE/R1+R2)

Answer: C
The equation for total current is obtained by applying Kirchhoff’s Voltage Law to the constant current circuit in a differential amplifier.

22. Constant current source in differential amplifier is also called as

A. Current Mirror
B. Current Source
C. Current Repeaters
D. All of the mentioned

Answer: A

The output current is a reflection or mirror of the reference input current. Therefore, the constant current source circuit is referred to as the Current Mirror.

23. When will be the mirror effect valid

A. β≫1
B. β=1
C. β<1
D. β≠1

Answer: A

If the value of β is used in the equation, IC=β/(β+2)×Iref. It almost becomes unity and the output current becomes equal to the reference current.

24. Calculate the value of reference current and input resistor for current mirror with IC=1.2μA & VCC=12v. Assume β=50.

A. 1.248mA, 9kΩ
B. 1.248mA, 9.6kΩ
C. 1.248mA, 9.2kΩ
D. 1.2mA, 9.6kΩ

Answer: A

We know that collector current, IC=β/(β+2)×Iref,

⇒ Iref=(β+2)/β×IC= (50+2)/50× 1.2μA = 1.248mA

⇒ Iref=(VCC-VBE)/R1

⇒ R1=(12v-07v)/1.248mA = 9.05kΩ.

25. Determine the early voltage, if the output resistance is 2.5×2kΩ and the input current is 2mA

A. 9.8v
B. 5.6v
C. 7.8v
D. 10v

Answer: D

Output resistance, Ro=VA/Iref

⇒ VA = Ro × Iref =2.5×2kΩ×2mA =10v.

26. In the practical application of current mirror, early voltage is assumed to be

A. Infinite
B. Zero
C. Unity
D. None of the mentioned

Answer: A

Early voltage is assumed to be infinity so that output resistance tends to infinity and the output current is constant.

27. A widlar current source is used

A. to get low value of current
B. to get high value of CMRR
C. to get low voltage of gain
D. to get high value of Output

Answer: A

In the widlar current source Re is added to emitter lead of transistor, which consequently results in smaller output current value.

28. What will be the value of emitter resistance in widlar current source for output current 10mA, having Iref=2.7A

A. 67/(1+1/β)Ω
B. 13/(1+1/β)Ω
C. 14/(1+1/β)Ω
D. 1.36/(1+1/β)Ω

Answer: C

Emitter resistor

RE=VT/(1+1/β) Iref × ln(Io/Iref)

⇒ RE= 0.025/(1+1/β)10mA × ln(12.7A/10mA. = 14/(1+1/β)Ω.

29. A current repeater having identical transistor has collector current, IC1 =0.39mA. Find IC2,IC4& IC6

A. 0.39mA, 0.39mA, 0.78mA
B. 0.78mA, 0.39mA, 0.39mA
C. 0.39mA, 0.78mA, 0.39mA
D. None of the mentioned

Answer: D

In current repeater, the current IC = IC1 =IC2 =⋯= IC N≅ Iref . Where, N – Number of transistors used in current repeater circuit.

30. If the reference and collector current is 0.539mA and 0.49mA respectively, how many transistors are used in the current repeater circuit? (Assume β =150)

A. 11
B. 14
C. 10
D. 8

Answer: B

The current equation is given as IC= Iref × β/(β+1+N)

⇒ 0.49mA= 0.539mA × 150/(150+1+N)

⇒ N=14.

Scroll to Top