This set of Linear Integrated Circuit Multiple Choice Questions & Answers (MCQs) focuses on “Operational Amplifier Internal Circuit – 2”.

1. How are the arbitrary signal represented, that are applied to the input of transistor? (Assume common mode signal and differential mode signal to be V_{CM} & V_{DM} respectively).

a) Sum of V_{CM} & V_{DM}

b) Difference of V_{CM} & V_{DM}

c) Sum and Difference of V_{CM} & V_{DM}

d) None of the mentioned

View Answer

Explanation: In practical situation, arbitrary signal are signal are represented as Sum and Difference of common mode signal and differential mode signal.

2. How the differential mode gain is expressed using ‘h’ parameter for a single ended output?

a) – h_{fe}RC/h_{ie}

b) 1/2×(h_{fe}RC)/h_{ie}

c) – 1/2×h_{fe}RC

d) None of the mentioned

View Answer

Explanation: Formula for differential mode gain using ‘h’ parameter model for a single ended output.

3. Find Common Mode Rejection Ration, given g_{m} =16MΩ^{-1}, RE=25kΩ

a) 58 db

b) 40 db

c) 63 db

d) 89 db

View Answer

Explanation: Formula for Common Mode Rejection Ration, CMRR= 1+2g

_{m}RE,

⇒ CMRR = 1+(2×16MΩ

^{-1}×25kΩ)

= 801 = 20log801 = 58.07 db.

4. In differential amplifier the input are given as V_{1}=30sinΠ(50t)+10sinΠ(25t) , V_{2}=30sinΠ(50t)-10 sinΠ(25t), β_{0} =200,RE =1kΩ and RC = 15kΩ. Find the output voltages V_{01}, V_{02} & g_{m}=4MΩ^{-1}

a) V_{01}=-60[10 sinΠ(25t) ]-6.637[30sinΠ(50t) ], V_{02}=60[10 sinΠ(25t) ]-6.637[30sinΠ(50t) ]
b) V_{01}=-6.637[10 sinΠ(25t) ]-60[30sinΠ(50t) ], V_{02}=6.637[10 sinΠ(25t) ]-60[30sinΠ(50t) ]
c) V_{01}=-60[30 sinΠ(50t) ]-6.637[10sinΠ(25t) ], V_{02}=60[30 sinΠ(50t) ]-6.637[10sinΠ(25t) ]
d) V_{01}=-6.637[30 sinΠ(50t) ]-60[10sinΠ(25t) ], V_{02}=6.637[30 sinΠ(50t) ]-60[10sinΠ(25t) ]
View Answer

Explanation: Differential mode gain, A

_{DM}= -g

_{m}RC,

⇒ A

_{DM}= -4MΩ

^{-1}×15kΩ = 60

⇒ r

_{Π}=β

_{0}/g

_{m}=200/4MΩ

^{-1}=50kΩ

Common mode gain, A

_{CM}=-β

_{o}×RC/r

_{Π}+(β

_{O}+1)×RE

⇒ A

_{CM}=-200×15kΩ/50kΩ+2(1+200)×1kΩ=-6.637

Common mode signal, V

_{CM}=(V

_{1}+V

_{2})/2= 30sinΠ(50t)

Differential mode signal, V

_{DM}=(V

_{1}-V

_{2})/2= 10 sinΠ(25t)

Output voltages are given as

⇒ V

_{01}=A

_{DM})× V

_{DM})+ A

_{CM}× V

_{CM}

= -60[10 sinΠ(25t)]-6.637[30sinΠ(50t)],

⇒ V

_{02}=-A

_{DM}× V

_{DM}+ A

_{CM}× V

_{CM}

= 60[10 sinΠ(25t)]-6.637[30sinΠ(50t)].

a) 0.036

b) -1.2

c) 4.8

d) 12

View Answer

Explanation: Common mode rejection ratio, CMRR =log

^{-1}×(40/20) = 100

⇒ CMRR =(∣A

_{DM}∣/ ∣A

_{CM}∣)

⇒ ∣A

_{DM}∣ =100×0.12 = 12.

6. To increase the value of CMRR, which circuit is used to replace the emitter resistance Re in differential amplifier?

a) Constant current bias

b) Resistor in parallel with Re

c) Resistor in series with Re

d) Diode in parallel with Re

View Answer

Explanation: Constant current bias offers extremely large resistor under AC condition and thus provide high CMRR value.

7. What is the purpose of diode in differential amplifier with constant current circuit?

a) Total current independent on temperature

b) Diode is dependent of temperature

c) Transistor is depend on temperature

d) None of the mentioned

View Answer

Explanation: The base emitter voltage of transistor (VBE) in constant current circuit by 2.5mv/

^{o}c, thus diode also has same temperature. Hence two variations cancel each other and total current I

_{Q}become in depend of temperature.

8. How to improve CMRR value

a) Increase common mode gain

b) Decrease common mode gain

c) Increase Differential mode gain

d) Decrease differential mode gain

View Answer

Explanation: For a large CMRR value, A

_{CM}should be small as possible.

9. Define total current (I_{Q}) equation in differential amplifier with constant current bias current

a) I_{Q}=1/R_{3}×(VEE/R_{1}+R_{2})

b) I_{Q} =(VEE×R_{2})/(R_{1}+R_{2})

c) I_{Q}=1/R_{3}×(VEE×R_{2}/R_{1}+R_{2})

d) I_{Q})=R_{3}×(VEE/R_{1}+R_{2})

View Answer

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

a) Current Mirror

b) Current Source

c) Current Repeaters

d) All of the mentioned

View Answer

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

11. When will be the mirror effect valid

a) β≫1

b) β=1

c) β<1
d) β≠1
[expand title="View Answer"]Answer: a
Explanation: If value of β is used in the equation, IC=β/(β+2)×I_{ref}. It almost become unity and the output current become equal to reference current. [/expand]

12. 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

View Answer

Explanation: We know that collector current, IC=β/(β+2)×I

_{ref},

⇒ I

_{ref}=(β+2)/β×IC= (50+2)/50× 1.2μA = 1.248mA

⇒ I

_{ref}=(VCC-VBE)/R

_{1}

⇒ R

_{1}=(12v-07v)/1.248mA = 9.05kΩ.

**Sanfoundry Global Education & Learning Series – Linear Integrated Circuit.**

Here’s the list of Best Reference Books in Electrical and Electronics Engineering.

__here is complete set of 1000+ Multiple Choice Questions and Answers__.