# Linear Integrated Circuit Questions and Answers – Operational Amplifier Internal Circuit – 2

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 VCM & VDM respectively).
a) Sum of VCM & VDM
b) Difference of VCM & VDM
c) Sum and Difference of VCM & VDM
d) None of the mentioned

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) – hfeRC/hie
b) 1/2×(hfeRC)/hie
c) – 1/2×hfeRC
d) None of the mentioned

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

3. Find Common Mode Rejection Ration, given gm =16MΩ-1, RE=25kΩ
a) 58 db
b) 40 db
c) 63 db
d) 89 db

Explanation: Formula for Common Mode Rejection Ration, CMRR= 1+2gmRE,
⇒ CMRR = 1+(2×16MΩ-1×25kΩ)
= 801 = 20log⁡801 = 58.07 db.

4. In differential amplifier the input are given as V1=30sin⁡Π(50t)+10sin⁡Π(25t) , V2=30sin⁡Π(50t)-10 sin⁡Π(25t), β0 =200,RE =1kΩ and RC = 15kΩ. Find the output voltages V01, V02 & gm=4MΩ-1
a) V01=-60[10 sin⁡Π(25t) ]-6.637[30sin⁡Π(50t) ], V02=60[10 sin⁡Π(25t) ]-6.637[30sin⁡Π(50t) ].
b) V01=-6.637[10 sin⁡Π(25t) ]-60[30sin⁡Π(50t) ], V02=6.637[10 sin⁡Π(25t) ]-60[30sin⁡Π(50t) ].
c) V01=-60[30 sin⁡Π(50t) ]-6.637[10sin⁡Π(25t) ], V02=60[30 sin⁡Π(50t) ]-6.637[10sin⁡Π(25t) ].
d) V01=-6.637[30 sin⁡Π(50t) ]-60[10sin⁡Π(25t) ], V02=6.637[30 sin⁡Π(50t) ]-60[10sin⁡Π(25t) ].

Explanation: Differential mode gain, ADM = -gm RC,
⇒ ADM = -4MΩ-1×15kΩ = 60
⇒ rΠ0/gm =200/4MΩ-1 =50kΩ
Common mode gain, ACM=-βo×RC/rΠ+(βO+1)×RE
⇒ ACM =-200×15kΩ/50kΩ+2(1+200)×1kΩ=-6.637
Common mode signal, VCM=(V1+V2)/2= 30sin⁡Π(50t)
Differential mode signal, VDM=(V1-V2)/2= 10 sin⁡Π(25t)
Output voltages are given as
= -60[10 sin⁡Π(25t)]-6.637[30sin⁡Π(50t)],
= 60[10 sin⁡Π(25t)]-6.637[30sin⁡Π(50t)].

5. If the value of Common Mode Rejection Ratio and Common Mode Gain are 40db and -0.12 respectively, then determine the value of differential mode gain
a) 0.036
b) -1.2
c) 4.8
d) 12

Explanation: Common mode rejection ratio, CMRR =log-1×(40/20) = 100

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

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

Explanation: The base emitter voltage of transistor (VBE) in constant current circuit by 2.5mv/oc, thus diode also has same temperature. Hence two variations cancel each other and total current IQ 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

Explanation: For a large CMRR value, ACM should be small as possible.

9. 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)

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

10. Constant current source in differential amplifier is also called as
a) Current Mirror
b) Current Source
c) Current Repeaters
d) All of the mentioned

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

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

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

Explanation: 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.

Sanfoundry Global Education & Learning Series – Linear Integrated Circuits.

To practice all areas of Linear Integrated Circuits, here is complete set of 1000+ Multiple Choice Questions and Answers.

If you find a mistake in question / option / answer, kindly take a screenshot and email to [email protected]