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

11. Find collector current IC2, given input voltages are V1=2.078v & V2=2.06v and total current IQ=2.4mA. (Assume α=1)

A. 0.8mA
B. 1.6mA
C. 0.08mA
D. 0.16mA

Answer: A

Collector current

I_C2=α_F×I_Q/(1+e^V_d⁄V_T),

V_T = Volts equivalent of temperature = 25mv,

⇒ V_d = V₁-V₂ =2.078v-2.06v

=0.018v (equ1)

Substituting equation 1,

⇒ V_d/V_T = 0.018v/25mv = 0.72v (equ2)

Substituting equation 2,

⇒ I_C2= 1×2.4mA/(1+e^0.72) =

2.4mA/(1+2.05) = 0.8mA.

12. A differential amplifier has a transistor with β0= 100, is biased at ICQ = 0.48mA. Determine the value of CMRR and ACM, if RE =7.89kΩ and RC = 5kΩ.

A. 49.54 db
B. 49.65 d
C. 49.77 db
D. 49.60 db

Answer: B

Differential mode gain, A_DM= -g_mRC and Common mode gain,

⇒ A_CM= -(g_mRC)/(1+2g_mRE)

(for β₀≫1).

Substituting the values,

⇒ g_m= I_CQ/V_T = 0.48mA/25mv=19.2×10^-3Ω^-1

⇒ A_DM= -g_m×RC= -19.2×10^-3Ω^-1×5kΩ= -96

⇒ A_CM= -(g_mRC)/(1+2g_mRE)

= -(19.2×10^-3Ω^-1×5kΩ) /(1+2×-⇒ 19.2×10^-3Ω^-1×7.89kΩ) = -0.3158

CMRR = -96/-0.3158= 303.976

=20log⁡303.976

=49.65db

13. How is the arbitrary signal represented, that is applied to the input of the transistor? (Assume the common-mode signal and the differential mode signal to be V_CM & V_DM respectively).

A. Sum of V_CM & V_DM
B. Difference between V_CM & V_DM
C. Sum and Difference of V_CM & V_DM
D. None of the mentioned

Answer: C

In practical situations, arbitrary signal signals are represented as the Sum and Difference of common-mode signal and differential mode signal.

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

A. – h_feRC/h_ie
B. 1/2 × (h_feRC./h_ie
C. – 1/2 × h_feRC
D. None of the mentioned

Answer: B

The formula for differential mode gain using ‘h’ parameter model for single-ended output.

15. Find Common Mode Rejection Ration, given g_m =16MΩ^-1, RE=25kΩ

A. 58 db
B. 40 db
C. 63 db
D. 89 db

Answer: A

Formula for Common Mode Rejection Ration, CMRR= 1 + 2g_mRE,

⇒ CMRR = 1+(2 × 16 MΩ^-1× 25kΩ)

= 801 = 20log⁡801 = 58.07 db.

16. In a differential amplifier the input is 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 & g_m=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) ].

Answer: A

Differential mode gain, A_DM = -g_m RC,

⇒ A_DM = -4MΩ^-1×15kΩ = 60

⇒ r_Π=β₀/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=(V₁+V₂)/2= 30sin⁡Π(50t)

Differential mode signal, V_DM=(V₁-V₂)/2= 10 sin⁡Π(25t)

Output voltages are given as

⇒ V₀₁=A_DM)× V_DM)+ A_CM× V_CM

= -60[10 sin⁡Π(25t)]-6.637[30sin⁡Π(50t)],

⇒ V₀₂=-A_DM× V_DM+ A_CM× V_CM

= 60[10 sin⁡Π(25t)]-6.637[30sin⁡Π(50t)].

17. If the value of Common Mode Rejection Ratio and Common Mode Gain is 40db and -0.12 respectively, then determine the value of differential-mode gain

A. 0.036
B. -1.2
C. 4.8
D. 12

Answer: D

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

⇒ CMRR =(∣A_DM∣/ ∣A_CM∣)

⇒ ∣A_DM∣ =100×0.12 = 12.

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

A. Constant current bias
B. Resistor in parallel with Re
C. Resistor in series with Re
D. Diode in parallel with Re

Answer: A

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

19. What is the purpose of the diode in a differential amplifier with a 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

Answer: A

The base-emitter voltage of the transistor (VBE) in the constant current circuit by 2.5mv/^oc, thus diode also has the same temperature. Hence two variations cancel each other and the total current I_Q becomes in depends on temperature.

20. 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

Answer: B

For a large CMRR value, A_CM should be small as possible.

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