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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) ].

I had been asked this question in a national level competition.

This interesting question is from Operational Amplifier Internal Circuit topic in chapter Operational Amplifier of Linear Integrated Circuits

1 Answer

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by (39.8k points)
The correct choice is (a) V01=-60[10 sin⁡Π(25t) ]-6.637[30sin⁡Π(50t) ], V02=60[10 sin⁡Π(25t) ]-6.637[30sin⁡Π(50t) ].

Delve into insight with: Differential mode gain, ADM = -gm RC,

⇒ ADM = -4MΩ^-1×15kΩ = 60

⇒ rΠ=β0/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

⇒ V01=ADM)× VDM)+ ACM× VCM

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


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

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