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In differential amplifier, reason of presence of noise at the output terminal

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Rahul Sharma

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Explanation is required. when small signal differential inputs are given to differential amplifier.

Two cases are given in attached figurewave_Page1.png
(A) When current source is ideal
(B) When current source is practical i.e. have one finite resistance.

If input is given as V1=1sinwt and V2=-1sinwt and Vcm= noise then what will be waveform of VA, VB, Vo1, Vo2 and Vo1-Vo2 in both the cases. I need explanation of each waveform generated.
 

Ata_sa16

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In A)

The differential gain is: gm*RL

gm=gm1=gm2=Is/vov

and the common mode gain is (gm*Rl )/2Rs



so your wave forms will be:

A)

common mode gain is zero Rs >> Rl

Va = Vcm+1sinwt /// Vb =Vcm-1sinwt /// Vo1 = -(gm*Rl)*sinwt /// Vo2 = +gm*Rl*sinwt vo1-vo2= -2*gm*Rl*sinwt

B)

common mode gain is (gm*Rl )/2Rs

Va = Vcm+1sinwt /// Vb =Vcm-1sinwt /// Vo1 = Vcm*(gm*Rl)/2Rs - (gm*Rl)*sinwt /// Vo2 = Vcm*(gm*Rl)/2Rs +(gm*Rl)*sinwt /// vo1-vo2= -2*gm*Rl*sinwt
 

Rahul Sharma

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In A)

The differential gain is: gm*RL

gm=gm1=gm2=Is/vov

and the common mode gain is (gm*Rl )/2Rs



so your wave forms will be:

A)

common mode gain is zero Rs >> Rl

Va = Vcm+1sinwt /// Vb =Vcm-1sinwt /// Vo1 = -(gm*Rl)*sinwt /// Vo2 = +gm*Rl*sinwt vo1-vo2= -2*gm*Rl*sinwt

B)

common mode gain is (gm*Rl )/2Rs

Va = Vcm+1sinwt /// Vb =Vcm-1sinwt /// Vo1 = Vcm*(gm*Rl)/2Rs - (gm*Rl)*sinwt /// Vo2 = Vcm*(gm*Rl)/2Rs +(gm*Rl)*sinwt /// vo1-vo2= -2*gm*Rl*sinwt
Thank you! Totally agree with you. Even i got the same answers. except according to me common mode gain is = -Rl*gm/(1 + 2gm*Rs) = Rl/2Rs. instead of (gm*Rl )/2Rs (you have written)

Am i correct here??
 

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