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

Rahul Sharma · Sep 28, 2016

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

Two cases are given in attached figure

(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 · Sep 28, 2016

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 · Sep 28, 2016

Ata_sa16 said:
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??

Ata_sa16 · Sep 28, 2016

yeah lol i forgot to temove gm ty

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

Rahul Sharma

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Ata_sa16

Full Member level 6

Rahul Sharma

Rahul Sharma

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Ata_sa16

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