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

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#### Rahul Sharma

##### Member level 3 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

##### Full Member level 6 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

### Rahul Sharma

points: 2

#### Rahul Sharma

##### Member level 3 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

##### Full Member level 6 yeah lol i forgot to temove gm ty

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