# Splitting of vd in differential amplifier

Discussion in 'General Electronics Chat' started by hamzahumayun, Sep 12, 2010.

1. ### hamzahumayun Thread Starter New Member

Sep 12, 2010
3
0
Hi,

I am struggling to understand why vd is split equally but phase is reversed at the two inputs of differential amplifier,
when vd is applied, the input at Q1 becomes: VBE + vd/2
and at Q2 it becomes: VBE-vd/2
why is that so?

http://www.ece.mtu.edu/faculty/goel/EE-4232/Diff-Amps.pdf for the diagram i am referring to

Thanks
humayun

2. ### marshallf3 Well-Known Member

Jul 26, 2010
2,358
201
Over to the far left it shows a + and a - Vd so it comes from the source, which I'm sure is somehow covered in the previous chapter(s)

3. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
783
This all pre-supposes matched transistors Q1 & Q2.

VBE is the mean of the two voltages VBE1 and VBE2 at a particular differential input condition. This mean value should be fairly constant over the common mode / linear range for a given differential input.

In the (narrow) linear region, the input voltages (w.r.t ground) lie symmetrically either side of VBE+Vx where Vx is the biased offset (w.r.t ground) at the common connected emitters. The difference between the inputs is vd. Or vd=V1-V2.

If the common connected emitters are at voltage Vx for the given input condition then ...

V1=Vx+VBE1
V2=Vx+VBE2

V1>V2

Also if VBE1=VBE+Vy and VBE2=VBE-Vy (assuming symmetry) what is the value of Vy?

V1=Vx+VBE1=Vx+VBE+Vy
V2=Vx+VBE2=Vx+VBE-Vy

vd=V1-V2=2Vy

Therefore Vy=vd/2

Hence

VBE1=VBE+Vy=VBE+vd/2

and

VBE2=VBE-Vy=VBE-vd/2

Which is a rather long-winded summary of what is meant to be "intuitive".