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?

    please refer to page 6 of following link

    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
    782
    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".
     
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