BJT Differential Pair

Discussion in 'General Electronics Chat' started by tingyu, Mar 8, 2005.

  1. tingyu

    Thread Starter New Member

    Mar 8, 2005
    Hi all,
    Microelectronic Circuits Sedra&Smith
    In the beginning of chapter6 'Differential and multistage Amplifier', it explains about differential signals. The part I dont understand is that
    why if VB2=0 and VB1=+1, Q2 is off and Q1 is on? and why if Vb1=-1 and VB2=0, Q2 is on and Q1 is off?
    (this is on page 488 and 489)
    Could anyone help me explain this?
    tingyu rolleyes.gif
  2. hgmjr

    Retired Moderator

    Jan 28, 2005
    See my answer in duplicate post.
  3. David Bridgen

    Senior Member

    Feb 10, 2005
    [attachmentid=461]This diagram supplements hgmjr's answer.
  4. krera_87

    New Member

    Oct 28, 2006
    in the firist case (VB2=0,and VB1=1) :
    you have pn junction ( Emitter Base Junction ) , and the pn junction has a v=0.7 , the input voltage like VB must have a voltage is more than the 0.7 , to be able to penetrate the ( Emitter Base Junction ) and go through, so in this case all the current IE will go through the firist transistor (VB1) , and nothing of the current of IE will go through the second transistor ; because the voltage (VB2=0) is less than 0.7 which the voltage of pn junction between emitter and base .