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

 

See my answer in duplicate post.

 

[attachmentid=461]This diagram supplements hgmjr's answer.

 

hiiiiiiiiiiiiii:
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 .