# BJT differential pair question plz help

#### devl82

Joined Sep 1, 2007
3
Hello ,
i'm reading Microelectronic Circuits Sedra&Smith's chapter about differential amplifiers and trying to understand how a BJT dfferential pair works...

In the book whenever Vb1>Vb2 ==> Q1 ON , Q2 OFF and
Vb1<Vb2 ==> Q1 OFF , Q2 ON
which agrees with hgmjr's answer in similar thread "Anytime the voltage difference between the two bases in an NPN transistor differential pair is greater than 0.7V (the forward voltage of a base-emitter junction) then the base with the higher voltage will conduct all of the available current leaving none for the remaining transistor. Thus the remaining transistor is in the off state since its base-emitter junction is reverse biased."
I understand how to calculate Ic and the other voltages but i dont really understand is WHY this is happening, i mean WHY the transistor's base with the higher voltage will conduct all of the available current...

thanks

#### hgmjr

Joined Jan 28, 2005
9,027
You may need to review the material in Sedra & Smith on the single transistor? It is basic transistor operation that is governing the differential pairs behavior. With the base of the npns Q1 and Q2 tied to the same dc voltage, the emitters are fixed at one Vbe below the bases. As the dc voltage on Q1 is taken more positive, its emitter will track the base always remaining one Vbe more negative. As Q1's emitter goes more positive, Q2 is turned off. Q2 is performing as a common base amplifier in this instance.

hgmjr

#### devl82

Joined Sep 1, 2007
3
It is the hypothesis about transistor's base with the higher voltage that conduct all of the available current that is giving me trouble...
Is it something fundamental that i should have known or something that i should understand by looking/solving the circuit?

#### hgmjr

Joined Jan 28, 2005
9,027
Study what Sedra and Smith say about NPN common base amplifiers. See if that helps clear things up for you.

hgmjr