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
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