In BJT diferential amplifier circuits why arent the bases of the BJTs DC biased???

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They are your inputs.

 

Its a matter of choice.
Normally, you would want to bias your BJTs/MOSFETs to have VO1 = VO2 = 0VDC.
So if you beforehand set VIN1=VIN2=0VDC, and still want to get VO1=VO2=0VDC, you will need to set:

1. RC in order to have VO1=VO2=0V.
RC = VCC / IC; (IC = IBIAS/2 since the BJTs are matched).

2. VBE in order to have the required VBE that corresponds to IB=IC/beta = IBIAS/(2*beta).
As said, you can set VB=0VDC beforehand, and set VE to the right value, according to the IB(VBE) exponential equation.
Normally, once you set IC, and VB=0Vdc, the current source IBIAS will automatically set its voltage (VE) to the required portion to give IB=IC/beta, that is if its output voltage swing is flexible enough.

By the way, i've also started learning about differential amplifiers as part of the course Linear Electronic Circuits (Also called Analog Circuits), which deals with single-stage, multi-stage, differential and feedback amplifers.

 

Hey alphacat,thnx for the reply.....am studyin Analog Circuits as well,which deals wth the same topics as u mentioned....But couldnt we just use coupling capacitors as we normally do in case of amplifiers????The output will be distorted in case the applied ac signal (our input) is small n fails to rise appreciably above the cut-in voltage of the B-E junction.....I guess using coupling capacitors would reduce the gain n hence the output,which is pretty small newaz,n so they arent used....

 

They are DC biased - positive with respect to the emitter exactly the same as any other NPN circuit.

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The bases in all your circuits are properly biased. The first two circuits have them set at 0V.

The transistors do not have a "cut-in voltage" because they are turned on all the time and the input signal simply varies their current.

Input coupling capacitors pass AC perfectly if their value is high enough but they block DC.

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Uncle $crooge

 

But all the circuits i hav chkd out do not show any dc bias for the B-E junction.....There is only a V_CC....No V__BB....Not even by using V_CC through a potential divider.....

 

Yes they do, the first circuit feeds DC bias via R1 and R2.

You're making the mistake of assuming voltages are absolute, and they aren't, they are relative - and (like I said before) the bases are biased positive with respect to the emitters.

__________________
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http://www.winpicprog.co.uk

 

Yup i got ur point...silly me....thnx......