HELP SOLVE BJT CIRCUIT QUESTION PLEASE
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@radar

Question

irishboy123 · Answer

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@radar

radar · Answer

Can't be much help. Q1 is an emitter follower with a typical voltage gain of one, feeding the input of Q2 a common base amplifier.  I have no idea what the voltage gain of Q2 while the current gain will be around 1, Q3 is acting as some kind of biasing circuit and it acting as emitter circuit of both Q1 and Q2.  The voltage out Vo is the same phase as voltage in Vi (no inversion)  Not up to snuff on the break down of the various biases and loads.  Review your study material, and emulate the circuit on SPICE.

ganeshie8 · Answer

Q3 is biased using a voltage divider.

Base voltage of Q3, $V_{BB,Q3} = \dfrac{10}{10+7.5}\times 12=6.9V$


Collector current of Q3 will be fixed at 
$I_C,Q3 = \dfrac{V_{BB,Q3} - 0.7}{4.3} = \dfrac{6.9-0.7}{4.3} =1.4A $

ganeshie8 · Answer

@518nad  that means the tail current of the differential amplifier is $1.4A$ and we didn't have to use $\beta$ so far. Rest should be easy ?

ganeshie8 · Answer

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