BJT Differential Amplifiers Question

vvkannan

Joined Aug 9, 2008
138
"-3+18Ic+0.7+11.5Ic-3=0"

The current through the 11.5 resistor is 2*Ic (as Ic1 and Ic2 flow through it) and hence the drop is 11.5*2Ic.
 

vvkannan

Joined Aug 9, 2008
138
Each Transistor contributes one half of the total emitter current .So to find the collector current of each transistor you need to divide the given Ie by 2 .
 
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