# BJT Differential Amplifiers Question

#### Digit0001

Joined Mar 28, 2010
100
Hi
I am having trouble with the following question
View attachment question4_BJTdiffamp.pdf
i don't understand why the value of Ic is 0.1293mA whilst i get 0.17966mA.

This i how i done it:

-3+18Ic+0.7+11.5Ic-3=0

6=18Ic+11.5Ic+0.7
6-0.7=(18+11.5)Ic
29.5Ic=5.3
Ic=0.17966mA

P.S

#### 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.

#### Digit0001

Joined Mar 28, 2010
100
i have another question that is similar in question 5
View attachment question5_BJTdiffamp.pdf
how would you find Ic which is needed for when i find r(pi)

this is what i get however the answer they got is 0.075
Ie=(beta+1)Ib
Ib=Ie/(beta+1) = 1.485micro
Ic=betaIb
Ic=0.01485mA

where is my mistake?

P.S

#### 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 .