# BJT Differential Amplifiers Question

Discussion in 'Homework Help' started by Digit0001, Mar 17, 2012.

1. ### Digit0001 Thread Starter Member

Mar 28, 2010
89
0
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

2. ### vvkannan Active Member

Aug 9, 2008
138
11
"-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.

3. ### Digit0001 Thread Starter Member

Mar 28, 2010
89
0
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

4. ### vvkannan Active Member

Aug 9, 2008
138
11
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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