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 .
     
    Digit0001 likes this.
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