Transistor Calculations

Discussion in 'Homework Help' started by Toshinatrix, Jan 21, 2010.

  1. Toshinatrix

    Thread Starter New Member

    Jan 21, 2010
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    Hi

    Basically I have been toying around with my electronics homework (which is one of my weaker subjects) and I have become stumped, I'l post a link of the circuit I am looking at further down.

    Basically the question asks: Calculate the Power Loss in Transistor T

    Information given:
    R1 = 400 ohms
    R2 = 500 ohms
    R3 = 50 ohms
    E = 10v
    Vbe = 0.6v
    Hfe = 10

    [​IMG]


    A formula I have worked out is(although not sure if correct):
    E= (Ir1xR1) + Vbe + (IeR3)..... where Ir1 = (Ib + Ir2)

    Basically I have been struggling with the initial phase of calculations this, ie.... E, Ib, Ir2, Ir1,
    I think if i had a formula to work out those then maybe I could get it.

    Thanks for taking the time to read this!
     
    Last edited: Jan 21, 2010
  2. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    Solve this
    E-I1*R1-I2*R2=0
    I1=Ib+I2
    I2*R2-Vbe-Ie*R3=0
    Ie=(hfe+1)*Ib

    Or use the Thévenin
    Et=E*R2/(R1+R2)=5.5V
    Rth=R1||R2=222Ω
    And then
    Ie=(Et-Vbe)/(Rb/(hfe+1)+R3=4.9V/70.18Ω=69.81mA
     
    Last edited: Jan 21, 2010
  3. Toshinatrix

    Thread Starter New Member

    Jan 21, 2010
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    0


    Soo from what I can get from this I can use thevenins to work out Ie, and from there Ib can be solved using ib= (hfe+1)/Ie and from there I should be able to work my way through it?

    slightly disappointed I didnt spot the thevenins way

    Thanks for the help
     
  4. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    To determine the power dissipated in bjt we usually ignore Ib*Vbe losses.
    So Ptot=Ic*Vce.
     
  5. Toshinatrix

    Thread Starter New Member

    Jan 21, 2010
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    Hi, the teacher has asked for us to use the formula:

    Ptot = Pc + Pb

    therefore I have used:
    Ie=(Et-Vbe)/(Rb/(hfe+1)+R3=4.95V/70.18Ω=70.5mA

    I then used:

    Ib = Ie / (hfe +1) = 6.5mA

    Therefore Ic = Ie - Ib = 64mA

    Vce = E - Ib(hfe+1)R3 = 6.425v

    Pb = Ib*Vbe = 3.9mW
    Pc = Ic*Vce = 411.2mW

    Therefore Ptot = 415.1mW

    Could you check through this working and see what you think please?
     
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