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BJT catastrophe!

  1. Jan 6, 2012 #1
    Now as you can see in the picture a common base configuration of BJT.
    The green markings in the picture are resistances. :p
    My question is:
    By KVL : VBB = VBE + IE*RB which means that IE is esentially a constant because
    VBB , VBE, RB are constants.
    Now If Voltage Vcb is increased Early effect states that depletion region
    of CB junction penetrates deeper into the Base region. Which means charge gradient in base increases and hence IE increases .

    What is the contradiction?
     

    Attached Files:

  2. jcsd
  3. Jan 6, 2012 #2
    There's no contradiction. Vbe isn't strictly constant. At constant current base, every BJT will show increasing colector current when Vce grows. Check any curve.
     
  4. Jan 6, 2012 #3
    Yes, but if at all IE increases, it would imply that VBE has increased. That will not fit
    the KVL equation.

    ?
     
  5. Jan 6, 2012 #4
    You need a higher Vbe (but very little) to increase Ie. It's the PN junction law. Perhaps you believe Vbe NEVER changes no matter the value of Ie. That's wrong.
     
  6. Jan 6, 2012 #5
    No i do believe. But what my question is that the KVL eqn says:
    VBB = VBE + IE*RB
    Do you agree that VBB and RB are constants?
    IE=function of VBE
    Now if VBE and IE were to simultaneously increase it would violate KVL.


    ?
     
  7. Jan 7, 2012 #6
    No i do believe. But what my question is that the KVL eqn says:
    VBB = VBE + IE*RB
    Do you agree that VBB and RB are constants?
    IE=function of VBE
    Now if VBE and IE were to simultaneously increase it would violate KVL.


    ?
     
  8. Jan 8, 2012 #7
    Hmm... trying my take on this.

    I don't think there's a catastrophe. Early effect increases Ic (not Ie) and increases the forward gain. One expects Ie to increase with an increase on Ic, but let's see. Ebers-Moll with Early corrections has for Ie

    Ie = i(Vbe)(a + 1/(Bf.a)) - a.i(-Vcb)
    i(Vx) = Is . ( exp( Vx/Vt ) - 1 )
    a = ( 1 + Vce/Va )

    i(-Vcb) is very small in forward active mode, so

    Ie = i(Vbe)(a + 1/(Bf.a))

    Now, Vee = Vbe + Re . Ie, so we get

    Vee = Vbe + Re.i(Vbe).(a+1/(Bf.a))

    This is identical to good old diode equation

    Vee = Vbe + Re*.i(Vbe)

    where Re* = Re.a(1 + 1/(Bf.a2))

    Therefore, the effect on Ie is the same as if Re increased by a factor "a" (plus a much smaller 2nd order factor)... so Ie should decrease, and because of the diode equation Vbe should decrease a tiny little bit too (albeit it will remain pretty much around 0.6-0.7).

    At least that's my take.
     
    Last edited: Jan 9, 2012
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