Common collector: How to calculate Ib

Discussion in 'General Electronics Chat' started by JStitzlein, Dec 8, 2010.

  1. JStitzlein

    Thread Starter Member

    Dec 6, 2010
    53
    0
    Hello,

    I am trying to calculate the base current for a common collector using KVL, but my calculations are wrong. Can someone please tell me what i'm doing wrong?

    here is my circuit:

    [​IMG]

    if IE = (1+B)Ib
    I am writing a KVL equation for the input loop like this:

    -10E3(101*Ib)-.7 + 3 = 0

    Ib = 2.27uA
    --

    At first i thought this may be plausible, but if i add resistors to my base my KVL equations don't even come close. Why is this?

    thanks for your help
     
  2. JStitzlein

    Thread Starter Member

    Dec 6, 2010
    53
    0
    Edit: i take back what i said about a base resistor having too much effect on the current, i miscalculated.
    --
    If i can save face and make something useful from this thread; given that Ic = B*Ib, i'm curious to know why the simulation is agreeable with my calculation of Ic, since my Ib was 2.27uA.
     
  3. steinar96

    Active Member

    Apr 18, 2009
    239
    4
    You are assuming beta of 100. 150 is more comon for small signal transistors today and i think circuit simulators simulate beta more around 150 then 100.

    The base voltage is 3V. You can assume that in steady state the voltage drop across the BE junction is ~0.7V. Which means that the emiter voltage is 2.3V.
    We notice then that there are 2.3V across the emitter resistor which is 10k.
    Using ohms law we calculate that the current Ic is aprox 230uA.

    If we assume beta of 150 instead of 100 we get Ib as aproximately 1.5uA.

    In practise it is really bad to design circuits assuming some value x for beta since it's production value varies from 100 to 200 easily, even for transistors of the same type
     
  4. JStitzlein

    Thread Starter Member

    Dec 6, 2010
    53
    0
    wow, thanks.
     
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