Widlar Circuit equation

Discussion in 'Homework Help' started by black_bear, Apr 14, 2010.

  1. black_bear

    Thread Starter New Member

    Apr 14, 2010
    2
    0
    I have the following equation from a widlar circuit:

    Vt[ln(.9927) -ln(Iout)] = 700(Iout-1)

    Vt = 26mV
    I(in) = 0.9927 mA
    R in the circuit = 700 Ohms
    I had a third transistor with V(be) ON = 700mV

    Can someone help me find I(out)

    Thanks
     
  2. syed_husain

    Active Member

    Aug 24, 2009
    61
    5
    as you know this type of equation is transcendental equation, u cant solve this like normal polynomial equation. if u want to do it manually, u have to use NEWTON's mehtod. it is an iterative process (quite tiresome:(). do some research u will able to figure it out.

    here i calculated the value of Iout using Mathemtica.

    NSolve[26*10^-3 (Log[0.9927*10^-3] - Log[Iout]) - 700 (Iout - 1) == 0,
    Iout] // N

    Iout-> 0.999743 A.

    hope this will help.
     
  3. black_bear

    Thread Starter New Member

    Apr 14, 2010
    2
    0
    Hello Syed
    (AoA)

    Thanks for the reply but unfortunately the answer to this problem is one of this 5 choices: I(out) equals

    1. 0.2 mA
    2. 0.5 mA
    3. 1 mA
    4. 1.7 mA
    5. 3.2 mA

    I think I have not set-up the equation properly. Please see the attached circuit diagram.

    WaS
    Faisal
     
  4. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    As a rough guess I'd say it's going to be about 1mA.

    I2*R2+VBE2=VBE1+VBE3

    If VBE1≈VBE2≈VBE3=0.7V

    I2≈0.7/700=1mA

    A more careful analysis is difficult for me due to the presence of Q1.

    If Q1 was omitted then normally one might try a closer solution using the relationship -

    I2=(Vt/R2)*Ln(I3/I2)

    - which is solved iteratively. I3 is the Q3 [&Q1 when included] emitter current which you work out as indicated in your analysis.

    An better 'approximate' guess with Q1 included might be obtained using

    I2=VBE1/700+(Vt/700)*Ln(I3/I2) - again solved iteratively

    I3 would be about 1mA (depending on your assumed VBE's for Q1 & Q3 - which I took as 0.65V)
     
  5. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    It's possible with the selected transistor that VBE at 1mA could be a lot less than 0.65V - perhaps as low as 0.57V

    In this case the current in Q1 & Q3 [I3] would be approx 1.01mA.

    The approx estimate for the resulting Q2 current would be determined from

    I=0.57/700+(Vt/700)*Ln(1.01e-3/I) where Vt=26mV

    When solved iteratively this would give

    I=822uA.

    Which isn't far removed from I=VBE/700=0.57/700=815uA.

    So the iterative method in this case doesn't add much to the most basic & simpler solution.
     
    Last edited: Apr 16, 2010
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