Simple problem about diode bias points

Discussion in 'Homework Help' started by ivanwind15, Mar 7, 2010.

  1. ivanwind15

    Thread Starter New Member

    Mar 7, 2010

    I'm having a little difficult how to solve the circuit that I attached. The question is: Find the bias points, Vd and Id, by using (i) an iterative approach, and (ii) assuming Vd is .7 V.

    I'm given: Is = 10^-15, n = 1, thermal voltage (Vt) = 25 mV

    I know the equation Id = Is(e^(Vd/(n*Vt))

    However, for part (i), I don't know how i can find Vd of the diode, since I would have assumed it was 0, since it would be like a short. If you plug 0 into the equation, you get 10^-15 amps, which I would not think makes sense, if it is acting like a short.

    Also, for part (ii), I would just plug in .7 V to the equation? That seems a little too easy, since he asks the same two questions on another circuit which I have not posted.

    Thanks for any help.
  2. Jony130

    AAC Fanatic!

    Feb 17, 2009
    Well in first step you assume Vd=0V so Id=1.5V/500=3mA (Thévenin equivalent)
    Next you substituted 3mA to this equation Id = Is(e^(Vd/(n*Vt)) and find Vd=718.240mV
    So in second step we find ID =(1.5V-718.240mV)/500=1.56351mA and again find Vd=701.948mV and so on. And the answer will be around 1.595mA