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Piecewise Linear Modeling (diode)

  1. Feb 24, 2013 #1
    1. The problem statement, all variables and given/known data
    A pn junction diode is connected to a Thevenin circuit consisting of a 0-to-1V, 1kHz square wave and 1kΩ resistor. Use piecewise linear modeling to estimate the diode voltage for the low (0V), and high (1V) states of the voltage source. Assume the diode to have an approximate turn-on voltage of 0.7V for small currents.

    2. Relevant equations
    rd = [itex]\frac{nV_{T}}{I_{D}}[/itex]
    iD = [itex]\frac{V_{th}-V_{f}}{R_{th}+r_{d}}[/itex]
    VD = Vf + iDrd

    3. The attempt at a solution
    I was thinking of just picking ID = 1mA, n = 1, and VT = 25mV so that:

    rd = 40Ω

    I don't know if I'm going about this right, or even where to go next.
  2. jcsd
  3. Feb 25, 2013 #2


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    For the 1V case...

    Are you familiar with the graphical method of finding the operating point? Plot a graph with two "curves" on it:

    First one for the voltage source/resistor: Plot open circuit voltage, short circuit current and join with straight line.

    Another for the diode, in particular the equation VD = iDrd + Vf is of the form y = mx + c ... a straight line.

    Where they cross is your answer.
  4. Feb 25, 2013 #3

    rude man

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    Picking iD = 1 mA is definitely not a good idea.

    You know VD will be close to 0.7V. So what is a logical choice for iD if the input to the 1K resistor is 1V?
  5. Feb 25, 2013 #4
    With that info, I found iD to be 0.3mA. Then, rd is ~83Ω with n=1 and VT=25mV. So, when the voltage source is 1V:

  6. Feb 25, 2013 #5

    rude man

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    Assuming the piecewise-linear approximation they have in mind is a straight line along the V axis (iD=0) up to V = 0.7V and then a straight line of slope 1/rd up to the operating point iD ~ 0.3 mA, then that would be correct.
  7. Feb 25, 2013 #6
    Alright, thanks :D.
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