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Diode - Resistor Combination

  1. Mar 25, 2017 #1
    Helo,

    I am in an electronic course and we just started studying diodes. I am stuck on the following problem:

    Place a 100 ohm resistance in series with a diode, and plot current versus voltage across the series combination.
    I am also given the following values for the diode:
    n = 1, Is = 10^-14A, Vt = 26mV


    I am trying to come up with an expression for current. I am using the following:

    iD = Is * exp(vD/(n*Vt)) ...... approximation of the current through the diode.

    and

    Vr = iR for the voltage drop across the resistor

    Then, the total voltage drop is:

    Vtot = vD + Vr

    I am then trying to solve for current, but I can't get it on its own. Can some one please help me. I have attached a picture of the original problem statement. Thank you very. much!

    Screen Shot 2017-03-25 at 7.33.08 PM.png
     
  2. jcsd
  3. Mar 25, 2017 #2
    I haven't checked to verify that your approximation equation is correct. But if it is, what do you get for the diode voltage if you plug in, say, 1 μA in as the current for the diode?
     
  4. Mar 25, 2017 #3

    phyzguy

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    Science Advisor

    Do you know how to take your expression for the current through the diode (iD = Is * exp(vD/(n*Vt))), and invert it to give vD as a function of iD? Also, how do the current through the diode and the current through the resistor compare? Then if you put these currents into your expression for Vtot, you will have a relation between Vtot and I which you should be able to solve on a computer. Do you know how to use Newton's method to solve a non-linear algebraic equation?
     
  5. Mar 25, 2017 #4
    Thank you for the reply! I do know how to solve the expression for the current through the diode and invert it to get vD. My problem is that I need to get an expression of the current in terms of voltage. Since it is not linear, I am not sure how to do that. I have not used Newton's method to solve non-linear algebraic equations.
     
  6. Mar 25, 2017 #5
    I think I get the fist part of the question, but I am stuck on getting an expression for the current in terms of voltage for the second situation where we add a resistor.
     
  7. Mar 25, 2017 #6
    I don't have the time right at the moment to try to come up with an expression; it doesn't look extremely trivial. But it didn't look to me like the problem statement was asking for an expression; it was asking for a plot. So if you pick a current - say, 1 μA - you can rearrange the diode equation and solve for the voltage drop for that particular current. And certainly you can calculate the voltage drop across a 100 ohm resistor for that same 1 μA current. So the total voltage would be VD + VR for that 1 μA current. You could pick a number of currents from 10 μA to 10 mA and come up with a plot. It would be fairly straightforward to do that in Excel, I would think.

    Maybe when I have a little more time on my hands, I might try to look at solving for current based on input voltage.
     
  8. Mar 25, 2017 #7
    Thank you very much! I will try that!
     
  9. Mar 27, 2017 #8

    Baluncore

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    Science Advisor

    Newtons method is used to solve the non-linear equation of R in series with D.
    Supply voltage is Vs. Assume initial diode voltage Vd = 0.
    Then resistor R, and diode have series current, I = Vs / R.
    An iterative solution for diode current should converge rapidly.
    Do.
    Given I, find Vd from the diode equation.
    Then I' = ( Vs – Vd ) / R
    Loop until value is stable.
     
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