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How to convert an I V curve for a diode into a straight line

Discussion in 'Electronic Basics' started by J. Payne, Nov 4, 2003.

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  1. J. Payne

    J. Payne Guest

    Having don some measuring of the I V characteristics of some diodes, I
    am stuck with the problem of converting them into a staight line, so
    that I can get some useful information out of them, and also to check
    that the effect that is caused is caused for the reason I expect.
    Now, before you go an say "plot V against ln(I)", I have tried that,
    and, getting another curve, am now fairly sure that this is due to an
    ohmic resistance (I have checked this on spice, and the approximate
    shapes look similar).
    Now, my question is, how do I show that this effect is due to the
    additional resistance and not anything else - ie - is there a function
    I can perform on the data to make it linear?
  2. I would think your best bet would be to produce an equation that closely
    fits your measured value (all sorts of software can do this fitting for
    you). Then coming up with an inverse function is easy, and your only task is
    to account for all the constants in your equation (or come up with an
    equation for which you can).

  3. There is an equation for the supply diode resister combination. I have
    it derived and shown at the end of the page at the following link

    Kevin Aylward
    SuperSpice, a very affordable Mixed-Mode
    Windows Simulator with Schematic Capture,
    Waveform Display, FFT's and Filter Design.
  4. Steve

    Steve Guest

    Draw a "line of best fit"
  5. J. Payne

    J. Payne Guest

    Cheers, thats the kind of thing I was looking for, however, its a
    little far from the scope of what I'm doing (this is A-level Physics).
    However, I had another idea today and am wondering whether it would
    work. As the I V curve converges towards becomming a staight lin for
    larger values of V, which is basically the resistor's characteristic
    curve, I could use the gradient of this to find the ohmic resistance.
    Then I calculate the voltage drop across this (V=IR) and subtract that
    from the drop I measured. This will give me the voltage across the
    perfect diode within. Then my new curve (V across ideal diode part vs
    I) should follow the standard exponential pattern. - Would this work?

    And steve, what exactly do you mean? My line is a curve, not straight,
    and drawing an arbitrary line through it will achieve nothing - the
    purpose of such lines is to smooth out random errors in data.
  6. Use a log/lin diagram and you will get a straight line.
  7. I think you need to do a curve fit for a function that has one term is
    proportional to the log of current, and one term that is proportional
    to the current, and solve for the coefficients that best fit your
    data. The definition of 'best' is also something you need to think
    about. The log curve is the ideal diode characteristic, and the
    proportional term is the resistive drop.

    If you have access to Mathcad, you can have it search for the
    coefficients, with your own definition of best fit.
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