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Linearizing tunnel diode equation

  1. Oct 1, 2008 #1
    1. The problem statement, all variables and given/known data
    Hi, I need to produce a linearized equation of the following,

    I = D. V. exp(-V/B)
    eq(1)​

    I is the current
    V is the voltage
    D is a constant
    B is a constant

    Data was collected in an experiment designed to investigate the characteristics of a tunnel diode. I didn't do the experiment myself, I just have to find the linearized form of eq(1) and determine the constants D and B. I have a set of I values and a set of V values given to me to allow me to calculate the constants.

    2. Relevant equations
    I = D. V. exp(-V/B)
    eq(1)​


    3. The attempt at a solution
    The problem I am having is the V term before the exponential term.

    Taking logs of both sides,
    Ln(I) = Ln(D) + Ln(V) - V/B
    eq(2)​

    I'm going to use excel to determine B and D. I'm just not sure how to graph it. If there was no V term before the exponential term then it would be,

    Ln(I) = Ln(D) - V/B
    eq(3)​

    and i could simply graph Ln(I) vs V, and 1/B would be my gradient and Ln(D) would be my y-intercept. I can't do this for eq(2) though because of the Ln(V) term.

    If I graphed Ln(I) vs Ln(v) - V, i can't determine B. I just can't seem to find a way to do it. I'm sure there's a simply way. I just need to work out how to bring the V's together.

    Any help would be greatly appreciated.
     
  2. jcsd
  3. Oct 1, 2008 #2

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Welcome to PF!

    Hi rjspurling! Welcome to PF! :smile:
    Stop trying to keep I and V separate …

    you're allowed to mix-and-match! :wink:

    Hint: you want an equation with B on the LHS, and a fraction on the RHS. :smile:
     
  4. Oct 1, 2008 #3
    argh mixing, of course!

    Ln(I) = Ln(D) + Ln(V) - V/B

    Ln(I) - Ln(V) = Ln(D) - V/B

    Ln(I/V) = Ln(D) - V/B

    Ln(I/V) vs V

    Ln(D) is the y-intercept.

    -1/B is the gradient.

    I think thats it.

    Thanks tiny-tim! much appreciated.
     
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