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Equation:I=Ioe^(eV/(kt)),so the graph to find e and Io is lnI against V?Nid c pic

  1. May 20, 2007 #1


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    Equation:I=Ioe^(eV/(kt)),so the graph to find e and Io is lnI against V?Nid c pic....

    1. The problem statement, all variables and given/known data
    SOLVED-I managed 2 call my teacher and he solved it
    [​IMG]-Zooms[tex]Equation\ is\ I=I_{o}e^\frac{eV}{kT}[/tex]
    Edit ** Yo .This is my question paper and I've written some working on it.I've 2 questions-part a)The graph needed to find e and Io is ln I against V is it? & for last part,e) to find the uncertainty of e I think I need the uncertainty of the y value,but they all are different at different pts,so I average them up in order to get e,correct way?!

    2. Relevant equations
    Uncertainty calculation:
    For getting the uncertainty,b in a,a[tex]\pm[/tex]b
    -If multiplication involved in calculating a then add fractional uncertainties then multiply by a to get b.
    -If addition or minus involved in calculating a just add uncertainties only to get b

    For natural logarithm: lne=1

    3. The attempt at a solution
    Well,I start by doing ln on both sides of the equation given ,ln I=lnIo + (ev)/(kt),then comparing it to y=mx+c the y is lnI,the x is V.Thus I think I need lnI against V to get the e and Io.I'm just checking the answer with u and at the same time doing my studies for my A lv(Which 2day is my exam for this type of questions).Hope the correct answer is answered here.So correct?
    Last edited: May 20, 2007
  2. jcsd
  3. May 20, 2007 #2
    Your question paper is unreadable.
    Your description is obscure. (for example: "The question the graph is needed ...")
    Your relevant equations are unclear.
    Your attempt at a solution not clear either.

    Take the time needed to ask properly your question, then I could take the time to answer.
    Also explain what your background is. Depending on this information, the way I could answer could be different. Also indicate what you have learned at school on this topic and the method you are probably expected to use.
  4. May 20, 2007 #3


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    I've redone the stuff u said,also added 1 more question needing to be answered,should be easy for pros out there.Pls do me a favour which is answering it?!
  5. May 21, 2007 #4
    Well, inv, real pros don't like to answer badly formulated questions.
    I hope that you realize that the symbol "e" in this formula:


    comes with two different meaning.
    Therefore, I would prefer to write the formula in this way:

    [tex]I=I_{o}exp(\frac{eV}{kT})[/tex] (problem equation)

    I have hidden the symbol "e" representing the base of the natural logarithm and replaced it by the symbol "exp".
    In this way I can avoid confusion with the symbol "e" appearing in the exponential. This symbol very probably represents the electric charge of the electron.

    Another problem for a pro, is that this equation very probably represents the I-V characteristic of an (ideal) diode, but it should then be corrected as follows:

    [tex]I=I_{o}(exp(\frac{eV}{kT})-1)[/tex] (correct equation for a diode)

    From the original (problem equation), you can write V as a function of I, as you did:

    [tex]V = a y + b[/tex]


    [tex]y = ln(I)[/tex]

    [tex]a = \frac{kT}{e}[/tex]

    [tex]b = -\frac{kT}{e} ln(I_{o})[/tex]

    The variables {V,y} are experimental data, and I assume you are asked to find a and b . (remember, I can not read or understand your problem statement)

    Of course, again, for a pro, this makes a problem since you probably know the temperature T, and since "k" and "e" are physical constants. Therefore, the only real unknown would be [tex]b = -\frac{kT}{e} ln(I_{o})[/tex].
    Anyway, assuming "a" and "b" are unknowns would allow you to check the physical theory that says [tex]a = \frac{kT}{e}[/tex].

    Assuming you need to determine a and b, the next question is then: have you learned linear regression, have you learned fitting a line to data points? If yes, you should go to your notes and use the theory. On this web site: http://people.hofstra.edu/faculty/Stefan_Waner/RealWorld/calctopic1/regression.html , you can find some explanations about how to calculate a and b, also called the "slope" and the "intercept".

    Assuming instead that a is known and that you need need only to find "b" and from b find Io, then the problem is easier since you can find b from the experimental data: b = V - a y.
    Since you will have an estimation of b for each data point, you will have to perform an average and evaluate the precision of your result. Here again, I should ask you: what have you learned at school?
    Indeed, it is clear that (in principle) data points with large error bars should have less weight in the average. If you have learned to take that into account, then use the weighting formula from your note. If you have not learned to do that, then use the usual average formula.
    For calculating the precision of you estimate, again refer either to your notes, or to the usual formula for the standard deviation of an average.

    You can also find on wiki the formula for a weighted average of unprecise data: http://en.wikipedia.org/wiki/Weighted_mean .
    The relevant formulas are:

    [tex]\bar{x} = \frac{ \sum_{i=1}^n x_i/{\sigma_i}^2}{\sum_{i=1}^n 1/{\sigma_i}^2}[/tex]


    [tex]\sigma_{\bar{x}}^2 = \frac{ 1 }{\sum_{i=1}^n 1/{\sigma_i}^2}[/tex]

    see the wiki page for the meaning of the symbols (x is the variable you average, and the sigmas are standard deviations)

    Finally, an important remark:

    If the parameter Io is what you need to find out, there is no need to express V as a function of I.
    You can write:

    [tex]I_{o} = I exp(-\frac{eV}{kT})[/tex]

    where I and V are the experimental data.
    For each data point {I,V}, an error bar on Io can easily be calculated from the error bars on I and V. Just use the rules of error calculations. The average and deviation on the calculated Io can proceed as explained above.
    Last edited: May 21, 2007
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