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Simple Diode Problem

  1. Feb 12, 2010 #1
    1. The problem statement, all variables and given/known data
    You have a series circuit consisting of a dc power supply, a 90 ohm resistor and a diode.
    If the measured current in the circuit is 136 mA, and the saturation current for the diode is 0.006 mA, what is the voltage drop across the diode? Assume the diode temperature is 300 K. Your answer should be given to three places after the decimal point.



    2. Relevant equations
    [tex]I = I_{o}(e^{\frac{qV}{k_{b}T}}-1)[/tex]

    [tex] q = 1.602x10^{-19}[/tex]
    [tex]k_{b} = 1.38x10^{-23}[/tex]


    3. The attempt at a solution

    Rearranging the above the equation I get

    [tex]\frac{k_{b}T}{q}ln(\frac{I}{I_{o}} + 1) = V[/tex]

    Plugging in the numbers I get 0.259V which isn't the right answer, I know this problem isn't difficult but I can't seem to wrap my head around why this doesn't work.
     
    Last edited: Feb 12, 2010
  2. jcsd
  3. Feb 12, 2010 #2

    berkeman

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    I believe you are missing a pair of parenthesis in your first equation...
     
  4. Feb 12, 2010 #3
    Oh yeah I missed that, fixed it in the post above
     
  5. Feb 12, 2010 #4

    berkeman

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    So does that fix your answer?
     
  6. Feb 12, 2010 #5
    no I formulated my equation from the correct expression, I just mistyped it in the original post
     
  7. Feb 13, 2010 #6

    ehild

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    Your result is correct.

    ehild
     
  8. Feb 13, 2010 #7
    Well that sucks, the problem is in an online assignment and it only accepts the right answer (we get 3 tries at it). It won't accept that one.
     
  9. Feb 13, 2010 #8

    ehild

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    Interesting that it gives the resistance of a series resistor and does not use its value in the question. Maybe, they wanted to ask the emf of the source.

    ehild
     
  10. Feb 13, 2010 #9
    Well there is a second part to this problem where the resistance of the resistor is needed, I just didn't post it since it had no bearing on this part of the problem. Here it is:

    What is the voltage across the diode-resistor combination?
     
  11. Feb 13, 2010 #10

    Redbelly98

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    I have a few ideas about what might be wrong.

    1. 0.006 mA seems pretty large for a diode saturation current. I think nA or pA range is more reasonable. Do they really say 0.006 mA, or could it have been 0.006 μA? Or something else?

    2. Perhaps there are too many significant figures in the answer "0.259 V". Try the calculation using saturation currents of 0.0055 and 0.0065 mA, to get a sense of how many sig figs are justified.

    3. Shouldn't the diode equation include an extra material-dependent ideality factor in the qV/kT term? Perhaps they expect the student to know what it should be, so they didn't provide the value in the problem statement.
     
  12. Feb 13, 2010 #11
    I just copied and pasted the problem directly from the problem set

    It says it wants my answer within 3 numbers after the decimal point

    I took that equation directly from our lab handout, this question is a for a prelab problem worth a small fraction of the lab mark, but I also found it mentioned in the textbook (however nothing really much was said about it)
     
  13. Feb 13, 2010 #12

    Redbelly98

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    Okay, sounds like those were three dead ends. Oh well, sorry.
     
  14. Feb 13, 2010 #13
    thanks anyways, I guess I'll just have to forget about it.
     
  15. Feb 13, 2011 #14
    You are doing it right, I had the same question but different values:
    120 ohm resistor, current in the circuit (I) = 73mA, saturation current (Io) = 0.010mA, and T=300K.
    My answer came out to be 0.230V which was accepted as correct. Just try it again and make sure all your calculations are correct.
     
  16. Feb 13, 2011 #15
    I'm not sure how to get the second part:
    What is the voltage across the diode-resistor combination?
    Any ideas?
     
  17. Feb 13, 2011 #16

    gneill

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    Have you considered VD + VR ?
     
  18. Feb 13, 2011 #17
    Ahh...yes. I had put in Vr-Vd.
    Vd + Vr worked, I got 8.99V/
    Thanks!
     
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