Find Vd across diode

Discussion in 'Homework Help' started by Heylow, Dec 21, 2010.

  1. Heylow

    Thread Starter New Member

    Dec 21, 2010
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    Me and my friends have been working on this for days, and none of us are even close to the answer.

    We are supposed to calculate the voltage drop (Vd) across the diode, while the saturation current is equal to 3x10^-16 A.
    R1 = 1kΩ and
    Is = 1 mA

    The circuit can be found at: [​IMG]

    Any help would be appreciated.
     
  2. Jony130

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    Feb 17, 2009
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  3. RRITESH KAKKAR

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    Jun 29, 2010
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    No idea..!!
     
  4. Heylow

    Thread Starter New Member

    Dec 21, 2010
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    But shouldn't Vs = (Ir + Id) * R
    where Ir is resistor current and Id is diode current.

    Let me clarify, in my image, Io is the current supply, while Is is meant as saturation current.
    Does your method apply under these conditions?
     
  5. Jony130

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    Vs = 1mA * 1K = 1V
     
  6. Jony130

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    No, Vs = 1mA * 1K Thévenin's equivalent circuit
     
  7. Jony130

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  8. Georacer

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    Nov 25, 2009
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    This is a diode specific problem, so simple circuit equations aren't enough. You need to take into account that V_d=n\cdot V_{\small{T}}\cdot ln\left(\frac{i_d}{I_{sat}}\right)
    That way you have that equation and V_d=V_r\\I_s=I_r+I_d

    You can solve that system with some recursions starting from an initial value. The solution will be approximate.
     
  9. Heylow

    Thread Starter New Member

    Dec 21, 2010
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    I understand now :), I just did not get the Vs part, but that now I get it.
    I answered my own question when I said Vs = (Ir + Id) * R, which is exactly what you did, but with the equivalent circuit.
    I guess I have to read up on my Thévenin's theorem.

    Again thank you very very very much!!! :D
     
  10. electric4891

    New Member

    Dec 21, 2010
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    i cant seem to get this exact number i keep getting 0.699
    even when i use goole calculator instead of my own.
    and when i increase the source current, Vd becomes greater than 0.7

    can anyone think of what im doing wrong?
     
  11. JoeJester

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    I got 0.699886 using excel ... from the listed information.
     
  12. electric4891

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    Dec 21, 2010
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    thats exactly what i got, at least i aint the only one. thanks
     
  13. Jony130

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    Feb 17, 2009
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    Well very strange
    When I put this on google
    Code ( (Unknown Language)):
    1. (ln( (1 - 0.6)/300E-15) -1 )) * 0.026
    I get
    0.699886283 (1)
    0.692416403 (2)
    0.693055624 (3)
    0.693001535 (4)
    Later on I check this with hand + calculator.
     
  14. Ron H

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    Apr 14, 2005
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    Interesting. I wrote a little Basic program to do the iterative calculation, and got VD=0.71689909. It checked out on my HP15C calculator.
    Code ( (Unknown Language)):
    1. Vt=.026
    2. Vs=1
    3. Is=3e-16
    4. R=1e3
    5. Vdnew=.6
    6. Do
    7.     Vd=Vdnew
    8.     Vdnew=Vt*log((Vs-Vd)/(R*Is)-1)
    9.     Print "vd=";Vd
    10. Print "Vdnew=";Vdnew
    11. Loop until Vdnew=Vd
    12.  
    13. end
    Note that "log" is the natural logarithm, at least in Just BASIC v1.01.
    What did I do wrong? Or what did you guys do wrong?
     
  15. Jony130

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    Feb 17, 2009
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    Well I think I know where I made the error.
    I forget one bracket in the google code

    Code ( (Unknown Language)):
    1. (ln(( (1 - 0.6)/300E-15) -1 )) * 0.026
    VD = 0.716899093V

    And I also use Mathematica software and I get this solution

    Vd = 0.7168990929699331
     
  16. JoeJester

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    Apr 26, 2005
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    with the corrected bracket

    I got 0.725886 in excel
    In google ... the result is (ln(( (1 - 0.6)/300E-15) -1 )) * 0.026 = 0.725886283


    The constant e in excel is 2.71828182845904

    I'll be checking it again later ... with my TI-83.
     
  17. Georacer

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    Nov 25, 2009
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    Take care not to mix the decical logarithm (log) with the natural logarithm (ln). I think the formula calls for the natural logarithm.
     
  18. Ron H

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    Apr 14, 2005
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    I didn't. From my previous post:

     
  19. JoeJester

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    "log" in basic is the natural log. Log10 in basic is the decimal log.

    Log10 is also used in excel for decimal log.

    Much to my surprise, the TI-83 agreed with google and excel at approximately 0.7258862829

    The TI-83 uses 2.718281828459 for e.

    I am at a loss for the error at this moment.
     
    Last edited: Dec 22, 2010
  20. Ron H

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    Apr 14, 2005
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    You didn't run the iterations. The answer you got, 0.7258862829, is the first step in the iteration, and you only get that if you use VD=0.6 as your first guess. You need to plug that number into VD in the right side of the equation, and re-evaluate. Plug the new answer into VD and evaluate again, ad nauseum, until the evaluated value doesn't change. See my Basic program in post #14,
    Code ( (Unknown Language)):
    1. Vt=.026
    2. Vs=1
    3. Is=3e-16
    4. R=1e3
    5. Vdnew=.6
    6. Do
    7.     Vd=Vdnew
    8.     Vdnew=Vt*log((Vs-Vd)/(R*Is)-1)
    9.     Print "vd=";Vd
    10. Print "Vdnew=";Vdnew
    11. Loop until Vdnew=Vd
    12.  
    13. end
    and the link posted by Jony130 in post #7.
     
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