Diode Voltage in Temperature

Discussion in 'Homework Help' started by matan, May 21, 2009.

  1. matan

    Thread Starter New Member

    May 21, 2009
    1
    0
    I'm a little bit confused about the Vd Vs. Temperature.

    The Diode Equation : [1] Id=Is(e^(Vd*q/n*k*T)-1),
    so => [2] Vd=[k*T*n*ln((Id/Is)+1)]/q
    Lets :
    n=1
    k=1.38e-23
    T(°k)=273+T(°c)
    q=1.6e-19
    Id=1mA
    Is=1e-12A

    Vd/T=86.25e-6*20.72=1.79e-3

    Vd(at 25°c)=1.79e-3*(273+25)=0.5326V
    Vd(at 26°c)=1.79e-3*(273+26)=0.5343V

    so ΔVd=Vd(26°c)-Vd(25°c)=0.5343-0.5326=1.7mV

    In the litterature and in all the world at high temperature the Diode Voltage shouled be -2mV/°c ! and i see "+".

    why the Vd at 26°c is higher then Vd at 25°c?
    In the formula i need to write "-", because "Electron Charge" (q) ?
    where is my misteke? :confused:
    someone have an articale or link that can explain in simple?

    10X
    :mad:
     
  2. Ron H

    AAC Fanatic!

    Apr 14, 2005
    7,050
    657
    I think the reason is that Is is also a function of temperature. Have a look at this site, and go down to "temperature dependence".
    See this site also. It is probably easier to understand.
     
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