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Q. on Diode Equation

Discussion in 'Electronic Basics' started by Blake, May 26, 2005.

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  1. Blake

    Blake Guest

    Recently I was looking into the use of a diode as a temperature sensor. I
    decided to start with the "Diode Equation":

    Id=Is*(exp(qVd/NKT)-1)

    This equation predicts that an increase in temperature T will result in an
    increase in diode voltage Vd, assuming constant current. But I know from
    experience that the opposite is true. Diode voltage drops as temperature
    increases.

    What gives? Is saturation current (Is) a function of temperature, strong
    enough to reverse the upward dVd/dT predicted by the diode equation? Or do I
    completely misunderstand the basics here?
     
  2. Yes, that's why. If you are interested, I'll post that equation, as
    well.

    Jon
     
  3. Ban

    Ban Guest

    Yes you misunderstand, look at the equation. The temperature T is appearing
    in the denominator making the *current* increase with constant Vd, or in
    other words you need less voltage for the same current. The relationship
    dVd/dT (partial d) is roughly -2mV/K.
     
  4. Spajky

    Spajky Guest

    If anybody is interested (to get maybe some further ideas) of analog
    converter for diode to thermistor read-outs:
    how to make a gadget that can do that for Cpu onDie diode for a PC
    MoBo that is natively not supporting that feature:
    my project is on my site under electronics ...
    may be of use for some other purpose for someone ... :)
     
  5. I think you may be wrong in your rationalizing, here. An increase in
    T would cause a _decrease_ in Id, not an increase. Thus, it would
    require an increase in Vd in order to keep Id unchanged.

    In this case, one needs to solve for Vd before taking the derivative
    with respect to T. Thus:

    Vd(T) = (kT/q) * ln( 1+Ic/Is )

    The derivative (assuming 'Is' is constant relative to T) is then
    trivially:

    d Vd(T) = (k/q) * ln( 1+Ic/Is ) dT

    and thus positive-going with respect to T.

    I believe the OP was not mixed up about the question. See my other
    reply.

    Jon
     
  6. Ignoring the emission coefficient (taken as 1, for now), the equation
    is:

    Id(T) = Is(T) * (e^(q*Vd/(k*T))-1)

    which becomes:

    Vd(T) = (kT/q) * ln( 1 + Ic/Is(T) )

    The derivative is trivially:

    d Vd(T) = (k/q) * ln( 1 + Ic/Is(T) ) dT

    which is a positive trend, very nearly +2mV/K for modest Ic... but
    positive.

    However:

    Is(T) = Is(Tnom) * (T/Tnom)^3 * e^(-(q*Eg/k)*(1/T-1/Tnom))

    which complicates things.

    The new derivative is a bit large.

    Assume:
    X = T^3 * Isat * e^(q*Eg/(k*Tnom))
    Y = Tnom^3 * Ic * e^(q*Eg/(k*T))

    Then the derivative, I think, is:

    X+Y
    k*Tnom*T*( (X+Y) * ln( -------- ) - 3*Y ) - q*Eg*( X*T+Y*T+Y*Tnom )
    Isat*T^3

    ---------------------------------------------------------------------

    q * Tnom * T * (X+Y)


    Tnom is the nominal temperature (Kelvin, of course) at which the
    device data is taken and Eg is the effective energy gap in electron
    volts for the semiconductor material. Of course, 'k' is Boltzmann's
    constant and T is the temperature of interest.

    Eg defaults to 1.11 eV in spice, I think. For an Ic=10uA and a stock
    Isat of about 1E-15, the figure comes out to about -2.07mV/K in the
    vicinity of 20 Celsius ambient.

    Jon
     
  7. Bob Masta

    Bob Masta Guest

    <snip>
    Check it out. Use an standard DMM on the 2 kOhm range to put
    a constant current through the diode, while reading the voltage
    across it. Now try warming and cooling the diode and you will
    see that the voltage is directly proportional to temperature.
    In fact, it is *so* proportional that this is one of the most linear
    temperature sensors you will find, from near absolute zero until
    the junction melts.

    Best regards,


    Bob Masta
    dqatechATdaqartaDOTcom

    D A Q A R T A
    Data AcQuisition And Real-Time Analysis
    www.daqarta.com
    Home of DaqGen, the FREEWARE signal generator
     
  8. I suggest you take your own advice and report back.
    Then read Mr. Kirwan's contribution to see why you
    are seriously mistaken with your assertion.
    I expect the leads will melt first. And long
    before that happens, the approximations
    underlying the diode equation will fail to
    be useful predictors of the junction voltage.
     
  9. Ban

    Ban Guest

    I throw in the formula for Is(T) = Isk * e(-Eg/mkT)
    Eg = 1.11eV bandgap of silicon; m=q=1.1; Isk= 1E6A; k=1.38E-23Ws/K
    Boltzmann

    for 300K we get Is = 11.2pA; for 310K Is= 39.3pA
    We put that into your formula and we get for 1mA current @300K 520.7mV and
    @310K 501.0mV
    -19.7mV/10K. The influence of Is is much bigger than the small change in
    Vt= from 25.9mV to 26,7mV.
    THX for the correction, Jon
     
  10. No problem. I'm just a hobbyist in electronics, so I rarely get to
    say all that much about it. But I'm okay in math, so I can work a
    model once in a while, here and there. It's nice when I can say
    something but it doesn't mean I know a darned thing about the reality,
    since my experience is so very limited.

    Jon
     
  11. Bob Masta

    Bob Masta Guest

    OOPS! My face is red now! (And not from
    high temperatures!)

    OK, to set the record straight, my point
    about the linearity still holds... I just got
    the direction backwards: The voltage
    goes *down* as the temperature goes
    up. (I've never actually pushed it to see
    what melts first, just assumed it was
    the junction. I seem to recall that linearity
    is typically good to well over 100C, though.)

    Hey, nobody's purfekt! ;-)

    Best regards,




    Bob Masta
    dqatechATdaqartaDOTcom

    D A Q A R T A
    Data AcQuisition And Real-Time Analysis
    www.daqarta.com
    Home of DaqGen, the FREEWARE signal generator
     
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