Question about diode temperature and forward voltage

I'm trying to understand how a diode can be used to detect changes in temperature. According to all the literature I've read, a diode's forward voltage, Vf, falls by 2 mV for every 1 degree C increase in temperature. This doesn't seem to agree with the diode equation, which suggests that Vf should increase with temperature.

Vf = k*T/q * ln( If / Is)

What am I missing?

Reply to
acataldo
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Where did you get this equation, and what do all of those other variables represent?

Thanks, Rich

Reply to
Rich Grise

Diode equation:

If = Is*exp( (Vf * q) / (k*T))

k = Boltzman constant T = Temperature in Kelvin q = charge of an electron, 1.6*10^-19 Coulombs If = forward current Is = saturation current Vf = forward voltage

Rich Grise wrote:

Reply to
acataldo

Reply to
acataldo

what your missing is the way it's being used in the circuit. diodes start conducting better as they get warmer. if you were to have a diode in a series circuit and measure voltage that way, then you would get the increase as indicated how ever what i think your looking at is the diode being used as a shunt load.. as the diode warms up it conducts better and there for dropping the voltage on the load. normally 2 diodes in series are used into a pot that drive's the base of a transistor. the pot, is to calibrate the bias point.

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Reply to
Jamie
** GG.

** Notice how only one significant digit is given.

Means actual tempco can be between -1.5 and - 2.5 mV per C

A diode needs to be calibrated to use as a thermometer.

** The "diode equation " has NOTHING to do with the matter.

That is a PHYSICS equation for a hypothetical IDEAL diode with NO forward voltage drop ( hence no forward drop tempco) and no internal resistance or other imperfections all real diodes have.

....... Phil

Reply to
Phil Allison

I think what is being missed is that in this application the diode is usually fed with a constant current source. When this is done everything except Vf and T become constant.

The bias current, in this case If, is selected to be much lower than Is to avoid self heating.

When If/Is

Reply to
G. Schindler

How do you get If/Is < 1? *Is* is REALLY small, isn't it (pico amps)?

Bob

Reply to
Bob

k is negative.

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Reply to
Jasen Betts

the diode equation is

i = i0((exp)(v/nVt)-1)

i=diode current v=applied voltage n=1(for germanium diode):2(for silicon diode) i0=reverse leakage current Vt=kt/q

k=boltzmann const t=temp of diode junction q=charge of electron

this eqn gives the relation between current and temp

Reply to
chrajesh911

the diode equation is

i = i0((exp)(v/nVt)-1)

i=diode current v=applied voltage n=1(for germanium diode):2(for silicon diode) i0=reverse leakage current Vt=kt/q

k=boltzmann const t=temp of diode junction q=charge of electron

this eqn gives the relation between current and temp

Reply to
chrajesh911

** But as any businessman knows, never f*ck around with your current " temp ".

..... Phil

Reply to
Phil Allison

Reply to
acataldo

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Reply to
John Fields

Good question, You are right with your idea, exept that you didn't consider the temperature dependence of Is, which doubles every 10 degrees and thus together they roughly decrease by 2mV/K

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Reply to
Ban

Reply to
acataldo

Ban,

Thank you for your insightful reply. Your approach seems correct. If I keep If constant and increase Is with temperature, dVf/dT is indeed negative for the rate of change that you mentioned.

I had considered that before, but I thought the change in Is would be too small to make a difference. How are you solving for Is? I used:

Is = q*A*(Dp/Lp*pn), for a p+ n diode

A = cross-sectional area Dp = Drift current coefficient Lp = recombination length pn = minority carrier concentration on the n side

Can you please tell me how you arrived at Is doubling every 10 degrees?

Thank you

Reply to
acataldo

I guess that would depend on what the definition of 'Is' is. ;-)

Sorry, couldn't help myself.

Reply to
Rich Grise

Yes, though none of the other respondents appear to know anything about it.

Here is the EM model equation for how Is varies with temperature, ignoring the emission coefficient (taken as 1, for now), the equation is, for the OP:

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

which becomes:

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

The derivative is then 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, that isn't the whole picture. Is also varies with temp:

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

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

Reply to
Jonathan Kirwan

Jon,

Great. I've never seen Isat in that form. I'll do some work in matlab and see if it works out. I've checked the spice results and it is about what you said. Thank you for taking the time to delve into this.

Anthony

Reply to
acataldo

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