Yes, that's why. If you are interested, I'll post that equation, as

well.

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