Yes, I meant to say the conductivity drops as the temperature drops (which is what was happening in the experiment).
Which would be normal for a semiconductor - but the conductivity was never measured. In order to measure the conductivity you need a piece of semiconductor without a junction! Then you will find that the conductivity is
\sigma = q ( µ
n * n + µ
p * p) // let's see if the forum can keep a \mu in place
where \mu is the mobility of either electrons (n) or holes (p), where n and p are the density of free electrons and holes. The mobility might have a slight temperature dependence as well, but the main part here is that the density of free electrons and holes increases exponentially with increasing temperature.
VT is proportional to temperature (i.e. equal to kT/q) so for a given voltage the current should increase, so for a constant current the voltage across the junction should drop if temperature is decreased (as in the experiment).
V
T is not the built-in potential, it is the thermal voltage, or rather the mean thermal energy of an electron divided by its elementary charge.
Assuming the ideality factor n is 1 (for ease of maths) then the diode equation is I0(eV/VT-1) so given VT = kT/q if T goes to half T then to keep the exponential term constant V also has to go to half its previous value.
You got the diode equation alright, but you are neglecting the temperature dependence of I
0 !
I
0 = I
00 e
-Eg/kTFor the analysis let's assume that the applied voltage is much higher than V
T which is 26 mV at room temperature, then we can neglect the -1 in the parenthesis and the diode equation becomes (assuming ideality factor n=1)
I = I
00 e
-Eg/kT e
qV/kT = I
00 e
(qV-Eg)/kTIn order to see in which direction the temperature coefficient goes, we have to look into (qV-Eg)/kT. Your interpretation is completely correct: to keep the exponent constant with changing temperature. I.e. for half T we need a voltage V which makes (qV-Eg) half the value as well - for all reasonable voltages over a diode, qV < Eg.
log I = log I
00 + (qV-Eg)/kT
V = kT/q * log I/I
00 + Eg/q
This seems to still indicate a positive temperature coefficient, but wait a second. How big is this strange I
00?
Let us take a standard 1N4148: NXPs datasheet states about 0.67 V at 5 mA, it's a silicon diode with a band gap of 1.1 eV
0.67 V = 0.026 V * log (5 mA/I
00) + 1.1 V
we see that the logarithm
must be negative!
-16.54 = log (5 mA/I
00) or then at 5 mA: V = 1.1 V - 16.54 * kT/q
which gives an I
00 = 76200 A
There is of course no current of 76200 A anywhere inside a small 1N4148, it is just a parameter in the equation. But it is the reason why the diode equation gives us a dropping forward voltage over a diode with increasing temperature.
Look again at the equations above. For T=0 K at absolute zero the diode equation predicts that the forward voltage drop independent of current will reach the value of the band gap in eV. But even before we reach these low temperatures with a real diode, other things will happen...