Shockley diode equation - temperature influence when looking at Shockley's ideal diode equation I wonder what "meaning" the temperature has in there?
If I assume $$I(T) = I_0 \left[\exp \left( \frac{U}{k_BT/q}\right)-1 \right]$$
and vary the temperature (250K, 300K, 350K -> blue, yellow, red) I'll see exactly the opposite trend of what I would've expected. I expect that at higher temperature the diode will start to conduct at lower voltages, but the opposite is given by Shockleys equation. What am I missing here?

 A: I was struggling with this same issue and finally found the answer!
The first reference is another post: Voltage across diode, Shockley equation. This references Ken Kuhn's Diode Characteristics whitepaper and the whitepaper references the diode_plots.xls workbook that he created to go along with the whitepaper. This workbook has all of the calculations (the ones we were missing that I will describe shortly) needed to properly model the forward voltage decrease with temperature.
If you simply plot the current vs. voltage per the Shockley equation, the result is that the voltage increases with increasing temperature, which is not what we know to be true. The missing piece is that the reverse saturation current, $I_s$ ($I_0$ in your formula) is also highly dependent on temperature and increases with temperature. Use the following formula to calculate the temperature-adjusted reverse saturation current. $$I_s(T_0)\cdot \exp\left[\frac{E_g\cdot q}{k}\cdot \left(\frac{1}{T_0}-\frac{1}{T}\right)\right]$$
Where,

*

*$I_S(T_0)$ is the reverse saturation current at 298.15°K (25°C)

*$E_g$ is the bandgap voltage for silicon (1.11V to 1.28V)

*$q$ is the elementary electric charge in Coulombs

*$k$ is Boltzmann's constant

If you're looking to plot BJT collector or emitter current vs. $v_{BE}$, there is an additional term used as shown in the post, Why does the base-emitter voltage of a BJT decrease with temperature?
