Why does the base-emitter voltage of a BJT decrease with temperature?

\$I_S \$ is highly temperature dependent. As the temperature of the material increases, more electron-hole pairs are thermally generated, increasing \$I_S \$. Here's a link that gives the formula SPICE uses for \$I_S \$

Temperature appears explicitly in the exponential terms of the BJT and diode model equations. In addition, saturation currents have a built-in temperature dependence. The temperature dependence of the saturation current in the BJT models is determined by:

\$ I_S(T_1)=I_S(T_0)\left[\dfrac{T_1}{T_0}\right]^{XTI}exp\left[{\dfrac{-E_gq(T_1T_0)}{k(T_1-T_0)}}\right] \$

I believe that \$I_S \$ is roughly cubic in \$ T \$

"So, how can vBE be inversely proportional to VT?"

I think, this leads to a false understanding of the effect to be observed. With other words: Base-emitter voltage does NOT decrease (automatically) with rising temperature.

The effect is as follows: For rising temperature the collector current Ic increases (because of Is temperature dependence). That means: To keep this current Ic on the same level the base-emitter voltage must be (externally !) decreased. Hence, the data sheet says that for constant Ic the well-known value −2mV/K applies.

EDIT: I like to add that in the following link the temperature dependence of Is is derived as well as a formula for the „magic“ value of -2mV/K .

web.mit.edu/klund/www/Dphysics.pdf

It is interesting to note that this derivation is based on transistor physics only - and without using the base current and the current gain at all.

In this context, I remember some - often controversal debated - questions whether the bipolar transistor is controlled by the current Ib or the voltage Vbe. For me, the derivation contained in the said document is a further clear evidence that the transistor - physically speaking - is controlled by the applied voltage Vbe.