Why does Vbe decreases with the increase in temperature for an n-p-n transistor?

Question

Microelectronics Circuits by Sedra/Smith says,

As in silicon diodes, the voltage across the emitter–base junction decreases by about 2 mV for each rise of 1°C in temperature, provided the junction is operating at a constant current.

$$i_c = I_s e^{\frac{V_{BE}}{V_T}}$$

$$\implies V_{BE} = V_T \ln\left( \dfrac{i_c}{I_s}\right)$$

$$\implies V_{BE} = \frac{kT}{q} \ln\left( \frac{i_c}{I_s}\right)$$

It can be clearly seen that when \$i_c\$ is constant, \$V_{BE}\$ is directly proportional to temperature.

Then why does the voltage decreases instead of increasing?

Answer 1

It decreases because Is is an exponential function of temperature, roughly doubling to quintupling for every 10°C, depending on \$\eta\$.

Much more from this chapter.

The equation for change is derived as follows:

\$dv/dT = \frac{v - (3V_T +V_G)}{T}\$

where Vt is the thermal voltage 0.0259V @ 300K (Vt = kT/q)

Vg is the bandgap voltage for silicon 1.11V

T is the absolute temperature say 300K

so at room temperature of 300K with v of 600mV dv/dT is about -1.96mV/°C