What happens to the Base emitter voltage of a transistor due to increase in temperature?

Explanation;

The schematic of the NPN and PNP transistors

Base emitter voltage means voltage drop across the diode.

We are considering the NPN configuration here for an explanation.

The intrinsic concentration is defined as:

\(n_i^2 = \frac{{{A_0}{T^3}{e^{ - \frac{{{E_G}_0}}{{kT}}}}}}{{{{\left( {c{m^3}} \right)}^2}}}\)

\({A_0} = 2.33 \times {10^{31}}{\left[ {\frac{{{m_n}{m_p}}}{{{m^2}}}} \right]^{\frac{3}{2}}}{e^{\frac{\beta }{k}}}\)

\(n_i^2 \propto {T^3}\)

The barrier potential of the diode is defined as:

\({V_0} = \frac{{kT}}{q}\ln \left( {\frac{{{N_A}{N_D}}}{{n_i^2}}} \right)\)

\({V_0} \propto \frac{1}{{n_i^2}} \Rightarrow {V_0} \propto \frac{1}{{{T^3}}}\)

The variation of the diode characteristics with the temperature is:

• Generally, when semiconductor devices are heated, there will be movement of Charge carriers, because these charge carriers are freed from the covalent bond.
• The Charge carriers add to the already moving Charge carriers and align with them, i.e., they possess more energy to move, this Energy reduces the conduction band and Valency band gaps.
• Hence the reduction in the forward-voltage. Typically, for every degree increase in temperature, there is a 2.5 mV reduction in the forward bias voltage.
• As the temperature increases there will decrease in the base-emitter voltage.

NOTE:

In Ge transistor VBE decreases by 2.1 mV/°C rise in temperature

\(\frac{{d{V_{BE}}}}{{dT}} = - 2.1mV/^\circ C\)

In Si transistor VBE decreases by 2.3 mV/°C rise in temperature

\(\frac{{d{V_{BE}}}}{{dT}} = - 2.3mV/^\circ C\)

In general VBE decreases by 2.5 mV/°C rise in temperature

\(\frac{{d{V_{BE}}}}{{dT}} = - 2.5mV/^\circ C\)