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

According to Sedra/Smith Microelectronic Circuits, $v_{BE}$ changes by $-2\text{mV}/\text{°C}$. I cannot understand how this could possibly be the case given the equations I am familiar with.

With all currents kept constant, we have:

$\large{i_E = \frac{I_s}{\alpha}e^{v_{BE}/V_T}}$

To keep $i_E$ constant, any change in $V_T$ would have to be accompanied by a change of the same factor in $v_{BE}$, otherwise $\alpha$ or $I_s$ would have to change, which as far as I understand is not possible.

So, how can $v_{BE}$ be inversely proportional to $V_T$?

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

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:

The corrected formula is:

$$I_S(T_1) = I_S(T_0) \left[\dfrac{T_1}{T_0}\right]^{XTI} \exp\left[ \dfrac{E_g q (1{\rm\,V})}{k} \left(\dfrac{1}{T_0}-\dfrac{1}{T_1} \right) \right],$$

where $$\E_g\$$ is in electron-Volts, remaining quantities are in SI units, and $$\XTI=3\$$ unless the transistor model specifies otherwise.

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

• Thanks very much, I knew I must have been missing something simple/obvious like that! Jul 13, 2012 at 17:01

"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.

As temperature increases , Ibo or reverse saturated current through emitter junction increases and hence overall current through Je reducess, and hence Vbe also get reduces proportionally.

• "....and hence..." Please can you explain why do you think that such an automatic Vbe reduction would take place? A current goes high - and a voltage goes down; is this logical? I think, this is NOT true. See my detailed answer below.
– LvW
Jan 20, 2017 at 15:25