Monday, 23 March 2015

transistors - 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 $V_T$?



Answer



$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 $


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