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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\$?

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\$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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  • \$\begingroup\$ Thanks very much, I knew I must have been missing something simple/obvious like that! \$\endgroup\$ – Gordon Bailey Jul 13 '12 at 17:01
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"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.

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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.

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  • \$\begingroup\$ "....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. \$\endgroup\$ – LvW Jan 20 '17 at 15:25

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