4
\$\begingroup\$

My questions pertains to the \$ \beta\$ value of a transistor. On page 162 of this book, it says that \$ \beta\$ increases as temperature of the transistor increases. But it doesn't explain why. Can anyone explain that bit more thoroughly?

\$\endgroup\$
2
  • 3
    \$\begingroup\$ The formula's in most books (including yours) are slllllllllllllllightly simplified. More accurate formulae are extremely hard to work with and unnecessarily complicated. For example a little more complicated version to calculate collector current already contains T for temperature: \$I_C = I_S \cdot e^{(\frac{q V_{BE}}{kT})}\$. Google will probably prompty you with more of this sort of formulae which can eventually lead to a formula for \$\beta\$. I think your better bet is to find a more descriptive answer than formulae. \$\endgroup\$
    – jippie
    Commented Mar 12, 2013 at 19:54
  • \$\begingroup\$ But then we know that saturation current increases with increasing temperature which would increase Ie and thus increase the denominator which in turn would decrease beta, how so ? Please explain! \$\endgroup\$ Commented Jul 12, 2020 at 12:29

1 Answer 1

4
\$\begingroup\$

If you look at the Ebers -Moll model, one can see that:

$$I_C = \dfrac{\beta_F}{\beta_F +1}I_E$$

which can be rewritten as

$$\beta_F +1 = \beta_F \dfrac{I_E}{I_C}$$

$$1+\dfrac{1}{\beta_F}=\dfrac{I_E}{I_C}$$

$$ \beta_F = \dfrac{1}{\frac{I_E}{I_C}-1}$$

Where

$$I_E = I_{ES}\{e^{\frac{V_{BE}}{V_T}}-1\}$$

where $$V_T=\dfrac{k_BT}{q}$$ is the thermal voltage.

Here it should be clear that the temperature cause a change in the thermal voltage which leads to a change in beta.

The equation is basically a statement about the distribution of energy per unit of charge, as the temperature increases, the exponential in the emitter current equation decreases, which causes the ratio of collector current to emitter current to increase, and since that ratio is in the denominator of the equation for beta, it has an inverse relationship, which causes beta to increase. So as temperature increases, beta increases.

\$\endgroup\$
1
  • 1
    \$\begingroup\$ Ther above explanation is correct - HOWEVER, the contribution of Vt is of less importance. More than that - an increase of Vt will cause a DECREASE of the exponential function (contrary to observations). The most important temperature influence comes from the saturation current Ies which strongly depends on temperatures. \$\endgroup\$
    – LvW
    Commented Mar 8, 2017 at 11:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.