I am having trouble understanding why is temperature coefficient for avalanche breakdown is positive? The explanation I found online says as the temperature increase the mean free path decreases making collision and generation of carrier pair easier. But the electric field require to accelerate charge carriers to point of collision remains constant . So why does avalanche breakdown voltage increases with temperature if the electric field remains same? Doesn't increase in voltage indicate it is slightly harder to obtain avalanche breakdown? If the mean free path is decreasing and charge carrier generation is getting easier shouldn't occur at same voltage and not increase further


1 Answer 1


In an avalanche breakdown minority carrier electrons are accelerated by the applied electric field and their kinetic energy increases until it becomes large enough to excite an electron from the valence to the conduction band.

But suppose that minority electron collides with and scatters off an atom while it is being accelerated by the applied field. Then it will lose some of its kinetic energy and it will take longer to reach a high enough energy to cause the breakdown. If it collides with atoms often enough it will never reach a high enough energy to cause a breakdown.

This is why increasing the temperature increases the breakdown voltage. The increased temperature increases the collision frequency of the minority electrons in the depletion layer so it decreases the maximum KE those electrons can reach in between each collision.

As a rough guide suppose the mean free path of the electrons is $\ell$ and the electric field strength in the depletion layer is $E$ volts per metre, then the maximum KE the electron can attain in between collisions is:

$$ KE = eE\ell $$

So if the band gap is $E_g$ for a breakdown to start we need:

$$ eE\ell \gt E_g $$


$$ E \gt \frac{E_g}{e\ell} $$

Increasing the temperature decreases the mean free path $\ell$ so it increases the required field strength. This is an oversimplification since the scattering is a random process so we get a random distribution of mean free paths not a single value. However hopefully it gives you the basic idea.


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.