0
\$\begingroup\$

I am currently reading Sze (Physics of Semiconductor Devices, 3e, Chapter 2.4.1) and am trying to understand his discussion of thermal instability breakdown/runaway in pn junctions. I have also consulted the reference he refers to (Strutt, Semiconductor Devices Vol 1, Chapter 2, Section 4.4) where a virtually identical discussion is given. I have given both discussions at the end of this post for reference, but I have two primary questions:

(1) I am confused about what exactly these hyperbolas (straight on log-log plot) actually are. Are we arguing as follows? For a diode to remain at steady state, in particular its temperature must remain at steady state. Then, for a given diode and ambient situation (i.e. thermal circuit), a fixed diode temperature implies a fixed power (call it \$P_0(T)\$) which the fixed diode can dissipate without changing its temperature. Then with reverse voltage as a variable we can write the constraint (I will use power and current rather than power density and current density under a tacit 1D assumption) $$J_0(V_R,T) = P_0(T)/V_R.$$ This would seem to make sense given the plots since we would expect the power \$P_0(T)\$ at steady state in the given thermal circuit to be greater at a higher temperature; therefore, \$J_0(V_R,T)\$ increases monotonically with temperature at fixed \$V_R\$.

We then have $$\log J_0 = \log P_0 - \log V_R$$ which gives the linear decrease with \$V_R\$ seen in the plots.

(2) Given that (1) is understood/true, I am still confused about why this implies thermal runaway. I can picture that continuing to drive up the reverse voltage on the junction will force us on the dark curve seen in the log-log plot since a steady state will only obtain when we have steady state both with respect to the IV characteristics (horizontal lines) and thermal circuit (sloped lines) at a given temperature. I am confused about what happens at the turnover voltage though. Why do we get runaway at this point? Both authors just make an allusion to "negative differential resistance" but can someone explain operationally why this means we get runaway?

Sze: enter image description here enter image description here

Strutt: enter image description here enter image description here

\$\endgroup\$

1 Answer 1

0
\$\begingroup\$

All the power dissipated occurs in the pn junction. There is some thermal resistance between the junction and the case and between the case and the environment. The power dissipated will increase the junction temperature to a temperature above ambient. The increase in internal temperature will cause a change in the characteristics of the semiconductor which will increase the conduction current and therefore an increase in power dissipation and internal temperature.

\$\endgroup\$
1
  • \$\begingroup\$ I kind of see, but can you explain what happens at the point at which we reach negative differential resistance? Why does this point define the runaway point? \$\endgroup\$
    – EE18
    Commented Dec 4, 2023 at 2:25

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.