0
$\begingroup$

When a pn-junction (e.g. an LED) is forward biased the height of the potential barrier is reduced...

forward biased pn junction

...unless you apply a huge potential $V$, in which case the barrier gets inverted and the band edges in the n-region go above their counterparts in the p-region. But I have never seen this in the literature. Is it impossible or just taboo?

$\endgroup$
  • 2
    $\begingroup$ Because you physically cannot get to that situation. Other physics beyond the simple diode equation come in to play. In other words, in this case, it will start acting like a resistor... $\endgroup$ – Jon Custer Aug 26 '15 at 16:58
  • $\begingroup$ @JonCuster I see... What phenomena are you referring to? e-hole recombination? $\endgroup$ – Rol Aug 26 '15 at 17:02
  • 1
    $\begingroup$ Not surprisingly, the SRH diode model is derived from a set of assumptions. For example, the resistance of the semiconductors away from the junction area do not appear in it. Yet, clearly, at high enough current they will become important. If you try to invert the bands, an infinite amount of current would try to flow. Since that is non-physical, it isn't happening, and the reason is that additional factors come in to play. $\endgroup$ – Jon Custer Aug 26 '15 at 22:32
2
$\begingroup$

In theory, yes, but remember that a diode is a balance between diffusion current in forward bias and drift current in reverse bias. Ideally $ J=exp^{(qV/kT)}-1$. Since the current is exponential in voltage, you easily get huge currents at forward bias (i.e. the diode turns on). In reality, there is some finite resistance in the structure (the bulk regions, the contacts), that add an IR voltage drop across the actual junction... $V_d=V_a-IR$. At small currents, even with ok resistances $V_d \approx V_a$, but for for large currents even at small resistances, it becomes important. Also, resistance is an indirect function of current through heating effects, so as you pass more current, omega increases too. Basically, thermodynamics will do everything in its power to make sure the situation you describe does not happen. Although, if you do something like put a heterostructure in the depletion region to act as a blocking layer to current, neat things can start happening.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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