I’m currently studying about Zener diodes. A line in my textbook says:

“Zener diode is fabricated by heavily doping both p-, and n- sides of the junction. Due to this, depletion region formed is very thin (<10^-6 m) and the electric field is extremely high (~5 x 10^6 V/m) even for a small reverse bias voltage of about 5V.”

I understand how the doping helps in decreasing the width of the depletion region, but I fail to understand why the electric field becomes exceptionally high. The Internet doesn’t help much either; all sources just directly state that the electric field increases, without providing a reason.

Any explanation would be appreciated.

  • 3
    $\begingroup$ The built in potential is a constant. The thinner the junction the higher the field. $\endgroup$
    – Jon Custer
    Dec 25, 2019 at 17:08
  • 1
    $\begingroup$ Does this have anything to do with the relation, E= V/d? And why is the potential constant? $\endgroup$
    – SG_27
    Dec 25, 2019 at 17:38
  • 1
    $\begingroup$ The potential across the junction is the difference in Fermi levels, so is essentially fixed for a given semiconductor. $\endgroup$
    – Jon Custer
    Dec 25, 2019 at 18:43
  • $\begingroup$ Oh alright, makes sense. Thanks. $\endgroup$
    – SG_27
    Dec 26, 2019 at 6:51

1 Answer 1


The built-in potential is not constant, as it depends on:

  • the doping (the higher these, both donor and acceptor impurities, the higher the built-in potential),
  • on the particular semiconductor and
  • on temperature.

In particular, built-in potential is: $$V_{\rm bi} = V_T \ln \frac{N_{\rm a}N_{\rm d}}{n_{\rm i}^2}$$ $n_{\rm i}$ (intrinsic carrier concentration) strongly depends on $T$, $V_T$ is the thermal voltage: $V_T= \frac{k_{\rm B} T}{e}$ (around 0,026 V at 300 K) and $N_{\rm a}, N_{\rm d}$ are acceptor and donor impurities concentrations at P and N sides, respectively (assuming null the other one).

Apart from that, the junction potential is not equal to the built-in potential, as it depends on bias (the applied voltage). At direct bias, junction potential is smaller than built-in potential, and depletion region (the central zone where there's net volume charge, due to ionized and non-neutralized impurities) narrows. At reverse bias, the opposite happens: junction potential is greater than built-in potential, and depletion region widens. The explanation to this is a bit long and you can find it in any PN junction university text (do not look at too-simple books, please); it's a matter of carriers injection or extraction, ionized impurities and net charges.

And about the electric field in the junction, the reasoning is similar: at thermodynamic equilibrium (when junction potential equals built-in potential), the electric field (typically it's maximum at the metallurgical junction, the transition surface between P and N zones) is high if doping is high. Also the electric field at the junction depends on biasing: at direct bias, as told above, depletion region narrows and electric field is reduced. At reverse bias, depletion region widens and electric field is greater.

In Zener diodes, the doping is very high (typically at least $10^{18}\,$cm$^{-3}$) and thus the "initial" (before bias) electric field is very high. At reverse bias, a small applied voltage (typically 6 V at most) increases the electric field even more, reaching a critical value (around $2·10^{5}\,$V/cm) that produces Zener effect (very narrow depletion region and some tunneling of electrons across it).

  • $\begingroup$ The formula you give for the built-in voltage is valid only for non-degenerate doping. In Zener diodes you usually very high degenerate doping. Degenerate doping limits the built-in voltage of a diode approximately to the voltage corresponding to the band gap! $\endgroup$
    – freecharly
    Apr 6 at 21:21

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.