It is my understanding that a highly doped, thin depletion region, reverse-biased P-N junction at low voltage can cause quantum mechanical tunneling. This (or sometimes avalanche breakdown) is used in some hardware random number generators. However, is the noise created in the junction truly quantum mechanically random, or is it chaotic but deterministic?

  • 1
    \$\begingroup\$ The only real difference between "random" and "deterministic" is the amount of knowledge of the observer. \$\endgroup\$ Apr 3, 2014 at 17:50
  • 1
    \$\begingroup\$ Are you sure? This white paper claims that their RNG products are superior because they use a truly random process, whether a photon is reflected by a semi-transparent mirror. They contrast this with chaotic but deterministic processes like the Zenner diode. I'm curious if it's true that the Zenner diode isn't truly random. In cryptography, you have to be concerned that your attacker will be able to predict a deterministic process even if you can't. Quantum processes are in theory unpredictable. \$\endgroup\$
    – bsamek
    Apr 3, 2014 at 18:13
  • 1
    \$\begingroup\$ I can't be absolutely certain, but only because we haven't yet reached a point where we can call quantum processes "deterministic". But absence of evidence is not the same as evidence of absence. \$\endgroup\$ Apr 3, 2014 at 18:16
  • 4
    \$\begingroup\$ @Spehro: I think Ignacio is coming out of older discussions about "hidden variables." A position taken by Einstein was that there may be hidden variables that govern quantum statistics and that if that underlying theory could be worked out, then determinism could be restored and what used to be "random" wouldn't be anymore. But Bell's inequality provides a very sensitive test which now experimentally appears to exclude all "local hidden variable" theories. \$\endgroup\$
    – jonk
    Apr 4, 2014 at 0:18
  • 3
    \$\begingroup\$ @jonk Exactly. And this means that there is a distinction between random processes with hidden variables and true quantum processes. In the former it might be a practical impossibility to determine the system's future states based on prior knowledge. In the latter case it is a theoretical impossibility. Hence my question. \$\endgroup\$
    – bsamek
    Apr 4, 2014 at 4:04

1 Answer 1


The noise that is created is truly random as it is generated by recombination of electrons with the atoms on the other side of the junction. The mean power density (RMS) can be calculated using the following formula:

$$i_n = \sqrt{2 I q \Delta B}$$

where \$q\$ is the electron charge, \$I\$ is the current that runs through the diode and \$\Delta B\$ is the bandwidth of the detector that measures the noise.

Apart from that, it is a stochastic process and therefore random. Read more about it here:


  • \$\begingroup\$ Does this apply to both avalanche and Zenner noise? \$\endgroup\$
    – bsamek
    Apr 3, 2014 at 18:14
  • \$\begingroup\$ Zener breakdown noise is shot noise if I remember that correctly. The diode type does not matter. A Zener diode will have an electron avalanche when it reaches its breakdown voltage (reverse biased). \$\endgroup\$
    – einball
    Apr 3, 2014 at 18:41
  • \$\begingroup\$ @einball: The noise generator and Zener diode articles seem to say that Zener diodes rated below 7 V produce mostly shot noise, Zener diodes rated above 7 V produce mostly avalanche noise, and Zener diodes near 7 V have a mixture of both. \$\endgroup\$
    – davidcary
    Aug 22, 2014 at 3:18
  • \$\begingroup\$ This answer seems to say that the shot noise is spectrally white, which is incorrect. \$\endgroup\$
    – DanielSank
    Dec 30, 2016 at 14:39
  • \$\begingroup\$ @davidcary Is there any correlation in that dividing line and the one about the direction of the thermal coefficient in relation to the reverse knee voltage (predominantly zener {quantum tunnelling} effect below 5V and avalanche breakdown above 7V)? Oh, hang on it is the same thing if the shot-noise IS from the zener effect. \$\endgroup\$
    – SlySven
    Jan 11, 2017 at 19:49

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.