I can understand that in the depletion region of a diode more covalent bonds are broken(more electrons goes to conduction band) with more temperature hence we say that the reverse saturation current is dependent upon the temperature i.e. thermal energy added.

But what is the argument which explains why the reverse saturation current is independent from the reverse voltage(at least until the zener breakdown)? (Until zener breakdown why does the increasing reverse voltage have no effect on the reverse current?)

But at some "zener" point something happens and the electrons starts to avalanche and create more electrons. But why doesnt this thing happen gradually? I mean the reverse saturation current is constant untill the avalanche why? Things start to happen at a particular point all of a sudden not gradually.

  • \$\begingroup\$ Whoever told you that the reverse current is constant? \$\endgroup\$ – Asmyldof Aug 26 '17 at 14:28
  • \$\begingroup\$ Isnt it almost constant between reverse voltage of 0 to zener breakdown? \$\endgroup\$ – atmnt Aug 26 '17 at 14:29
  • \$\begingroup\$ @Asmyldof This says that it's virtually constant regardless of voltage if temperature stays constant: en.wikipedia.org/wiki/Reverse_leakage_current \$\endgroup\$ – horta Aug 26 '17 at 14:34
  • \$\begingroup\$ @horta it is virtually constant. To pretend it doesn't depend on voltage at all, or that it suddenly goes from nothing to full at break-down like a binary switch over is not helping understanding of the actual physics involved. \$\endgroup\$ – Asmyldof Aug 26 '17 at 21:22

This current is due solely to carrier generation within the depletion region due to phonon interactions creating an electron-hole pain every once in awhile. Once this hole and electron are generated in this region of high electric field, they are swept in opposite directions. Since there's so few carriers being generated due to thermal interactions generally, your reverse current is very little and isn't at all dependent upon the reverse voltage until you get to to the zener voltage which allows tunneling to occur or until you get to the avalanche voltage where the e-field is adding so much energy to the carriers that they can knock other electrons free thereby creating more electron-hole pairs. Up to either of these two points, you've got the trickle of current due solely to thermally generated carriers within the depletion region of the p-n junction. This is also why the reverse current is highly dependent upon the temperature of the junction. Without any temperature, you'd have no phonons to generate the e-h pairs.

  • \$\begingroup\$ "phonon interactions" do you mean thermal energy? \$\endgroup\$ – atmnt Aug 26 '17 at 14:45
  • \$\begingroup\$ there's no phonon in my book \$\endgroup\$ – atmnt Aug 26 '17 at 14:46
  • \$\begingroup\$ @user134429 yes, thermal energy. Phonons are lattice vibrations, i.e. thermal motion. \$\endgroup\$ – horta Aug 26 '17 at 14:53

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