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I'm only talking about current conducted by the majority carriers.

First things first, barrier potential/height is different than the built-in voltage of a diode. The built-in voltage is a constant. The external bias changes the barrier height. In forward bias, the external biasing voltage decreases the barrier potential/height. If the external biasing voltage is connected under reverse bias, the opposite happens. The barrier height equals the built-in voltage when the external biasing voltage is 0V.

Now, for any amount of current to be established between the two terminals of the external biasing voltage connected under forward bias, the voltage of the external bias must be greater than the barrier potential. Otherwise, no current would be able to be established between the two terminals of the external biasing voltage. Am I correct?

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  • \$\begingroup\$ You don't provide a detailed glossary of your terms. What is meant by "built-in voltage of a diode," for example, as opposed to a "barrier potential?" I've few clues. I can suggest that a diode may conduct current with any forward biasing. (Even reverse-biasing.) I can't parse the detailed meaning of your writing. Perhaps someone else can do better. \$\endgroup\$
    – jonk
    Jun 13 at 5:31
  • \$\begingroup\$ I tried to make it clearer; do you understand it now? \$\endgroup\$
    – user545735
    Jun 13 at 5:38
  • \$\begingroup\$ There exists also a reverse leakage and junction capacitance decreases with reverse bias . \$\endgroup\$ Jun 13 at 5:38
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I think your description ignores two things which anyway lead to current: temperature gives a fraction of electrons sufficient energy to cross the "barrier". This gives the Shockley equation its general exponential dependence. Even the carriers with insufficient energy can tunnel through the barrier if the other side of the barrier is favorable (there is some forward bias.) The likelihood of this also increases exponentially the smaller the remaining barrier is.

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    \$\begingroup\$ Okay, so, what you are trying to say is, even if the external biasing voltage is less than the barrier height/potential, some electrons can regularly gain the remaining energy from the heat of the system (as they exist on an energy distribution) to overcome the barrier. Once they have diffused to the p side, they are now attracted to the positive terminal of the battery and head there. Thus current is established. Did I understand you correctly? \$\endgroup\$
    – user545735
    Jun 13 at 6:15
  • \$\begingroup\$ yes after they are past the barrier, either overcoming it thermally or penetrating it via tunneling, they carry on towards to positive voltage source in a regular way. \$\endgroup\$
    – tobalt
    Jun 13 at 6:29
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    \$\begingroup\$ In normal pn junction diodes the barrier is far too thick to allow any measurable amount of tunnel current. \$\endgroup\$
    – Matt
    Jun 13 at 11:27

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