We know that there will always be a net current in the circuit between the two terminals of the forward biasing voltage source, however small the forward biasing voltage is (see @jonk 's comment).

There will be a net forward current under forward bias, no matter how small the forward biasing voltage (read the conversation between me and @Matt in the comment section) is. When no voltage is applied, $I_{diff}=I_{drift}$, but when a forward bias is applied, no matter how small the biasing voltage is, $I_{diff}$ becomes larger than $I_{drift}$ by an amount corresponding to the biasing voltage. As electrons exist on an energy distribution, no matter how small the biasing voltage is, some electrons regularly can acquire the required energy to overcome the barrier height of the depletion region from the heat of the system. So, for current to be established biasing voltage doesn't have to be equal to the voltage across the depletion region. Current can be established in the forward direction for much less than the voltage across the depletion region.

Granted, the current will also be extremely small if the forward biasing voltage is very small, but the voltage drop due to the barrier height/ potential across the depletion region will be big if the forward biasing voltage is small, and the carriers must go through this voltage drop for current to be established. So, is it possible for the voltage drop to be bigger than the actual external forward biasing voltage?

  • $\begingroup$ You have been asking a lot of questions that should be covered by any intro to semiconductor device physics book. An in depth answer to this question, and many others you asked, will rely on a good understanding of band diagrams. I suggest reading about them ecee.colorado.edu/~bart/book/book/chapter4/ch4_2.htm (if you have an external voltage all that voltage must be across various parts of the internal diode structure, but it doesnt have to be across junction itself, a lot of other stuff goes on getting from the junction itself to the metal wires) $\endgroup$ – Matt Jun 13 at 17:11
  • $\begingroup$ Are you familiar with Kirchhoff's voltage law (KVL)? $\endgroup$ – Matt Jun 13 at 17:16

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