let's answer this question using the example below:
Let's imagine an ideal condition, where a 5V ideal voltage source with no internal resistance is connected to a diode, the voltage source can only supply max of 1A of current, how will the circuit behave? what will be the voltage and current in the diode? (ignoring heating, resistive and capacitive variables within the diode)
My thinking which is probably wrong:
In Fermi band diagram, to put it simply, you need to increase the fermi level of electrons in the n region to overcome the reverse bias, so that some electrons can be higher than the conduction band in the P region and spill over, thus conduction follows. The more you can lift the fermi level of electrons in the N region the more of them will spill over, as shown in picture below:
My misunderstanding is probably surrounding the cause and effect of the fermi level change.
I thought the elevation of fermi level governs the maximum
I that can be allowed to flow thru the junction. The actual
I is determined by the how much the circuit allows. In a sense the voltage across diode is like opening an adjustable valve in a water pipe, if you only have trickle of flow in the pipe, opening a valve any further won't get your more water.
Please comment on this if you can
Now my thinking is: once conducting, the fermi level on the N region needs constant replenishment of electrons to maintain the fermi level. If not enough electrons can be supplied to the N region, voltage across the diode will drop. That's why V and I's relationship is not "valve opening governs the max allowable water flow", but more like "water flowing over a weir, and the junction V is how high the water can be above the weir".
So the answer to this question is: It is impossible to have 5V junction voltage while the electron flow is only 1A, 1A is not enough to sustain fermi level of that high on the N region.