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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:
enter image description here

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.

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2 Answers 2

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After a diode start conducting, can you increase voltage across the diode without current change in the diode?

No, actually the opposite is true.

Consider Shockley's diode equation $$I = I_S \left(e^{qV/kT} - 1\right).$$

Shockley diode chararteristic
(Image taken from here)

For voltages in forward direction much larger than $kT/q$ ($\approx 0.025$ V) you need only a small increase in voltage ($V$) to get a large increase in current ($I$).

And for voltages in reverse direction much larger than $kT/q$ ($\approx 0.025$ V), you get almost no increase in current even for a large increase in voltage.

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  • $\begingroup$ I am aware of diode's "exponential current increase" circuit behavior, but I can't reconcile with the the fermi diagram. I guess band diagram can illustrate a diode junction clearly when there is no conduction. Afterall the band digram illustrates if conduction can or can not happen. Once conduction happens, the band digram ceases to be useful. $\endgroup$
    – eliu
    Commented Jan 31, 2020 at 16:25
  • $\begingroup$ If hypothetically, there are infinite holes and eletrons generation, no resistance in P and N, and recombination is snappy and smooth. a diode can be forever 0.7v across and conduct infinite amount of current. Fermi diagram in this case tells us that once recombination can happen, current is limitess, the reason for real world voltage across diode is purely resistive after conduction. $\endgroup$
    – eliu
    Commented Jan 31, 2020 at 16:32
  • $\begingroup$ In a way, my question was almost asking what if you put a 4.3v voltage source across a short circuit. $\endgroup$
    – eliu
    Commented Jan 31, 2020 at 18:10
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Consider the following analogous question:

A 5V ideal voltage source with no internal resistance is connected to a perfectly conducting wire. What is the current in the wire?

If you apply $V = IR$, you find that the current in the wire is infinity. That clearly can't be true.

What actually happens is that non-idealness of the various components takes over. The current $I(t)$ will be determined by a combination of the self-inductance of the wire loop, various stray capacitances throughout the circuit, radiation resistance, the internal resistances of the wire and battery, and the current the battery can produce.

At the very least, you need to model the current with a differential equation which accounts for some of these effects. Just writing down $V = IR$ implicitly assumes that a reasonable steady state exists, which the system can actually reach, which isn't the case here.

None of these complications are captured in the $I(V)$ characteristic, so looking at it alone won't tell you what will happen in your situation: it depends on precisely how nonideal various features are. That said, you can make stories about what could happen, depending on the circumstances.

For example, a conventional battery has a limited current because its emf is produced by positive and negative ions being pushed to opposite sides of it, through a chemical reaction. If you take out the ions faster than the diffusion of ions within the battery can replenish it, then the emf of the battery will go down. In other words, when we say this kind of battery has a maximum current 1 A, we really mean that the steady state voltage it produces starts to fall rapidly above currents of this size. Thus, in this case both the battery and the diode have nontrivial $I(V)$ characteristics, and the steady state solution is their intersection. That is one possible story for what could happen.

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  • $\begingroup$ I am very sorry that this does not answer my concern, which is more related to fermi level in a semi conductor. Or why my "water valve" analogy is wrong. But, the battery example has some parallels. As in: When a battery is rated 1.5V, what is maintaining that 1.5V. When it is supplying more than rated current, why the voltage would go down. $\endgroup$
    – eliu
    Commented May 1, 2020 at 19:11
  • $\begingroup$ @eliu "When it is supplying more than rated current, why the voltage would go down." That's addressed in my final paragraph, does it help at all? $\endgroup$
    – knzhou
    Commented May 1, 2020 at 19:12
  • $\begingroup$ I don't understand it because I don't fully understand how 1.5V was there the first place. Especially how it translate from microscopic eV in eletron energy level differences between 2 materials into macroscopic battery. Besides, I don't think a battery can supply more electrons than the ion diffusion rate at all. But again I can't predict what happens when you short circuit a battery, as in real life, heat will cause internal resistance of the battery to rise. $\endgroup$
    – eliu
    Commented May 1, 2020 at 19:19
  • $\begingroup$ However, the "parallel" I mentioned is more akin to: "voltage is the result of accumulation of charges". you can't have arbitrary voltage across something unless you have the reservior of charges to back it up. Fermi level and battery volts all seem to follow this rule. $\endgroup$
    – eliu
    Commented May 1, 2020 at 19:22

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