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I have just finished reading about biasing in pn junctions. I learnt that in forward biasing, the external electric field applied slowly neutralizes the depletion layer and a current is established by majority charge carriers. However consider this graph:

Why does this not follow ohm's law when the external voltage completely neutralizes the depletion layer? When the depletion layer is absent shouldn't it act like any other conducting device?

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  • $\begingroup$ You should take a look at some other books (Haliday, Resnick, Walker or HC Verma Volume 2) to properly understand this stuff. But here it is in short. The current when the p-n junction is forward biased is the diffusion current responsible for the electricity flowing in the circuit, and the diffusion current is actually holes moving from p side to n side, so once the voltage supplied by external cell is more than the barrier potential the rate of flow of holes to the n side increases rapidly and the current increases rapidly. $\endgroup$ Mar 7, 2021 at 17:11

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Ohm's law holds for linear devices. Linear devices are those in which the change in current is proportional to the change in voltage or in other words the resistance is constant. A diode or a pn junction is a non-linear device. It's resistance varies with the voltage applied. Hence, after a value of voltage (called the threshold voltage), the resistance drops to a minimum and the junction basically acts like a piece of wire. In ideal scenario the voltage drop is zero across the diode after the threshold voltage is reached and hence we can call it a switch. But practically, there's always a drop in voltage, about 0.7 V in case of Silicon semiconductor used and 0.3 for Germanium. An additional link just in case. https://learningaboutelectronics.com/Articles/Diode-resistance.php

Hope it helps.

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All electrical devices (with the exception of superconductors) always have some effective series resistance. For example, the doped Si in the diodes has some resistance. The graph you're showing is for an ideal diode, where this resistance is ignored. This is often a useful approximation because the effective series resistance is probably small, so it only matters at voltages you typically don't subject a diode to.

To capture the behavior of a real diode, you probably need to add in a lot of things including the series resistance, the leakage (parallel resistance), and especially for AC circuits, the capacitance between the p and n layers. These additions get complicated really fast. The exact solution to a diode and resistor in series requires a special function.

But, to a decent approximation, the graph you're showing should become linear at high voltages because the series resistance will become the limiting factor for the current.

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  • $\begingroup$ "But, to a decent approximation, the graph you're showing should become linear at high voltages because the series resistance will become the limiting factor for the current." Pardon me but what series resistance are you referring to here ? $\endgroup$ Mar 7, 2021 at 17:14
  • $\begingroup$ The one I mention at the start of the post. It stems from the fact that a typical diode is made of doped semiconductors, which have a non-zero resistance (as does everything other than superconductors). $\endgroup$
    – lnmaurer
    Mar 7, 2021 at 21:40

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