If a PN junction diode is connected to a battery, the P part must have a higher potential than the N part(because these points are connected to the battery) and so there should be a potential drop across the diode. However there seems to be potential RISE (because of the potential barrier) Why?
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1$\begingroup$ Refer this book : Semiconductor Physics and Devices Basic Principles Donald A. Neamen $\endgroup$– Sahan ManodyaCommented Sep 8, 2021 at 15:35
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2 Answers
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I think the term potential drop is just a convention: it could be called potential step or potential jump or potential difference.
A similar question could be asked about a resistor, which is a simpler device than a diode. The resistor also presents a barrier and is causing a potential drop across it.
The preference for this term may have something to do with the way we draw circuits, with the negative terminal of a battery often on the bottom and, at least, mentally associated with the ground or zero potential, and the positive terminal on the top, associated with the actual battery voltage.
Since the direction of the conventional current is from positive negative, we tend to think that the current somehow starts on the top and flows down, toward the ground, "dropping" voltages on various barriers, like resistors and diodes, along the way.
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At zero bias, a pn junction has a built in potential (0.3V - 0.7V) that depends on junction temperature, doping concentrations, and the semiconductor material. This is due to the depletion region created at the pn junction interface. A forward bias greater than the built in voltage is required to narrow the depletion region enough for current to flow.
answered Jun 5, 2022 at 22:57