Is resistance what makes a diode have a voltage drop?

Or is it impedence?

If it is electrical energy being converted to light, please tell me what this effect is called.

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    \$\begingroup\$ Generally when you have electrical energy absorbed you are creating heat. This often generates IR, but unless you are buying something that converts to another medium, heat is the side affect of dissipating power. \$\endgroup\$ – Kortuk Feb 19 '13 at 19:00

The forward biased semiconductor junction takes a voltage level to be able to push the electronic charges over the P-N zone. Think of it as similar to how you have to "lift" each marble up to the table top from on the floor. In addition to the energy level difference needed to transport charge across the junction there is also a resistive part of the diode that drops some voltage as well. The resistive drop in the diode will be dependent upon the amount of current flow allowed through the junction.

  • \$\begingroup\$ I assume that the "lifting" slows down the current flow... does this "lifting" act like resistance in that sense except for generating heat? \$\endgroup\$ – Vial Feb 19 '13 at 18:34
  • \$\begingroup\$ @Vicky resistors are one of the simplest elements, yes, you can think of a diode as having a resistance to current flow, but it is not that simple resistance you think of an ideal resistor having. The voltage drop is a logarithmic function where every time you multiply your diode current by 10 you only lightly increase your voltage(on the order of 70mV in silicon). It is you pushing current across the small "depletion layer" of the diode. \$\endgroup\$ – Kortuk Feb 19 '13 at 18:59
  • \$\begingroup\$ It's more like a drop in the physical analogy - moving towards ground. Lifting inputs energy. \$\endgroup\$ – Nick Johnson Feb 20 '13 at 10:23

This is due to the inherent Depletion Region in the PN Junction. http://en.wikipedia.org/wiki/Depletion_region

It's a little confusing, but essentially it is due to a loss of carriers (electrons and holes). Holes are simply the place where an electron "Could" be, but think of it like a small zone where there is No charge carriers, or like a small strip of insulation.

You need voltage(force) to push the electrons over this area.

When you reverse bias a diode, this depletion region becomes even BIGGER.


All semiconductor junctions (PN) have such voltage drop (about 0,7V for silicon). Not only in diodes, also in BJT transistors, JFETs, etc.

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    \$\begingroup\$ OP seems to know there's a voltage drop, he wants to know why there's one. \$\endgroup\$ – Johan.A Aug 11 '13 at 13:21

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