1

I was studying diodes. I understand the depletion region.

I am wondering if I tie together the P and N end of the diode, there will be a depletion region around the junction. But there should also be a depletion region around the wire connection due to majority carriers combining by flowing around the wire, and the depletion region should be opposite to the depletion region around the junction, thereby cancelling out the built-in voltage. This must be the case so measuring exactly the negative of the built-in voltage. Can someone comment on my thoughts?

Another way to express my question is if I connect a P type semiconductor with an N type semiconductor together using wires, will there be a PN juncton?

clarification n is tied to the ground..

  • 1
    You claim that, but *why* would there be a depletion region at the wire connections? That's not correct, as far as I can tell. – mmmm May 06 '21 at 08:10
  • A PN junction: yes, as the difference in doping (N-region vs P-region) still exists. The depeletion region will exist around the junction, there is no need to short the N and P side of the diode. As the wikipedia page states: "A depletion region forms instantaneously across a p–n junction" read: https://en.wikipedia.org/wiki/Depletion_region A common misunderstanding is that there will be a voltage (the built-in voltage) across the diode. This voltage is only present **inside** the junction, you cannot measure it with a multimeter. – Bimpelrekkie May 06 '21 at 08:22
  • thanks for the response. u misunderstood my question. my point is that if you short circuits a diode, there should be two depletion regions. One centered around the pn junction, the other around the wire connection, they carry opposite voltage and cancel each other out – hardware noob May 06 '21 at 08:30
  • if I measure the voltage across the pn junction, do I read zero? if zero is read, then there must be another voltage across pn junction that cancel out the built-in voltage across the pn junction. – hardware noob May 06 '21 at 08:31
  • No, there should *not* be two depletion regions. Why do you think that would be the case? It's not the case. – Marcus Müller May 06 '21 at 09:16
  • it has to be two depletion regions... otherwise, hooking up the p and n junction with wire and ground it, you will not get a zero voltage... there must be a inverse built-on voltage somewhere to cancel it out!!! – hardware noob May 06 '21 at 09:18
  • There will be zero voltage across the depletion region and that is no problem. the voltage across the central depletion regions doesnt magically appear. It needs current. No current, no voltage across the depletion region – tobalt May 06 '21 at 09:38
  • you are outta ur mind, there is certainnly a built-in voltage across the depletion region!!! it is delta v!!! but u can't use it as current source!! – hardware noob May 06 '21 at 09:49
  • Looks like you are right. look in this topic: https://electronics.stackexchange.com/questions/106496/why-isnt-there-a-potential-difference-across-a-disconnected-diode?rq=1 Consensus there seems to be that there is actual current through a shorted diode, which is fueled by the thermal junction drift, if I understand correctly. – tobalt May 06 '21 at 10:05
  • The built-in potential of the diode's depletion zone is exactly cancelled by the built-in potentials of each metal-semiconductor contact found on the ends of the silicon block. (That's true if it's all at the same temp.) PN depletion zones don't provide zero voltage, they provide a "balanced" voltage which is canceled out by the pair of metal-to-p and metal-to-n semiconductor contact zones. – wbeaty May 07 '21 at 00:33

1 Answers1

4

A JUNCTION-POTENTIAL forms at the wire connections to the diode's silicon surface. Not a depletion region, not a new diode. It's called a "non-rectifying junction" (google search.) Also see "Ohmic Contact" https://en.wikipedia.org/wiki/Ohmic_contact

A built-in junction-potential is not a depletion zone. We can have built-in potentials without depletion zones, but we can't have depletion zones without built-in potentials.

Remember, a junction-potential also forms whenever copper touches solder, or where iron touches nickel, or where Chromel touches Alumel. These built-in potentials are also called Volta potentials, or Galvanic potentials. Touching two different conductors together always causes them to become opposite-charged, as Alessandro Volta discovered, back in the "static electricity" era. (This is not battery chemistry. This is the metal-metal physics of thermocouples. It was controversial during Volta's era, because Volta was certain that these built-in potentials appearing between differing metals were the long-sought perpetual-motion machine! Heh.)

And back in the 1910 "catswhisker diode" days, it was a challenge to form diodes by touching sharp metal wires against Galena semiconductor. Most of the time your little gold wire would just form a junction-potential, a short circuit but no diode. You had to find a naturally low-doped region on the Lead-sulfide surface, since high-doped regions won't have a depletion zone. (Hence it was a great discovery at Bell Labs to learn the trick of zone-refining of long semiconductor rods, where all the PPM parts-per-million contaminants could be swept out, producing low-doped semiconductors which don't normally occur in nature.)

Think: if diodes always formed at every metal contact, then transistors would be impossible, because all three wire connections would always create inwards-pointing rectifiers on the semiconductor surface, and one of these would always be reverse-biased and turned off. (These are EXTRA junctions, not part of the transistor's two base-junctions.) Heh, even diodes themselves would be impossible, since diodes always have three separate junctions: the PN junction, plus two metal/semiconductor junctions where the wires are connected.

A good example of this effect is the TE module, the Peltier device. It's made of 100 to 200 junctions, where all the junctions are metal/semiconductor, yet none of them act like diodes. This happens whenever we use high-doped semiconductor, labeled as p++ or n++ materials. They're also called "metallic" or "degenerate" semiconductor. It becomes "safe" to make lead-connections against p++ or n++, since they only form thermocouples, not diodes. To instead create metal-semiconductor diodes, or "Shottky diodes," we must always use low-doped materials.

If you look closely at IC doping diagrams, or even transistor crystal diagrams, or read up on diode manufacturing, you'll see the p- and the n- regions which form the PN junction. But out on the surface of the crystal block, we deposit small patches of n++ or p++ regions, and that's where the aluminum conductors are welded. No diode is formed between a metal and a heavy-doped semiconductor. (We call the heavy-doped material by the name "metallic" semiconductor, as if it had the electron-sea of metals, rather than the very sparse electron-gas of low-doped semiconductor.)

https://physics.stackexchange.com/questions/176547/how-does-annealing-improves-the-ohmic-contact

Official explanation: heavy doping makes the depletion region contract to nearly zero thickness. It becomes so thin that it's thinner than the DeBroglie quantum wavelength of electrons, and constantly produces quantum tunneling in both directions, even at extremely low voltage. So, your metal contact is actually forming a tunnel-diode! But if it always turns on at far below 0.1VDC, then we simply ignore it. (Although maybe we shouldn't, at least when writing electronics textbooks. I think we should be teaching about "metallic" doped semiconductors and their non-rectifying junctions right at the start, at the same time that we teach about PN junctions. It would end much confusion. Heh, also I think we should teach about transistors without PN junctions, as Bardeen and Brattain originally invented, before William Shockley strode in and took the project away, inventing PN junction theory and modern semiconductor physics in the process!)

wbeaty
  • 10,761
  • 23
  • 39
  • two metals touching each other should not cause junction potential because they both own infinite free electrons – hardware noob May 06 '21 at 23:14
  • metal is similar to n type of semiconductor, I think there is depletion region around a metal/ p semiconductor junction – hardware noob May 06 '21 at 23:15
  • agree with me that you can't use diode as a voltage source :) – hardware noob May 06 '21 at 23:23
  • @hardwarenoob read answer, LOOK UP: voltaic potentials between metals, been known since early 1800s, it's how thermocouples work. The junction potential is caused by mismatched "work functions" of two different metals. Touch iron to copper, you get some voltage. But it's not measurable with voltmeters, because the metal probes of the voltmeter will produce new voltages in opposite direction, which perfectly cancel the voltage you're trying to measure. – wbeaty May 07 '21 at 00:19
  • @hardwarenoob Touch metal to p-type, there is no depletion region, unless the surface of the p-type is extremely new and ultra-clean, otherwise the surface will be p++ contaminated, and only produces an "Ohmic" contact. Also, we can create diodes by touching metal to n-type, since n-type produces a depletion zone when contacting metal. Shottky diodes don't use PN junctions, they use metal-semiconductor junctions, either metal-N or metal-P will work. – wbeaty May 07 '21 at 00:25
  • @hardwarenoob Right, cannot use junction-potentials as power supplies, UNLESS we apply heat or cold to some junctions and not to others. A Peltier-module powering a fan is using junction potentials (but they are *mismatched* junction potentials, with imbalanced voltage caused by non-uniform heating. Normally all the junction-potentials sum to zero around the loop. Heat one of them up, and a net voltage appears, which then can power a motor etc. But without the heating, the voltages are still there, but in perfect balance, so the motor terminals see zero volts.) – wbeaty May 07 '21 at 00:29