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Consider a PN junction diode

enter image description here

Now, this system will naturally form a depletion region due to diffusion and will convert into this:

enter image description here

Now, I wish to connect the ends of this diode with a conducting wire and resistance like this:

enter image description here

Thus, I expect the electric field in this space to be aligned as follows (indicated dim green:)

enter image description here

This will make the charges in the wire accumulate in the following way:

enter image description here

Finally, this system will end up neutralizing all the charges so that no electric field remains inside and outside the diode; creating equilibrium state for the free charges. So, we are going to be back on a simple P-N junction diode with no depletion region and no electric fields.

However, this will again create a difference in electron concentration between N and P sides; these high-speed free electron will naturally flow from N-side to P-side creating an electric field which would later be balanced by the conducting wire.

I thus expect this cycle to go on and on leading to continuous charge flow through the resistance. Overall, the diode will be taking up heat from the surroundings, while the resistance would be releasing heat there.

Obviously I expect this analysis to be wrong, but I am not able to figure out where did I made any mistake.

PS: I asked this question originally at the Physics StackExchange, but they consider it to be similar to (https://physics.stackexchange.com/questions/108314/why-isnt-there-a-potential-difference-across-a-disconnected-diode). However, I am still not satisfied because I am not finding where I am going wrong. Simply stating that there is no potential difference across the diode is of no help.

Here, I have used basic core principles of electrodynamics only. I would be grateful if someone points out the exact step where I am making the blunder.

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This circuit has three dissimilar materials: two doped semiconductor regions, and a wire (we can allow the resistor to be a wirewound type, so it's just... part of the wire).

That means there are three junctions. While the metal joints are perhaps not semiconductor diodes, they ARE subject to the same charge-diffusion boundaries at the joint, and even between metals, there is a known effect (thermocouple or Seebeck effect) due to charge migration at that boundary.

So, the total circular sum of voltages across those three junctions might be zero, but none of the individual junctions is exactly zero voltage. In fact, if the joints are at different temperatures, you'll get a net thermocouple effect, and current may flow through the resistor. It's a heat engine, though, and not a perpetual motion solution.

The metallurgy of making a metal joint to a semiconductor is a significant problem, and the easy solutions (for silicon, an aluminum interface layer) are often unappreciated.

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    \$\begingroup\$ by 'unappreciated' do you mean 'not mentioned', causing this question to be asked by OP? \$\endgroup\$ – quetzalcoatl Jul 6 at 8:51
  • \$\begingroup\$ The actual PN diode-and-wire circuit will (for technical reasons) have at least one additional metal alloy - maybe several- in the circuit. The question doesn't require a discussion of those (though they ARE present). \$\endgroup\$ – Whit3rd Jul 6 at 10:52
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    \$\begingroup\$ @quetzalcoatl No, I think it's just a side note saying these junctions are surprisingly hard to make. Like: Everyone thinks about silicon physics, but nobody thinks about how to connect metal wires to it. \$\endgroup\$ – user253751 Jul 6 at 12:56
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The process you are describing equalizes the Fermi level across the junction. Connecting wires equalizes the Fermi levels of the wires. The relative Fermi level is what a voltmeter reads, so a voltmeter connected in place of your resistor would read zero volts. There is no voltage to drive a current through your resistor.

The detailed physics here is a bit mind bending: even those of us who do things like chip design get confused sometimes. Wikipedia has an excellent article on this.

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If you take the field from any point-like charge -- an electron or hole -- and integrate it around any loop, the total field along the loop will add up to zero.

The field from any distribution of such charges, integrated around a loop, is the sum of the field from all of the point charges. They all sum to 0 individually, so you get 0 when you add them all up.

So the real answer to your question is easy: It doesn't matter what kind of charge distribution you have -- a static charge distribution doesn't push electrons around any loop.

In this case, you forgot to count the voltage drop across the PN junction.

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  • \$\begingroup\$ What voltage drop across the PN junction? What causes that? \$\endgroup\$ – user253751 Jul 6 at 12:57
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    \$\begingroup\$ Also I think you've just restated Kirchoff's Voltage Law, which obviously doesn't prohibit the existence of batteries or solar panels that produce useful energy. \$\endgroup\$ – user253751 Jul 6 at 12:58
  • \$\begingroup\$ See that green arrow you drew at the top? It opposes all the other ones. Causing an electron/hole recombination across the depletion region get costs energy. In a solar panel, that is provided by the incident light. \$\endgroup\$ – Matt Timmermans Jul 6 at 13:18
  • \$\begingroup\$ The electric field around a circuit containing a battery and a lightbulb is also zero, since the electric field from the lightbulb opposes the electric field from the battery. \$\endgroup\$ – user253751 Jul 6 at 13:27
  • \$\begingroup\$ Batteries are powered by chemical reactions, not charge distributions. It's a more complicated topic, but you can google "redox reactions" if you want to learn about it. \$\endgroup\$ – Matt Timmermans Jul 6 at 13:29

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