simulate this circuit – Schematic created using CircuitLab

People are taught how to calculate this very simple thing, and the calculation goes from source plus to minus, and the diode is there to make sure the electricity flows in one direction.

However, we also know that electrons actually flow from the negative to the positive.

How does this actually work in real life?

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    \$\begingroup\$ "How does *this" work in real life? " What exactly? Electron flow through a diode? Electricity in general? I think there's no real question here, you're just forgetting that everything we know about reality and physics really is just a model, and the closer to reality you get, the less practical these models become. So, the actual flow direction of electrons doesn't matter. The model you've learned works for this circuit. But it's just a model, which doesn't care about in which directions charges migrate. \$\endgroup\$ Commented Jun 18, 2017 at 5:18
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    \$\begingroup\$ The flow of electrons and the calculations seem to be in opposite directions. I'm pretty sure the practical side matters, to the manufacturers of the diode at least. Also, I believe that there is a question, you just aren't answering it. \$\endgroup\$ Commented Jun 18, 2017 at 5:25
  • \$\begingroup\$ @user3635998 You aren't ready for an answer, I suspect. Let me ask you a much simpler question involving fewer bits to worry about. There cannot be any excess charges in a wire, so the number of mobile electrons must everywhere equal the number of positive atomic cores and the copper is electrically neutral, as it must be. Electrons cannot "push" each other, as any repulsion from one direction is always compensated by a positive attraction from another. So how is it that electrons move in a wire, at all? Answer that correctly and I may have a shot at a better explanation about diodes. \$\endgroup\$
    – jonk
    Commented Jun 18, 2017 at 5:39
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    \$\begingroup\$ We generally talk about "conventional current" which flows from the positive terminal of the battery, through the external circuit, and returns to the negative terminal. Atoms and electrons hadn't been invented when the early scientists were studying electricity so they arbitrarily declared this "conventional current" direction. Although we now know that negatively charged electrons carry the charge in most materials, we rarely need to deal with the actual charge carriers so we almost always deal with conventional (positive) current. \$\endgroup\$ Commented Jun 18, 2017 at 5:41
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    \$\begingroup\$ Unfortunate that the people here dont know the answer and only know how to vote down. \$\endgroup\$ Commented May 17, 2018 at 3:54

1 Answer 1


Conventional current is the total flow of charged carriers.

In an electrolyte, these are both positively and negatively charged ions. In a plasma, electrons and positively charged ions. In ice, it's protons that flow to conduct electricity. In N and P type semiconductors, it's mostly electrons and holes respectively, but in intrinsic (undoped) semiconductors, it's a more even mix. In metals, it's just electrons.

It's rather metal-chauvinistic to concentrate on metals and get uptight about electrons having a negative charge, when there's such a diversity of current conduction mechanisms available, using both negative and positive charge carriers. The pioneers of electrical theory and measurement chose one way, but they could have just as easily picked the other. Then I guess people would have objected to the centres of atoms being negative!

The diode has three junctions in it, metal-N, N-P, and P-metal. Different things happen at each junction, only the middle one is what we'd call a rectifying diode.

Use your favourite search engine to find out how doped semiconductors conduct electricity, and how metals do, and how the junctions work. There's much too much information in those topics for a simple answer here.

The schematic you've shown models at the 'conventional current' level, and is not concerned with what particular charge carriers are involved at various parts of the circuit.


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