3
\$\begingroup\$

As I understand it, electrons (which are negatively charged) flow towards positive.

As a convention, we say that current flows from positive to negative. Is this to say that what you connect the positive terminal of a battery to, say, VDD of an IC, and the negative terminal to the IC's GND, what is actually called the "positive terminal" is really ground/negative? Are the electrons flowing from "positive" to VDD, or "negative" to GND, etc?

Are the flowing electrons themselves responsible for the work done, or is it the "holes" moving in the opposite direction?

|improve this question
\$\endgroup\$
  • \$\begingroup\$ physics.stackexchange.com/a/17131/74763 \$\endgroup\$ – Solomon Slow Jul 23 '18 at 15:52
  • \$\begingroup\$ In what, a conductor, a diode, PNP transistor? The whole concept of an electron (and hole) flowing to create current, in itself is an abstraction. Both of them, in this context, are known as quasiparticles, which are an approximation of what the real electrons are doing. The confusing part is that we call the physical particle and the quasiparticle "electrons". \$\endgroup\$ – user71659 Jul 23 '18 at 16:31
  • \$\begingroup\$ it doesn't really matter, both are just a simplified human model of a complex process \$\endgroup\$ – dandavis Jul 24 '18 at 17:03
10
\$\begingroup\$

Conventional current goes from positive to negative, which is the opposite of electron flow which goes from negative to positive.

Neither the electrons nor the holes are responsible for doing the work, it's simply the flow of charge that does the work. The direction of the flow of charge, the electrons, or the conventional current, is not relevant to the amount of work that is being done.

In electronics, when we say "current", we refer to conventional current. If for any reason would we would want to refer to electron flow, we make sure to say so.

The reasons for this are mostly historical.

|improve this answer
\$\endgroup\$
  • \$\begingroup\$ Is the flow of charge in the same direction as electron current, and is there "anything" tangible which flows in the opposite direction? electromagnetism? \$\endgroup\$ – 19172281 Jul 23 '18 at 15:28
  • 5
    \$\begingroup\$ Electrons carry negative charge. Protons carry positive charge. If any of them move, then charge is flowing. Whether positive or negative charge is flowing to the right or to the left, depends on whether it's being carried by an electron or a proton, and it depends on whether the particle is moving right or left. In copper wires and most other electronic circuits, it is mostly the electrons that are moving, and they are moving in the opposite direction of what we call conventional current. \$\endgroup\$ – Dampmaskin Jul 23 '18 at 15:30
  • \$\begingroup\$ A few thoughts, why would you have a ground plane at the end of the circuit? Wouldn't that mean that all the charge/electrons would flow first through the ground plane, and then through the circuit's elements? \$\endgroup\$ – 19172281 Jul 23 '18 at 16:09
  • 2
    \$\begingroup\$ @B4039, a wide ground plane provides much less resistance to electrons flowing than a skinny wire. \$\endgroup\$ – The Photon Jul 23 '18 at 16:10
  • 1
    \$\begingroup\$ @user71659 Inside batteries and electrolytic cells, for example, charge can be carried by protons or other positive ions. \$\endgroup\$ – user253751 Jul 23 '18 at 22:42
3
\$\begingroup\$

Positive is positive.

If a pin is "positive" in relation to another pin, then the voltage difference between that pin and the reference pin is positive. In your example VDD is connected to the "positive" terminal, which means the voltage at that pin is higher with respect to GND.

Electrons "flow" from negative to positive, so if a pin is at higher voltage, they will flow towards it. Electron current is therefore from GND to VDD. Conventional current is from VDD to GND.

|improve this answer
\$\endgroup\$
  • \$\begingroup\$ Is conventional current what actually "powers" the IC? \$\endgroup\$ – 19172281 Jul 23 '18 at 15:21
  • \$\begingroup\$ And is it equal to Electron current? \$\endgroup\$ – 19172281 Jul 23 '18 at 15:22
  • \$\begingroup\$ @B4039 conventional current is just electron current, but negated. Mr franklin got it wrong. \$\endgroup\$ – Tom Carpenter Jul 23 '18 at 15:22
  • 1
    \$\begingroup\$ Electromagnetism doesn't flow. Electromagnetism is just a field of study. Charge flows. \$\endgroup\$ – Dampmaskin Jul 23 '18 at 15:28
  • 4
    \$\begingroup\$ @B4039 Everything is the same regardless of whether you use conventional or electron current. You really don't need to worry about which way electrons flow. \$\endgroup\$ – Tom Carpenter Jul 23 '18 at 16:30
1
\$\begingroup\$

Voltage is electrical potential energy per charge, i.e. EPE/q. If moving an electron from Point A to Point B takes 1 electron volt of work, then the voltage difference from A to B is 1eV/(-1e) = -1V (note that the notation is a bit confusing, as the 'e' in 'eV' stands for "electron", while the 'e' on the RHS stands for "elementary charge"). Conversely, if an electron travels between two points with a voltage difference of +1V, then it will perform 1eV of work.

If we define current as the signed flow of charge, then a positive charge traveling from the positive terminal to the negative terminal is a positive current from the positive terminal to negative. A negative charge flowing from negative to positive is also a positive current; the negative sign from going the opposite direction and the negative sign from opposite charge combine to give a positive current.

"Positive" and "negative" are a bit misleading, as there's no "absolute zero" of voltage. The "positive" terminal is simply the one with the higher voltage, and the "negative" one is the one with lower voltage.

If one Coulomb of electrons travels from one terminal to another, and the second terminal has a voltage that is 5V higher than the first, then we have -1C of going across -5V of voltage, giving (-1C)(-5v), or 5J, of work. The electrons do work by being negative charge going across a negative voltage difference.

As something that's somewhat analogous, if you release a helium balloon (negative weight), it will float upwards, and the work done will be the weight of the balloon times the signed change in height. Both the weight and the displacement will be negative, giving a positive amount of work done by the balloon.

|improve this answer
\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.