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If current is just the movement of charged particles, why do the all have to move in the same direction?

For example, if you reverse-bias a diode (connect the positive terminal to the n-type side and the negative terminal to the p-type side), the positive "holes" are attracted to the negative terminal and the electrons are attracted to the positive terminal.

Firstly, if positive holes moving towards the negative terminal corresponds to electrons moving the opposite way (since "holes" aren't real, they're just a lack of electrons). So on both sides of the diode, electrons are moving in the same direction. I don't quite understand how this doesn't correspond to a current flowing.

Not even looking that deep into it, if positive charges are moving one way and negative charges the other, why does it matter if they cross the PN junction? Moving charges = electricity, right?

On top of all this, the battery creates an electric field that goes through all the wires, so why is there no current in the circuit? Electrons don't even move that fast (I've heard drift speed is on the order of cm/s), so "current" is localized in the sense that an electron on one side of a circuit may never even reach the other side. So why aren't localized electric fields enough to create a circuit?

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    $\begingroup$ Current tends to flow in the direction from high to low potential. But I'm willing to say the average current does. No doubt in my mind some electrons swim up stream. But most know the right way $\endgroup$ – docscience Mar 16 '17 at 19:15
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All charges don't move in the same direction. It's the net effect that we see. I think you're missing the fact that conventially current was thought to be the flow of positive charges.

Let's consider an example (something less complex than the diode example you've mentioned)

Consider an area element of a conductor and view it in a direction along its plane. Let there be both positive and negative charges (yes, these are electrons (for a metallic conductor) but for the time being let them be positive and negative charges) to the left and right of the element. If a charge +q flows from the left to right we say that a current is flowing from left to right. If it were flowing from the right to the left a current would be flowing from right to left. For a charge -q a current is said to flow from the left to right if it moves from right to left (transport of a negative charge from right to left can be visualised as a transport of an equal amount of positive charge from left to right) and similarly the other case. When we say that a current I is flowing from the left to the right it is due to a net charge Q=+q-q flowing across the area element from left to right (in the time frame we have chosen). It could be that no negative charge is flowing from the right to left. In such a case the current would be due to the positive charges solely. Similarly it could be that no positive charge flows from left to right. In this case the current would be due to negative charges flowing from right to left only. The more general case assumes the net +q-q flowing from left to right (if the current is flowing from left to right).

Now getting to the part of your question in which you address a diode being reverse biased. Who says there isn't a current when you reverse bias it? There is a current! But it's so negligibly small (in microamperes) compared to the current we get in forward bias (milliamperes) that we can neglect it and say that there is no current in reverse bias. Think about it. Both differ by a factor of 10³. Current in the reverse bias is limited by the number of minority charge carriers present in either wafers of a junction diode.

Hope that clears your doubt.

EDIT

Causes of the miniscule reverse current.

enter image description here

(Sorry for the poor quality picture. I've cropped it from a screenshot of one of the pages of an e-book I had. Ignore the W, that's was there in another context).

You can see that the diode is reverse biased. The external biasing provided by the battery sets up an electric field which is in the same direction as that produced by the space charge regions in the depletion region. Now any hole on the n-side (remember that both holes and electrons exist in any type (p or n) of a doped semiconductor, it's just that one is in excess) would move towards the depletion region because of the field due to the battery. If it gains enough kinetic energy to just enter the barrier region, the field there will push it to its majority charge region (I.e. the p-side). Similar is the case with an electron on the p-side. A natural doubt would be why does that make the current small? The answer to that is the fact that the concentration of minority charge carriers (electrons in p-type and holes in n-type) is much much smaller than the concentration of majority charge carriers (holes in p-type and electrons on n-type). Since their (minority charge carriers') concentration is much much smaller their flow is much much slower.

I hope that that's a bit intuitive to you. If it isn't, here's a weird analogy: Ten people jumping on a trampoline have more chances of breaking it than a single person jumping on it. The ten people here may be thought of as those causing the large (~milliamps) current in forward bias while the single person may be thought of the person causing a negligible current (~microamps) in the reverse bias.

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  • $\begingroup$ That all makes sense, but what are minorty charge carriers? And how do they contribute to the (albeit tiny) reverse current? $\endgroup$ – rcplusplus Mar 17 '17 at 17:24
  • $\begingroup$ @rcplusplus Whenever a p-type or n-type semiconductor is formed it is doped with a particular impurity. P-type semiconductors are formed by doping the pure (or intrinsic) semiconductor with an acceptor atom (an impurity from group 13 of the periodic table). While N-type semiconductors are formed by doping the pure semiconductor with a donor impurity (an impurity from group 15 of the periodic table). Basically P-type or N-type semiconductors have on type of charges in excess. It is holes for p-type and electrons for the n-type. If you're not satisfied with that I'd be happy to elaborate $\endgroup$ – Kunal Pawar Mar 17 '17 at 17:29
  • $\begingroup$ @rcplusplus I have edited my answer. $\endgroup$ – Kunal Pawar Mar 17 '17 at 17:50
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The battery supplies a part of its chemical energy to the electrons that emerge out of it. These energized electrons lose this energy in the outer circuit, when they go through various electronic components. The electrons in a simple conductor collide due to their thermal energy.

Electrons move from the -ve terminal of the battery to the +ve terminal. In a reverse-biased diode, the electrons move from minority region (the p side) to the majority region (the n side) due to the electric field inside the diode which forces the charges from their minority region into their majority region.

Current is defined as the number of charges passing unit area of conductor per sec. The electrons inside a conductor without a battery, suffer a lot of collisions but their velocity on an average can be considered $0$ and hence, they are unable to produce current.

But when a battery is joined across the conductor an electric field is set up in it. Even in the presence of this electric field, the free electrons suffer a lot of collisions but now their movement gets a certain direction. The real motion is still zig-zag but with a certain direction. The high number of collisions are responsible for their slow drift speed.

As the electric field starts to accelerate the electron, the electron collides and loses its kinetic energy in form of heat. Then the field accelerates the electron again but the electron again collides and this goes on.

The electrons drift through the conductor and their speed ,in presence of E.field, is called drift speed. Although the drift speed of electrons is very low $( −0.000023 m/s)$ a large current is produced because of their high quantity crossing per unit area of the conductor per sec, which is about $10^{23}$ electrons.

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  • $\begingroup$ Yes, as you apply a reverse bias you get a little current for a bit. Then net current stops when the depletion width is fully established for the applied bias. $\endgroup$ – Jon Custer Mar 16 '17 at 22:52

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