I'm trying to understand a reverse biased PN junction, conceptually. For reference, I'll adopt the convention that the P-type is on the left, and N-type is on the right. So without a bias, the depletion region forms and there should be a positive electric-potential step crossing from P to N. This potential step deters more holes from diffusing into the N side, and electrons from diffusing into the P side. I understand that applying a forward bias will reduce this barrier, allowing carriers to cross the junction quite easily.

So applying a reverse bias increases the potential step, making it even harder for holes to migrate to the right, and for holes to migrate to the left. But here's where my confusion comes in: who's asking them to? If the positive side of the voltage source is applied to the N-side, shouldn't the electron current be pointed left to right, and the hole-current be pointed right to left? In that case, the potential "step" should be a "downhill", which the carriers should be happy to cross.

So what am I missing here?


I think I figured it out, can anyone confirm that the following understanding is correct? The N-side could, in theory, provide the electron current, and the P-side could provide the hole current. but to complete the circuit, a electron current will need to flow out of the voltage source's anode, and all the way through the P-side. Electrons entering the P-side should be able to fill in holes in acceptors, and migrate towards the right as holes drift left. However, when the electrons get to the depletion region, all (or most) of the acceptors have already been filled by the electrons which diffused over from the N-side. So not only are there few (if any) holes left to be filled in the depletion region, but there is also a net negative charge there due to the acceptor ions, so there will be a force pushing back on the electrons anyway. I guess if any electrons manage to get past this wall of acceptor ions, then they would get to slide down the potential hill towards the n-side, but the issue is getting that far in the first place.

  • \$\begingroup\$ Here's a reasonable video that may help a bit. \$\endgroup\$
    – Oli Glaser
    Commented Dec 3, 2012 at 20:32
  • \$\begingroup\$ Thanks, that's a decent video. It at least got me thinking in a direction that allowed me to figure it out, but was still far from addressing what I was really after. \$\endgroup\$ Commented Dec 4, 2012 at 23:14
  • \$\begingroup\$ Here is another excellent visualization of p-n junction that answers this question fairly well youtube.com/watch?v=7ukDKVHnac4 \$\endgroup\$
    – akhmed
    Commented Aug 19, 2016 at 0:49

3 Answers 3


Diffusion current

When a p-n junction is formed, a diffusion phenomena causes electrons from the n-doped region to diffuse to the p-doped region. At the same time (even if it's an abstraction) holes diffuse from the p-type region to the n-type one. The atoms that lose a carrier (electron or hole) become ions, which means that instead of being neutral, they have a positive or negative net charge. This happens because the ideal equilibrium would have the same concentration of mobile carriers equal all over the region.

Ohmic current

However, this diffusion causes the growth of a region, populated by ions, called depletion region, because all atoms have lost their carrier. These ions, as we said, are electrically charged, and cause an electric field directed from the n-region to the p-region, pushing carriers in the opposite way than diffusion. Therefore an equilibrium is reached in which the current (movement of carriers) caused by diffusion is perfectly balanced by the current caused by the electric field (ohmic current).

Effect of biasing

Applying a potential to the junction causes a perturbation on this equilibrium, making one of the currents dominant on the other. Reverse biasing the junction causes the ohmic current to prevail, while forward biasing increases the diffusion current.

Now, the diffusion current is a much stronger phenomena, from which derives the exponential growth of the forward bias current with the bias voltage. Ohmic current, on the other side, is much weaker, and saturates quite soon (neglecting avalanche effect) because the width of the depletion region (which determines the resistivity) is proportional to the reverse bias voltage.

  • 1
    \$\begingroup\$ +1 to this! I'm currently taking a semiconductor physics course in my 3rd-year program, and this is a great summary of the drift-diffusion model we used to learn about the PN junction. Wish I had read this a month or two ago =) \$\endgroup\$
    – Shamtam
    Commented Dec 5, 2012 at 5:35
  • \$\begingroup\$ Shamtam, regarding your semiconductor course: The above description of the pn junction properties is an excellent basis for understanding the working principle of the bipolar transistor (which contains two pn junctions). \$\endgroup\$
    – LvW
    Commented Aug 27, 2016 at 8:10

Honestly, I found the other answers here fairly incomplete. I've researched this some more and the key difference between forward and reverse bias is whether you're trying to move majority or minority carriers across the depletion zone.

Under forward bias, you're trying to move electrons from the n-side to the p-side. Since the electrons on the n-side are majority carriers, there are a large number of free electrons, meaning they are not attached to any nucleus, and they are relatively easy to move, once you apply a sufficient bias voltage to give them the energy they need to make it over the potential barrier at the junction (alternatively, once the applied bias voltage lowers the potential barrier sufficiently for the electrons to cross it easily).

Under reverse bias, you're trying to move electrons from the p-side to the n-side. But electrons are minority carriers in the p-side, so there are very few that are free. Most electrons in the n-side are bound to atoms/ions as part of covalent bonds, and can only be moved by filling in an adjacent hole. There are plenty of holes in the quasi-neutral region of the p-side, so the electrons can move around, but once they reach the depletion region, there are very few (if any) available holes, which is what defines the depletion region. Thus the bound electrons cannot readily be moved into the depletion zone and from the p-side to the n-side, thus no current can flow (until breakdown is reached, of course).

Note that the increased magnitude of the potential energy change at the junction that is associated with reverse bias is not the reason that current can't flow. In fact, because of the orientation of the reverse bias voltage, current is actually trying to move "downhill" across the junction, so any free charges that actually do make it to the junction will readily be swept down this large drop in potential energy.

In fact, this must be the case in order for BJTs to work in common-emitter mode, where the base-collector junction is reverse biased, but the large potential energy change does not prevent, or even significantly hinder, current from flowing through the junction. In this case, the forward biased base-emitter junction provides ("emits") a large supply of free electrons into the p-type region which make up the current flowing through the B-C junction.


The electric field generated by the space charge that tends to counteract the diffusion at equilibrium at zero bias. The electric field collapses as it becomes more forward biased.

The diffusion zone is only a few microns, but this animations shows the Electric field gradients and animated with bias changes.

enter image description here (Image taken from Wikipedia)

With a negative bias the holes and electron move away from the junction with greater strength in the depletion zone, thus resisting current flow and visa versa. The junction also reducing capacitance with voltage. (e.g. varicap is a more precise diode voltage control capacitance, but all diodes do this)

Both polarities are non-linear.

In the reverse direction when the breakdown threshold is reached with a series resistance current flows again with more power dissipation from the VI drop.

In the forward direction once conduction takes place, the voltage increases according to the bulk series resistance or ESR.

The series resistance is not the same in both polarities due to doping and material process factors.

  • \$\begingroup\$ Can you elaborate on "thus resisting current flow"? This is precisely the part that I don't understand. If electrons are being pulled towards the positive voltage, and holes are pulled towards the negative voltage, isn't that a current? \$\endgroup\$ Commented Dec 3, 2012 at 20:46
  • \$\begingroup\$ not unless they flow across the junction its just a strong electric field with holes and mobile electrons stuck on either side moving away from the barrier. No flow just a charge build up and the capacitance of the charge reduces as the charges move away from the barrier. \$\endgroup\$ Commented Dec 3, 2012 at 20:57
  • \$\begingroup\$ Ok, that's more clear, but then why can't electrons move from the circuit, into the P-side, and across the junction? The potential difference across the junction is in the correct direction for that, isn't it? If the P-side is what blocks the electrons, why doesn't it do that for forward bias as well? \$\endgroup\$ Commented Dec 3, 2012 at 21:01
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    \$\begingroup\$ "Without a bias there is no electric feld." Not true. The diffusion of the majority carriers to the other side of the junction causes a electric field. In fact, it is exactly this field that causes the diffusion to stop at a certain level, since the field is trying to push the carriers back to where they were. \$\endgroup\$ Commented Dec 3, 2012 at 21:21
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    \$\begingroup\$ If you copy images from Wikipedia or other sources, remember to cite them \$\endgroup\$
    – clabacchio
    Commented Dec 3, 2012 at 22:42

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