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When free electrons from P side drift towards N side,it constitutes drift current.But more free electrons are generated in P side due to thermal energy from temperature,so again it drifts towards N side.So i think continuous drift current flows in unbiased diode.But i don't know about diffusion current.

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    \$\begingroup\$ I don't think monetary is the word you're looking for lol. \$\endgroup\$ – KingDuken Jul 18 '17 at 20:26
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    \$\begingroup\$ cut the ESL jokes \$\endgroup\$ – Sunnyskyguy EE75 Jul 19 '17 at 1:22
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When you form a pn junction, mobile carriers from each side of the junction flow to the other side due to diffusion. I believe you mistakenly called this drift in your question. Diffusion is the process by which things move from a region of high concentration to a region of low concentration. So in a pn junction, electrons from the n side, and holes from the p side, will tend to diffuse to the other side of the junction.

This diffusion of free carriers leads to a depletion of those carriers on each side of the junction. (Diffusion current) The depletion of electrons and holes leaves behind ionized donors and acceptors which gives you the charged depletion region. This charged region creates an electric field which causes charge carriers to move in the other direction. This component of current is the drift current: the current that flows due to the electric field near the junction. These two currents must balance out in order to satisfy KCL, and you have your diode in equilibrium.

There are cases where the currents are not balanced, such as if you apply a voltage to the diode, but in that case, the behavior is probably more intuitive. In your question you brought up thermal carrier generation. This plays a role in the diffusion current, since thermal generation is where the carriers come from to support the diffusion current. However, it does not grow out of control like you expect since it is limited by the drift current. If you were to increase the carrier generation rate, you would have either an increase in the potential between the p and n sides of the junction which increases the diffusion current to cancel out the increased drift current, or there would indeed be some net current if there is a circuit for it to flow through. This is precisely how a solar cell works. Photons create electron-hole pairs and increase the diffusion current.

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  • \$\begingroup\$ Could you fix the sentence in your second paragraph? I'm trying to understand the fragmented sentences there: "The current that flows due to the electric field in near the junction These" \$\endgroup\$ – horta Aug 14 '17 at 4:46
  • \$\begingroup\$ @horta Should be fixed \$\endgroup\$ – Matt Aug 14 '17 at 4:50
  • \$\begingroup\$ @horta I re-read that section and I don't think i mentioned phonons, but I did say photons can generate e-h pairs. Anyway, there is thermal generation of carriers, but this just leads to the normal carriers we think of when working with semiconductors. As you increase temp you get more carriers, but this doesnt cause current to flow in a diode. It would however change the built in voltage. There is always a potential difference between the sides if a diode, but you cant really see this from the outside since in order to connect wires to the semiconductor you form metal-semiconductor contact \$\endgroup\$ – Matt Aug 14 '17 at 5:02
  • \$\begingroup\$ Right, I get that part of it, but to exaggerate what I'm trying to get at, with a PIN diode, there's a large region that has an electric field on it. Given the fact that thermally generated carriers exist, if they exist within the I-region of this kind of diode, and if the PIN diode were connected end to end with a wire, would there not be current flow from thermally generated carriers within the electrostatic field? \$\endgroup\$ – horta Aug 14 '17 at 5:06
  • \$\begingroup\$ Is there a difference besides the amount of energy of photon generated e-h pairs and thermally generated carriers? Are e-h pairs significantly different than thermally generated carriers? In k-space they're represented orthogonally, but that doesn't quite make sense to me because band gap thicknesses vary due to the temperature (due to thermally generated carriers). \$\endgroup\$ – horta Aug 14 '17 at 5:15

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