Can anyone tell me the typical paths of holes and electrons in an unbiased PN junction at equilibrium? In terms of diffusion currents, is that only holes (majority carriers in the the P-Type) traveling exclusively within the valence band and electrons (majority carriers in the N-type) traveling exclusively within the conduction band, both remaining in their respective 'lanes' even across the PN junction and both due to concentration of charges (the coulomb effect)? And in terms of drift currents in the opposite direction, is that minority carrier holes traveling exclusively in the valence band and electrons traveling exclusively in the conduction band with both traveling due to the charge imbalance created by the diffusion current?
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Yes that's correct, the drift current is due to minority carriers in distance from the depletion area smaller then their diffusion length being drift across the depletion area (due to the built in field created by the diffusion).
Here is a band diagram
Note that the drift current under reversed bias is negligible (if it has something with what you asked)
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\$\begingroup\$ Many thanks for that. But my understanding is that the drift current is just as 'negligible' as the diffusion current, both being equal at equilibrium in all bias states. Also, when a conductor injects electrons and holes, do those migrate to the depletion region? If so, in response to what force? If not, then does the depletion region actually extend across the entirety of the diode, becoming significant only near the junction? Thanks. \$\endgroup\$ Commented Jul 22, 2020 at 17:58
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\$\begingroup\$ How can you say equilibruim at all bais state? Bais means the device is not at equilibruim. At equilibruim, the whole point is that the drift current and diff current cancle each other both for holes and electrons, thus no current flows in the device (as you would expect under no voltage drop) \$\endgroup\$ Commented Jul 23, 2020 at 8:44
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\$\begingroup\$ I agree that drift and diffusion currents are equal and opposite at any bias (perhaps short of full conduction). Their equality MEANS equilibrium and they were unequal in the brief time it took to achieve that equilibrium (that intrinsic depletion voltage), perhaps after being joined in the first place. You seem to suggest that no net current through the device means no equilibrium. Maybe we're just defining equilibrium differently. \$\endgroup\$ Commented Jul 23, 2020 at 22:19
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\$\begingroup\$ By equilibrium I mean no external force on the device. "But my understanding is that the drift current is just as 'negligible' as the diffusion current, both being equal at equilibrium in all bias states" those two current are the same under no bais at all, they are not the same under any bais, for example, under reverse bais the depletion region extends, which means it takes more enegry for elctron and holes to diffuse, on the other hand, the drift current is only due to minority carriers so it might be negligible (this state is closed diode). forward bais is when the opossite occurs \$\endgroup\$ Commented Jul 24, 2020 at 7:41
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\$\begingroup\$ I can direct you to a book where every thing is explained perfectly \$\endgroup\$ Commented Jul 24, 2020 at 7:45