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In a forward-biased PN junction, the potential barrier decreases, allowing more majority carriers from one side to diffuse to the other side where they are minority carriers. After they cross the potential barrier, they form a diffusion current, the drift current of minority carriers is insignificant, then they recombine with majority carriers and form a drift current under the effect of the applied electric field.

Why do minority carriers form a diffusion current not a drift current after they cross the potential barrier? It is counter-intuitive that the main current is diffusion when there is an applied electric field.

This is according to all the microelectronics book I'm currently reading. There is one which says this can be proved but without providing anything. Can someone please provide a proof for this.

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2 Answers 2

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Why do minority carriers form a diffusion current not a drift current after they cross the potential barrier?

Because the electric field is zero outside the depletion region, so a drift current cannot be driven.

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  • $\begingroup$ Could you please look at page 24 at the link below? From the current density plot, it is clear that there are both drift and diffusion on the quasi-neutral region even no electric field. Could you explain why? solar.udel.edu/ELEG620/04_pnjunction.pdf $\endgroup$
    – emnha
    Commented Mar 10, 2017 at 20:55
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The unbiased PN junction is in an equilibrium where the diffusion current is cancelled by the drift current.

The applied electric field in a forward biased PN junction "cancels" the electric field in the depletion region, eliminating the potential barrier and allowing diffusion current to flow.

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