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My Textbook reads:

During forward biasing, holes move towards n side and electrons move towards p side. The diffusion of electrons and holes into the depletion layer decreases its width.

If more holes and electrons will diffuse, wouldn't they form more immobile ions i.e. increase the width of depletion region. How come this decreases its width?

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You can easily show that the width of the depletion layer is given by $$W=\sqrt{\frac{2\varepsilon}{q}\left(\frac{N_a +N_d}{N_a N_d}\right)V_{bi}},$$ where $N_i$ is the acceptors/donors atoms density and $V_{bi}$ is the built-in voltage. If you forward bias the p-n junction, equation above modifies as follows: $$W=\sqrt{\frac{2\varepsilon}{q}\left(\frac{N_a +N_d}{N_a N_d}\right)\left(V_{bi}-V\right).}$$ So you can see that high forward bias voltage $V$ means decreasing of $W$.

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  • $\begingroup$ So depletion region disappears when $V>V_{b}i$? $\endgroup$ – Stacy Barrymore Feb 15 '18 at 12:17
  • $\begingroup$ Yes, something new happens in this case (saturation). $\endgroup$ – Stig Feb 15 '18 at 12:49
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I think you missunderstand the meaning of "diffusion" in this context. It simply says that holes and electrons are pushed further into depletion region due to electromagnetic forces acting on them.

They are not dissappearing, which the unfortunate phrasing of the sentence from your book might suggest.

See this wikipedia article about Diffusion as a term used in mechanical engineering context.

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  • $\begingroup$ No. Diffusion is due to a concentration gradient and random walks. Not due to a Coulomb force. $\endgroup$ – Pieter Feb 15 '18 at 15:48

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