How does the doping concentration affect the depletion width?

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SUMMARY

The depletion width in semiconductor junctions inversely correlates with doping concentration. As doping concentration increases, the likelihood of electron-hole recombination rises, leading to a reduction in the depletion region. This phenomenon occurs because excessive electrons from the n-type material diffuse towards the p-type material, where they encounter holes more frequently due to the higher doping levels. The asymmetry in the depletion region is attributed to the differing effective masses of electrons and holes, which affects their recombination dynamics.

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Kaushik
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TL;DR
In my book, it is given that the depletion width increases with a decrease in doping concentration.
I do not understand why that is the case. Is there any intuitive explanation for it? Thanks
 
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Stating it the other way might give a better intuition: Depletion region decreases with increasing doping concentration.

The definition of the depletion region is there are no free charge carriers there. Because they are "recombined". Recombination is the process of an electron meeting with a hole and terminating each other. This is how it works:

When you make a junction the excessive electrons in the n-type is going to diffuse towards the p-type. (And vice versa.) If your doping concentration is high the probability of an electron to meet with a hole is going to be enhanced. So all the electrons diffused to the p-side are going to find their mathces in a shorter path.

The exact details are somehow mathematical involving the "distribution functions","effective masses" and "mean free paths". That is the reason that even if you get the same concentrtion of electrons and holes on both sides the depletion region will be asymetric becausse the effective mass of a hole is bigger. Hope that helps.
 

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