I was studying about semiconductors and I came up with a doubt in my mind . Let us assume a p-n junction diode is in its steady state in unbiased condition that is no charge is flowing through the depletion region. In this situation the electrons in the n type semiconductor are facing repulsion among them but the potential barrier in the depletion region is resisting them to cross the region and go to the p side. But when we make it forward biased with a constant DC voltage source then we are able to reduce the potential barrier a bit and the electrons from the n side will again start flowing. After a few of them has crossed the region the charge difference across the region will increase and the Barrier potential will again neutralize the repulsion between the remaining electrons of the n side and the external voltage source also. Now in my book it is written that when this will happen the width of the depletion region will decrease . But why should it decrease ?


1 Answer 1


What you are missing from this analysis is the charge on the dopant atoms.

So, the n side of a silicon p-n junction might be doped with Phosphorus atoms which have 5 valence electrons, only four of which are involved in bonding to silicons. The remaining electron is easily ionized and travels through-out the n-side, but this leaves a small positive charge on the Phosphorus dopant atoms. The sum of the positive Phosphorus charges keeps the electrons on the n side of the depletion region. (A similar situation is going on with holes on the p side of the depletion region.) The depletion region is the area of the junction where electrons and holes meet each other (because the doping switches from n to p) so that neither can exist for long (beyond the intrinsic level - there is always some thermal creation of electrons and holes).

When the diode is forward biased the voltage on the n-side starts to compensate for the phosphorus atoms' positive charge and favors electrons venturing away from the phosphorus charge on the n-side and into the depletion region. (The opposite happens on the p side driving holes into the region.) It takes a while for an electron to find a hole in the depletion region so the region begins to shrink as idle electrons seeking a hole venture into it and a little current begins to flow as the electron-hole recombination rate begins to rise. With a little more voltage the region disappears entirely as the voltage overwhelms the attaction of the dopant atoms' charge and electron-hole recombination becomes rapid and the current across the junction increases exponentially.

  • $\begingroup$ What do you mean by "takes a while" ? Why does it take longer? $\endgroup$ Commented Feb 21, 2021 at 16:22
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    $\begingroup$ This is like a chemistry re-combination equation. The product of: (the density of holes) times (the density of electrons) times (a recombination rate) controls how rapidly they recombine. $\endgroup$
    – Paul Young
    Commented Feb 25, 2021 at 3:22

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