A p-n junction has a depletion layer of the order of
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A depletion region is a natural feature of doping a semiconductor n-type and right next to it p-type. At the p-n junction (the border of the two types), the mobile electrons in the n-type (this is due to the excess electron provided by the n-dopant) diffuse across the junction, into the p-type area. This then leaves the n-type area with a net positive charge, because the electron left.
Holes (the majority charge-carriers in the p-type, created through doping with an element such as Indium (3 valence electrons)) can be said to 'diffuse' toward the n-type. (I think this is because of the electrons from the n-type causing the electrons in the p-type to repel away, and thus travel away from the n-type, filling existing holes and thus creating holes on their way, however this is not important.)
So we have already established that the n-type is left with a positive charge. The electrons which moved to the p-type area from the n-type combined with the holes in the p-type area, and thus leave the p-type with a negative charge.
So, p-type = net negatively charged and
n-type = net positively charged
(The real answer starts here)
When the electron combined with the hole in the p-type, the outer shell of the atom which previously had the hole is now filled, and thus the atom becomes very unreactive, as it has a stable full outer shell. Therefore, when many electrons fill holes, this happens, and so the area becomes very reactive. ALSO, since the p-type is negative, electrons are repelled away.
In the n-type, atoms have always had a full outer shell, but have gotten rid of their free electron. So now, the p-type region and the n-type region both are charged, so repel charge-carriers, and also are unreactive because of their full outer shells. This -freezes- this region, and no mobile charge-carriers exist here. This is called the depletion zone.
NOTE^^^ This only happens towards close to the boundary of the p-n junction, and so the depletion zone is not infinite, it has a defined width/size.
Holes (the majority charge-carriers in the p-type, created through doping with an element such as Indium (3 valence electrons)) can be said to 'diffuse' toward the n-type. (I think this is because of the electrons from the n-type causing the electrons in the p-type to repel away, and thus travel away from the n-type, filling existing holes and thus creating holes on their way, however this is not important.)
So we have already established that the n-type is left with a positive charge. The electrons which moved to the p-type area from the n-type combined with the holes in the p-type area, and thus leave the p-type with a negative charge.
So, p-type = net negatively charged and
n-type = net positively charged
(The real answer starts here)
When the electron combined with the hole in the p-type, the outer shell of the atom which previously had the hole is now filled, and thus the atom becomes very unreactive, as it has a stable full outer shell. Therefore, when many electrons fill holes, this happens, and so the area becomes very reactive. ALSO, since the p-type is negative, electrons are repelled away.
In the n-type, atoms have always had a full outer shell, but have gotten rid of their free electron. So now, the p-type region and the n-type region both are charged, so repel charge-carriers, and also are unreactive because of their full outer shells. This -freezes- this region, and no mobile charge-carriers exist here. This is called the depletion zone.
NOTE^^^ This only happens towards close to the boundary of the p-n junction, and so the depletion zone is not infinite, it has a defined width/size.
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