Chemistry, asked by mamoni3, 1 year ago

if Phosphorus is doped with silicon and silicon is doped with Boron identify the type of semiconductors

Answers

Answered by kanika58
1
Doping means the introduction of impurities into a semiconductor crystal to the defined modification of conductivity. The conductivity of a deliberately contaminated silicon crystal can be increased by a factor of 10⁶.

1.  Elements with 5 valence electrons are used for n-doping. The 5-valent dopant has an outer electron more than the silicon atoms. Four outer electrons combine with ever one silicon atom, while the fifth electron is free to move and serves as the charge carrier. The dopants are positively charged by the loss of negative charge carriers and are built into the lattice, only the negative electrons can move. Doped semimetals whose conductivity is based on free (negative) electrons are n-type or n-doped.

2. Boron has atomic number 5. It has 3 electrons in its outer orbit, i.e. 3 valence electrons. Elements with 3 valence electrons are used for p-type doping. The 3-valent dopants can catch an additional outer electron, thus leaving a hole in the valence band of silicon atoms. Therefore the electrons in the valence band become mobile. Due to the presence of positive holes, these semiconductors are called p-conductive or p-doped. Hence when Boron is doped with Silicon, it creates p-type doping.
Answered by definitelyslayer
1
First of all I'm assuming you mean Silicon doped with Boron (ie a small amount of Boron added to Silicon), not the other way around. Also this is the sort of question that's MUCH easier to answer with animations or at least diagrams but I'll do my best (if anyone knows how to attach powerpoint slides to an answer please let me know).

So the short answer: p-type

The explanation as to why:

Silicon atoms have 4 electrons in their outer shell (4 valence electrons in their valence shell if you want to use fancy words) and in a sample of pure silicon the atoms link up with neighbouring atoms by forming 4 covalent bonds (a covalent bond is a shared pair of electrons, one from each atom, so this brings the electrons in each atom's outer shell up to 8).

If a small amount of boron is added throughout the material then statistics tells us that the overwhelming majority of Boron atoms will be separate from each other and surrounded by Silicon.

Boron atoms have 3 electrons in their outer shell (boron's column on the periodic table tells us that). This gives them the ability to support 3 covalent bonds by sharing these electrons, however they cannot manage a 4th bond. If an extra electron were added then a 4th bond would form with a neighbouring silicon atom, but there just aren't any spare outer shell electrons around, they're all already part of a bond.

It's this place where an electron is missing from the 4th bond that makes the 'hole', a place where an electron would happily go (becoming the other half of a covalent bond) if one was available.

The important thing to remember is that the forces holding electrons in bonds aren't enough to keep them locked in place. Electrons from nearby bonds can move over and fill the hole, however that creates a new hole where the moving electron used to be.

A good analogy to use is imagine looking down on a crowded assembly hall with nearly all the seats filled but a few here and there still empty. People can move into empty seats next to them and then someone else can move into their old seat and so forth. The net affect of all this is that the empty seats seem to move around the room.

The same thing happens in silicon doped with boron, the holes actually move through the material as the electrons shuffle around. In fact they act exactly as if they were real particles with a positive charge, hence this type of semiconductor is a p-type (positive type).

Incidentally the explanation for n-types is pretty similar. If we dope silicon with an element with 5 valence electrons then 1 electron from each atom is not needed. The doping atoms use 4 of their valence electrons to form covalent bonds and the 5th is left to move around freely (hence the negatively charged free electrons making it an n-type)

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