I know it depends upon the concentration of electrons and holes . So if we add donor atoms, even though electron concentration in conduction band increases , hole concentration decreases due to more recombination such that np=ni^2. Why do we say that conductivity increases?
Electronic – Why conductivity increases in semiconductor when doped
semiconductorssolid-state-devices
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The electron states in the band gap are localised, whereas the states which contribute to the bands are not.
The electron states in a solid are not simple, there's a lot of non-trivial quantum mechanics going on. The electron states in a free atom are localised around the atom - the electrons in those states can't leave the atom without a lot of energy, so they can't conduct anything. When you pack lots of atoms together, the surrounding electron states overlap and mix. Which states mix with each other is dictated by the energy: similar energies means more mixing. You end up with a new set of states which extend over the whole block of material. If the material has a periodic lattice, these electron states group together into bands.
Every state in a band has some velocity (called a Fermi velocity) associated with it, and an electron in that state can be thought of as moving through the material with that velocity. The Fermi velocity of electrons in the conduction band is very large, but because the electrons are all going in different directions, there is no net current. An applied electric field moves some electrons from states which were going in one direction, to states going in the other. In a metal, one of the bands is part full, so there are plenty of nearby states to move electrons into. In a semiconductor, there is a gap between a full band and an empty one so it's much harder to push electrons into the higher band.
When dopants are added, they don't form a nice periodic lattice and they are much more spread out than the silicon atoms that host them. This means that the electron states around the dopant can't mix with states from other dopants to form a band. Since the energy levels of the dopant states are different from the silicon states, they don't mix (much) with them either. Instead, the electron states are localised around the dopant, much like the states around the free atom. An electron in that state can't conduct in the way one in a band can. It either has to jump up into the band, or jump to another nearby dopant. The former happens in semiconductors, the later is known as incoherent transport, and appears in some other materials.
I'm not sure how well I've explained this, but if you don't get a clear answer here, you could try the physics stack exchange. This definitely feels more like condensed matter physics than electrical engineering!
In an intrinsic semiconductor at thermal equilibrium, carrier generation and recombination rate will be balanced so that net carrier density is constant. Also the concentrations n and p, of the electrons and holes are equal.
For any semiconductor,
$$ np = n_i^2 $$
Where ni is known as intrinsic career concentration of the semiconductor. This relation is called mass action law.
When an intrinsic semiconductor is doped with a dopor impurity like Phosphorus, electron concentration will increase, due to the surplus electron provided by each of the doper atom. Hole concentration remains the same. The net effect is -> \$ np> n_i^2 \$ . But mass action law has to be obeyed in the doped semiconductor too. So when it is doped, recombination rate will increase from its earlier rate to reduce the hole concentration. It drives the semiconductor back to the thermal equilibrium. i.e,\$ np = n_i^2 \$ .
So yes, Recombination Process is the sole reason for reducing the hole concentration while doping. Recombination Processes are of different types : band to band , auger recombination etc.
Best Answer
Conductivity can be dominated by holes (in P-type material), or by electrons (in N-type material), but the generation and recombination of hole/electron pairs must be equal (or it's not in equilibrium, and will soon have different carrier concentrations).
The generation rate (set by temperature) isn't a variable, but the recombination depends on the concentration of the (relatively rare) electrons and holes. It is proportional to both. So, we have an equation $$N_e \times N_{hole} = constant $$ One cannot manipulate the electron concentration without affecting the hole concentration.
Conductivity is proportional to a weighted sum of the hole and electron concentrations.
Holes, in silicon, are less effective, but we'll ignore that.
Now, visualize a rectangle, where the vertical sides are the hole concentration, and the horizontal sides are the electron concentration. The product (the rectangle area) is fixed because the product of concentrations is a constant.
The perimeter length of that rectangle is the conductivity; either extreme P doping (hole concentration) or N doping (electron concentration) yields a large-perimeter rectangle. Similar electron and hole concentrations make the rectangle a square, with minimum perimeter for the given area.
The formulae insist that high conductivity results from either extreme of doping.