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Photovoltaics: Carrier Concentration in Doped Semiconductors




Now, that we know what doping is, let’s see how we can determine the charge carrierconcentration in doped semiconductors. Once we do that, I will also show to you how tocalculate the position of the Fermi levels of doped semiconductors as well.As we saw in a previous post, the product of electron and hole concentrations in asemiconductor is independent on the position of the Fermi-energy level. This same equationcan be applied for doped semiconductors.You can remember that ‘n’ is the concentration of electrons that occupy energy states in theconduction band.

‘P’ is the concentration of holes that occupy energy states in the valenceband. N_i is the intrinsic carrier concentration. N_C and N_V are the effective densities ofstates in the conduction and valence band, respectively. ‘E_G’ is the bandgap. ‘k_B is theBoltzmann constant and ‘T’ is the temperature. Such relation is called the “law of massaction for semiconductors”.Though the values of ‘n’ and ‘p’ will change under doping, their product will still be equal tothe square of the intrinsic carrier concentration, n_i.Let’s work out how to calculate the charge carrier concentration from a given dopantconcentration.

We will use an example of an n-type silicon at room temperature that has a donorconcentration of 1 times 10 to the power of 18 phosphorus atoms per cubic centimeter.Since we are at room temperature we can assume that all the dopants are ionized.We can recall that the intrinsic carrier concentration of undoped silicon at roomtemperature is 1 times 10 to the power of 10 carriers per cubic centimeter.It is clear to see that the dopant concentration here is orders of magnitude higher than theintrinsic carrier concentration. Therefore, the intrinsic concentration essentially has noeffect on the electron concentration.We can equate the electron concentration to the donor concentration.

‘n’ is then equal to 1times 10 to the power of 18 electrons per cubic centimeter since each phosphorus atomdonates one mobile electron.We now have ‘n’, but how do we calculate ‘p’?With this relationship we can easilycalculate that the concentration of holes, ‘p’, is equal to 100 hole per cubic centimeter.When we compare the concentration of electrons to the concentration of holes we clearlynotice the enormous difference in the numbers of electrons and holes that are present inthis example of n-type Si.

We can do the same calculation for a p-type semiconductor with a given acceptorconcentration. I won’t step through the entire solution here. The only key difference now isthat we can set ‘p’ equal to the acceptor concentration and solve ‘n’ through the massaction law. This is because we can assume that at room temperature, all boron atoms areionized and each boron atom contributes with a hole.We have seen how doping can manipulate the concentration of electrons and holes insemiconductor. We also know that due to the mass-action law when one type of the chargecarrier is increased the other type is decreased.We call the carriers whose concentration is much larger than that of the other type majoritycarriers.

These are holes in p-type materials and electrons in n-type materials. Minoritycharge carriers are the carriers with much lower concentration than the majority carriers.These are electrons in in p-type materials and holes in n-type materials.From the concentrations of charge carriers we can calculate the position of the Fermi level.Here we see the band diagram of an n-type semiconductor.We are interested in the difference between the conduction band and the Fermi level. Wecalculate that by recalling this equation from a previous post. We can see that the quantity,E_F - E_C is present in the equation.We expect that the Fermi level is below the edge of the conduction band E_C, therefore thisquantity should be negative. 

Or in other words the position of the Fermi level is negativewith respect to the position of the edge of the conduction band.We know that the effective density of states in the conduction band of Si. From here, solvingfor the Fermi level is quite straight forward and we end up with a value of minus 0.09electron volts. This means that the Fermi level is 0.09 electron volts below the edge of theconduction band. We can use the same process to solve for the Fermi level in a p-typesemiconductor.

This time we are interested in the distance between the edge of the valenceband and the Fermi level. We expect that the Fermi level is close to the valence band.We use the relationship to calculate the concentration of holes in which the effectivedensity of states in the valence band N_V is present. The value for the position of Fermi levelwith respect to the edge of the valence band is plus 0.19 electron volts, which means that inthis case the Fermi level is above the valence band.

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