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CONCENTRATION OF CARRIERS IN SEMICONDUCTORS AND METALS

We are now ready to evaluate the densities of carriers in the bands of semiconductors which form one of the main factors of their classical conductivity. Let us consider first the number of carriers per unit volume in the conduction band of a semiconductor. If Pfd> the Fermi-Dirac probability of occupation of a state in a conduction band is, say, 0.7, that means that on average there are 0.7 electrons in that state, so it is correct to write that the number of electrons per unit volume, n, in that band is

where V is the volume of the semiconductor. The problem with equation 2.50 is that while Ek is a discrete quantity, there are roughly 1023 energies per mole to be counted so that it would be beneficial to convert the sum in 2.50 into an integral. To achieve that we only need to count the number of energy levels AN(E) between E and E+AE. Then the sum can be converted into an integral as follows

where

is the number of states per unit volume between E and E+AE in the conduction band or alternatively g„ (£) is the density of states per unit energy per unit volume in the conduction band.

In the limit of AE —» 0, g(E) can be thought as the derivative of N(E), where the latter is defined as the number of states up to the energy E. Although the energies Ek are discrete, as the subscript к signifies, they are so dense that they can be treated as a continuum. The task is then to find an expression for g(E). We have seen in the previous paragraph that most semiconductors of interest can have their conduction band expanded about its minimum as (cf. equation 2.48 which we reproduce here)

that is, the constant energy surfaces in к space are in general ellipsoids. We remind the reader that in Si the conduction band minima Ec occur (see figure 2.16) along the ГХ axis. Two effective masses are equal in 2.48 and are labelled m,, and the third one is labelled trip see equation 2.46. In GaAs with Ec occurring at k= 0 all effective masses are equal. However, for the sake of generality we will proceed with the calculation of g(E) assuming all m' are different.

The volume of an ellipsoid of the form

4

is V = —nabc. So the volume in к space of an ellipsoid of the form of equation 2.48, enclosed by a surface of energy E is

Therefore, an incremental volume dV enclosed between ellipsoids of energy E and E+dE is (taking the derivative)

But we have seen in section 2.2 (see equation 2.12) that in 1 dimension, the distance 2л (2kV

between к points is —, and in 3 dimensions, it is — for a cubic sample of size L. Hence L L )

the number of states AN(E) between E and E+dE will be given by the elemental volume dV(E) divided by the above spacing

so that

where in 2.52 we have divided by the volume V=L3 and have simplified the denominator, noting that (2л)3 Й3 = h}. Equation 2.52 is not the final result for the density of states (usually abbreviated DOS). There are two additional multiplicative factors: first as we have noted earlier, there are six such ellipsoids in Si so we have to multiply by this number. We assign the symbol/to refer to the number of minima in any semiconductor. Second, we have to take care of spin, every Bloch state can carry two electrons. Hence the final result is

Therefore, from 2.51 we get the concentration of electrons in the conduction band

where m,) = . We remind the reader that К is Boltzmann’s constant. If the 1 in

the denominator of the above integral can be ignored, then the integral can be calculated analytically and the result is

/ , 3/2

where Nc = 2/ -^-I is called the effective density of states. If, on the other hand,

the Fermi level is very close to the conduction band edge Ec, the 1 in the denominator can’t be ignored and the integral has to be evaluated numerically. Equation 2.54 can then be rewritten as

where the so-called Fermi integral F,A of order Vi is defined as

The above formulae 2.56 and 2.56a are necessary only when the Fermi energy EF moves inside the conduction band.

A completely parallel argument shows that the concentration of holes p in the valence band

where gp{E), the density of states in the valence band, is (again provided that EF is not near Ev)

The degeneracy of the band edge is taken as/=l in this case because Ev is situated at k= 0 for Si and most other semiconductors.

Consider now an intrinsic semiconductor (i.e. no doping present). For every electron that is excited into the conduction band, a hole is left in the valence band. Hence n=p. Substituting into n=p the expression for n from equation 2.55 and the expression for p from equation 2.58 and solving for EF, we get

The second term in 2.59 is negligible compared to the first, (remember at room temperature KT=26meV). So to a very good approximation £f,=the Fermi level of the intrinsic

(Ес + Ev)

semiconductor is EF. = ----We note that Ec - EF. > KT and therefore the approximate expressions 2.55 and 2.58 were correctly used to obtain EFj.

By multiplying 2.55 and 2.58 we get for an intrinsic semiconductor

where Eg is the band-gap Ec—Ev or

Since the product np is independent of the Fermi level, which (as we shall see) is sensitive to the doping of a semiconductor, the above equation holds for any level of doping of a semiconductor. Equation 2.61 is usually called the law of mass action. Equations 2.55 and 2.58 can now be rewritten in terms of EF. and и,

and

Now consider an N-type semiconductor. As explained in section 2.4, for every donor atom in the crystal a new energy level ED is created just below Ec (see figure 2.14), which is occupied by the extra electron of the donor atom at 7=0 and which is progressively emptied (the donor atom is being ionized) as the temperature T increases. The electrons at ED jump into the conduction band. What can we say about the Fermi level EF of such a system if the states at Ed are occupied at T= 0 and the ones in the conduction band are empty? The Fermi level at T= 0 must lie in between ED and Ec since EF separates the full from the empty states (at T=0). In fact it lies midway at (Ed+Ec)/2. As the temperature increases EF moves down the gap since the levels Ed are emptied. An exactly parallel argument holds for a P-type semiconductor. The Fermi level EF lies below EA at (Ev + EA)/2 and as the levels at EA are being filled and holes are created in the valence band, EF moves up the gap. In both cases of donor and acceptor doping if the temperature is raised at such levels that the number of electrons and holes created by thermal excitation from the valence to the conduction band exceeds the number created by doping, the Fermi level returns to the mid-point of the energy gap. The movement of EF with temperature in both N- and P-type semiconductors is shown in figures 2.17a,b.

Since any semiconductor N- or P-type must remain neutral we must have

Movement of the Fermi level Ep as the temperature increases in an N-type (a) and P-type (b) semiconductor

FIGURE 2.17 Movement of the Fermi level Ep as the temperature increases in an N-type (a) and P-type (b) semiconductor.

The Fermi level E,: of any kind of semiconductor can be calculated as follows: divide equation 2.62 by 2.63, we then get

When we have an N-type semiconductor, и » p and its Fermi level EFn lies in the upper half of the band-gap and conversely for a P-type semiconductor /)»и and its Fermi level EFp lies in the bottom half of the band-gap. The ionization process of both types of impurities is complete at room temperature because the impurity levels ED or EA happen to be very close to the respective band-edge Ec or Ev. Therefore, at room temperature

The doping of most semiconductors is such that и, No or n, NA. Using the relation 2.61

we also have for an N- and P-type semiconductor respectively (i.e. for the minority carriers)

Hence we get by virtue of 2.64 that at room temperature for an N-type semiconductor to a very good approximation

and for a P-type semiconductor

Substitution of the above values for n and p in the general relation 2.65 for the Fermi levels of an N-doped and a P-doped semiconductor respectively leads to

Now that we have dealt with semiconductors, the evaluation of the corresponding quantities in metals is straightforward. The number of electrons per unit volume n in the conduction band of a metal can be evaluated by integrating the density of states (DOS) multiplied by the Fermi-Dirac probability from the bottom of the band to infinity. The DOS is the same for metals as for semiconductors to the extent that both can be treated as parabolic (i.e. °c k2). Furthermore, we can choose as our zero of energy the bottom of the conduction band so that

where gm(E) is the DOS of metals at E. Then

The above integral is easily evaluated at T = 0 when the denominator of the integrand is 1 and the upper limit of the integral will be Ep. We get

The above formula may also be used for T = room temperature because the Fermi level of a metal does not move significantly compared to the width of the conduction band of a metal. We remind the reader that KT ~ 26meV and the width of the bands in metals are of the order ofeV.

 
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