Question

The Fermi energy level in a silicon sample at T = 300 K is as (Eg/4) close to the top of the valence band. (a) Is the material n type or p type? (b) Calculate the values of n0 and p0.

Answer 1

Answer 2

(a) The given information implies that the Fermi level is located in the valence band and is closer to the top by an amount of Eg/4, where Eg is the bandgap energy of silicon. This indicates that the majority carriers are holes, and the material is p-type.

(b) At thermal equilibrium, the product of the electron and hole concentrations, n0p0, is given by:

n0p0 = ni^2

where ni is the intrinsic carrier concentration of silicon, which is approximately equal to 1.5 x 10^10 cm^-3 at T = 300 K. Since the material is p-type, the hole concentration is higher than the electron concentration, i.e., p0 > n0.

The position of the Fermi level can also be related to the carrier concentrations by the following equation:

n0/Nc * p0/Nv = exp[(Ef - Ev)/kT]

where Nc and Nv are the effective densities of states in the conduction and valence bands, respectively, k is the Boltzmann constant, and T is the temperature in Kelvin.

Since the Fermi level is closer to the top of the valence band by an amount of Eg/4, we have:

Ef - Ev = -Eg/4

Substituting the appropriate values and simplifying, we get:

n0 * p0 = (Nc*Nv/4) * exp(Eg/4kT)

Using the values for silicon, we have:

Nc = 2.8 x 10^19 cm^-3 Nv = 1.04 x 10^19 cm^-3 Eg = 1.12 eV k = 8.62 x 10^-5 eV/K

Substituting these values, we get:

n0 * p0 = 4.62 x 10^19 cm^-6

Since p0 = n0 * exp(Eg/4kT), we can substitute for p0 to obtain:

n0^2 * exp(Eg/4kT) = 4.62 x 10^19 cm^-6

Taking the natural logarithm of both sides, we get:

2ln(n0) + Eg/4kT = ln(4.62 x 10^19)

Solving for n0, we get:

n0 = sqrt(4.62 x 10^19 / exp(Eg/2kT))

Substituting the values, we get:

n0 = 1.26 x 10^15 cm^-3

Therefore, the hole concentration is:

p0 = n0 * exp(Eg/4kT) = 5.21 x 10^16 cm^-3