Consider silicon at T300 K Assume Eg112 eV Nc28 1019 cm3 N

Consider silicon at T=300 K. Assume Eg=1.12 eV, Nc=2.8 × 10^19 cm^3 , Nv=1.04 × 10^19 cm^3 . (a) If the hole concentration is p0=5 × 10^5 cm^3 , find the electron concentration n0. Is it an n-type or p-type semiconductor? (b) If the ratio of hole to electron concentrations is 1 × 10^11, what is the position of Fermi energy with respect to (1) intrinsic Fermi level, (2) conduction band?

Solution

(a)

the intrinsic concentration is,

     ni^2 = N_cN_vexp[-E_g/kT]

            = [2.8 × 10^19][1.04 × 10^19 cm^3]*exp[- 1.12*1.6x10^-19/1.38x10^-23*300]

ni = 6.95x10^9 cm^-3

the electron concentration n0 is.

n0 = ni^2/p0 = [6.95x10^9]^2/[5 × 10^5 ] = 9.66e+13 cm^-3

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(1) intrinsic Fermi level:

the position of Fermi energy:

Ef - Ei = kT * ln(n0/p0)

=  (0.0259) * ln(1x10^-11)

= -0.656 eV

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conduction band:

Calculate the value of n0 as follows

n0p0 = ni^2 [since, p0/n0 = 1 × 10^11]

n0*[1 × 10^11*n0] = [6.95x10^9]^2

hence, n0 = 2.198e+4. Thus, the osition of Fermi energy:

    Ef - Ei = kT * ln(Nv/n0)

= (0.0259)*ln[1.04 × 10^19 / 2.198e+4.]

= 0.875 eV