ENE 311

Homework # 3

Due: 23Jan 2007 (Group 21)

26Jan 2007 (Group 1)

  1. At room temperature the effective density of states in the valence band is 2.66 x 1019 cm-3 for silicon. Find the corresponding effective mass of holes. Compare this mass with the free-electron mass. Ans: 1.03m0
  2. Calculate the location of Ei in silicon at liquid nitrogen temperature at 77K and at 100ºC. Is it reasonable to assume that Ei is in the center of the forbidden gap? Given mh = m0 and me = 0.19m0.Ans: 0.583 eV and 0.557 eV
  3. A silicon sample at T = 300 K contains an acceptor impurity concentration of NA = 1016 cm-3. Determine the concentration of donor impurity atoms that must be added so that the silicon is n-type and the Fermi energy is 0.20 eV below the conduction band edge. Ans: 2.26 x 1016 cm-3
  4. Find the electron and hole concentration and Fermi level in silicon at 300 K for 1 x 1015 boron atoms/cm3. Ans:1015 cm-3, 9.3 x 104 cm-3, 0.26 eV

Extra problem:(not required for grading)

  1. Calculate the Fermi level of silicon doped with 1015, 1017, and 1019 phosphorus atoms/cm3 at room temperature, assuming complete ionization. From the calculated Fermi level, check if the assumption of complete ionization is justified for each doping. Assume that the ionized donors is given by Ans: valid for the case of 1015 cm-3
  2. Repeat the problem 4 by changing from boron impurities alone to be 3 x 1016 boron atoms/cm3 and 2.9 x 1016 arsenic atoms/cm3.Ans:1015 cm-3, 9.3 x 104 cm-3, 0.26 eV
  3. For an n-type silicon sample with 1016 cm-3 phosphorus donor impurities and a donor level at ED = 0.045 eV, find the ratio of the neutral donor density to the ionized donor density at 77 K where the Fermi level is 0.0459 eV below the bottom of the conduction band. Ans: 0.876

Solutions

Homework 3

NV = 2(2πmp kT / h2)3/2

The effective mass of holes in Si is

mp = (NV / 2) 2/3 ( h2 / 2πkT )

=

= 9.4 × 10-31 kg = 1.03 m0.

Calculate the location of Ei in silicon at liquid nitrogen temperature at 77K and at 100ºC. Is it reasonable to assume that Ei is in the center of the forbidden gap? Given mh* = m0 and me* = 0.19m0.

Soln

We may find Ei by measuring it from the top of the valence band as Ei – EVOR we may assume EV = 0 to be our reference. Therefore, we have

(1)

Actually, Eg is the function of temperature T. Since the problem did not give Eg(T), therefore, we may assume Eg(T) is about the same as Eg(300K) = 1.12 eV.

At 77 K

At 373 K

Because the second term on the right-hand side of the Eq.1 is much smaller compared to the first term, over the above temperature range, it is reasonable to assume that Ei is in the center of the forbidden gap.

  1. From Ec – EF= kT ln [NC / (ND– NA)]

which can be rewritten as ND – NA = NC exp [–(EC – EF) / kT ]

Then ND– NA = 2.86 × 1019 exp(–0.20 / 0.0259) = 1.26 × 1016 cm-3

or ND= 1.26 × 1016 + NA = 2.26 × 1016 cm-3

A compensated semiconductor can be fabricated to provide a specific Fermi energy level.

  1. The ionization energy for boron in Si is 0.045 eV. At 300 K, all boron impurities are ionized. Thus pp = NA = 1015 cm-3

np = ni2 / nA= (9.65 × 109)2 / 1015 = 9.3 × 104 cm-3.

The Fermi level measured from the top of the valence band is given by:

EF – EV= kT ln(NV/ND) = 0.0259 ln (2.66 × 1019 / 1015) = 0.26 eV

For ND = 1015 cm-3,

For ND = 1017 cm-3,

For ND = 1019 cm-3,

The number of electrons that are ionized is given by

nND[1 – F(ED)] = ND / [1 + e]

We obtain the number of ionized donors as

n = 1015 cm-3 for ND = 1015 cm-3

n = 0.975 × 1017 cm-3 for ND = 1017 cm-3

n = 0.275 × 1019 cm-3 for ND = 1019 cm-3

Therefore, the assumption of complete ionization is valid only for the case of 1015 cm-3.

  1. The boron atoms compensate the arsenic atoms; we have

pp = NA – ND = 3 × 1016 – 2.9 × 1016 = 1015 cm-3

Since pp is the same as given in (a), the values for np and EF are the same as

in problem 4. However, the mobilities and resistivities for these two samples are different.

  1. ND+ = =

= = 5.33 × 1015 cm-3

The neutral donor = 1016 – 5.33 ×1015 cm-3 = 4.67 × 1015 cm-3

∴Theratio of = = 0.876 .