CH274 - Electrons in Solids and Materials Exercise 4.3
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CH274 - Electrons in Solids and Materials
Exercise 4.3 – electron density in saturation regime
Demonstrate that the temperature of 300 K is in the saturation regime for an n-doped Si crystal, with a donor concentration of 1ppb. Assume an energy gap Eg = 1.1 eV, an effective density of states NC = 6.01024 m一3, a density pSi = 2.33 g.cm一3, and an atomic mass mSi = 28.09 for silicon.
Model answer
In order to determine whether the temperature of 300 K corresponds to the saturation regime, we need to compare the density of intrinsic electrons in the conduction band at this temperature, ni, with the density of donor atoms, nd.
The former is given by the usual expression:
ni = NC e −
By substituting the values for Si we obtain
ni = 6.0 × 1024 e − 2×1.38×10−23 ×300 = 6.0 × 1024 e − 2×1.38×3× 10−21 = = 6.0 × 1024 e −0.2126×102 = 6.0 × 1024 e −21.26 = 6.0 × 1024 × 5.87 × 10 −10 = = 6.0 × 1024 e −0.2126×102 = 6.0 × 1024 e −21.26 = 6.0 × 1024 × 5.87 × 10 −10 =
= 3.5 × 1015 m一3
The density of dopant is obtained by first calculating the numeric density of Si atoms and then diving this by 10一9 (since the donor concentration is of 1ppb).
The numeric density of silicon (i.e. the number of silicon atoms per unit volume) can be obtained by considering the mass density (pSi = 2.33 g.cm一3 = 2.33 × 10 −3 × 106 = 2.33 103 kg.m一3) and diving this by the mass of a single Si atom (equal to mSi mproton):
pSi 2.33 × 103 2.33 103
nSi = mSi × mproton = 28.09 × 1.67 × 10 −27 = 28.09 × 1.67 × 10 −27 =
= 5.0 × 10 −2 × 1030 = 5.0 × 1028 m一3
Thus, the donor density is equal to:
nd = nSi × 10 −9 = 5.0 × 1019 m −3
Finally, since nd >> ni, we can conclude that for this doped semiconductor, the temperature of 300 K is within the saturation regime.
2022-09-19