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ECSE-2210 Microelectronics Technology
Homework 2 – Solution
1. Always, n p ≈ ni 2
At 300 K, n = 1017 cm–3, ni = 1010 cm–3, p = 103 cm–3
At 200 K, n = 1017 cm–3, ni = 105 cm–3, p = 10-7 cm–3
That is, the hole concentration at 200 K is extremely small. You will find only one hole in a
volume of 107 cm3 of Si !
2. a) At room temperature in Si, ni = 1010 cm-3. Thus here ND >> NA , ND >> ni and
n = ND = 1015 cm–3
p = ni 2/ ND = 105 cm–3
b) Since ND << NA, NA >> ni
p = NA = 1016 cm–3
n = ni 2/ NA = 104 cm–3
c) Here we must retain both ND and NA, but ND-NA >> ni
n = ND – NA = 1015 cm–3
p = ni 2/ (ND – NA) = 105 cm–3
d) We deduce from Figure 2.20 that, at 450 K, ni (Si) ≈ 5 × 1013 cm–3. Clearly, ni is
comparable to ND and we must use Eq. 2.29a
n = ND /2 + [ (ND /2) + ni 2] 1/2 = 1.21 × 1014 cm–3
p = ni 2 /n = 2.07 × 1013 cm–3
e) We conclude from figure 2.20, that at 650 K, ni = 1016 cm–3 Here ni >> ND
p = ni = 1016 cm–3
n = ni = 1016 cm–3
