# A find ec ef b determine ef ev c calculate

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Unformatted text preview: = 300 K is doped with arsenic atoms such that the concentration of electrons is n0 = 7 × 1015 cm- 3. (a) Find Ec – EF. (b) Determine EF – EV. (c) Calculate p0. (d) Which carrier is the minority carrier? (e) Find EF – EFi. (a) From the next to last equation on p.125, we have that "n % " 7 × 1015 % EC − EF = −kBT ln \$ 0 ' = −kBT ln \$ = 0.215eV 19 ' # 2.8 × 10 & # NC & (b) Using the gap from B4, (c) Then, EF − EV = EF − EC + EC − EV = −0.215 + 1.12 = 0.91eV 2 10 ni2 (1.5 × 10 ) p0 = = = 3.0 × 10 4 cm −3. n0 7 × 1015 (d) The holes are the minority carrier. (e) From (4.65), we have " n0 % " 7 × 1015 % EF − EFi = kBT ln \$ ' = kBT ln \$ = 0.338eV . 10 ' # 1.5 × 10 & # n1 & 4.19 The electron concentration in silicon at T = 300 K is n0 = 2 × 105 cm-...
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## This document was uploaded on 01/28/2014.

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