KaSap Electronic Materials_Chapter_5_problem_solutions

KaSap Electronic Materials_Chapter_5_problem_solutions -...

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5.1 Chapter 5 solutions 5.2 Intrinsic Ge Using the values of the density of states effective masses m e * and m h * in Table 5.1, calculate the intrinsic concentration in Ge. What is n i if you use N c and N v from Table 5.1? Calculate the intrinsic resistivity of Ge at 300 K. Solution From Table 5.1, we get and e e m m 56 . 0 = e h m m 4 . 0 = Now, 2 / 3 2 34 1 23 31 2 / 3 2 ) s J 10 626 . 6 ( ) K 300 )( K J 10 38 . 1 )( kg 10 1 . 9 56 . 0 ( 2 2 2 2 × × × × = = π π h kT m N e c = 1.05×10 25 m -3 or 1.05 ×10 19 cm -3 2 / 3 2 34 1 23 31 2 / 3 2 ) s J 10 626 . 6 ( ) K 300 )( K J 10 38 . 1 )( kg 10 1 . 9 4 . 0 ( 2 2 2 2 × × × × = = π π h kT m N v v = 6.35×10 24 m -3 or 6.35 ×10 18 cm -3 The intrinsic concentration is = kT E N N n g v c i 2 exp ) ( 2 / 1 [ ] × × × = ) K 300 )( K eV 10 62 . 8 ( 2 eV 66 . 0 exp ) cm 10 33 . 6 )( cm 10 049 . 1 ( 1 5 2 / 1 3 18 3 19 i n = 2.33 ×10 13 cm -3 2.3 ×10 13 cm -3 From Table 5.1, we get N c = 1.04×10 19 cm -3 and N v = 6.0×10 18 cm -3 [ ] × × × = ) K 300 )( K eV 10 62 . 8 ( 2 eV 66 . 0 exp ) cm 10 0 . 6 )( cm 10 04 . 1 ( 1 5 2 / 1 3 18 3 19 i n = 2.26 × 10 13 cm -3 2.3 ×10 13 cm -3 The intrinsic conductivity is ) ( h e i h e en ep en μ μ μ μ σ + = + = = 0.022 1 1 2 3 13 19 s V cm ) 1900 3900 )( cm 10 34 . 2 )( C 10 6 . 1 ( + × × = σ -1 cm -1 And the intrinsic resistivity is ρ = 1/ σ = 45.45 cm
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5.2 5.3 Fermi level in intrinsic semiconductors Using the values of the density of states effective masses m e * and m h * in Table 5.1, find the position of the Fermi energy in intrinsic Si, Ge and GaAs with respect to the middle of the bandgap ( E g /2). Solution + = h e g v Fi m m kT E E E ln 4 3 2 1 For Si, and e e m m 08 . 1 = e h m m 6 . 0 = × + = e e g v Fi m m E E E 6 . 0 08 . 1 ln ) K 300 )( eVK 10 62 . 8 ( 4 3 2 1 1 5 eV 011 . 0 2 1 + = g v Fi E E E So intrinsic Fermi level of Si is 0.011 eV below the middle of the bandgap ( E g /2). For Ge, and e e m m 56 . 0 = e h m m 4 . 0 = × + = e e g v Fi m m E E E 4 . 0 56 . ln ) K 300 )( eVK 10 62 . 8 ( 4 3 2 1 1 5 eV 0065 . 0 2 1 + = g v Fi E E E So intrinsic Fermi level of Ge is 0.0065 eV below the middle of the bandgap ( E g /2). For GaAs, and e e m m 067 . 0 = e h m m 5 . 0 = × + = e e g v Fi m m E E E 5 . 0 067 . 0 ln ) K 300 )( K eV 10 62 . 8 ( 4 3 2 1 1 5 eV 039 . 0 2 1 + + = g v Fi E E E So intrinsic Fermi level of GaAs is 0.039 eV above the middle of the bandgap ( E g /2). 5.5 Extrinsic Si Find the concentration of acceptors required for a p -type Si crystal to have a resistivity of 1 Ω cm. Solution The resistivity, h a eN μ ρ 1 = ) cm 1 )( s V cm 450 )( C 10 6 . 1 ( 1 1 1 1 2 19 Ω × = = ρ μ h a e N = 1.38×10 16 cm -3
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5.3 It is assumed that the doping concentration does not affect drifty mobility. At this concentration, from Figure 5.19, the hole drift mobility is approximately 350 cm 2 V -1 s -1 rather than 450 cm 2 V -1 s -1 . Using 350 cm 2 V -1 s -1 we find, ) cm 1 )( s V cm 350 )( C 10 6 . 1 ( 1 1 1 1 2 19 Ω × = = ρ μ h a e
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