Chapter03

Microelectronic Circuit Design

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CHAPTER 3 3.1 φ j = V T ln N A N D n i 2 = 0.025 V () ln 10 19 cm 3 10 18 cm 3 ( ) 10 20 cm 6 = 0.979 V w do = 2 ε s q 1 N A + 1 N D j = 211.7 8.854 x 10 14 F cm 1 1.602 x 10 19 C 1 10 19 cm 3 + 1 10 18 cm 3 0.979V w do = 3.73 x 10 6 cm = 0.0373 μ m x n = w do 1 + N D N A = 0.0373 m 1 + 10 18 cm 3 10 19 cm 3 = 0.0339 m | x p = w do 1 + N A N D = 0.0373 m 1 + 10 19 cm 3 10 18 cm 3 = 3.39 x 10 -3 m E MAX = qN A x p s = 1.60 x 10 19 C 10 19 cm 3 3.39 x 10 7 cm 11.7 8.854 x 10 14 F / cm = 5.24 x 10 5 V cm 3.2 p po = N A = 10 18 cm 3 | n po = n i 2 p po = 10 20 10 18 = 10 2 cm 3 n no = N D = 10 15 cm 3 p no = n i 2 n no = 10 20 10 15 = 10 5 cm 3 j = V T ln N A N D n i 2 = 0.025 V ln 10 18 cm 3 10 15 cm 3 ( ) 10 20 cm 6 = 0.748 V w do = 2 s q 1 N A + 1 N D j = 8.854 x 10 14 F cm 1 1.602 x 10 19 C 1 10 18 cm 3 + 1 10 15 cm 3 0.748V w do = 98.4 x 10 6 cm = 0.984 m 3.3 p po = N A = 10 18 cm 3 n po = n i 2 p po = 10 20 10 18 = 10 2 cm 3 n no = N D = 10 18 cm 3 p no = n i 2 n no = 10 20 10 18 = 10 2 cm 3 j = V T ln N A N D n i 2 = 0.025 V ln 10 18 cm 3 10 18 cm 3 ( ) 10 20 cm 6 = 0.921 V w do = 2 s q 1 N A + 1 N D j = 8.854 x 10 14 F cm 1 1.602 x 10 19 C 1 10 18 cm 3 + 1 10 18 cm 3 0.921V w do = 4.881 x 10 6 cm = 0.0488 m 34
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3.4 p po = N A = 10 18 cm 3 | n po = n i 2 p po = 10 20 10 18 = 10 2 cm 3 n no = N D = 10 18 cm 3 p no = n i 2 n no = 10 20 10 18 = 10 2 cm 3 φ j = V T ln N A N D n i 2 = 0.025 V () ln 10 18 cm 3 10 20 cm 3 ( ) 10 20 cm 6 = 1.04 V w do = 2 ε s q 1 N A + 1 N D j = 211.7 8.854 x 10 14 F cm 1 1.602 x 10 19 C 1 10 18 cm 3 + 1 10 20 cm 3 1.04V w do = 0.0369 μ m 3.5 p po = N A = 10 16 cm 3 n po = n i 2 p po = 10 20 10 16 = 10 4 cm 3 n no = N D = 10 19 cm 3 p no = n i 2 n no = 10 20 10 19 = 10 cm 3 j = V T ln N A N D n i 2 = 0.025 V ln 10 19 cm 3 10 16 cm 3 ( ) 10 20 cm 6 = 0.864 V w do = 2 s q 1 N A + 1 N D j = 8.854 x 10 14 F cm 1 1.602 x 10 19 C 1 10 19 cm 3 + 1 10 16 cm 3 0.864V w do = 0.334 m 3.6 w d = w do 1 + V R j | (a) w d =2 w do requires V R = 3 j = 2.55 V w d = 0.4 m 1 + 5 0.85 =1.05 m 3.7 w d = w do 1 + V R j w d =3 w do requires V R = 8 j = 4.80 V w d = 1 m 1 + 10 0.6 = 4.20 m 3.8 j n = σ E , = 1 ρ = 1 0.5 Ω⋅ cm = 2 cm E = j n = 1000 A cm 2 2 cm 1 = 500 V cm 35
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3.9 j p = σ E | E = j n = j n ρ = 5000 A cm 2 () 2 Ω⋅ cm = 10.0 kV cm 3.10 j j n = qnv = 1.60 x 10 19 C 4 x 10 15 cm 3 10 7 cm s ⎟ = 6400 A cm 2 3.11 j j p = qpv = 1.60 x 10 19 C 5 x 10 17 cm 3 10 7 cm s ⎟ = 800 kA cm 2 3.12 j p = q μ p pE qD p dp dx = 0 E =− D p p 1 p kT q dp dx 1 p dp dx p ( x ) = N o exp x L 1 p E= V T L 0.025 V 10 4 cm 250 V cm dp dx = 1 L The exponential doping results in a constant electric field. 3.13 j p = qD n dn dx = q n V T dn dx dn dx = 2000 A / cm 2 1.60 x 10 19 C 500 cm 2 / V s 0.025 V = 1.00 x 10 21 cm 4 3.14 10 = 10 4 10 16 exp 40 V D 1 [] + V D and the solver yields V D = 0.7464 V 3.15 f = 10 10 4 I D 0.025ln I D + I S I S f ' 10 4 0.025 I D + I S I D ' = I D f f ' Starting the iteration process with I D = 100 μ A and I S = 10 -13 A: I D f
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Chapter03 - CHAPTER 3 3.1(1019 cm-3(1018 cm-3 = 0.979V NA...

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