34 0175 134 0175 i d1 sat 0175 152 ma 1

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Unformatted text preview: From the data, we find that eN d a 2 1.6 × 10 −19 ⋅ 2 × 10 22 ( 3.5 × 10 −7 )2 Vp 0 = = = 1.69V 2ε s 2 ⋅ 13.1 ⋅ 8.854 × 10 −12 ⎛ 5 × 1018 ⋅ 2 × 1016 ⎞ Vbi = 0.0259 ln ⎜ = 1.34V ⎝ (1.8 × 10 6 )2 ⎟ ⎠ . VG 1.69 ⋅ 2.69 × 10 −3 I p1 = p 0 01 = = 1.52 mA 3 3 Vp = Vbi − Vp 0 = −0.35V Now, we have (c) Now, we use (13.35) to find the two cases: ⎧ ⎪ ⎪ ⎛ 1.34 − 0 ⎞ ⎡ 2 1.34 − 0 ⎤ ⎫ I D1 ( sat; 0 ) = 1.52 mA ⎨1 − 3 ⎜ ⎥ ⎬ = 51µ A ⎟ ⎢1 − ⎝ 1.69 ⎠...
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