8 solution

8 solution - Ω/sq width = 2 spacing =2 2 2 38 303 3 2 3 2 5 23 20 um K A = × × ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ × × = Area of CMOS process Area of

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Solution HW8 Problem 1: AA’ cross section: Problem 2: DD’ cross-section
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Problem 3: 1. resistor # squares from corners 2.75 = 440 # squaresfrom horizontal + vertical 22.25 =3560 Total Resistor 4000 2. Capacitor: V D = 4V, φ B = 0.6V, n = 0.5 fF C 3 . 94 6 . 0 4 1 ) 12 12 ( 2 1 . 3 6 . 0 4 1 ) 12 12 ( 78 . 0 5 . 0 5 . 0 = + + × × + + × × = Problem 4: Problem 5: Assume BJT works in Forward Active I = 1mA - I C = I B1 + I B2 -> 1mA = I C + I B1 + I B2 = 100 I B1 + I B1 + 4I B1 =105 I B1 -> I B1 = 1mA / 105 = 9.5 uA -> I C2 = 100* 9.5u = 0.95 mA Problem 6: 2 2 1 2 1 2 2 2 1 2 1 2 1 1 1 1 ) 1 1 ( ) 1 1 ( C C C C C B B C IN I A A A A I I A A I I I I I I + + = + + = + + = + + = β + + = > 1 1 2 1 2 1 2 A A A A I I IN C
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Problem 7: In a bipolar process: p – base diffusion: 160 /sq, width = 3 λ , spacing =3 λ 2 1 25 . 101 ) 3 . 0 3 ( ) 3 . 0 3 ( 160 20 um K A = × × × × = In a CMOS process: Poly: 23.5
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Unformatted text preview: Ω /sq, width = 2 λ , spacing =2 λ 2 2 38 . 303 ) 3 . 2 ( ) 3 . 2 ( 5 . 23 20 um K A = × × ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ × × = Area of CMOS process / Area of Bipolar process = 3 Problem 8: In a bipolar process: n+ emmitter difussion: 4.5 Ω /sq, width = 3 λ , spacing =3 λ 2 3 3600 ) 3 . 3 ( ) 3 . 3 ( 5 . 4 20 um K A = × × ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ × × = In a CMOS process: N active: 83.5 Ω /sq, width = 3 λ , spacing =3 λ 2 4 01 . 194 ) 3 . 3 ( ) 3 . 3 ( 5 . 83 20 um K A = × × ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ × × = Area of Bipolar process / Area of CMOS process= 18.5...
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8 solution - Ω/sq width = 2 spacing =2 2 2 38 303 3 2 3 2 5 23 20 um K A = × × ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ × × = Area of CMOS process Area of

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