Fundamentals of Microelectronics

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12.1 (a) Y = A 1 ( X KA 2 Y ) Y (1 + KA 1 A 2 ) = A 1 X Y X = A 1 1 + KA 1 A 2 (b) Y = X KY A 1 ( X KY ) Y (1 + K A 1 K ) = X (1 A 1 ) Y X = 1 A 1 1 + K (1 A 1 ) (c) Y = A 2 X A 1 ( X KY ) Y (1 A 1 K ) = X ( A 2 A 1 ) Y X = A 2 A 1 1 A 1 K (d) Y = X ( KY Y ) A 1 [ X ( KY Y )] Y = X KY + Y A 1 X + KA 1 Y A 1 Y Y [ A 1 (1 K ) + K ] = X (1 A 1 ) Y X = 1 A 1 A 1 (1 K ) + K
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12.5 The loop gains calculated in Problem 4 are used. (a) A OL = A 1 A loop = KA 1 p R 2 R 1 + R 2 P Y X = A 1 1 + KA 1 ± R 2 R 1 + R 2 ² (b) A OL = A 1 A loop = g m 3 R D A 1 p R 2 R 1 + R 2 P Y X = A 1 1 + g m 3 R D A 1 ± R 2 R 1 + R 2 ² (c) A OL = A 1 A loop = g m 3 R D A 1 Y X = A 1 1 + g m 3 R D A 1 (d) A OL = A 1 p g m 1 R 2 1 + g m 1 R 2 P A loop = A 1 p g m 1 R 2 1 + g m 1 R 2 P Y X = A 1 ± g m 1 R 2 1+ g m 1 R 2 ² 1 + A 1 ± g m 1 R 2 1+ g m 1 R 2 ²
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12.8 A OL = g m r o = r 2 W L μ n C ox I D 1 λI D = 1 λ I D r 2 W L μ n C ox V out V in = A OL 1 + KA OL We want to look at the maximum and minimum deviations that V out V in will have from the base value given the variations in λ and μ n C ox . First, let’s consider what happens when λ decreases by 20 % and μ n C ox increases by 10 %. This causes A OL to increase in magnitude by a factor of 1 . 1 0 . 8 = 1 . 311. We want V out V in to change by less than 5 % given this deviation in A OL . 1 . 311 A OL 1 + 1 . 311 KA OL < 1 . 05 A OL 1 + KA OL KA OL > 3 . 982 Next, let’s consider what happens when λ increases by 20 % and μ n C ox decreases by 10 %. This causes A OL to decrease in magnitude by a factor of 0 . 9 1 . 2 = 0 . 7906. We want V out V in to change by less than 5 % given this deviation in A OL . 0 . 7906 A OL 1 + 0 . 7906 KA OL < 0 . 95 A OL 1 + KA OL KA OL > 4 . 033 Thus, to satisfy the constraints on both the maximum and minimum deviations, we require KA OL > 4 . 033 .
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12.10 A OL = g m p r o b 1 sC L P = g m r o 1 + sr o C L V out V in = A OL 1 + KA OL = g m r o 1+ sr o C L 1 K g m r o 1+ sr o C L = g m r o 1 + sr o C L Kg m r o Setting the denominator equal to zero and solving for s gives us the bandwidth B . B = Kg m r o 1 r o C L K = 1 + Br o C L g m r o
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12.11 (a) Feedforward system: M 1 and R D (which act as a common-gate ampli±er) Sense mechanism: C 1 and C 2 (which act as a capacitive divider) Feedback network: C 1 and C 2 Comparison mechanism: M 1 (which ampli±es the di²erence between the fed back signal and the input) (b) A OL = g m R D A loop = g m R D p C 1 C 1 + C 2 P v out v in = g m R D 1 + g m R D ± C 1 C 1 + C 2 ² (c) R in,open = 1 g m R in,closed = 1 + g m R D ± C 1 C 1 + C 2 ² g m R out,open = R D R out,closed = R D 1 + g m R D ± C 1 C 1 + C 2 ²
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12.15 (a) v in + g m 1 R D v in 1 /g m 2 v out (b) i in 1 g m 1 + R D i in 1 /g m 2 v out (c) v in g m v in r o (d) i in 1 g m 1 i in
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12.18 (a)
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This note was uploaded on 05/03/2009 for the course ECSE 2050 taught by Professor Monahella during the Spring '08 term at Rensselaer Polytechnic Institute.

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ch12sol - 12.1 (a) Y = A1 (X KA2 Y ) Y (1 + KA1 A2 ) = A1 X...

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