HW12_Solutions

HW12_Solutions - 1.58 v I-V TN ( 29 = V DD-v I + V TP a (...

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Homework Solutions ECE302 Spring 2007 HW 12 1 7.6 (a) V H = 2.5 V, V L = 0 V (b) Same as (a). V H and V L don't depend upon W/L in a CMOS gate. 7.16 For the NMOS device +5 V 1.5 mA 0.6 V + - ( 29 1 10 . 6 10 5 . 1 6 . 0 2 6 . 0 6 . 0 5 10 100 3 6 = = - - - - n n L W x L W x For the PMOS device +5 V 60 μ A 2.6 V + - 40 x 10 - 6 ( 29 W L p 5 - 0.6 - 2.6 2 2.6 = 6 x 10 - 5 W L p = 1 5.37 7.25 t r 2245 3.6 R onP C = 3.6 C K p V GS - V TP = 3.6 0.2 pF ( 29 54 x 10 - 5 ( 29 - 3.3 + 0.75 = 1.41 ns t f 2245 3.6 R onN C = 3.6 C K n V GS - V TN ( 29 = 3.6 0.2 pF ( 29 2 10 - 4 ( 29 3.3 - 0.75 ( 29 = 1.41 ns t PLH = 1.2 R onP C = t r 3 = 0.470 ns t PHL 2245 1.2 R onN C = t f 3 = 0.470 ns t P = 0.470 + 0.470 2 = 0.470 ns
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Homework Solutions ECE302 Spring 2007 HW 8 2 7.33 ( a ) 5 W 2 x 10 6 gates = 2.5 m W / gate P = CV DD 2 f ; C = 2.5 x 10 -6 3.3 2 5 x 10 6 ( 29 = 45.9 fF ( b ) C = 2.5 x 10 -6 2.5 2 5 x 10 6 ( 29 = 80.0 fF 7.36 Peak current occurs for v O =v I . Assume both transistors are saturated since v O = v I . 20 1 100 x 10 - 6 2 v I - V TN ( 29 2 = 20 1 40 x 10 - 6 2 v I - V DD - V TP ( 29 2
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Unformatted text preview: 1.58 v I-V TN ( 29 = V DD-v I + V TP a ( 29 v O = v I = V DD + 1.58 V TN + V TP 2.58 = 3.3 + 1.58 0.6 ( 29 + -0.6 ( 29 2.58 = 1.414 V i D = 20 1 100 x 10-6 2 1.414-0.6 ( 29 2 = 663 m A Checking the current : i D = 20 1 40 x 10-6 2 1.414-3.3 + 0.6 ( 29 2 = 662 m A | Within roundoff error. b ( 29 v O = v I = 2.5 + 1.58 0.6 ( 29 + -0.6 ( 29 2.58 = 1.104 V | i D = 20 1 100 x 10-6 2 1.104-0.6 ( 29 2 = 254 m A Checking the current : i D = 20 1 40 x 10-6 2 1.104-2.5 + 0.6 ( 29 2 = 253 m A | Within roundoff error....
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This note was uploaded on 03/30/2008 for the course ECE 302 taught by Professor Barlage during the Spring '07 term at N.C. State.

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HW12_Solutions - 1.58 v I-V TN ( 29 = V DD-v I + V TP a (...

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