Hw2_sol-16- - hw2_sol.mcd Page 1 of 3 Reference:C\Users\Bernhard\Lib\MathCAD\Default\defaults.mcd 240 Homework 2 Problem 1 L:= 0.35m fu = 2 CL C s

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hw2_sol.mcd Page 1 of 3 C gd 13.2 fF = C gd 0.24 fF μ m W 1 = I D1 680 μ A = W 1 55 μ m = I D 513.761 μ A = I D ID_over_W W 1 = neglecting versus Cgs probably not too accurate need about 10% more W and ID C gs 63.535 fF = C gs 2 3. W 1 L C ox = W 1 51.376 μ m = W 1 root 1 π 1 Vstar ID_over_W 1 C L W 1 2 3 W L C ox C s + f u - W , = W 50 μ m = f u 100MHz = C s 100fF = f u 1 π 1 Vstar I D W 1 C L W 1 2 3 W L C oxx C s + = C L 5pF = Vstar 200mV = f u 1 2 π g m I D I D C L C s C s C GS + = ID_over_W 10 μ A μ m = neglect Cgd . .. check later if ok f u 1 2 π g m C L C s C s C GS + = C ox 5.3 fF μ m 2 = L 0.35 μ m = Problem 1 240 Homework 2 Reference:C:\Users\Bernhard\Lib\MathCAD\Default\defaults.mcd
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hw2_sol.mcd Page 2 of 3 Vo in dB $ AC simulation for HW2 problem 1. ******************** .param wn1=55u
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This note was uploaded on 08/01/2008 for the course EECS 240 taught by Professor Boser during the Spring '04 term at University of California, Berkeley.

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Hw2_sol-16- - hw2_sol.mcd Page 1 of 3 Reference:C\Users\Bernhard\Lib\MathCAD\Default\defaults.mcd 240 Homework 2 Problem 1 L:= 0.35m fu = 2 CL C s

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