ECE559_Fall2009_Exam2_Solution

V 5 2 d d v v 5 2 d d v u 1 x m v 3 n u v i m 1 m v n

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Unformatted text preview: 0.5)( −0.5) − L 3 2 2 n I D, M 1 + I D , M 3 = I D, M 2 . VDS, M3 = (VX – 2.5) = -0.5 V > VGS,M3 - Vtp = (-2.5 + 0.5) V = -2 V. So C Exam 2 Solution ECE 559 (Fall 2009), Purdue University = 0 D D V V n e s a i When Vin = VM- = 1 V, VX = VM- = 1 V. VGS,M1 = (1 – 2.5) V = -1.5 V < Vtp = -0.5 V. So M1 is ON. o i g e r n o i t a r u t a n s o n i i e g s r i 1 n M o i t a r t u a s n i s i 2 M VGS,M2 = (1 – 0) V = 1 V > Vtn = 0.5 V. So M2 is ON. n VDS, M1 = (VX – 2.5) = -1.5 V < VGS,M1 - Vtp = -1 V. So the transistor . since VDS, M2 = (VX – 0) = 1 V > VGS,M2 - Vtn = (1 – 0.5) V = 0.5 V. We will assume sharp transition of the output, Vout. Since we have started from Vout = 2.5 V, we will assume that Vout is still at 2.5 V. n o i g e r r a e n i l n i s i 4 M VGS,M4 = (2.5 – 0) V = 2.5 V > Vtn = 0.5 V. So M4 is ON. since VDS, M4 = (VX – 0) = 1 V < VGS,M4 - Vtn = (2.5 – 0.5) = 2 V. I D, M 1 = I D,M 2 + I D, M 4 ( −1.5 + 0.5) ⇒ 50*2* 2 2 (1 − 0.5) = 100*1* 2 2 2 (1) W + 100* * ( 2.5 − 0.5)(1) − 2 L 4 W ⇒ 0.5 = 0.125 + *1.5 L 4 W 0.375 0.75 ⇒ = = 1.5 3 L 4 0.75*300 ⇒ W4 = nm = 75 nm 3 W4 = 75 nm W4 has come out to be less than L = 300 nm. It would not be quite allowable by a technology library. So for proper operation of the Schmitt trigger circuit, we need to make the width of the transistor M1 higher so that the W4 conforms at least to the specification of transistor widths provided by a technology library. m u m i n i m Page 9 of 9 C...
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This note was uploaded on 02/19/2014 for the course ECE 559 taught by Professor Staff during the Fall '08 term at Purdue.

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