1x10 7 fcm2 2 si0 kt n a 103 nm 2 ln q qn a ni cd si

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Unformatted text preview: 0 kT N a =103 nm 2 ln q qN a ni '' Cd si tox =0.146 '' Cox ox xd ∆ V (1 ) kT 0.0292 V q Q' kT V V ∆ V kT W '' qvDS qvDS kT kT iDsub I n 1 e Cox ∆ V exp GS T n 1 e L q q ∆V L qvDS kT =2.58x10-9 1 e Assume vDS >> kT/q, iDsub 2.58 109 A , nearly independent of vDS C Vinitial ∆ Q iDsub t 2 Then, the time required to drop voltage to 50% is t CVinitial =1.94 10-4Vinitial s 2iDsub 3. J&B P7.55 1⁄ , is proportion a3nal to is proportional to , so when capacitance is scaling ∗ with factor of , ∝ 4. J&B P7.83 In order to be twice faster compared with the reference inverter with a load of capacitance of 2C, , . The worst path contains 3 NMOS _ and 2 PMOS, hence 5. J&B P8.6 _ , Because the bitlines a...
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This note was uploaded on 06/28/2013 for the course EE 331 taught by Professor Taicheng during the Fall '08 term at University of Washington.

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