454Fall07HW4_Solution

454Fall07HW4_Solution - ECEN 454 Digital Integrated Circuit...

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ECEN 454 Digital Integrated Circuit Design Fall 2007 Homework 4 Solutions 1.
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2. Solution:
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V IH : load in saturation, driver in linear. ) ) ( 2 ( 2 ) ( 2 2 0 2 0 out out T in driver T out DD load V V V V k V V V k = (2) Take derivative with respect to V in on both sides and use 1 = IH V in out dV dV We have + = IH IH IH V in out out V in out in out V in out out dV dV V dV dV V V dV dV V 2 ) 8 . 0 ( 2 2 8 ) 2 . 4 ( 2 out in out out V V V V 8 ) 8 . 0 ( 8 8 ) 2 . 4 ( + = ) ( 129 . 0 471 . 0 V V V IH out = when in IH V V = . Plug back to (2), 0 06 . 15 286 . 3 539 . 5 2 = IH IH V V ) ( 06 . 2 V V IH = V V V V V NM IH OH H 14 . 2 06 . 2 2 . 4 = = = (b) When input logic is “1”, V V V OH in 2 . 4 = = , it’s the same condition when we calculated V OL : A V k I I OL load load driver μ 353 ) 2 . 4 ( 2 2 = = = When input logic is “0”, the driver transistor is in cut-off therefore the current is zero. So the average power consumption is
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This note was uploaded on 04/17/2008 for the course ELEN 454 taught by Professor Jianghu during the Spring '08 term at Texas A&M.

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454Fall07HW4_Solution - ECEN 454 Digital Integrated Circuit...

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