week_6_&_7 - 39 Integrated Source Follower VDD ....

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39 Integrated Source Follower A simpler version: out R . . ~ . DD V in v DD V . 1 R L R out v . . . M2 DD V S R . . g R ~ in v DD V . L R out v . . . + 1 R . . _ Common Gate Amplifier Non-integrated version
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40 out R ~ . s R in v 3 C . D R L R out v . . 2 C . . DD V 1 C C R in R 1 R DD V 2 R from you can find and check for ideal MOS saturation condition: D I DS V n t GS DS V V V > . Equivalent Circuit D C DD S G GS I R V R R R V V V + = = 2 1 2 D C S DD G t GS ox n D D C D DD DS I R V V R R R V V V L W C I R R V V n = + = = + = 2 1 2 2 ) ( 2 ) ( µ I 2 equations and 2 unknowns: V , . GS D I C R s s v g ds r gs m v g g v gs v + _ out v s v . D R L R . out R ~ s R in v in R .
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41 To calculate in R lso: test i C R s s v g ds r ) ( s m v g g v d v s v . L D R R || . . + _ ~ test v C test R v . C test test R v i + = 1 || || . C L D test L D test ds R R R v R R i v s d ds v v v = test s d L D C test test v v v R R R v i = = || ) ( A + = + + + + = + + + = + ds L D test C ds C L D C m s test C test test ds L D test ds C L D C test test m s C test test ds ds s m s s r R R i R r R R R R g g v R v i r R R i r R R R R v v g g R v i r v v g v g || 1 1 . || || || ) ( C ds C L D C m s ds L D ds test test in R r R R R R g g r i v R 1 . || + + + + = = R R r || +
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You can simplify the equation; assuming is large Æ ds r C L D ds s m ds R R R r g g r , || 1 >> + >> Similar to of source-follower o calculate α out R T out R : ds test s ds m s test ds s test s m s s test S C test s r v v r g g i r v v v g v g i R R i v = + + + + = = ) 1 ( ) ( ) || ( C in R 1 1 = s m R g g + + C s m in R g g R || 1 + = Disconnect, now whatever test i S C R s s v g ds r R || s m v g g v s v D R + _ ~ test v . test
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This note was uploaded on 02/19/2012 for the course ECE 455 taught by Professor Mohammadi during the Fall '10 term at Purdue University-West Lafayette.

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week_6_&_7 - 39 Integrated Source Follower VDD ....

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