MOS Inverer = 0 = = = 0 r SDP r SDP r DSN r DSN
MOS Inverer r SDP = 0 = Q P : Triode small v SD Q N : uoff = r SDP r DSN
MOS Inverer r SDP = 0 = Q P : Triode small v SD Q N : uoff = r SDP
MOS Inverer r SDP = = 0 Q P : uoff Q N : Triode small v DS = r DSN r DSN
MOS Inverer = = 0 Q P : uoff Q N : Triode small v DS = r DSN r DSN
MOS Inverer = = 0 Q P : uoff Q N : Triode small v DS = r DSN = 0 = Q P : Triode small v SD Q N : uoff = r SDP
MOS Inverer
MOS Inverer
MOS Inverer
MOS Inverer
MOS Inverer = = 0 Q P : uoff Q N : Triode small v DS = r DSN = 0 = Q P : Triode small v SD Q N : uoff = r SDP
MOS Inverer = 0 = Q P : Triode small v SD Q N : uoff = r SDP = = 0 Q P : uoff Q N : Triode small v DS = r DSN
MOS Inverer = 0 = = = 0
(), () =? =0 ()=? =0 + ()=? =0 -
(), () =? =0 (0)= =0 + ()=? =0 -
(), () ()/ = () =0 () = A1 exp(-/( )) + A 2 (0)=, ( )=0 () = exp(-/( )) (0)= =0 + () =0 -
(), () =0 ()=, < 0 () = exp(-/( )) > 0 (0) =0 + () =0 -
(), () () = exp(-/( )) i r = / = ( / ) exp(-/( )) P PDN = 2 / P PDN = (2 / ) exp(-2/( )) (0) =0 + () =0 -
(), () =0 =0 - ()=0 (0)=? =0 +
(), () =0 () = 0, < 0 () = exp(-/( )) > 0 =0 - ()=0 () =0 +
(), () =0 () = 0, < 0 () = exp(-/( )) > 0 =0 - ()=0 () =0 +
(), () () = exp(-/( )) i r = ( )/ = ( / ) exp(-/( )) P PUN = (2 / ) exp(-2/( )) P SUPPLY = (2 / ) exp(-/( )) =0 - ()=0 () =0 +
(), () ()=? ()=?
(), () i (), i SUPPLY () P SUPPLY, P PUN,PDN
(), () P SUPPLY = (2 / ) exp(-/( )) P AVG = (1/T) T (2 / ) exp(-/( ) d P AVG = - (1/T) (2 / ) exp(-/( ) 0 o T P AVG = (1/T) (2 ) if << T P SUPPLY, P PUN,PDN P AVG = f 2
(), () 0.9 0.5 0.1 90 50 10 0.9 0.5 0.1 10 50 90 τ PHL = 50 on op plo τ PLH = 50 on boom plo τ THL = 10 90 on op plo τ TLH = 90 10 on boom plo
(), () =0 ( 50 ) = exp(- 50 /( )) = /2-50 /( ) = ln(1/2) 50 = ln(2) - 10 /( ) = ln(1/10) 10 = ln(10) - 90 /( ) = ln(9/10) 90 = ln(10/9)
(), () =0 τ PHL = 50 = ln(2) τ THL = 10 90 = ln(10) ln(10/9) τ THL = (ln(10) ln(10/9)) = ln(10 9/10) τ PHL = ln(2) τ THL = ln(9)
(), () 0.9 0.5 0.1 90 50 10 0.9 0.5 0.1 10 50 90 τ PHL,τ PLH =,PUN ln(2) τ THL,τ TLH =,PUN ln(9) This is for he simple model of a resisior and capacior.
(), () =0 5 4 v IN v O U T () = exp(-/( )) k n = 130 ma/v V n = 1.62 V = 5 V = 200 nf = r DS = 1/(.130*(5-1.62)) = 2.27 Ω = 454 nsec () = 5 exp(-/(454x10-9 )) v 3 2 1 0-5 0 5 10 ime (usec)
5 4 3 k n = 130 ma/v V n = 1.62 V = 5 V = 200 nf v 2 1 0-5 0 5 10 ime (usec)
(), () P AVG = f 2 τ PHL,τ PLH =,PUN ln(2) τ THL,τ TLH =,PUN ln(9) For his model he minimum clock period is proporional o τ THL or τ THL. f max ~ 1/τ T ~ 1/(,PUN ) r SD,DS = 1/(k (W/L)( -V ))
(), () P AVG = N f 2 f max ~ (2/) k (W/L) ( -V ) A = N W L Performance (f max ) vs. Power Dissipaion (P D ) vs. Area/Number (A,N) (assume always operaing a f max ) 1. Increase (Power vs f max ) f max f max goes up by -V Power goes up by ( -V ) 2 2. Increase W/L (f max vs A) f max goes up by W Power goes up by W Uses more area on he I 3. Find a beer process ha has Larger k P D Smaller, bu k =µ ox, so maybe higher mobiliy. Smaller V A
MOS Inverer
(), () =? =0 For his model, he oupu volage would be a 2 nd order differenial equaion. a. overdamped. b. underdamped (ringing) c. criically ()=? =0 + ()=? =0 -