Distributed RC InterconnectsGSI ClassECE6458Two types of devices: transistors and wires!silicon wafer surfaceM1M2M3M4Interconnections between transistors are stacked on top!!IBM microprocessor micrographTop view IBM Copper Processviax-y orthogonal pairInterconnect Devicesx-y orthogonal pairSide view of 4-metal level stackSide view (and zoom) of IBM Copper ProcessIBM microprocessor micrographLocalInterconnectsSemi-Global InterconnectsGlobal InterconnectsCLASS OUTLINEGOAL: Understand Interconnect Device and Circuit Models•Common Mathematical Techniques to Deriving Transient Models•Example: Simple Lumped RC Model•Example: Distributed RC Model (without transistor driver effects)•Complete single line model with real drivers (Reading)•Coupled Line SolutionsInterconnect Modeling: Lumped RC Modelelectrical modeldriverpowerAssume for the moment that driver transistors are perfect switches!+-Vdd uo(t)RCreceiverAssume for the moment that transistor receivers have negligible load capacitanceModels for Wire Resistance and CapacitanceWHρHεR LWH!!=r oWC LH!! !=•parallel plate approximation•parallel plate plus fringing field approximationModels for Wire Resistance and Capacitance1.15 2.8r oHWC LH H!" "" "# $% &= +' () *+ ,- .•resistance per unit length and capacitance per unit length defined1RrL WH!!= =1.15 2.8r oHC WcL H H!" "" "# $% &= = +' () *+ ,- .Lumped RC Model for Transient Expression+-Vdd uo(t)RC( ) ( ) ( )dd c cV u t V t dV tCR dt!=int( )cdV tI Cdt=Iint(t)Vc(t)-+Laplace Transforms•Single-Sided Laplace Transforms0( ) ( )stF s f t e dt!"=#( )( )( ) 001 1( )t a sa sF s e dt e ea s s a!" "! " "= = " =" " +#( )atf t e!=•Example:( ) ( )f t u t=•Example:( )( ) 001 1( )st sF s e dt e es s!" ! "= = " ="#•Integration in the time domain….0( )stdf t e dtdt!"# $% &' ()[ ]( ) ( ) ( ) '( ) '( ) ( )tD u t v t u t v t u t v t= +( ) ( ) ( ) '( ) '( ) ( )u t v t u t v t dt u t v t dt= +! !( ) '( ) ( ) ( ) '( ) ( )u t v t dt u t v t u t v t dt= !" "00 0( ) ( ) ( )( )st st stdf t e dt f t e f t s edt! !!" " "# $= " "% &' () )00( ) (0) ( )s stf e f e s f t e dt!" ! "# $= ! " +% &'( ) ( ) ( 0)df t sF s fdt! "Laplace Transforms( ) ( ) ( )dd c cV u t V t dV tCR R dt! =( )( )dd ccV V sCsV ssR R! =1( )1ddcVV sRCs sRC=! "+# $% &Lumped RC Model Example: Laplace Domain( )1( )A Bs s a s s a= ++ +•Partial-Fraction Expansion Technique( )1 A B s aA= + +( )0A B+ =1 aA=A B= !1Aa=1Ba= !( )1 1 1 1( )s s a a s s a! "= #$ %+ +& 'Inverse-Laplace Transforms1( )1ddcVV sRCs sRC=! "+# $% &Transient Expression using Lumped RC Model1 1( )1c ddV s VssRC! "# $= %# $# $+# $& '( ) 1 ( )tRCc dd oV t V e u t!" #= !$ %& 'Interconnect Modeling: Lumped RC Model( ) 1tRCc ddV t V e!" #= !$ %& '1tRCdd ddvV V e!" #= !$ %& 'v =• fraction of supply voltage (Vc/Vdd)1ln1vt RCv! "=# $%& '• time delay (50% and 90%)0.50.693t RC=0.92.3t RC=• using parallel plate model, shows the problem with global interconnects20.50.693 0.693r o r oLW LLtWH H H H! " " !" " !" "!# $ # $# $= =% & % &% &% & % &' (' ( ' (+-x=0ΔxrΔxcΔxx=LVdd uo(t)Interconnect Modeling: Distributed RC LineV(x,t) = voltage along the lineI(x,t) = current along the linerΔxcΔ xI(x,t)V(x+Δx,t)I(x+ Δx,t)V(x+Δx,t)•KCL( , ) ( , ) ( , )dI x t I x x t c x V x tdt! + " = "•KVL0( , ) ( , ) ( , ) 1lim ( , )xV x x t V x t V x tI x tr x x r! "+ ! # $= # =! $0( , ) ( , )lim ( , ) ( , )xI x t I x x tI x t c V x tx x t! "# + ! $ $= # =! $ $Distributed RC Line( , ) 1( , )V x tI x tx r!" =!( , ) ( , )I x t c V x tx t! !" =! !22( , ) ( , )V x t rc V x tx t! !=! !•Find solutions to this partial differential equation!!Distributed RC Line22( , ) ( , ) 0V x s rcsV x sx!" =!• Use Laplace Transform• General Solution1 1( , ) sinh( ) cosh( )V x s A x src B x src= +( )1 1( , ) 1( , ) cosh( ) sinh( )V x t srcI x s A x sr c B x srcx r r!= " = " +!Distributed RC Line• Boundary Conditions (Ideal Driver and Load)+-x=0ΔxrΔxcΔxx=LVdd uo(t)( 0, ) ( )ddV x t V u t= =( , ) 0I x L t= =•time domain•Laplace domain( 0, )ddVV x ss= =( , ) 0I x L s= =Distributed RC Line1 1( 0, ) sinh(0) cosh(0)ddVV x s A Bs= = + =• Use Boundary Conditions to solve for B11ddVBs=( )1 1( , ) cosh( ) sinh( ) 0srcI x L s A L src B L srcr= = ! + =1 1cosh( ) sinh( ) 0A L src B L src+ =1sinh( )cosh( )ddVL srcAsL src= !• Use Boundary Conditions to solve for A1Distributed RC Linesinh( )( , ) sinh( ) cosh( )cosh( )dd ddV VL srcV x s x src x srcs sL src= ! +( )2sinh( )( , ) cosh( )cosh( )dd ddL srcV VV x L s L srcs sL src= = ! +( ) ( )2 2cosh( ) sinh( )( , )cosh( ) cosh( )ddL src L srcVV x L ssL src L src! "# $= = %# $# $& '( ) ( )2 2cosh( ) sinh( ) 1L src L src! =1( , )cosh( )ddVV x L ssL src! "= =# $% &• Transfer function for any position x• Transfer function at end of lineLaplace Transform for Distributed RC Line• Assuming that T(s) can be approximated by a power series expansion• This can be transformed into a time domain expansion• To find the K coefficients, multiply both sides of (28) by sT(s)• Setting s=0 gives• Where δk is a complex root of T(s) is a solution to ( ) 0ks T s!= =Time Domain Expansion Technique• Using L’hopital’s rule gives• Solving for Kk gives:• To solve for the kth coefficient let s approach δkTime Domain Expansion Technique1( , )cosh( )ddVV x L ssL src! "= =# $% & is a solution to ( ) 0ks T s!= =( ) cosh( ) 0T s L src= =( )2 1 ; 0,1, 2,3...2kL rc j m m!"= + =( )212 1 ; 1; 0,1, 2,32km k m mrc L!"# $= % + = + =& '( )( )212 1 ; 1, 2,32kk krc L!"# $= % % =& '( )Time Domain Expansion for Distributed RC Line• Solving for δk gives:1( , )cosh( )ddVV x L ssL src! "= =# $% &1 11(0) cosh(0)oKT= = =1( ) cosh( …