超大规模集成电路秋段成华老师第四次作业.docx
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超大规模集成电路秋段成华老师第四次作业
超大规模集成电路2017年秋段成华老师第四次作业
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1.Shownbelowarebuffer-chaindesigns.
(1)Calculatetheminimumdelayofachainofinvertersfortheoveralleffectivefan-outof64/1.
(2)UsingHSPICEandTSMC0.18umCMOStechnologymodelwith1.8Vpowersupply,designacircuitsimulationschemetoverifythemwiththeircorrespondentparametersofN,f,andtp.
(1)
所以最佳反相器数目约为3
通过仿真可以得到tphl=1.3568E-11tplh=1.7498E-11tp0=1.5533E-11
(2)N=1时,tphl=5.2735E-10tplh=8.1605E-10tpd=6.7170E-10
N=2时,tplh=2.2478E-10tphl=2.5567E-10tpd=2.4023E-10
N=3时,tphl=2.0574E-10tplh=2.1781E-10tpd=2.1178E-10
N=4时,tplh=2.1579E-10tphl=2.2189E-10tpd=2.1884E-10
从仿真结果可以看出N=3或者N=4时延迟时间最优,且N=2、3、4得到的仿真延迟时间与理论推导的时间比较接近,比例基本上是18、15、15.3,而N=1时仿真得到的延迟时间远小于理论推导的时间,但是最优结果依旧是N=3,f=4,tp=15。
*SPICEINPUTFILE:
Bsim3demo1.sp--achainofinverters
.paramSupply=1.8
.lib'C:
\synopsys\Hspice_A-2007.09\tsmc018\mm018.l'TT
.optioncaptab
.optionlistnodepostmeasout
.tran10p6000p
************************************************************
.paramtdval=10p
.meastrantplhtrigv(in)val=0.9td=tdvalrise=2
+targv(out)val=0.9rise=2
.meastrantphltrigv(in)val=0.9td=tdvalfall=2
+targv(out)val=0.9fall=2
.meastpdparam='(tphl+tplh)/2'
*macrodefinitions
************************************************************
*
*nmos1
*
.subcktnmos1n1n2n3
mnn1n2n3Gndnchl=0.2uw=0.4uad=0.2p^2pd=0.4uas=0.2p^2ps=0.4u
.endsnmos1
*
*pmos1
*
.subcktpmos1p1p2p3
mpp1p2p3Vccpchl=0.2uw=0.8uad=0.4p^2pd=0.8uas=0.4p^2ps=0.8u
.endspmos1
*
.subcktinv1inout
xmnoutinGndnmos1
xmpoutinVccpmos1
vccVccGndSupply
.endsinv1
*
*nmos2
*
.subcktnmos2n1n2n3
mnn1n2n3Gndnchl=0.2uw=1.12uad=0.56p^2pd=1.12uas=0.56p^2ps=1.12u
.endsnmos2
*
*pmos2
*
.subcktpmos2p1p2p3
mpp1p2p3Vccpchl=0.2uw=2.24uad=1.12p^2pd=2.24uas=1.12p^2ps=2.24u
.endspmos2
*
.subcktinv2inout
xmnoutinGndnmos2
xmpoutinVccpmos2
vccVccGndSupply
.endsinv2
*
*nmos3
*
.subcktnmos3n1n2n3
mnn1n2n3Gndnchl=0.2uw=3.2uad=1.6p^2pd=3.2uas=1.6p^2ps=3.2u
.endsnmos3
*
*pmos3
*
.subcktpmos3p1p2p3
mpp1p2p3Vccpchl=0.2uw=6.4uad=3.2p^2pd=6.4uas=3.2p^2ps=6.4u
.endspmos3
*
.subcktinv3inout
xmnoutinGndnmos3
xmpoutinVccpmos3
vccVccGndSupply
.endsinv3
*
*nmos4
*
.subcktnmos4n1n2n3
mnn1n2n3Gndnchl=0.2uw=9.04uad=4.52p^2pd=9.04uas=4.52p^2ps=9.04u
.endsnmos4
*
*pmos4
*
.subcktpmos4p1p2p3
mpp1p2p3Vccpchl=0.2uw=18.08uad=9.04p^2pd=18.08uas=9.04p^2ps=18.08u
.endspmos4
*
.subcktinv4inout
xmnoutinGndnmos4
xmpoutinVccpmos4
vccVccGndSupply
.endsinv4
*maincircuitnetlist
xinv1inout1inv1
xinv2out1out2inv2
xinv3out2out3inv3
xinv4out3outinv4
cloutGnd154.24f
VininGnd0.9pulse(0.01.8219p40p40p1100p2400p)
.printtranv(in)v(out)
.end
2.Considerthelogicnetworkbelow,whichmayrepresentthecriticalpathofamorecomplexlogicblock.Theoutputofthe。
networkisloadedwithacapacitancewhichis5timeslargerthantheinputcapacitanceofthefirstgate,whichisaminimum-sizedinverter.TheeffectivefanoutofthepathhenceequalsF=CL/Cg1=5.
UsingHSPICEandTSMC0.18umCMOStechnologymodelwith1.8Vpowersupply,designacircuitsimulationschemetoverify
(1)theOPTIMAZATIOMparametersofg,f,andsforeachoftheinverterandgatesand
Thepathlogicaleffort
Thetotalpatheffort
Theoptimalgateeffort
在所有的nmos和pmos均采用最小尺寸晶体管的情况下
tplh=1.9047E-10tphl=2.2742E-10tpd=2.0895E-10
在所有所有的mos管尺寸均是前一个mos管尺寸的2倍的情况下
tplh=1.8353E-10tphl=2.4356E-10tpd=2.1355E-10
在参数最优的情况下tplh=1.7151E-10tphl=2.2853E-10tpd=2.0002E-10
所以最优参数为上面的推到过程。
*SPICEINPUTFILE:
Bsim3demo1.sp--achainofinverters
.paramSupply=1.8
.lib'C:
\synopsys\Hspice_A-2007.09\tsmc018\mm018.l'TT
.optioncaptab
.optionlistnodepostmeasout
.tran10p6000p
************************************************************
.paramtdval=10p
.meastrantplhtrigv(in)val=0.9td=tdvalrise=2
+targv(out)val=0.9rise=2
.meastrantphltrigv(in)val=0.9td=tdvalfall=2
+targv(out)val=0.9fall=2
.meastpdparam='(tphl+tplh)/2'
*macrodefinitions
************************************************************
*
*nmos1
*
.subcktnmos1n1n2n3
mnn1n2n3Gndnchl=0.2uw=0.4uad=0.2p^2pd=0.4uas=0.2p^2ps=0.4u
.endsnmos1
*
*pmos1
*
.subcktpmos1p1p2p3
mpp1p2p3Vccpchl=0.2uw=0.8uad=0.4p^2pd=0.8uas=0.4p^2ps=0.8u
.endspmos1
*
.subcktinv1inout
xmnoutinGndnmos1
xmpoutinVccpmos1
vccVccGndSupply
.endsinv1
*
*nmos2
*
.subcktnmos2n1n2n3n4
mnn1n2n3n4nchl=0.2uw=0.772uad=0.386p^2pd=0.772uas=0.386p^2ps=0.772u
.endsnmos2
*
*pmos2
*
.subcktpmos2p1p2p3p4
mpp1p2p3p4