超大规模集成电路2017秋段成华老师第四次作业

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1、1.Shownbelowarebuffer-chaindesigns.(1)Calculatetheminimumdelayofachainofinvertersfortheoveralleffectivefan-outof64/1.(2)UsingHSPICEandTSMC0.18umCMOStechnologymodelwith1.8Vpowersupply,designacircuitsimulationschemetoverifythemwiththeircorrespondentparametersofN,f,andtp.(1)γ=1F=64∴f=NF=3.6∴N=3.246所以最佳

2、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-10N=2时，tplh=2.2478E-10tphl=2.5567E-10tpd=2.4023E-10N=3时，tphl=2.0574E-10tplh=2.1781E-10tpd=2.1178E-10N=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

3、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:synopsysHspice_A-2007.09tsmc018mm018.l'TT.optioncaptab.optionlistnodepostmeasout.tran10p6000p*****************************************************

4、*******.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*.subcktnmos1n1n2n3mnn1n2n3Gndn

5、chl=0.2uw=0.4uad=0.2p^2pd=0.4uas=0.2p^2ps=0.4u.endsnmos1**pmos1*.subcktpmos1p1p2p3mpp1p2p3Vccpchl=0.2uw=0.8uad=0.4p^2pd=0.8uas=0.4p^2ps=0.8u.endspmos1*.subcktinv1inoutxmnoutinGndnmos1xmpoutinVccpmos1vccVccGndSupply.endsinv1**nmos2*.subcktnmos2n1n2n3mnn1n2n3Gndnchl=0.2uw=1.12uad=0.56p^2pd=1.12uas=0.5

6、6p^2ps=1.12u.endsnmos2**pmos2*.subcktpmos2p1p2p3mpp1p2p3Vccpchl=0.2uw=2.24uad=1.12p^2pd=2.24uas=1.12p^2ps=2.24u.endspmos2*.subcktinv2inoutxmnoutinGndnmos2xmpoutinVccpmos2vccVccGndSupply.endsinv2**nmos3*.subcktnmos3n1n2n3mnn1n2n3Gndnchl=0.2uw=3.2uad=1.6p^2pd=3.2uas=1.6p^2ps=3.2u.endsnmos3**pmos3*.sub

7、cktpmos3p1p2p3mpp1p2p3Vccpchl=0.2uw=6.4uad=3.2p^2pd=6.4uas=3.2p^2ps=6.4u.endspmos3*.subcktinv3inoutxmnoutinGndnmos3xmpoutinVccpmos3vccVccGndSupply.endsinv3**nmos4*.subcktnmos4n1n2n3mnn1n2n3Gndnchl=0.2