集成電路設(shè)計-05-反相器動靜態(tài)特性

1、EIS-WUHAN UNIVERSITY1集成電路設(shè)計第五章第五章 CMOS反相器反相器EIS-WUHAN UNIVERSITY2Outlinen電路特性電路特性n反相器反相器nCMOS反相器電壓傳輸特性反相器電壓傳輸特性n噪聲容限噪聲容限n傳輸延遲傳輸延遲n驅(qū)動大電容負(fù)載驅(qū)動大電容負(fù)載q功耗及低功耗設(shè)計功耗及低功耗設(shè)計EIS-WUHAN UNIVERSITY35-1 特性n成本q復(fù)雜性和面積n完整性和穩(wěn)定性q靜態(tài)（穩(wěn)態(tài)）特性n性能q動態(tài)（瞬態(tài)）特性n能量效率q能耗和功率EIS-WUHAN UNIVERSITY45-2 反相器（VinVoutCLVDDCMOS InverterPolysili

2、conInOutVDDGNDPMOS2l lMetal 1NMOSContactsN WellEIS-WUHAN UNIVERSITY5Two InvertersConnect in MetalShare power and groundVDDEIS-WUHAN UNIVERSITY6CMOS 反相器基本特點n輸出輸出q電源和電源和GNDGNDq噪聲容限大噪聲容限大n邏輯電平與尺寸無關(guān)，可以采用最小尺寸邏輯電平與尺寸無關(guān)，可以采用最小尺寸n穩(wěn)態(tài)輸出時，穩(wěn)態(tài)輸出時，VDDVDD或或GNDGND與輸出之間總存在有限電阻的通路與輸出之間總存在有限電阻的通路q低輸出阻抗低輸出阻抗q對噪聲和干擾不敏感對

3、噪聲和干擾不敏感n極高的輸入阻抗（極高的輸入阻抗（input resistanceinput resistance）n穩(wěn)態(tài)下穩(wěn)態(tài)下 V Vdddd 和和 GND GND 間無直流通路間無直流通路q無靜態(tài)功耗無靜態(tài)功耗n傳輸延遲（傳輸延遲（Propagation delayPropagation delay）q是負(fù)載電容和晶體管電阻的函數(shù)。是負(fù)載電容和晶體管電阻的函數(shù)。EIS-WUHAN UNIVERSITY7CMOS InverterFirst-Order DC AnalysisVOL = 0VOH = VDDVM = f(Rn, Rp)VDDVDDVin=VDDVin= 0VoutVoutR

4、nRpEIS-WUHAN UNIVERSITY8CMOS Inverter: Transient Response tpHL = f(Ron.CL)= 0.69 RonCLVoutVoutRnRpVDDVDDVin=VDDVin=0(a) Low-to-high(b) High-to-lowCLCLEIS-WUHAN UNIVERSITY95-3 Voltage Transfer CharacteristicnNMOS+PMOSn圖解法EIS-WUHAN UNIVERSITY10I-V NMOSID (A)VDS (V)X 10-4VGS = 1.0VVGS = 1.5VVGS = 2.0VV

5、GS = 2.5VLinear dependenceNMOS transistor, 0.25um, Ld = 0.25um, W/L = 1.5, VDD = 2.5V, VT = 0.4VEIS-WUHAN UNIVERSITY11I-V Plot (PMOS)ID (A)VDS (V)X 10-4VGS = -1.0VVGS = -1.5VVGS = -2.0VVGS = -2.5VPMOS transistor, 0.25um, Ld = 0.25um, W/L = 1.5, VDD = 2.5V, VT = -0.4VAll polarities of all voltages an

6、d currents are reversedEIS-WUHAN UNIVERSITY12PMOS Load LinesVoutIDnVin = VDD+VGSpIDn = - IDpVout = VDD+VDSpVinVoutCLVDDEIS-WUHAN UNIVERSITY13PMOS Load LinesVDSpIDpVGSp=-2.5VGSp=-1VDSpIDnVin=0Vin=1.5VoutIDnVin=0Vin=1.5Vin = VDD+VGSpIDn = - IDpVout = VDD+VDSpEIS-WUHAN UNIVERSITY14CMOS Inverter Load Ch

7、aracteristics IDnVoutVin = 2.5Vin = 2Vin = 1.5Vin = 0Vin = 0.5Vin = 1NMOSVin = 0Vin = 0.5Vin = 1Vin = 1.5Vin = 2Vin = 2.5Vin = 1Vin = 1.5PMOSEIS-WUHAN UNIVERSITY15CMOS Inverter VTC VoutVin0.511.522.50.511.522.5NMOS resPMOS offNMOS satPMOS satNMOS offPMOS resNMOS satPMOS resNMOS resPMOS satEIS-WUHAN

8、UNIVERSITY16噪聲容限logic 1logic 0unknownVDDVSSVHVLn反映了對噪聲的敏反映了對噪聲的敏感程度；感程度；n電路電路0 0，1 1電平允許電平允許的輸入范圍；的輸入范圍；n越大越好；越大越好；q高電平噪聲容限高電平噪聲容限q低電平噪聲容限低電平噪聲容限EIS-WUHAN UNIVERSITY17Logic level matchingnLevels at output of one gate must be sufficient to drive next gate.EIS-WUHAN UNIVERSITY18Transfer characteristic

9、snTransfer curve shows static input/output relationshiphold input voltage, measure output voltage.EIS-WUHAN UNIVERSITY19反相器噪聲容限的三種求法n求法求法1q最低輸出高電平、最高最低輸出高電平、最高輸出低電平；輸出低電平；q找到對應(yīng)的輸入；找到對應(yīng)的輸入；q求差；求差；VNL=Voff VilVNH=Vih VonVolVol,maxVoh,minVohVonVihVoffVilEIS-WUHAN UNIVERSITY20n求法求法2q單位增益點（斜率為單位增益點（斜率為1，

10、-1）；）；q找到對應(yīng)的輸入；找到對應(yīng)的輸入；q求差；求差；VNL=Voff VilVNH=Vih VonVolVol,maxVoh,minVohVonVihVoffVilEIS-WUHAN UNIVERSITY21n求法求法3q工作中心點；工作中心點；nVin = VoutnVgs = Vdsq找到對應(yīng)的輸入；找到對應(yīng)的輸入；q求差；求差；EIS-WUHAN UNIVERSITY22Noise Margins Determining VIH and VILVinVoutVOH = VDDVMBy definition, VIH and VIL are where dVout/dVin = -

11、1 (= gain)VOL = GNDA piece-wise linear approximation of VTC NMH = VDD - VIH NML = VIL - GNDApproximating: VIH = VM - VM /g VIL = VM + (VDD - VM )/gSo high gain in the transition region is very desirableEIS-WUHAN UNIVERSITY23CMOS Inverter VTC from SimulationVin (V)Vout (V)0.25um, (W/L)p/(W/L)n = 3.4(

12、W/L)n = 1.5 (min size)VDD = 2.5VVM 1.25V, g = -27.5VIL = 1.2V, VIH = 1.3VNML = NMH = 1.2(actual values are VIL = 1.03V, VIH = 1.45VNML = 1.03V & NMH = 1.05V)Output resistance low-output = 2.4khigh-output = 3.3kEIS-WUHAN UNIVERSITY24VM與PMOS及NMOS的寬長比(W/L)p/(W/L)nVM (V)qIncreasing the width of the

13、PMOS moves VM towards VDDq Increasing the width of the NMOS moves VM toward GNDq決定因素：寬長比寬長比q近似為等效電阻之比。近似為等效電阻之比。.1工藝因子：工藝因子： k = Cox導(dǎo)電因子：導(dǎo)電因子： n = k（W/L）3.4Rn1/n (Vgs Vt)EIS-WUHAN UNIVERSITY25Gain DeterminatesVingainGain is a strong function of the slopes of the currents in the saturation region, fo

14、r Vin = VM (1+r)g - (VM-VTn-VDSATn/2)(ln - lp )Determined by technology parameters, especially channel length modulation (l). Only designer influence through supply voltage and VM (transistor sizing).EIS-WUHAN UNIVERSITY26Gain as a function of VDD00.000.0Vin (V)Vout (V)00.511.5

15、22.500.511.522.5Vin (V)Vout(V)Gain=-1n100mv時，時，VTC變差；變差；n過渡區(qū)增益接近過渡區(qū)增益接近-1n一般，為達(dá)到足夠的增益，電源應(yīng)大于熱電勢的兩倍一般，為達(dá)到足夠的增益，電源應(yīng)大于熱電勢的兩倍qVDDmin 2, 4 KT/qqKT/q室溫下約為室溫下約為26mvEIS-WUHAN UNIVERSITY27Simulated VTC 00.511.522.500.511.522.5Vin (V)Vout(V)EIS-WUHAN UNIVERSITY28Impact of Process Variations00.511.522.500.511.5

16、22.5Vin (V)Vout(V)Good PMOSBad NMOSGood NMOSBad PMOSNominalEIS-WUHAN UNIVERSITY295-4 傳輸延遲（Propagation Delay）EIS-WUHAN UNIVERSITY30DelaynAssume ideal input (step), RC load.EIS-WUHAN UNIVERSITY31tpHL = f(Ron.CL)= 0.69 RonCLVoutVoutRnRpVDDVDD(a) Low-to-high(b) High-to-lowCLCLn上升時間（rise time）, pullup on

17、;n下降時間（fall time）, pullup off.EIS-WUHAN UNIVERSITY32Current through transistornTransistor starts in saturation region, then moves to linear region.nVout增大增大q充電電流減小。充電電流減小。q Vds 減小。減小。EIS-WUHAN UNIVERSITY33Resistive approximation可使用積分求解可使用積分求解等效電阻平均值等效電阻平均值VGS VTRonSDEIS-WUHAN UNIVERSITY34Req求求VDD/2，

18、VDD區(qū)間的電阻平均值區(qū)間的電阻平均值EIS-WUHAN UNIVERSITY35Gate delaynDelay: 傳輸延遲nVDD 50% VDD n50% VDD VDDnTransition time: 轉(zhuǎn)換時間qtime required for gates output to reach 10% (logic 0) or 90% (logic 1) of final value.n10% 90%n90% 10%EIS-WUHAN UNIVERSITY36Inverter delay circuitnLoad is resistor + capacitor, driver is re

19、sistor.EIS-WUHAN UNIVERSITY37Inverter delay with t modelnt model: qgate delay based on RC time constant t.nVout(t) = VDD exp-t/(Rn+RL)CLn90% (logic 1) 10% (logic 0) qtf = 2.2 R CLn100% (logic 1) 50%qtD= 0.69 R CLnFor pullup time, use pullup resistance.EIS-WUHAN UNIVERSITY38t model inverter delay n0.

20、5 micron process: qRn = 3.9 kqCL= 0.68 fFn延遲時間qtd = 0.69 x 3.9 x 0.68E-15 = 1.8 ps.n上升延遲qtf = 2.2 x 3.9 x 0.68E-15 = 5.8 ps.EIS-WUHAN UNIVERSITY39Quality of RC approximationEIS-WUHAN UNIVERSITY40VDDVoutVin = VDDRonCLtpHL = f(Ron.CL)= 0.69 RonCLtVoutVDDRonCL10.5ln(0.5)0.36EIS-WUHAN UNIVERSITY4100.511

21、.522.5x 10-10-0.500.511.522.53t (sec)Vout(V)傳播延遲50%平均延遲時間tp = 0.69 CL (Reqn+Reqp)/2tpLHtpHLVOUT = 0.5VDD時時EIS-WUHAN UNIVERSITY42nRn1/n1/n n ( (Vgs Vt)qRn1/n1/n nn導(dǎo)電因子qn n = k（W/L）qk = n Coxn CoxqCgCg = CoxCox *(W*L）EIS-WUHAN UNIVERSITY43Delay as a function of VDD0.81.822.22.411.522.533.544

22、.555.5VDD(V)tp(normalized)EIS-WUHAN UNIVERSITY44n等效電阻與等效電阻與W/LW/L成反比；成反比；n當(dāng)當(dāng)V VDDDDVt+VVt+VDDDD/2/2時，等效電阻與電源無關(guān)；時，等效電阻與電源無關(guān)；n當(dāng)當(dāng)V VDDDD=Vt zero delayCLtp = k RWCLRWRWWunit = 1k is a constant, equal to 0.69EIS-WUHAN UNIVERSITY51輸出端電容構(gòu)成輸出端電容構(gòu)成nCout = CFET + CLntf = 2.2 Rn ( CFET + CL )ntr = 2.2 Rp ( CFE

23、T + CL )nCFET 由幾何圖形決定由幾何圖形決定EIS-WUHAN UNIVERSITY52Inverter with LoadLoadDelayCintCLDelay = kRW(Cint + CL) = kRWCint + kRWCL = kRW Cint(1+ CL /Cint)= Delay (Internal) + Delay (Load)CN = CunitCP = 2Cunit2WWEIS-WUHAN UNIVERSITY53Delay Formula/1/10intftCCCkRtCCRDelaypintLWpLintWCint = Cgin with 1f = CL/

24、Cgin - effective fanoutR = Runit/W ; Cint =WCunittp0 = 0.69RunitCunitEIS-WUHAN UNIVERSITY54Apply to Inverter ChainCLInOut12Ntp = tp1 + tp2 + + tpNjginjginunitunitpjCCCRt,1,1LNginNijginjginpNjjppCCCCttt1,1,1,01, ,1EIS-WUHAN UNIVERSITY55Optimal Tapering for Given NDelay equation has N - 1 unknowns, Cg

25、in,2 Cgin,NMinimize the delay, find N - 1 partial derivativesResult: Cgin,j+1/Cgin,j = Cgin,j/Cgin,j-1Size of each stage is the geometric mean of two neighbors- each stage has the same effective fanout (Cout/Cin)- each stage has the same delay1,1,jginjginjginCCCEIS-WUHAN UNIVERSITY56延遲時間及級數(shù)優(yōu)化1 ,/gin

26、LNCCFfWhen each stage is sized by f and has same eff. fanout f:NFf /10NppFNttMinimum path delayEffective fanout of each stage:EIS-WUHAN UNIVERSITY57ExampleCL= 8 C1InOutC1283fCL/C1 has to be evenly distributed across N = 3 stages:EIS-WUHAN UNIVERSITY58級數(shù)優(yōu)化For a given load, CL and given input capacita

27、nce CinFind optimal sizing ffffFtFNttpNpplnlnln1/0/100ln1lnln20fffFtftppfFNCfCFCinNinLlnln with EIS-WUHAN UNIVERSITY59級數(shù)的近似收斂解收斂解：For = 0, f = e, N = lnFff1expCint = Cgin此時，忽略自載。此時，忽略自載。f = e =2.71828,N = lnFEIS-WUHAN UNIVERSITY60Optimum Effective Fanout fOptimum f for given process defined by ff1ex

28、pfopt = 3.6for =1EIS-WUHAN UNIVERSITY61Buffer Design111186464646442.881622.6Nftp164652818341542.815.3EIS-WUHAN UNIVERSITY625-6 功耗（Power Dissipation） Lead microprocessors power continues to increaseP6Pentium 486386286808680858080800840040.1110100197119741978198519922000YearPower (Watts)EIS-WUHAN UNIV

29、ERSITY63Chip Power Density40048008808080858086286386486PentiumP611010010001000019701980199020002010YearPower Density (W/cm2)Hot PlateNuclearReactorRocketNozzleSunsSurfacechips might become hotSource: Borkar, De Intel EIS-WUHAN UNIVERSITY64Chip Power Density DistributionnPower density is not uniforml

30、y distributed across the chipnSilicon is not a good heat conductornMax junction temperature is determined by hot-spotsqImpact on packaging, coolingPower MapOn-Die TemperatureEIS-WUHAN UNIVERSITY65Power DissipationSource: Borkar, De Intel n來源：來源：q動態(tài)功耗（動態(tài)功耗（Dynamic Power Consumption）nCharging and Disc

31、harging Capacitorsq短路電流（短路電流（ Short Circuit Currents ）nCircuit Path between Supply Rails during Switchingq漏電流（漏電流（Leakage）nLeaking diodes and transistorsEIS-WUHAN UNIVERSITY66Power consumption circuitnInput is square wave.EIS-WUHAN UNIVERSITY67驅(qū)動電路ni(t)=dQ/dt ，i=c*dV/dtq電壓不能突變，柵電壓的變電壓不能突變，柵電壓的變化有延遲時

32、間。化有延遲時間。nQ=CVqC大，意味著延遲時間加長大，意味著延遲時間加長n影響影響C的因素？的因素？P=V（C*dV/dt）=d （ 0.5CV2 ） /dtE= 0.5C V2輸入從0到VDD時，E= 0.5C VDD2每次開關(guān)消耗能量。EIS-WUHAN UNIVERSITY68動態(tài)功耗動態(tài)功耗VinVoutCLVddnA single cycle qE = CL(VDD - VSS)2 .nClock frequency f = 1/t.nEnergyq E = CL(VDD - VSS)2.nPower q E * f = f CL(VDD - VSS)2.n影響因素q f qCL

33、qVDD 其中負(fù)載消耗其中負(fù)載消耗1/2。EIS-WUHAN UNIVERSITY69Energy/transition = CL * VDD2 * P01Pdyn = （Energy/transition） * f = CL * VDD2 * P01 * fPdyn = CEFF * VDD2 * f where CEFF = P01 CL f01Data dependent - a function of switching activity!EIS-WUHAN UNIVERSITY70nConsider a 0.25 micron chip, 500 MHz clock, average

34、 load cap of 15fF/gate (fanout of 4), 2.5V supply. qDynamic Power consumption per gate is ?q46.875uw ? nWith 1 million gates (assuming each transitions every clock) qDynamic Power of entire chip = ?.n46.875w ?EIS-WUHAN UNIVERSITY71Lowering Dynamic PowerPdyn = CL VDD2 P01 fCapacitance:Function of fan

35、-out, wire length, transistor sizesSupply Voltage:Has been dropping with successive generationsClock frequency:IncreasingActivity factor:How often, on average, do wires switch?EIS-WUHAN UNIVERSITY72Speed-power productnPower-delay product (PDP ）nSP = P/f = CV2EIS-WUHAN UNIVERSITY73短路電流（短路電流（Short Cir

36、cuit Current）IV DD (mA)5Vin (V)5.04.03.02.01.00.0VinVoutCLVddEIS-WUHAN UNIVERSITY74nDuration and slope of the input signal, tscnIpeak determined by qthe saturation current of the P and N transistors which depend on their sizes, process technology, temperature, etc.qstrong function of the

37、ratio between input and output slopesna function of CLEsc/transition = tsc VDD Ipeak P01Psc = tsc VDD Ipeak f01EIS-WUHAN UNIVERSITY75Ipeak as a Function of CLIpeak (A)time (sec)x 10-10 x 10-4CL = 20 fFCL = 100 fFCL = 500 fFEIS-WUHAN UNIVERSITY76Impact of CL on PscVinVoutCLIsc 0VinVoutCLIsc ImaxLarge capacitive loadSmall capacitive loadEIS-WUHAN UNIVERSITY77Psc as a Function of Rise/Fall TimesP normalizedtsin / tsoutVDD= 3.3 VVDD = 2.5 VVDD = 1.5VWhen load capacitance is small (tsin/tsout 2 for VDD 2V) the power is dominated by PscW/Lp = 1.125 mm/0.25 mmW/Ln = 0.375 mm/0.25 mmCL