t r,f =t r.fo +α p,n *C L Ru=[k’(W/L)(Vdd-Vt )]-1 C GU =Cox(WL)u

tr,f=tr.fo+αp,n*CL Ru=[k’(W/L)(Vdd-Vt)]-1 CGU=Cox(WL)u CDU=(CGD+CDB)u CSU=(CGS+CSB)u (W/L)m=m(W/L)u Rm=Ru /m C(G,D,S)m=mC(G,D,S)u RmCm=RuCu=常數

tr,f3=tr,f,o+(1/3)αp,nuCLtr,f---m大,tr,f變小

tr=[(N+1)/2]tro+(αpu/m)CL tf=(N+1)tfo+(Nαnu/m)CL Cin=mCmin tr,f因m大而變小

tr=(N+1)tro+(Nαpu/m)CLtf=[(N+1)/2]tfo+(αnu/m)CL Cin=mCmin tr,f因m大而變小

td=tnot|m=1+tNAND2|m=2+tNOR2|m=3 =tfo+αnu(2Cmin)+ (3/2)tro+αpu(3Cmin)+ (3/2)tfo+αnu(4Cmin)

Wn=Wmin Wp=rWmin r=Un/Up βn=βp Cin=(1+r)Cu=Cinv ts=to+(α/m)CL td=(A+Bn) τmin n=CL/Cmin τmin=Rmin Cmin ,

Time Delay • td=(A+Bn) τmin :單一inv • td,N=(x1)(N-1) (A+Bn) τmin N:扇入 • EX: x1=1.17 N=1----N=2 • tmd,N=(x1)(N-1) (A+(Bn/m)) τmin m:縮放比 • tmd,N=x2(x1)(N-1) (A+(Bn/m)) τmin x2:複雜N • EX8.1

8-2驅動大電容負載 (W/L)p=r*(W/L)n r=Un/Up=kn’/kp’>1 Vtn=|Vtp|=Vt Rn=Rp=R VM=0.5VDD τ=RCout=R(CFET+CL) ts=tr=tf Cin=CGn+CGp =Cox(AGn+AGp)= CoxL(Wn+Wp)=(1+r)(CoxLWn) =(1+r)CGn

CL1=Cin ts1=to+αCin :單位負載降低ts1降低αβ大 β’= Sβ, R’=R/S α’=α/S ts=to+(α/S)CL ts變小

CL大 CL=SCin Wn’=SWn Cin’=SCin

β1<β2<β3……………………..<βN-1<βN β2=Sβ1 β3=Sβ2 ……… βj+1=Sβj β3=Sβ2=S2 β1

βj=S(j-1)β1 Cj=S(j-1)C1 Rj=Rj/S(j-1)

τj=RjCj+1 τd=τ1+τ2+τ3+ …………+τN-1+τN (假設Cj+1>>CFET,j)=R1C2+R2C3+R3C4+……+RN-1CN+RNCL (CL=CN+1=SNC1) =R1SC1+(R1/S)S2C1+(R1/S2)S3C1+……. ………. +(R1/SN-2)SN-1C1 +(R1/SN-1)SNC1

τd • τd=SR1C1+SR1C1+SR1C1+………SR1C1+SR1C1 • =NSR1C1 (8.71)<----有bug • =NSτr • CL=SNC1 ln(SN)=ln(CL/C1)=NlnS • N= [ln(CL/C1)]/lnS • τd=τr [ln(CL/C1)][S/lnS] • …S=e EX8.2 Real Case S>e

多CF,j=SJ-1CF,1 ……… τx=R1CF,1 S(lnS-1)=τx/ τr EX8.3

8.3 邏輯效力 g=Cin/Cref gNOT=1 gNand=(2+r)/(1+r) gNor=(1+2r)/(1+r)

h=Cout/Cin (電性效力) d=h+p p:寄生電容延遲

D=d1+d2=(h1+p1)+(h2+p2) (反相器g=1) H=h1*h2 D=(h1+p1)+(H/h1+p2) P8-27;28 公式(8.122;123;124) h1=h2, DMin.

8.3.2 推廣 Cin=CGn(1+r) gNAND2=(2+r)/(1+r) ; gNOR2=(1+2r)/(1+r)

D=Σdi (8.134 p.8-30) G=Πgi=g1*g2*g3….. H=Πhi=h1*h2*h3…. =500fF =20fF F=GH=f1*f2*f3…..=fN ; F(1/N)=f=gh(每一級皆相等Dmin.) D=NF(1/N)+P (8.137-142)

範例 8.4 • G=gNOTgNOR2gnand2= 1*[(1+2r)/(1+r)][(2+r)/(1+r)] = 1(6/3.5)(4.5/3.5)=2.2 (r=2.5) • H=C4/C1=500/20=25 • F=GH=55 f=F(1/N)=55(1/3)=3.8 • D=3(3.8)+P=11.4+P • P=pNOT+pNOR2+pNAND2 • h3=f/g3=3.8/1.29=2.95=C4/C3; C3=500/2.95= 169.5fF=S3CGn(2+r)=S3(4.5CGn); S3:縮放因數 • h2=3.8/g2=3/8/1.71=2.22=C3/C2

C2=169.5/2.22=76.35fF=S2CGn(1+2r) =S2CGn(1+2r)=S2(6CGn) h1=3.8/1=C2/C1; C1=76.35/3.8=20fF CGn=20/3.5=5.71fF S2=76.35/(6*CGn)=2.23 (Bug?) (8.159) S3=169.5/(4.5*CGn)=6.60 (Bug?) (8.159) (另外假設Cref=8fF 即Cgn=8/3.5fF=2.2857fF) P.8-33 最下面一行 S2=1.59*3.5=5.565 (Bug?) S3=4.71*3.5=16.485 (Bug?)

範例8.5 • F=200 • N=3 NF(1/N)=3*(200)(1/3)=17.54 • N=4 NF(1/N)=4*(200)(1/4)=15.04 • N=5 NF(1/N)=5*(200)(1/5)=14.43 • N=10 NF(1/N)= 10*(200)(1/10)=16.99 • N=5 Optimum Value

8.3.5 分枝 B=Πbi b=CTotal/Cpath Example 8.6

8.4 BiCMOS 驅動器

Ic≒ISe(VBE/Vth)Vth=kT/q=26mV27℃=300 0K

Ic=αIE

XB<=電子游離半徑

Vdd,0 SAT OFF H L OFF SAT 0,Vdd Vdd 0 0 Vdd

VOH=VDD-VBE,SATVOL=VBE,SAT=0.8 VEX8.7

CL>Cx BiCMOS有優勢

Exercise • 習題…..1,2-9 • 題目8.3中第一行”1被切換至一個0”改成”0被切換至一個1” (解答有bug?) • td=tfo+αnu(Cmin)+3tro+2αpu(4Cmin) +3tfo+(2/2)αnu(3Cmin)+tro+(α/3)pu(10cmin)=4tfo+4tro+4αnu(Cmin)+(34/3) αpu(Cmin) [紅色解答有bug?] • 習題8.4第一行加Wn=2.2μm • 習題8.6第一行c’=0.86改成 8.6pF/cm • 習題8.9 解答有問題把b改成 • b=(2+r+3+r)/(2+r)=2.22

t r,f =t r.fo +α p,n *C L Ru=[k’(W/L)(Vdd-Vt )]-1 C GU =Cox(WL)u