超大規(guī)模集成電路第四次作業(yè)2016秋-段成華

1、1. Shown below are buffer-chain designs.（1） Calculate the minimum delay of a chain of inverters for the overall effective fan-out of 64/1.Solution:由題可知：F 64根據(jù)經(jīng)驗 1 3.6為最合適的值，所以f N下 N64 3.6, 所以N 3.24,但是級數(shù)必須 為整數(shù)所以取 N 3 ,又因為 1 ,所以: f <64 4,tp/ 3 (1 <64) 15,所以 f 4時最合適。tp0 Using HSPICE and TSMC 0.18

2、 um CMOS technology model with 1.8 V power supply, design a circuit simulation scheme to verify them with their correspondent parameters of N, f, andptSolution:根據(jù)（1）中計算知道三級最合適，所以驗證如下：A）、一級無負(fù)載測本征延時代碼如下：.title buffer-chain 1.lib 'C:synopsysHspice_D-2010.03-SP1tsmc018mm018.l' TT * set 0.18um li

3、brary.opt scale=0.1u * set lambda.options post=2 list.temp 27 .global vddVdd vdd gnd 1.8vin vin 0 0.9 pulse 0 1.8 25n 5p 5p 49.99n 100n 頻率為$10MhzCl vout gnd 0f $Cg1=2.46fF,負(fù)載為 CL=157.44fF.subckt inv in out wn=3.5 wp=10 t=7.5mn out in gnd gnd NCH l=2 w=wn ad='wn*t' pd='wn+2*t' as='

4、;wn*t' ps='wn+2*t'mp out in vdd vdd PCH l=2 w=wp ad='wp*t' pd='wp+2*t' as='wp*t' ps='wp+2*t' .endsX1 vin vout inv wn=3.5 wp=10 t=7.5.op.tran 5p 5n.meas tran voutmax max v(vout) from=5p to=5n.meas tran voutmin min v(vout) from=5p to=5n$一級.meas tran tphl1+tr

5、ig v(vin)+val=0.9+rise=1+targ v(vout)+val='0.5*(voutmax-voutmin)+voutmin'+fall=1.meas tran tplh1+trig v(vin)+val=0.9+fall=1 +targ v(vout)+val='0.5*(voutmax-voutmin)+voutmin'+rise=1.end1) 一級無負(fù)載測得本征延時約為17ps；$DATA1 SOURCE-, HSPICEJ VERSION=，D-2010. 03Tpi 32-B1T7,TITLE J .title buffer-ch

6、ain 1'voutinaKvoutinintphlltplhlteinperal t er#2. 2321-S 24721.725e-U k 703e-ll.27- 000012)帶上64倍Cg1大小的負(fù)載測得延時為750.35ps,是本征延時的44倍$DATA1 SOURC拉'HSPICE' VERSION D-2010.03-SF1 32-BIT'TITLE J title buffer-chainvoutmaxvoutmintphlltplhltsinperalter#1. S083T, 476-031. 059e-107. 94叫TO27,00001B

7、）、三級帶負(fù)載測延時代碼如下: .title buffer-chain 3 .lib 'C:synopsysHspice_D-2010.03-SP1tsmc018mm018.l' TT * set 0.18um library.opt scale=0.1u * set lambda.options post=2 list.temp 27.global vdd.param fan=4Vdd vdd gnd 1.8vin vin 0 0.9 pulse 0 1.8 25n 5p 5p 49.99n 100nCl vout gnd 0f $Cg1=2.46fF,負(fù)載為 CL=157.

8、44fF.subckt inv in out wn=3.5 wp=10 t=7.5 mn out in gnd gnd NCH l=2 w=wn ad='wn*t' pd='wn+2*t' as='wn*t' ps='wn+2*t' mp out in vdd vdd PCH l=2 w=wp ad='wp*t' pd='wp+2*t' as='wp*t' ps='wp+2*t' .endsX1 vin 2 inv wn=3.5 wp=10 t=7.5X2 23 in

9、v wn='fan*3.5' wp='fan*10' t=5X3 3 vout inv wn='fan*fan*3.5' wp='fan*fan*10' t=5.op.tran 50p 500n.meas tran voutmax max v(vout) from=50p to=500n.meas tran voutmin min v(vout) from=50p to=500n$三級.meas tran tphl3+trig v(vin)+val=0.9+rise=1+targ v(vout)+val='0.5*(vou

10、tmax-voutmin)+voutmin'+fall=1.meas tran tplh3+trig v(vin)+val=0.9+fall=1+targ v(vout)+val='0.5*(voutmax-voutmin)+voutmin'+rise=11)帶上64倍Cg1大小的負(fù)載測得延時為174.6ps,是本征延時的10.27倍32-BIT1tplh31, 794e-10$ DAT Al SOURCE'' HSP1CE1 VERSION" D-2 01k 03-SPl.TITLE ' . title bufferchainvout

11、inaxiroutjpintphl3teipperalter#L 8449-6. U9e-021. 693e-102- 00001總結(jié)如下：經(jīng)過調(diào)整參數(shù)近似時每一級的1 ,所以經(jīng)過手工計算得到一級帶負(fù)載和三級帶負(fù)載的延時比值為：tp3 空收 0.2344,而仿真得到的結(jié)果為tp1 65tp0同理其他級的延時代碼也是如上的旦立0.2327,所以符合手工計算的比值,750.35寫法，經(jīng)過仿真得到三級延時最小。.end2. Consider the logic network below, which may represent the critical path of a more complex

12、 logic block. The output of the network is loaded with a capacitance which is 5 times larger than the input capacitanceof the first gate, which is a minimum-sized inverter. The effective fanout of the path hence equals F = C/Cg1 =5.Using HSPICE and TSMC 0.18 um CMOS technology model with 1.8 V power

13、 supply, design a circuit simulation scheme to verify the OPTIMAZATION parameters ofg, f, and s for each of the inverter and gates.Solution:由題得到路徑邏輯努力 G 1 5 5 1 25 ,由于沒有分支B=1 ,所以3 39H GFB衛(wèi)5,所以使延時最小的邏輯努力為h N日4陛1.93,得到如下 9, 92.6的扇出系數(shù)：f1 1.93, f2 1.16, f3 1.16, f4 1.93,利用書上公式6.18計算得到尺寸系數(shù) § 1,S2 f1

14、gl/g2 1.16,S3 f1f2gl/g3 1.34,s4f1f2f3gl/g4電路仿真代碼如下:.title INV 2NAND 2NOR.lib 'C:synopsysHspice_D-2010.03-SP1tsmc018mm018.l' TT * set 0.18um library .options post=2 list.temp 27.global vddVdd vdd gnd 1.8vin vin 0 0.9 pulse 0.0 1.8 150p 5p 5p 290p 600pC1 vout gnd 12.3f $Cg1=2.46fF,所以負(fù)載為 12.3fF

15、.subckt inv1 in out wn=0.35u wp=1u t=0.75umn out in gnd gnd NCH l=0.2u w=wn ad='wn*t' pd='wn+2*t' as='wn*t' ps='wn+2*t'mp out in vdd vdd PCH l=0.2u w=wp ad='wp*t' pd='wp+2*t' as='wp*t' ps='wp+2*t' .ends.subckt nand3 NAND-A1 NAND-D1 NAND

16、-B1 NAND-C1 wn='0.35u*1.16' wp='1u*1.16't=0.5u $優(yōu)化尺寸系數(shù)S2*.subckt nand3 NAND-A1 NAND-D1 NAND-B1 NAND-C1 wn=0.35u wp=1ut=0.5u$未優(yōu)化尺寸系數(shù)S2mn3 NAND-S2 NAND-C1 gnd gnd NCH l=0.2u w=wn ad='wn*t' pd='wn+2*t' as='wn*t' ps='wn+2*t'mn2 NAND-S1 NAND-B1 NAND-S2 gnd

17、NCH l=0.2u w=wn ad='wn*t' pd='wn+2*t' as='wn*t' ps='wn+2*t'mn1 NAND-D1 NAND-A1 NAND-S1 gnd NCH l=0.2u w=wn ad='wn*t' pd='wn+2*t' as='wn*t' ps='wn+2*t'mp1 NAND-D1 NAND-A1 vdd vdd PCH l=0.2u w=wp ad='wp*t' pd='wp+2*t' as=&

18、#39;wp*t' ps='wp+2*t'vdd PCH l=0.2u w=wp ad='wp*t' pd='wp+2*t'mp2 NAND-D1 NAND-B1 vdd as='wp*t' ps='wp+2*t'vdd PCH l=0.2u w=wp ad='wp*t' pd='wp+2*t'mp3 NAND-D1 NAND-C1 vdd as='wp*t' ps='wp+2*t'.ends .subckt nor2 NOR-A1 NOR-D

19、1 NOR-B1 wn='0.35u*1.34' wp='1u*1.34' t=0.5u$優(yōu)化尺寸系數(shù)S3t=0.5u*.subckt nor2 NOR-A1 NOR-D1 NOR-B1 wn=0.35u wp=1u$未優(yōu)化尺寸系數(shù)S3mn2 NOR-D1 NOR-B1 gnd as='wn*t' ps='wn+2*t'mn1 NOR-D1 NOR-A1 gnd as='wn*t' ps='wn+2*t'mp1 NOR-S1 NOR-A1 vdd as='wp*t' ps='

20、wp+2*t'gnd NCH l=0.2u w=wn ad='wn*t' pd='wn+2*t'gnd NCH l=0.2u w=wn ad='wn*t' pd='wn+2*t'vddPCH l=0.2u w=wp ad='wp*t' pd='wp+2*t'mp2 NOR-D1 NOR-B1 as='wp*t' ps='wp+2*t'NOR-S1 vdd PCH l=0.2u w=wp ad='wp*t' pd='wp+2*t'

21、.ends.subckt inv2 in out wn='0.35u*2.6' wp='1u*2.6' t=0.5u$優(yōu)化尺寸系數(shù)S4*.subckt inv2 in out wn=0.35u wp=1u t=0.5u$未優(yōu)化尺寸系數(shù)S4mn out in gnd gnd NCH l=0.2u w=wn ad='wn*t' pd='wn+2*t' as='wn*t' ps='wn+2*t'mp out in vdd vdd PCH l=0.2u w=wp ad='wp*t' pd=&

22、#39;wp+2*t' as='wp*t' ps='wp+2*t'.endsX1 vin 2inv1X2 23vdd vdd nand3X3 34gnd nor2X4 4 voutinv2.op.tran 5p 3000p.meas tran voutmax max v(vout) from=5p to=3000p.meas tran voutmin min v(vout) from=5p to=3000p.meas tran tphl+trig v(vin)+val=0.9+rise=2+targ v(vout)+val='0.5*(voutmax-voutmin)+voutmin'+rise=2.m