超大规模集成电路第五次作业2016秋-段成华

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1、1. Shown below is a level restore circuit of pass transistor. (1) Without transistor Mr, verify by using HSPICE and TSMC 0.18 um CMOS technology model with 1.8 V power supply that the high input to the signal- restoring inverter only charges up to VDD - VTn.(2) After inserting Mr, verify that the hi

2、gh input to the signal-restoring inverter can charge up to VDD.Solution:(1)不带电平恢复管的代码:.title LEVEL NO RESTROE .lib C:synopsysHspice_D-2010.03-SP1tsmc018mm018.l TT * set 0.18um library.options post=2 list.temp 27.global vdd Vdd vdd gnd 1.8vin vin 0 0.9 pulse 0.0 1.8 250p 5p 5p 490p 1000p*C1 2 gnd 0.0

3、1f.subckt no-restore A s1 wn=0.4u t=0.75umn A vdd s1 gnd NCH l=0.2u w=wn ad=wn*t pd=wn+2*t as=wn*t ps=wn+2*t.ends.subckt inv in out wn=0.4u wp=1.2u t=0.75um1 out in gnd gnd NCH l=0.2u w=wn ad=wn*t pd=wn+2*t as=wn*t ps=wn+2*tm2 out in vdd vdd PCH l=0.2u w=wp ad=wp*t pd=wp+2*t as=wp*t ps=wp+2*t.endsX1

4、 vin x no-restore X2 x vout inv .op.tran 5p 4000p.meas tran voutmax max v(x) from=50p to=4000p.meas tran voutmin min v(x) from=50p to=4000p.meas tran tphl+trig v(vin) +val=0.9 +rise=2+targ v(vout) +val=0.5*(voutmax-voutmin)+voutmin +fall=2.meas tran tplh+trig v(vin) +val=0.9 +fall=2+targ v(vout) +va

5、l=0.5*(voutmax-voutmin)+voutmin +rise=2.print dc v(vin) v(vout) v(x).end(2) 带电平恢复管的代码:.title LEVEL RESTROE .lib C:synopsysHspice_D-2010.03-SP1tsmc018mm018.l TT * set 0.18um library.options post=2 list.temp 27.global vdd Vdd vdd gnd 1.8vin vin 0 0.9 pulse 0.0 1.8 250p 5p 5p 490p 1000p*C1 x gnd 0.01f.

6、subckt restore A s1 out wn=0.4u wp=1u t=0.75umn A vdd s1 gnd NCH l=0.2u w=wn ad=wn*t pd=wn+2*t as=wn*t ps=wn+2*tmr s1 out vdd vdd PCH l=0.2u w=wp ad=wp*t pd=wp+2*t as=wp*t ps=wp+2*t.ends.subckt inv in out wn=0.4u wp=1.2u t=0.75um1 out in gnd gnd NCH l=0.2u w=wn ad=wn*t pd=wn+2*t as=wn*t ps=wn+2*tm2

7、out in vdd vdd PCH l=0.2u w=wp ad=wp*t pd=wp+2*t as=wp*t ps=wp+2*t.endsX1 vin x vout restore X2 x vout inv .op.tran 5p 4000p.meas tran voutmax max v(x) from=50p to=4000p.meas tran voutmin min v(x) from=50p to=4000p.meas tran tphl+trig v(vin) +val=0.9 +rise=2+targ v(vout) +val=0.5*(voutmax-voutmin)+v

8、outmin +fall=2.meas tran tplh+trig v(vin) +val=0.9 +fall=2+targ v(vout) +val=0.5*(voutmax-voutmin)+voutmin +rise=2.print dc v(vin) v(vout) v(x).end二者分析如下不带电平恢复管的测量数据:带电平恢复管的测量数据:得到的电压波形如下:上方的波形为加入电平恢复管,下方为未加入电平恢复管。根据测量数据可知道:未加入电平恢复管最大v(x)=1.1446,加入电平恢复管v(x)=1.8003,但是根据延时数据和图像可以知道加入电平恢复管延时增加,但是对称性变好了

9、,同时上方图像的v(x)的图形有一个弯曲,经过实验可以知道这个弯曲和x点的电容值有关。2. Figure 2-1 shows a cascade of dynamic n-type inverters and Figure 2-2 a cascade of domino logic of inverters. (1) Verify by HSPICE that the straightforward cascading of dynamic gates to create more complex structures (Figure 2-1) does not work. (2) Verif

10、y by HSPICE that the cascade of domino logic of inverters (Figure 2-2) works.Figure 2-1Figure 2-2Solution:(1) 、串联N型动态门代码如下:.title N-TYPE INV .lib C:synopsysHspice_D-2010.03-SP1tsmc018mm018.l TT * set 0.18um library.options post=2 list.temp 27.global vdd Vdd vdd gnd 1.8vclk vclk 0 0.9 pulse 0.0 1.8 2

11、50p 5p 5p 490p 1000pvin vin 0 0.9 PWL 0p 0,150p 0,155p 1.8*C1 x gnd 0.01f.subckt N-TYPE-INV clk in out wn=0.4u wp=1.2u t=0.75umn d1 clk gnd gnd NCH l=0.2u w=wn ad=wn*t pd=wn+2*t as=wn*t ps=wn+2*tmin out in d1 gnd NCH l=0.2u w=wn ad=wn*t pd=wn+2*t as=wn*t ps=wn+2*tmp out clk vdd vdd PCH l=0.2u w=wp a

12、d=wp*t pd=wp+2*t as=wp*t ps=wp+2*t.endsX1 vclk vin out1 N-TYPE-INV X2 vclk out1 out2 N-TYPE-INV .op.tran 5p 2000p.meas tran out1max max v(out1) from=50p to=2000p.meas tran out1min min v(out1) from=50p to=2000p.meas tran out2max max v(out2) from=50p to=2000p.meas tran out2min min v(out2) from=50p to=

13、2000p.print dc v(vclk) v(vin) v(out1) v(out2).end得到测量数据如下:波形图如下:从测量数据和波形图可以看出out1下降到VTn以下之后,out2停留在中间电平电平约为1.0V,因为毛刺的原因使整个电路的最大电压和最小电压偏高和偏低。所以整个电路不能正常工作。(2) 、多米诺反相器逻辑代码如下:.title DOMINO LOGIC OF INVERTERS.lib C:synopsysHspice_D-2010.03-SP1tsmc018mm018.l TT * set 0.18um library.options post=2 list.tem

14、p 27.global vdd Vdd vdd gnd 1.8vclk vclk 0 0.9 pulse 0.0 1.8 250p 5p 5p 490p 1000pvin vin 0 0.9 PWL 0p 0,400p 0,405p 1.8,550p 1.8,555p 0*C1 x gnd 0.01f.subckt N-TYPE-INV clk in out wn=0.4u wp=1.2u t=0.75umn d1 clk gnd gnd NCH l=0.2u w=wn ad=wn*t pd=wn+2*t as=wn*t ps=wn+2*tmin out in d1 gnd NCH l=0.2u w=wn ad=wn*t pd=wn+2*t as=wn*t ps=wn+2*tmp out clk vdd vdd PCH l=0.2u w=wp ad=wp*t pd=wp+2*t as=wp*t ps=wp+2*t.ends.subckt inv in out wn=0.4u wp=1.2u t=0.75um1 out in gnd gnd NCH l=0.2u w=wn ad=wn*t pd=wn+2*t as=wn*t ps=wn+2*tm2 out in vdd vdd PCH l=0.2u w=wp ad=wp*t pd=wp+2*t a

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