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PTAT电流

时间:10-02 整理:3721RD 点击:



请大家帮我看下这个仿真,无论怎么设置MOS的长宽都还是在截止区,是不是应该调BJT的参数,请大家教教我,谢谢拉.
.MODEL nmos NMOS (LEVEL= 49
+VERSION = 3.1TNOM= 27TOX= 7.8E-9
+XJ= 1E-7NCH= 2.2E17VTH0= 0.4730643
+K1= 0.6129988K2= 7.542666E-4K3= 100
+K3B= -10W0= 3.02827E-5NLX= 4.058176E-7
+DVT0W= 0DVT1W= 0DVT2W= 0
+DVT0= 0.4689888DVT1= 0.2917777DVT2= -0.3
+U0= 355.1321831UA= -8.40947E-10UB= 2.261637E-18
+UC= 2.996492E-11VSAT= 1.875162E5A0= 0.8883904
+AGS= 0.1167968B0= 1.155842E-6B1= 5E-6
+KETA= 3.56905E-3A1= 0A2= 0.3740786
+RDSW= 1.131183E3PRWG= -0.1100856PRWB= -0.2
+WR= 1WINT= 1.481909E-7LINT= 3.346531E-10
+XL= -5E-8XW= 1.5E-7DWG= -4.274339E-9
+DWB= 5.243879E-9VOFF= -0.0876189NFACTOR = 2.0725615
+CIT= 0CDSC= 2.4E-4CDSCD= 0
+CDSCB= 0ETA0= 1ETAB= -0.0732914
+DSUB= 0.7823743PCLM= 1.676784PDIBLC1 = 1.543456E-4
+PDIBLC2 = 4.733841E-3PDIBLCB = 0.1DROUT= 3.345392E-4
+PSCBE1= 7.163825E8PSCBE2= 1E-3PVAG= 3.072725E-3
+DELTA= 0.01RSH= 78.2MOBMOD= 1
+PRT= 0UTE= -1.5KT1= -0.11
+KT1L= 0KT2= 0.022UA1= 4.31E-9
+UB1= -7.61E-18UC1= -5.6E-11AT= 3.3E4
+WL= 0WLN= 1WW= 0
+WWN= 1WWL= 0LL= 0
+LLN= 1LW= 0LWN= 1
+LWL= 0CAPMOD= 2XPART= 0.5
+CGDO= 2.69E-10CGSO= 2.69E-10CGBO= 1E-12
+CJ= 9.015359E-4PB= 0.8MJ= 0.3608969
+CJSW= 2.782744E-10PBSW= 0.8MJSW= 0.1801287
+CJSWG= 1.82E-10PBSWG= 0.8MJSWG= 0.1824357
+CF= 0PVTH0= -0.05PRDSW= -121.7614848
+PK2= 4.477143E-5WKETA= -1.015121E-3LKETA= -0.0120304)
.MODEL pmos PMOS (LEVEL= 49
+VERSION = 3.1TNOM= 27TOX= 7.8E-9
+XJ= 1E-7NCH= 8.52E16VTH0= -0.6954453
+K1= 0.4304724K2= -0.0112912K3= 86.4300172
+K3B= -5W0= 6.598922E-6NLX= 2.06513E-7
+DVT0W= 0DVT1W= 0DVT2W= 0
+DVT0= 0.3128439DVT1= 0.8240817DVT2= -0.2517458
+U0= 150.785921UA= 1E-10UB= 1.789607E-18
+UC= -1.8798E-11VSAT= 1.999188E5A0= 1.1641922
+AGS= 0.3099053B0= 2.120097E-6B1= 5E-6
+KETA= -4.16221E-3A1= 4.218536E-3A2= 1
+RDSW= 3.913702E3PRWG= -0.1079321PRWB= -4.46657E-3
+WR= 1WINT= 1.500604E-7LINT= 0
+XL= -5E-8XW= 1.5E-7DWG= -1.858462E-8
+DWB= 1.143752E-8VOFF= -0.140743NFACTOR = 2
+CIT= 0CDSC= 2.4E-4CDSCD= 0
+CDSCB= 0ETA0= 0.032156ETAB= 5.85316E-3
+DSUB= 0.2940371PCLM= 4.7751455PDIBLC1 = 2.143338E-3
+PDIBLC2 = -2.300289E-6PDIBLCB = -6.805486E-4DROUT= 2.413411E-4
+PSCBE1= 7.927378E10PSCBE2= 5.007084E-10PVAG= 15
+DELTA= 0.01RSH= 150.7MOBMOD= 1
+PRT= 0UTE= -1.5KT1= -0.11
+KT1L= 0KT2= 0.022UA1= 4.31E-9
+UB1= -7.61E-18UC1= -5.6E-11AT= 3.3E4
+WL= 0WLN= 1WW= 0
+WWN= 1WWL= 0LL= 0
+LLN= 1LW= 0LWN= 1
+LWL= 0CAPMOD= 2XPART= 0.5
+CGDO= 2.05E-10CGSO= 2.05E-10CGBO= 1E-12
+CJ= 1.397158E-3PB= 0.99MJ= 0.5773462
+CJSW= 3.176388E-10PBSW= 0.99MJSW= 0.3570517
+CJSWG= 4.42E-11PBSWG= 0.99MJSWG= 0.3570517
+CF= 0PVTH0= 0.0253723PRDSW= -76.9871264
+PK2= 2.04063E-3WKETA= 4.61596E-3LKETA= -8.540344E-3)
.model PBJT pnp
+level= 1
+is= 1.0982e-17
+bf= 100
+nf= 1.007
+vaf= 188.5529
+ikf= 2.9138e-04
+ise= 4.4775e-16
+ne= 1.672
+br= 8.9525e-03
+nr= 1.0138
+var= 16.297
+ikr= 2.50e-04
+isc= 2.0329e-13
+nc= 1.592
+rb= 0
+irb= 1.60e-04
+rbm= 0
+re= 0
+rc= 0
+xti= 4.8129
+eg= 1.1502
+cje= 2.1155e-14
+vje= 8.3140e-01
+mje= 4.7169e-01
+tvje= 1.8766e-03
+cte= 1.1810e-03
+tf= 1.00e-10
+xtf= 1.0
+vtf= 10.0
+itf= 0.1
+cjc= 1.4063e-14
+vjc= 7.5000e-01
+mjc= 2.4000e-01
+tvjc= 3.0000e-03
+ctc= 1.0000e-05
+xcjc= 0.5
+fc= 0.75
+cjs= 0.00
+vjs= 0.75
+mjs= 0.5
+tr= 0.00
+ptf= 0.00
+kf= 0.00
+af= 1.0
+cbcp= 0.00
+cbep= 0.00
+ccsp= 0.00
+nkf= 0.4795
+tref= 25.0
+tlev= 0
+tlevc= 1
+xtb= 1.792
q1 1 0 0 PBJT
q2 6 0 0 PBJT m=8
m1 2 2 1 0 nmos w=6u l=2u
m2 3 2 7 0 nmos w=6u l=2u
m3 2 3 vdd vdd pmos w=15 l=2u
m4 3 3 vdd vdd pmos w=15 l=2u
m5 6 3 vdd vdd pmos w=3u l=2u
r1 5 4 4.3k
vdd vdd 0 dc 3.3v
.op
.dc vdd 0 3.3v 0.1
.dctemp -20 100 5
.print i(m5)
.end

贴个电路吧

netlist 跟电路不对应啊,比如图上有2个R,netlist只有r1

那个无所谓啊,我看的是输出电流

两种解释:
1. M1~M4组成的电路为与电源无关的偏置,其中M2带负载,这种电路具有稳定的非零偏置点条件为M2的宽长比大于M1,具体原理看参见拉扎为书第11章(page 311)
2. 假设电路有PTAT电流,根据电路架构,应该有结论VX=VY 并且 I1=I2,由于BJT的Vbe与I为exp形曲线,当I1=I2时,Vbe,Q1=Vbe,Q2,那么有VX=Vbe,Q1, VY=Vbe,Q2+I2*R,根据VX=VY,可知,R=0;从上述推理可知,该电路没有非零偏置点
解决方法:修改网表,加大M2宽长比,务必保证M2宽长比大于M1宽长比

谢谢楼上,我现在就去试试,还有想问问BJT有需要调整的参数什么的吗?还有PNP的模型没问题吧,我是用的tsmc035工艺

PNP的模型应该还好
如果是网上下载的一般都还能用
不过level=1基本上只能仿真PNP本身
有些PNP的模型会附带有寄生的BJT模型,在仿真漏电的时候和饱和的时候比较准确

我加了个启动电路后M1~M4都在饱和区了,但问题是I1不等 I2,subckt
element0:ms20:ms30:m10:m20:m30:m4
model0:cmosn0:cmosn0:cmosp0:cmosp0:cmosn0:cmosn
regionLinearCutoffSaturatiSaturatiSaturatiSaturati
id1.8110m9.4658n -670.9874u -707.7647u670.9874u707.7553u
ibs0.0.0.0.0.0.
ibd0.0.0.0.0.0.
vgs3.8779255.8578m-1.4608-1.46081.06471.1968
vds255.8578m1.9017-1.4608-5.59835.20221.1968
vbs0.0.0.0.-836.9606m -704.9079m
vth396.9978m394.7592m -396.6578m -392.2850m611.2485m584.5114m
vdsat3.544533.0013m -830.0189m -833.5606m360.4754m469.2796m
vod3.4809-138.9014m-1.0642-1.0686453.4539m612.2437m
beta2.2221m446.4001u1.5058m1.5072m7.3531m4.5882m
gam eff576.7999m576.1427m530.7678m530.7678m575.9592m576.0443m
gm0.245.6912n1.1336m1.1687m2.6748m2.1081m
gds6.8541m830.0623p10.8416u8.5942u9.2924u12.4827u
gmb323.8627u76.1138n400.7434u407.3719u567.1733u470.5713u
cdtot319.6411f2.5602f39.1422f38.8061f42.6531f27.4250f
cgtot730.0629f18.4159f737.0476f736.6577f744.7557f478.3115f
cstot165.6529f3.0059f399.9598f398.2655f380.6537f245.2594f
cbtot171.9375f12.7379f187.4853f187.9369f153.7285f99.9063f
cgs387.2755f3.3558f687.0519f683.2093f672.0755f433.3995f
cgd343.9679f2.5494f38.8007f38.4037f42.3319f27.2460f
所以结果电流只在一段温度范围内是正温度系数的,现在该怎么调呢,请求指教?

这个这个不就是每个书上的经典电路么。

1. 从仿真角度说,这个电路不加启动电路也应该有非零的仿真结果
2. 观察启动电路在启动后又没有关掉,换句话说,电路在正常工作时,启动电路应该是不工作的,估计这是I1不等于I2的原因,否则按照电路架构,M3和M4为镜像,工作在饱和区的话电流就应该相等
3. 贴个启动电路上来哦


这样电流就应该不是PTAT成分了吧,应该VY-VBE,Q2中含有delta(Vgs)的成分了?

他这个电路貌似只用M1M2来钳位Vx Vy,靠下面的BJT来提供△BE吧?所以个人认为你的第一个说法有点问题。

是不是没有加启动电路的原因?希望LZ后续把原因及解决办法贴出来!

,学习中

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