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下面的程序是我基于SOI例子改的,但程序老是出错,求大神指教。
# (c) Silvaco Inc., 2015 go atlas tiTLE SOI device simulation # # 0.2um of silicon on 0.4um oxide substrate # mesh space.mult=1.0 # x.mesh loc=0.00 spac=0.50 x.mesh loc=1.15 spac=0.02 x.mesh loc=1.5 spac=0.1 x.mesh loc=1.85 spac=0.02 x.mesh loc=3 spac=0.5 # y.mesh loc=-0.025 spac=0.02 y.mesh loc=0.00 spac=0.005 y.mesh loc=0.1 spac=0.02 y.mesh loc=0.2 spac=0.01 y.mesh loc=0.6 spac=0.25 # region num=1 y.min=0.2 oxide region num=2 y.max=0.2 y.min=0.199 silicon region num=3 y.min=0.0 y.max=0.199 silicon region num=4 y.min=-0.001 y.max=0.0 oxide region num=5 y.max=-0.001 y.min=-0.025 oxide # #*********** define the electrodes ************ # #1-GATE #2-SOURCE #3-DRAIN #4-SUBSTRATE(below oxide) # electrode name=gate x.min=1 x.max=2 y.min=-0.025 y.max=-0.025 electrode name=source x.max=0.5 y.min=0 y.max=0 electrode name=drain x.min=2.5 y.min=0 y.max=0 electrode substrate # #*********** define the doping concentrations ***** # doping uniform conc=2e17 p.type reg=3 doping gauss n.type conc=1e20 char=0.2 lat.char=0.05 reg=2 x.r=1.0 doping gauss n.type conc=1e20 char=0.2 lat.char=0.05 reg=2 x.l=2.0 # # define ferroelectric material # model region=5 ferro material region=3 ferro.ps=0.5e-6 ferro.pr=0.4e-6 ferro.ec=100000.0 ferro.eps=10.0 # # set workfunction of gate contact name=gate n.poly material mun=800.0 # # set interface charge separately on front and back oxide interfaces interf qf=3e10 y.max=0.1 interf qf=1e11 y.min=0.1 # # select models models conmob srh auger bgn fldmob print save outf=lee03_0.str tonyplot lee03_0.str -set lee03_0.set # solve init # # do ID-VG characteristic # method newton trap solve prev solve vgate=-0.2 solve vdrain=0.05 solve vdrain=0.1 # # ramp gate voltage log outf=lee03_1.log master solve vgate=0.1 vstep=0.1 name=gate vfinal=1.5 # # plot resultant ID-VG threshold voltage curve tonyplot lee03_1.log -set lee03_1.set # # plot resultant IDVG subthreshold slope curve tonyplot lee03_1.log -set lee03_2.set # # extract name="subvt" 1.0/slope(maxslope(curve(v."gate",log10(abs(i."drain"))))) # # extract name="vt" (xintercept(maxslope(curve(v."gate",abs(i."drain")))) - abs(ave(v."drain"))/2.0) # quit |
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2个回答
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没有人吗
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可以加个qq吗576238308。我也正在解决这个问题
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