I overleaped the netlist here.........following is the controlled description.
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V_VSA vsa 0 DC=0
V_VDA vda 0 DC=3
V_vip vip 0 DC=1
I_IIN 0 iin DC=2.5uA
v_vim vim 0 1 AC=1 $ ac source
R_AC_COUPLE vim vimz 1e9 ac=0 .
R_self-bias vout vimz 1m ac=1e30
c_load vout vsa cval $ varying load capacitance
.TEMP 25
.options nomod numdgt=6 acout=0 runlvl=0 accurate post probe
.prot
.lib 'mm06.l' tt
.unprot
.op
.param cval=0
.AC DEC 10 1m 10MEG
.print AC vdb(vout) vp(vout) v(*)
.alter
.param cval=1p
.alter
.param cval=10p
.alter
.param cval=100p
.end
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I'm fresh ,pls see my questions below:
1,Through viewing the phase waveform,it has two turnings,whether means it has 2-pole response?and how did you calculate it?
2,How does change in load capacitance affect pole location(s)?
Need your helps, thanks in advance!
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