Hspice simulation of an circuit problem!

Hi dear all!
I trying to simulate a circuit by HSPICE but there is a some problem. Firstly, I don't know how to get gm of the whole circuit, however I know how to calculate gm of a single stage transistor but when the circuit is complicated I get confused!

here is the circuit:

it's original link is: IEEE Xplore - Linearly Tunable CMOS OTA With Constant Dynamic Range Using Source-Degenerated Current Mirrors

HERE is my incomplete code:
vdd 1 40 2.5
ibias 1 2 30u
vtun1 1 11 .2
vtun2 1 15 .2
vtunmin 1 31 .2
vop 1 37 1
von 1 38 1
vcm 1 35 1.25
vv 1 22 1
vv2 1 41 1
vinp 1 6 2
vinn 1 8 2

************* FIRST BLOCK ***************

M5 2 2 1 0 N1
M6 4 3 1 0 N1
M7 5 3 1 0 N1
M8 7 3 1 0 N1
M9 9 3 21 0 N1
M1 4 6 5 0 N1
M2 9 8 7 0 N1
M4 5 6 7 0 N1
M3 5 8 7 0 N1
M2a 4 13 12 0 P1
M1a 12 11 40 0 P1
M2b 9 9 10 0 P1
M1b 10 11 40 0 P1

************** SECOND BLOCK ****************

M11 23 21 1 0 N1
M12 24 21 1 0 N1
M13 19 22 23 0 N1
M14 20 22 24 0 N1
C1 19 21
C2 20 21
M15 19 41 18 0 P1
M16 20 41 17 0 P1
M4b 18 9 16 0 P1
M4a 17 13 14 0 P1
M3b 16 15 40 0 P1
M3a 14 15 40 0 P1

************ CM FEEDBACK ********************

M17 25 21 1 0 N1
M18 21 22 25 0 N1
M19 21 41 26 0 P1
M4c 26 27 29 0 P1
M3c 29 15 40 0 P1
M1C 30 31 40 0 P1
M2C 28 27 30 0 P1
M20 27 41 28 0 P1
M21 21 37 36 0 N1
M25 36 3 1 0 N1
M22 33 35 36 0 N1
M23 33 35 39 0 N1
M24 27 38 39 0 N1
M26 39 3 1 0 N1
M29 32 31 40 0 P1
M28 35 33 32 0 P1
M27 33 34 35 0 P1

*********************************

*model = bsim3v3
*Berkeley Spice Compatibility
*Lmin= .35 Lmax= 20 Wmin= .6 Wmax= 20
.model N1 NMOS
+Level= 8
+Tnom=27.0
+Acnqsmod=1 elm=3
+Capmod=3
+Nch= 2.498E+17 Tox=9E-09 Xj=1.00000E-07
+Lint=9.36e-8 Wint=1.47e-7
+Lintnoi=1e-9
+Vth0= .6322 K1= .756 K2= -3.83e-2 K3= -2.612
+Dvt0= 2.812 Dvt1= 0.462 Dvt2=-9.17e-2
+Nlx= 3.52291E-08 W0= 1.163e-6
+K3b= 2.233
+Vsat= 86301.58 Ua= 6.47e-9 Ub= 4.23e-18 Uc=-4.706281E-11
+Rdsw= 650 U0= 388.3203 wr=1
+A0= .3496967 Ags=.1 B0=0.546 B1= 1
+Dwg = -6.0E-09 Dwb = -3.56E-09 Prwb = -.213
+Keta=-3.605872E-02 A1= 2.778747E-02 A2= .9
+Voff=-6.735529E-02 NFactor= 1.139926 Cit= 1.622527E-04
+Cdsc=-2.147181E-05
+Cdscb= 0 Dvt0w = 0 Dvt1w = 0 Dvt2w = 0
+Cdscd = 0 Prwg = 0
+Eta0= 1.0281729E-02 Etab=-5.042203E-03
+Dsub= .31871233
+Pclm= 1.114846 Pdiblc1= 2.45357E-03 Pdiblc2= 6.406289E-03
+Drout= .31871233 Pscbe1= 5000000 Pscbe2= 5E-09 Pdiblcb = -.234
+Pvag= 0 delta=0.01
+ Wl = 0 Ww = -1.420242E-09 Wwl = 0
+ Wln = 0 Wwn = .2613948 Ll = 1.300902E-10
+ Lw = 0 Lwl = 0 Lln = .316394
+ Lwn = 0
+kt1=-.3 kt2=-.051
+At= 22400
+Ute=-1.48
+Ua1= 3.31E-10 Ub1= 2.61E-19 Uc1= -3.42e-10
+Kt1l=0 Kt1=-0.1 Prt=764.3


.model P1 PMOS
+Level= 8
+Tnom=27.0
+Acnqsmod=1 elm=3
+Nch= 3.533024E+17 Tox=9E-09 Xj=1.00000E-07
+Lint=6.23e-8 Wint=1.22e-7
+Lintnoi=1e-9
+Vth0=-.6732829 K1= .8362093 K2=-8.606622E-02 K3= 1.82
+Dvt0= 1.903801 Dvt1= .5333922 Dvt2=-.1862677
+Nlx= 1.28e-8 W0= 2.1e-6
+K3b= -0.24 Prwg=-0.001 Prwb=-0.323
+Vsat= 103503.2 Ua= 1.39995E-09 Ub= 1.e-19 Uc=-2.73e-11
+ Rdsw= 460 U0= 138.7609
+A0= .4716551 Ags=0.12
+Keta=-1.871516E-03 A1= .3417965 A2= 0.83
+Voff=-.074182 NFactor= 1.54389 Cit=-1.015667E-03
+Cdsc= 8.937517E-04
+Cdscb= 1.45e-4 Cdscd=1.04e-4
+ Dvt0w=0.232 Dvt1w=4.5e6 Dvt2w=-0.0023
+Eta0= 6.024776E-02 Etab=-4.64593E-03
+Dsub= .23222404
+Pclm= .989 Pdiblc1= 2.07418E-02 Pdiblc2= 1.33813E-3
+Drout= .3222404 Pscbe1= 118000 Pscbe2= 1E-09
+Pvag= 0
+kt1= -0.25 kt2= -0.032 prt=64.5
+At= 33000
+Ute= -1.5
+Ua1= 4.312e-9 Ub1= 6.65e-19 Uc1= 0
+Kt1l=0

**********************************************************
*? how i can get gm of this circuit (gm Amp) ?

.END

HAIDE said:

how i can get gm of this circuit (gm Amp) ?

Click to expand...

Inject an ac unit voltage vin between the differential inputs and measure the ac output current iout between the differential outputs, then run an ac analysis. gm = δ(iout)/δ(vin). If vin=1V_ac then gm = iout[A].

Dear erikl, thanks for your reply but how i can measure the iout? could you please explain more about writing its code?

HAIDE said:

how i can measure the iout? could you please explain more about writing its code?

Click to expand...

For g_m measurement, the output current should be measured in short-circuit mode. So connect a 0Ω resistor or a CCVS across the outputs and .PROBE its current resp. the CCVS output voltage.

I'm not used any more to HSPICE syntax, sorry. It's something like
.PROBE i(R0)

Check the HSPICE User Guide for the exact syntax!

then gm = δ(iout)/δ(vin) is an iout that measured by r=0 resistance?

HAIDE said:

then gm = δ(iout)/δ(vin) is an iout that measured by r=0 resistance?

Click to expand...

Yes, if you inject vin=1v_ac between the inputs.

This works only with an ac analysis; ac signals are calculated in a linearized manner around the selected operation point, so their absolute values don't matter.

.ac type step start stop sweep MCcommand
for example
.ac dec 10 1k 10Meg interval=10
or
.ac dec 1 10 1G
thats for ac sweep
this is for measuring
.probe ac gm=par('vm(out)/vm(in)')
or something like vp(out)-vp(in)
it depends on what do you want to do
hspice also provides .measure command... refer to hspice references

DEAR erikl, i updated my code, but it didnt work , do you have any other way?


vdd 1 40 2.5
ibias 1 2 30u
vtun1 1 11 1
vtun2 1 15 1
vtunmin 1 31 .2
vop 1 37
von 1 38
vop1 1 19
von1 1 20
vcm 1 35 1.25
vv 1 22 1
vv2 1 41 1
vinp 1 6 1 AC 1
vinn 1 8 1 AC 1

************* FIRST BLOCK ***************

M5 2 2 1 0 N1 L=.35u W=2.8u
M6 4 3 1 0 N1 L=.35u W=2.8u
M7 5 3 1 0 N1 L=.35u W=2.8u
M8 7 3 1 0 N1 L=.35u W=2.8u
M9 9 3 21 0 N1 L=.35u W=2.8u
M1 4 6 5 0 N1 L=.35u W=2.8u
M2 9 8 7 0 N1 L=.35u W=2.8u
M4 5 6 7 0 N1 L=.35u W=2.8u
M3 5 8 7 0 N1 L=.35u W=2.8u
M2a 4 13 12 0 P1 L=.35u W=2.8u
M1a 12 11 40 0 P1 L=.35u W=2.8u
M2b 9 9 10 0 P1 L=.35u W=2.8u
M1b 10 11 40 0 P1 L=.35u W=2.8u


************** SECOND BLOCK ****************

M11 23 21 1 0 N1 L=.35u W=2.8u
M12 24 21 1 0 N1 L=.35u W=2.8u
M13 19 22 23 0 N1 L=.35u W=2.8u
M14 20 22 24 0 N1 L=.35u W=2.8u
C1 19 21 100u
C2 20 21 100u
M15 19 41 18 0 P1 L=.35u W=2.8u
M16 20 41 17 0 P1 L=.35u W=2.8u
M4b 18 9 16 0 P1 L=.35u W=2.8u
M4a 17 13 14 0 P1 L=.35u W=2.8u
M3b 16 15 40 0 P1 L=.35u W=2.8u
M3a 14 15 40 0 P1 L=.35u W=2.8u

************ CM FEEDBACK ********************

M17 25 21 1 0 N1 L=.35u W=2.8u
M18 21 22 25 0 N1 L=.35u W=2.8u
M19 21 41 26 0 P1 L=.35u W=2.8u
M4c 26 27 29 0 P1 L=.35u W=2.8u
M3c 29 15 40 0 P1 L=.35u W=2.8u
M1C 30 31 40 0 P1 L=.35u W=2.8u
M2C 28 27 30 0 P1 L=.35u W=2.8u
M20 27 41 28 0 P1 L=.35u W=2.8u
M21 21 37 36 0 N1 L=.35u W=2.8u
M25 36 3 1 0 N1 L=.35u W=2.8u
M22 33 35 36 0 N1 L=.35u W=2.8u
M23 33 35 39 0 N1 L=.35u W=2.8u
M24 27 38 39 0 N1 L=.35u W=2.8u
M26 39 3 1 0 N1 L=.35u W=2.8u
M29 32 31 40 0 P1 L=.35u W=2.8u
M28 35 33 32 0 P1 L=.35u W=2.8u
M27 33 34 35 0 P1 L=.35u W=2.8u


*********************************

*model = bsim3v3
*Berkeley Spice Compatibility
*Lmin= .35 Lmax= 20 Wmin= .6 Wmax= 20
.model N1 NMOS
+Level= 8
+Tnom=27.0
+Acnqsmod=1 elm=3
+Capmod=3
+Nch= 2.498E+17 Tox=9E-09 Xj=1.00000E-07
+Lint=9.36e-8 Wint=1.47e-7
+Lintnoi=1e-9
+Vth0= .6322 K1= .756 K2= -3.83e-2 K3= -2.612
+Dvt0= 2.812 Dvt1= 0.462 Dvt2=-9.17e-2
+Nlx= 3.52291E-08 W0= 1.163e-6
+K3b= 2.233
+Vsat= 86301.58 Ua= 6.47e-9 Ub= 4.23e-18 Uc=-4.706281E-11
+Rdsw= 650 U0= 388.3203 wr=1
+A0= .3496967 Ags=.1 B0=0.546 B1= 1
+Dwg = -6.0E-09 Dwb = -3.56E-09 Prwb = -.213
+Keta=-3.605872E-02 A1= 2.778747E-02 A2= .9
+Voff=-6.735529E-02 NFactor= 1.139926 Cit= 1.622527E-04
+Cdsc=-2.147181E-05
+Cdscb= 0 Dvt0w = 0 Dvt1w = 0 Dvt2w = 0
+Cdscd = 0 Prwg = 0
+Eta0= 1.0281729E-02 Etab=-5.042203E-03
+Dsub= .31871233
+Pclm= 1.114846 Pdiblc1= 2.45357E-03 Pdiblc2= 6.406289E-03
+Drout= .31871233 Pscbe1= 5000000 Pscbe2= 5E-09 Pdiblcb = -.234
+Pvag= 0 delta=0.01
+ Wl = 0 Ww = -1.420242E-09 Wwl = 0
+ Wln = 0 Wwn = .2613948 Ll = 1.300902E-10
+ Lw = 0 Lwl = 0 Lln = .316394
+ Lwn = 0
+kt1=-.3 kt2=-.051
+At= 22400
+Ute=-1.48
+Ua1= 3.31E-10 Ub1= 2.61E-19 Uc1= -3.42e-10
+Kt1l=0 Kt1=-0.1 Prt=764.3


.model P1 PMOS
+Level= 8
+Tnom=27.0
+Acnqsmod=1 elm=3
+Nch= 3.533024E+17 Tox=9E-09 Xj=1.00000E-07
+Lint=6.23e-8 Wint=1.22e-7
+Lintnoi=1e-9
+Vth0=-.6732829 K1= .8362093 K2=-8.606622E-02 K3= 1.82
+Dvt0= 1.903801 Dvt1= .5333922 Dvt2=-.1862677
+Nlx= 1.28e-8 W0= 2.1e-6
+K3b= -0.24 Prwg=-0.001 Prwb=-0.323
+Vsat= 103503.2 Ua= 1.39995E-09 Ub= 1.e-19 Uc=-2.73e-11
+ Rdsw= 460 U0= 138.7609
+A0= .4716551 Ags=0.12
+Keta=-1.871516E-03 A1= .3417965 A2= 0.83
+Voff=-.074182 NFactor= 1.54389 Cit=-1.015667E-03
+Cdsc= 8.937517E-04
+Cdscb= 1.45e-4 Cdscd=1.04e-4
+ Dvt0w=0.232 Dvt1w=4.5e6 Dvt2w=-0.0023
+Eta0= 6.024776E-02 Etab=-4.64593E-03
+Dsub= .23222404
+Pclm= .989 Pdiblc1= 2.07418E-02 Pdiblc2= 1.33813E-3
+Drout= .3222404 Pscbe1= 118000 Pscbe2= 1E-09
+Pvag= 0
+kt1= -0.25 kt2= -0.032 prt=64.5
+At= 33000
+Ute= -1.5
+Ua1= 4.312e-9 Ub1= 6.65e-19 Uc1= 0
+Kt1l=0

**********************************************************
.OP
.ac dec 10 1k 1meg
R0 19 20 0
R1 37 38 0

.probe ac I(R0)
.probe ac I(R1)

.END

---------- Post added at 10:28 ---------- Previous post was at 10:27 ----------

gm=par('vm(out)/vm(in)') not works for finding gm, what is vm?

HAIDE said:

... it didnt work ...

Click to expand...

It can't work; there are a lot of errors in your code, e.g.:

• The connection between node#1 & node#0 (GND) is missing
• All bulk connections are connected to GND (0), even the PMOS bulks !!
• Your AC stimuli are common mode, you can't expect any output from them.

Anyway, before you analyze the AC behavior, you should check if your DC biasing generates a reasonable operating point.

HAIDE said:

gm=par('vm(out)/vm(in)') not works for finding gm, what is vm?

Click to expand...

That's what you should ask yourself, you wrote it!
The equation is wrong anyway: gm can never be a voltage ratio!

Hello, erikl
Thanks for your reply, I fix some of my code's errors, but how I can change the common mode AC stimuli
to the correct form?

HAIDE said:

how I can change the common mode AC stimuli to the correct form?

Click to expand...

Add the AC voltage to just one of the inputs!

Or: add a minus sign or a 180° phase to one of them -- see the HSPICE manual -- and then only ½V each -- this will add up to 1Vac , which is practical for the gm calculation.

hi, erikl
I put vin as you told
vinp 0 6 AC 0.5 180
vinn 0 8 AC 0.5
but it didnt work yet, I also consider bulk effect for transistors to HSPICE
BUT in the output I(R0) is 0 yet for all!


vdd 0 40 2.5
ibias 0 2 30u
vtun1 0 11 1
vtun2 0 15 1
vtunmin 0 31 .2
vop 0 37
von 0 38
vop1 0 19
von1 0 20
vcm 0 35 1.25
vv 0 22 1
vv2 0 41 1
vinp 0 6 AC 0.5 180
vinn 0 8 AC 0.5

************* FIRST BLOCK ***************

M5 2 2 0 0 N1 L=.35u W=2.8u
M6 4 3 0 0 N1 L=.35u W=2.8u
M7 5 3 0 0 N1 L=.35u W=2.8u
M8 7 3 0 0 N1 L=.35u W=2.8u
M9 9 3 21 0 N1 L=.35u W=2.8u
M1 4 6 5 5 N1 L=.35u W=2.8u
M2 9 8 7 7 N1 L=.35u W=2.8u
M4 5 6 7 7 N1 L=.35u W=2.8u
M3 5 8 7 7 N1 L=.35u W=2.8u
M2a 4 13 12 12 P1 L=.35u W=2u
M1a 12 11 40 0 P1 L=.35u W=2u
M2b 9 9 10 10 P1 L=.35u W=2u
M1b 10 11 40 0 P1 L=.35u W=2u


************** SECOND BLOCK ****************

M11 23 21 0 0 N1 L=.35u W=2.8u
M12 24 21 0 0 N1 L=.35u W=2.8u
M13 19 22 23 23 N1 L=.35u W=2.8u
M14 20 22 24 24 N1 L=.35u W=2.8u
C1 19 21 100u
C2 20 21 100u
M15 19 41 18 18 P1 L=.35u W=2u
M16 20 41 17 17 P1 L=.35u W=2u
M4b 18 9 16 16 P1 L=.35u W=2u
M4a 17 13 14 14 P1 L=.35u W=2u
M3b 16 15 40 0 P1 L=.35u W=2u
M3a 14 15 40 0 P1 L=.35u W=2u

************ CM FEEDBACK ********************

M17 25 21 0 0 N1 L=.35u W=2.8u
M18 21 22 25 25 N1 L=.35u W=2.8u
M19 21 41 26 26 P1 L=.35u W=2u
M4c 26 27 29 29 P1 L=.35u W=2u
M3c 29 15 40 0 P1 L=.35u W=2u
M1C 30 31 40 0 P1 L=.35u W=2u
M2C 28 27 30 30 P1 L=.35u W=2u
M20 27 41 28 28 P1 L=.35u W=2u
M21 21 37 36 36 N1 L=.35u W=2.8u
M25 36 3 0 0 N1 L=.35u W=2.8u
M22 33 35 36 36 N1 L=.35u W=2.8u
M23 33 35 39 39 N1 L=.35u W=2.8u
M24 27 38 39 39 N1 L=.35u W=2.8u
M26 39 3 0 0 N1 L=.35u W=2.8u
M29 32 31 40 0 P1 L=.35u W=2u
M28 35 33 32 32 P1 L=.35u W=2u
M27 33 34 35 35 P1 L=.35u W=2u
R0 19 20 0
R1 37 38 0


*********************************

*model = bsim3v3
*Berkeley Spice Compatibility
*Lmin= .35 Lmax= 20 Wmin= .6 Wmax= 20
.model N1 NMOS
+Level= 8
+Tnom=27.0
+Acnqsmod=1 elm=3
+Capmod=3
+Nch= 2.498E+17 Tox=9E-09 Xj=1.00000E-07
+Lint=9.36e-8 Wint=1.47e-7
+Lintnoi=1e-9
+Vth0= .6322 K1= .756 K2= -3.83e-2 K3= -2.612
+Dvt0= 2.812 Dvt1= 0.462 Dvt2=-9.17e-2
+Nlx= 3.52291E-08 W0= 1.163e-6
+K3b= 2.233
+Vsat= 86301.58 Ua= 6.47e-9 Ub= 4.23e-18 Uc=-4.706281E-11
+Rdsw= 650 U0= 388.3203 wr=1
+A0= .3496967 Ags=.1 B0=0.546 B1= 1
+Dwg = -6.0E-09 Dwb = -3.56E-09 Prwb = -.213
+Keta=-3.605872E-02 A1= 2.778747E-02 A2= .9
+Voff=-6.735529E-02 NFactor= 1.139926 Cit= 1.622527E-04
+Cdsc=-2.147181E-05
+Cdscb= 0 Dvt0w = 0 Dvt1w = 0 Dvt2w = 0
+Cdscd = 0 Prwg = 0
+Eta0= 1.0281729E-02 Etab=-5.042203E-03
+Dsub= .31871233
+Pclm= 1.114846 Pdiblc1= 2.45357E-03 Pdiblc2= 6.406289E-03
+Drout= .31871233 Pscbe1= 5000000 Pscbe2= 5E-09 Pdiblcb = -.234
+Pvag= 0 delta=0.01
+ Wl = 0 Ww = -1.420242E-09 Wwl = 0
+ Wln = 0 Wwn = .2613948 Ll = 1.300902E-10
+ Lw = 0 Lwl = 0 Lln = .316394
+ Lwn = 0
+kt1=-.3 kt2=-.051
+At= 22400
+Ute=-1.48
+Ua1= 3.31E-10 Ub1= 2.61E-19 Uc1= -3.42e-10
+Kt1l=0 Kt1=-0.1 Prt=764.3


.model P1 PMOS
+Level= 8
+Tnom=27.0
+Acnqsmod=1 elm=3
+Nch= 3.533024E+17 Tox=9E-09 Xj=1.00000E-07
+Lint=6.23e-8 Wint=1.22e-7
+Lintnoi=1e-9
+Vth0=-.6732829 K1= .8362093 K2=-8.606622E-02 K3= 1.82
+Dvt0= 1.903801 Dvt1= .5333922 Dvt2=-.1862677
+Nlx= 1.28e-8 W0= 2.1e-6
+K3b= -0.24 Prwg=-0.001 Prwb=-0.323
+Vsat= 103503.2 Ua= 1.39995E-09 Ub= 1.e-19 Uc=-2.73e-11
+ Rdsw= 460 U0= 138.7609
+A0= .4716551 Ags=0.12
+Keta=-1.871516E-03 A1= .3417965 A2= 0.83
+Voff=-.074182 NFactor= 1.54389 Cit=-1.015667E-03
+Cdsc= 8.937517E-04
+Cdscb= 1.45e-4 Cdscd=1.04e-4
+ Dvt0w=0.232 Dvt1w=4.5e6 Dvt2w=-0.0023
+Eta0= 6.024776E-02 Etab=-4.64593E-03
+Dsub= .23222404
+Pclm= .989 Pdiblc1= 2.07418E-02 Pdiblc2= 1.33813E-3
+Drout= .3222404 Pscbe1= 118000 Pscbe2= 1E-09
+Pvag= 0
+kt1= -0.25 kt2= -0.032 prt=64.5
+At= 33000
+Ute= -1.5
+Ua1= 4.312e-9 Ub1= 6.65e-19 Uc1= 0
+Kt1l=0

**********************************************************
.OP

.probe ac I(R0)
.probe ac I(R1)
.END

You applying an AC voltage but no DC bias to the inputs. Obviously the amplifier doesn't support an input voltage range down to ground.

HAIDE said:

... it didnt work yet, I also consider bulk effect for transistors to HSPICE

Click to expand...

No, you still have several PMOS bulks connected to GND: M1a,b,c M3a,b,c M29 -- may be more?

Moreover:

• Your bias current connection is wrong (also in schematic)
• vop, von are short circuits (0V by default). Is this what you intent?

May be there are more errors. All your transistors have min. length and only 2 different widths, and your bias voltages seem to be selected arbitrarily. How can you expect that such a circuit will ever work, without checking for a reasonable operating point?
You can't create a working design with this method, sorry!

Hi again erikle
about M1a,b,c M3a,b,c M29 their source is grounded in small signal then I choose their bulk zero, isn't true?
about Bias current, it is original schematic of the IEEE journal paper, and I somehow agree with you but I have to simulate this schematic for my course.



Sorrily the paper that I downloaded from IEEE dont define many things and I want to get an answer with try and error!
even I dunno W/L of each transistor, I just make it to each transistor be ON (triode or saturation)

[/COLOR]thanks for your help, dear erikle!

HAIDE said:

about M1a,b,c M3a,b,c M29 their source is grounded in small signal then I choose their bulk zero, isn't true?

Click to expand...

No, it is not true: Their sources are connected to Vdd, so should their bulks!
In small signal representation, Gnd & Vdd are equivalent, but of course not in real circuit!

HAIDE said:

about Bias current, it is original schematic of the IEEE journal paper, and I somehow agree with you but I have to simulate this schematic for my course.

Click to expand...

But it's wrong and it won't work like this :-( Try to understand where from the current should be sourced!

HAIDE said:

I want to get an answer with try and error!
even I dunno W/L of each transistor, I just make it to each transistor be ON (triode or saturation)

Click to expand...

Fine, if you think you can achieve this. But it will get you a long try and error rallye. Comprehension is the right way!

And don't try to simulate g_m before you aren't sure to have found a reasonable OP (operating point) and ICMR (input common mode range).