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Biasing of Multiplier Circuit

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IntuitiveAnalog

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Hi Everyone,

I am designing a multiplier circuit which has to act as a voltage squarer with Vout = (Vin^2)/25. The input waveform is sinusoidal with zero dc level ,100 mV peak to peak voltage and 10 kHz frequency.
I am having problem in adjusting biasing of the circuit .
I will be thankful if anybody can help.

--------------------------


Code: 
***Multiplier***
Vdd 100 0 1.5
*.option scale=1u
.option post
.probe
.tran 0.1u 600u


Vp1 1 0 SIN(0.8V 0.2V 10k 0 0 0)
Vp2 7 0 SIN(0.8V 0.2V 10k 0 0 180) 
Vp3 8 0 SIN(0.8V 0.2V 10k 0 0 0) 
Vp4 12 0 SIN(0.8V 0.2V 10k 0 0 180) 
Vn1 3 0 SIN(1.5V 0.1V 10k 0 0 0) 
Vn2 5 0 SIN(1.5V 0.1V 10k 0 0 180) 
Vn3 13 0 SIN(1.5V 0.1V 10k 0 0 180) 
Vn4 14 0 SIN(1.5V 0.1V 10k 0 0 0) 
Mp1 2 1 100 100 pmos W=0.8u L=0.35u
Mp2 6 7 100 100 pmos W=0.8u L=0.35u
Mp3 9 8 100 100 pmos W=0.8u L=0.35u
Mp4 11 12 100 100 pmos W=0.8u L=0.35u
Mn1 4 3 2 0 nmos W=0.8u L=0.35u
Mn2 4 5 6 0 nmos W=0.8u L=0.35u
Mn3 10 13 9 0 nmos W=0.8u L=0.35u
Mn4 10 14 11 0 nmos W=0.8u L=0.35u
M1 2 100 0 0 nmos W=0.8 L=0.35
M2 6 100 0 0 nmos W=0.8u L=0.35u
M3 9 100 0 0 nmos W=0.8u L=0.35u
M4 11 100 0 0 nmos W=0.8u L=0.35u

.MODEL nmos NMOS (                                LEVEL   = 49
+VERSION = 3.1            TNOM    = 27             TOX     = 7.7E-9
+XJ      = 1E-7           NCH     = 2.3579E17      VTH0    = 0.5048265
+K1      = 0.5542796      K2      = 0.0155863      K3      = 2.3475646
+K3B     = -3.3142916     W0      = 4.145888E-5    NLX     = 1.430868E-7
+DVT0W   = 0              DVT1W   = 0              DVT2W   = 0
+DVT0    = -0.0150839     DVT1    = 1.51022E-3     DVT2    = 0.170688
+U0      = 415.8570638    UA      = 5.057324E-11   UB      = 1.496793E-18
+UC      = 2.986268E-11   VSAT    = 1.237033E5     A0      = 0.9098788
+AGS     = 0.2120181      B0      = 1.683612E-6    B1      = 5E-6
+KETA    = -4.011887E-4   A1      = 0              A2      = 1
+RDSW    = 1.156967E3     PRWG    = -8.468558E-3   PRWB    = -7.678669E-3
+WR      = 1              WINT    = 5.621821E-8    LINT    = 1.606205E-8
+XL      = -2E-8          XW      = 0              DWG     = -6.450939E-9
+DWB     = 6.530228E-9    VOFF    = -0.1259348     NFACTOR = 0.3344887
+CIT     = 0              CDSC    = 1.527511E-3    CDSCD   = 0
+CDSCB   = 0              ETA0    = 1.21138E-3     ETAB    = -1.520242E-4
+DSUB    = 0.1259886      PCLM    = 0.8254768      PDIBLC1 = 0.4211084
+PDIBLC2 = 6.081164E-3    PDIBLCB = -5.865856E-6   DROUT   = 0.7022263
+PSCBE1  = 7.238634E9     PSCBE2  = 5E-10          PVAG    = 0.6261655
+DELTA   = 0.01           MOBMOD  = 1              PRT     = 0
+UTE     = -1.5           KT1     = -0.11          KT1L    = 0
+KT2     = 0.022          UA1     = 4.31E-9        UB1     = -7.61E-18
+UC1     = -5.6E-11       AT      = 3.3E4          WL      = 0
+WLN     = 1              WW      = -1.22182E-15   WWN     = 1.137
+WWL     = 0              LL      = 0              LLN     = 1
+LW      = 0              LWN     = 1              LWL     = 0
+CAPMOD  = 2              XPART   = 0.4            CGDO    = 3.5E-10
+CGSO    = 3.5E-10        CGBO    = 0              CJ      = 8.829973E-4
+PB      = 0.7946332      MJ      = 0.3539285      CJSW    = 2.992362E-10
+PBSW    = 0.9890846      MJSW    = 0.1871372      PVTH0   = -0.0148617
+PRDSW   = -114.7860236   PK2     = -5.151187E-3   WKETA   = 5.687313E-3
+LKETA   = -0.018518       )
*
.MODEL pmos PMOS (                                LEVEL   = 49
+VERSION = 3.1            TNOM    = 27             TOX     = 7.7E-9
+XJ      = 1E-7           NCH     = 8.52E16        VTH0    = -0.6897992
+K1      = 0.4134289      K2      = -5.342989E-3   K3      = 24.8361788
+K3B     = -1.4390847     W0      = 2.467689E-6    NLX     = 3.096223E-7
+DVT0W   = 0              DVT1W   = 0              DVT2W   = 0
+DVT0    = 1.3209807      DVT1    = 0.4695965      DVT2    = -8.790762E-4
+U0      = 150.6275733    UA      = 2.016943E-10   UB      = 1.714919E-18
+UC      = -1.36948E-11   VSAT    = 9.559222E4     A0      = 0.9871247
+AGS     = 0.3541967      B0      = 3.188091E-6    B1      = 5E-6
+KETA    = -0.0169877     A1      = 0              A2      = 1
+RDSW    = 2.443009E3     PRWG    = 0.0260616      PRWB    = 0.141561
+WR      = 1              WINT    = 5.038936E-8    LINT    = 1.650588E-9
+XL      = -2E-8          XW      = 0              DWG     = -1.535456E-8
+DWB     = 1.256904E-8    VOFF    = -0.15          NFACTOR = 1.5460516
+CIT     = 0              CDSC    = 1.413317E-4    CDSCD   = 0
+CDSCB   = 0              ETA0    = 0.3751392      ETAB    = 2.343374E-3
+DSUB    = 0.8877574      PCLM    = 5.8638076      PDIBLC1 = 1.05224E-3
+PDIBLC2 = 3.481753E-5    PDIBLCB = 2.37525E-3     DROUT   = 0.0277454
+PSCBE1  = 3.013379E10    PSCBE2  = 3.608179E-8    PVAG    = 3.9564294
+DELTA   = 0.01           MOBMOD  = 1              PRT     = 0
+UTE     = -1.5           KT1     = -0.11          KT1L    = 0
+KT2     = 0.022          UA1     = 4.31E-9        UB1     = -7.61E-18
+UC1     = -5.6E-11       AT      = 3.3E4          WL      = 0
+WLN     = 1              WW      = -5.22182E-16   WWN     = 1.125
+WWL     = 0              LL      = 0              LLN     = 1
+LW      = 0              LWN     = 1              LWL     = 0
+CAPMOD  = 2              XPART   = 0.4            CGDO    = 3.02E-10
+CGSO    = 3.02E-10       CGBO    = 0              CJ      = 1.397645E-3
+PB      = 0.99           MJ      = 0.5574537      CJSW    = 3.665392E-10
+PBSW    = 0.99           MJSW    = 0.3399328      PVTH0   = 0.0114364
+PRDSW   = 52.7951169     PK2     = 9.714153E-4    WKETA   = 0.0109418
+LKETA   = 7.702974E-3     )  

.END

Thanks
 
Last edited by a moderator:

If the signal is always sine and the same frequency convert it to DC first.
 
Last edited:

zulqar said:
I am having problem in adjusting biasing of the circuit .
Vdd 100 0 1.5
Click to expand...
Click to expand...

I think your power supply voltage is too low for this design in a 0.35µm process with |Vth| ≈ 0.5V ; you'd better use Vdd=3V.

I'd suggest to set VP to about Vdd/3 by appropriate W/L values of M1..M4, respectively VO to about Vdd/2 by appropriate W/L values of N1..N4 . With the VTH0 values from the model files you can then calculate the X±x and Y±y values, where X resp. Y should be the mean (quiescent) values of the input nodes, x resp. y the half peak (sin) values of the applied signals.
 
Erikl , First of all I am thankful to your response.
Actually I have taken this design from an IEEE transaction. Author has taken Vdd as 1.5V in 0.35um process.
 

zulqar said:
Erikl , First of all I am thankful to your response.
Actually I have taken this design from an IEEE transaction. Author has taken Vdd as 1.5V in 0.35um process.
Click to expand...

Hi,

Do you know what is the technology the author did use? Normally the mininum voltage for 0.35um process is 1.8V. What is value of Vth?
 

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