ECE422_HW#2_Solutions - ECE 422/522 HW #2 Solution 1 2 2b)...

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ECE 422/522 HW #2 Solution 1
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3 2b) SPICE simulation : *ECE422/522 *Sooch Configuration *HW#2 Problem 2b) .options post node $ These are HSPICE options, just include them .options runlvl=0 $ (Check the manual for more info) *Sources VD Vdd 0 12 *FETS *M# D G S B M1 Vo2 Vo1 0 0 NMOSM W=2e-6 L=1e-6 M2 Vo1 Vo1 Vo2 Vo2 NMOSM W=8e-6 L=1e-6 M3 Vo1 Vo1 Vdd Vdd PMOSM W=18e-6 L=1e-6 .MODEL NMOSM NMOS (LEVEL=1 KP=100e-6 VTO=1.0 ) .MODEL PMOSM PMOS (LEVEL=1 KP=50e-6 VTO=-1.0 ) .op .end *Results ***** operating point status is all simulation time is 0. node =voltage node =voltage node =voltage +0:vo2 = 3.3223 0:vo1 = 8.8606 0:vdd = 12.0000 **** voltage sources subckt element 0:vd volts 12.0000 current -2.0596m power 24.7157m Given I M3 = I M2 = I M1 = Id K N ’ = 2K P (W/L) 1 = 2μm ; (W/L) 2 = 4(W/L) 1 ; (W/L) 3 = 9(W/L) 1 NOTE: M1 is in non saturation, and M2,M3 are in saturation Vdd = 12 volts Vt N = | Vt P | = 1 volts Assume Vo 1 = X , Vo 2 = Y, and Id = Z I M3 = ½ K P ’ (W/L) 3* ((V dd – Vo 1 ) - |Vt P |) 2 = ½ K P ’ 9* (W/L) 1* ((12 –X)-1) 2 I M2 = ½ K N ’ (W/L) 2* ((Vo 1 – Vo 2 ) - Vt N ) 2 = K P ’ 4* (W/L) 1* ((X-Y)-1) 2 I M1 = K N ’ (W/L) 1* ((Vo 1 – Vt N ) * Vo 2 - (Vo 2 ) 2 /2) =2* K P ’ (W/L) 1* ((X-1)*Y-(Y 2 )/2) Using TI-89 calculator, Solve (Z =50e-6* 9* ((12 –X)-1) 2 and Z= 50e-6 *8* ((X-Y)-1) 2 and Z = 4*50e-6 ((X-1)*Y-(Y 2 )/2) , {X,Y,Z}) X = Vo 1 = 8.034144 Volts Y = Vo 2 = 3.888379 Volts Z = Id = 3.958335 mA % Matlab solution by John Guerber Ans = solve('Id = (.000450)*((12-V01 - 1)^2)','Id = (.000400)*((V01- V02-1)^2)','Id = (.000200)*(((V01 -1)*V02 + (V02^2)/2))') Ans.Id Ans.V01 Ans.V02 Id = 5.061 mA, Vo1 = 7.646v, and Vo2 = 3.089v.
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6 ECE422/522 HW2 1a M3 3 3 2 2 NCH L=1u W=20u M2 2 2 1 1 NCH L=1u W=20u M1 1 1 0 0 NCH L=1u W=20u Vdd 3 0 15 .model NCH NMOS Level=1 VT=1 KP= 25u .option spice .op .end RESULTS element 0:m3 0:m2 0:m1 model 0:nch 0:nch 0:nch region Saturati Saturati Saturati id 4.000e-03 4.000e-03 4.000e-03 ibs 0. 0. 0. ibd -5.000e-14 -5.000e-14 -5.000e-14 vgs 5.000e+00 5.000e+00 5.000e+00 vds 5.000e+00 5.000e+00 5.000e+00 vbs 0. 0. 0. vth 1.000e+00 1.000e+00 1.000e+00 vdsat 4.000e+00 4.000e+00 4.000e+00 ECE422/522 HW2 1b M3 3 3 2 0 NCH L=1u W=20u M2 2 2 1 0 NCH L=1u W=20u M1 1 1 0 0 NCH L=1u W=20u Vdd 3 0 15 .model NCH NMOS Level=1 VT=1 KP=25u gamma=.5 phi=.6 .option spice .op .end RESULTS element 0:m3 0:m2 0:m1 model 0:nch 0:nch 0:nch region Saturati Saturati Saturati id 2.820e-03 2.820e-03 2.820e-03 ibs -9.444e-14 -4.359e-14 0. ibd -1.500e-13 -9.444e-14 -4.359e-14 vgs 5.556e+00 5.085e+00 4.358e+00 vds 5.556e+00 5.085e+00 4.358e+00 vbs -9.443e+00 -4.358e+00 0. vth 2.197e+00 1.726e+00 1.000e+00 vdsat 3.358e+00 3.358e+00 3.358e+00 ECE422/522 HW2 1d M3 2 2 3 3 PCH L=1u W=40u M2 1 1 2 3 PCH L=1u W=40u M1 0 0 1 3 PCH L=1u W=40u Vdd 3 0 15 .model PCH PMOS Level=1 VTO=-1 KP=10u .option spice
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7 .op .end RESULTS element 0:m3 0:m2 0:m1 model 0:pch 0:pch 0:pch region Saturati Saturati Saturati id -3.200e-03 -3.200e-03 -3.200e-03 ibs 0. 5.000e-14
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This note was uploaded on 01/18/2011 for the course ALS 161 taught by Professor Sfuiaf during the Spring '10 term at American InterContinental University Los Angeles.

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ECE422_HW#2_Solutions - ECE 422/522 HW #2 Solution 1 2 2b)...

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