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"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter10: Digital CMOS Logic Circuits"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex10.1:pg-962"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Example 10.1 : To determine t_PHL, t_PLH and t_P\n",
"# Consider CMOS inverter\n",
"import math\n",
"C_ox=6*10.0**-15; # (F/um**2)\n",
"uC_n=115.0*10**-6; #uC_n=u_n*C_ox (A/V**2)\n",
"uC_p=30.0*10**-6; #uC_p=u_p*C_ox (A/V**2)\n",
"V_tn=0.4; # (V)\n",
"V_tp=-0.4; # (V)\n",
"V_DD=2.5; # (V)\n",
"W_n=0.375*10**-6; # W for Q_N\n",
"L_n=0.25*10**-6; # L for Q_N\n",
"W_p=1.125*10**-6; # W for Q_P\n",
"L_p=0.25*10**-6; # L for Q_P\n",
"C_gd1=0.3*W_n*10**-9; # (F)\n",
"C_gd2=0.3*W_p*10**-9; # (F)\n",
"C_db1=10**-15; # (F)\n",
"C_db2=10**-15; # (F)\n",
"C_g3= 0.375*0.25*6*10**-15+2*0.3*0.375*10**-15; # (F)\n",
"C_g4=1.125*0.25*6*10**-15+2*0.3*1.125*10**-15; # (F)\n",
"C_w=0.2*10**-15; # (F)\n",
"C=2*C_gd1+2.0*C_gd2+C_db1+C_db2+C_g3+C_g4+C_w; # (F)\n",
"i_DN0=uC_n*W_n*(V_DD-V_tn)**2/(2*L_n); # i_DN0 = i_DN(0) (A)\n",
"i_DNtPHL=uC_n*W_n*((V_DD-V_tn)*V_DD/2-((V_DD/2.0)**2)/2.0)/L_n; # i_DNtPHL = i_DN(t_PHL) (A)\n",
"i_DNav=(i_DN0+i_DNtPHL)/2; # i_DN|av (A)\n",
"t_PHL=C*(V_DD/2)*1e12/i_DNav;\n",
"print math.ceil(t_PHL),\"= t_PHL (picoseconds)\"\n",
"t_PLH=1.3*t_PHL; # Since W_p/W_n=3 and u_n/u_p=3.83 thus t_PLH is greater than t_PHL by 3.83/3\n",
"print round(t_PLH,-1),\"= t_PLH (picoseconds)\"\n",
"t_P=(t_PHL+t_PLH)/2; \n",
"print round(t_P),\"= t_P (picoseconds)\"\n",
"\n",
" # the answer in the textbook is slightly dirfferent due to approximation"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"23.0 = t_PHL (picoseconds)\n",
"30.0 = t_PLH (picoseconds)\n",
"26.0 = t_P (picoseconds)\n"
]
}
],
"prompt_number": 17
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex10.2:pg-972"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Example 10.2 : W/L ratios for the logic circuit\n",
"#For basic inverter\n",
"n=1.5;\n",
"p=5;\n",
"L=0.25*10**-6; # (m)\n",
"WbyL=2*n; # W/L for Q_NB , Q_NC , Q_ND\n",
"print WbyL,\"= W/L ratio for Q_NB\"\n",
"print WbyL,\"= W/L ratio for Q_NC\"\n",
"print WbyL,\"= W/L ratio for Q_ND\"\n",
"WbyL=n; # W/L ratio for Q_NA\n",
"print WbyL,\"= W/L ratio for Q_NA\"\n",
"WbyL=3*p; # W/L for Q_PA, Q_PC , Q_PD\n",
"print WbyL,\"= W/L ratio for Q_PA\" \n",
"print WbyL,\"= W/L ratio for Q_PC\"\n",
"print WbyL,\"= W/L ratio for Q_PD\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"3.0 = W/L ratio for Q_NB\n",
"3.0 = W/L ratio for Q_NC\n",
"3.0 = W/L ratio for Q_ND\n",
"1.5 = W/L ratio for Q_NA\n",
"15 = W/L ratio for Q_PA\n",
"15 = W/L ratio for Q_PC\n",
"15 = W/L ratio for Q_PD\n"
]
}
],
"prompt_number": 18
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex10.3:pg-981"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Example 10.3 : To determine the parameters of pseudo NMOS inverter\n",
"# Consider a pseudo NMOS inverter\n",
"uC_n=115.0*10**-6; #uC_n=u_n*C_ox (A/V**2)\n",
"uC_p=30*10.0**-6; #uC_p=u_p*C_ox (A/V**2)\n",
"V_tn=0.4; # (V)\n",
"V_tp=-0.4; # (V)\n",
"V_DD=2.5; # (V)\n",
"W_n=0.375*10**-6; # W for Q_N (m)\n",
"L_n=0.25*10**-6; # L for Q_N (m)\n",
"r=9.0;\n",
"\n",
"# 10.3a\n",
"V_OH=V_DD;\n",
"print round(V_OH,2),\"= V_OH (V)\"\n",
"V_OL=(V_DD-V_tn)*(1-math.sqrt(1.0-1/r));\n",
"print round(V_OL,2),\"= V_OL (V)\"\n",
"V_IL=V_tn+(V_DD-V_tn)/math.sqrt(r*(r+1.0));\n",
"print round(V_IL,2),\"=V_IL (V)\"\n",
"V_IH=V_tn+2*(V_DD-V_tn)/(math.sqrt(3.0*r));\n",
"print round(V_IH,2),\"= V_IH (V)\"\n",
"V_M=V_tn+(V_DD-V_tn)/math.sqrt(r+1.0);\n",
"print round(V_M,2),\"= V_M (V)\"\n",
"NM_H=V_OH-V_IH;\n",
"NM_L=V_IL-V_OL;\n",
"print round(NM_H,2),\"=The highest and the lowest values of allowable noise margin (V)=\",round(NM_L,2)\n",
"\n",
"# 10.3b\n",
"WbyL_p=uC_n*(W_n/L_n)/(uC_p*r); # WbyL_p=(W/L)_p\n",
"print round(WbyL_p,2),\"=(W/L)_p\"\n",
"\n",
"#10.3c\n",
"I_stat=(uC_p*WbyL_p*(V_DD-V_tn)**2)/2;\n",
"print round(I_stat*1e6,1),\"=I_stat (microA)\"\n",
"P_D=I_stat*V_DD;\n",
"print round(P_D*1e6),\"=Static power dissipation P_D (microW)\"\n",
"\n",
"#10.3d\n",
"C=7*10**-15;\n",
"t_PLH=1.7*C/(uC_p*WbyL_p*V_DD);\n",
"print round(t_PLH*1e9,2),\"=t_PLH (ns)\"\n",
"t_PHL=1.7*C/(uC_n*(W_n/L_n)*math.sqrt(1-0.46/r)*V_DD);\n",
"print round(t_PHL*1e9,2),\"= t_PHL (ns)\"\n",
"t_p=(t_PHL+t_PLH)/2.0;\n",
"print round(t_p*1e9,2),\"= t_p (ns)\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"2.5 = V_OH (V)\n",
"0.12 = V_OL (V)\n",
"0.62 =V_IL (V)\n",
"1.21 = V_IH (V)\n",
"1.06 = V_M (V)\n",
"1.29 =The highest and the lowest values of allowable noise margin (V)= 0.5\n",
"0.64 =(W/L)_p\n",
"42.3 =I_stat (microA)\n",
"106.0 =Static power dissipation P_D (microW)\n",
"0.25 =t_PLH (ns)\n",
"0.03 = t_PHL (ns)\n",
"0.14 = t_p (ns)\n"
]
}
],
"prompt_number": 24
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex10.4:pg-985"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Example 10.4 : To determine parameters for NMOS transistor\n",
"# Consider NMOS transistor switch\n",
"uC_n=50.0*10**-6; #uC_n=u_n*C_ox (A/V**2)\n",
"uC_p=20.0*10.0**-6; #uC_px `=u_p*C_ox (A/V**2)\n",
"V_tO=1.0; # (V)\n",
"y=0.5; # (V**1/2)\n",
"fie_f=0.6/2; # (V)\n",
"V_DD=5; # (V)\n",
"W_n=4*10.0**-6; # (m)\n",
"L_n=2*10.0**-6; # (m)\n",
"C=50*10.0**-15; # (F)\n",
"\n",
"# 10.4a\n",
"V_t=1.6; # (V)\n",
"V_OH=V_DD-V_t; # V_OH is the value of v_O at which Q stops conducting (V)\n",
"print V_OH,\"= V_OH (V)\"\n",
" \n",
"# 10.4b\n",
"W_p=10.0*10**-6; # (m)\n",
"L_p=2*10.0**-6; # (m)\n",
"i_DP=uC_p*W_p*((V_DD-V_OH-V_tO)**2)/(2*L_p);\n",
"print round(i_DP*1e6),\"= Static current of the inverter (microA)\"\n",
"P_D=V_DD*i_DP;\n",
"print round(P_D*1e6),\"= Power dissipated (microW)\"\n",
"V_O=0.08; # Output voltage (V) found by equating the current of Q_N=18uA\n",
"print round(V_O,2),\"= The output voltage of the inverter (V) \"\n",
"\n",
"# 10.4c\n",
"i_D0=uC_n*W_n*((V_DD-V_tO)**2)/(2*2*10**-6); # i_D0=i_D(0) (A) current i_D at t=0\n",
"v_O=2.5; # (V)\n",
"V_t=V_tO+0.5*(math.sqrt(v_O+2*fie_f)-math.sqrt(2*fie_f)); # at v_O=2.5V\n",
"i_DtPLH=(uC_n*W_n*(V_DD-v_O-V_t)**2)/(2*L_n); # i_DtPLH=i_D(t_PLH) (A) current i_D at t=t_PLH\n",
"i_Dav=(i_D0+i_DtPLH)/2; # i_Dav=i_D|av (A) average discharge current\n",
"t_PLH=C*(V_DD/2)/i_Dav;\n",
"print round(t_PLH*1e9,2),\"t_PHL (ns)\"\n",
"\n",
"# 10.4d\n",
"# Case with v_t going low\n",
"i_D0=uC_n*W_n*((V_DD-V_tO)**2)/(2*2*10**-6); # i_D0=i_D(0) (A) current i_D at t=0\n",
"i_DtPHL=uC_n*W_n*((V_DD-V_tO)*v_O-(v_O**2)/2.0)/(L_n); # i_DtPHL=i_D(t_PHL) (A) current i_D at t=T_PHL\n",
"i_Dav=(i_D0+i_DtPHL)/2; # i_Dav=i_D|av (A) average discarge current\n",
"t_PHL=C*(V_DD/2)/i_Dav;\n",
"print round(t_PHL*1e9,2),\"= t_PHL (ns)\"\n",
"\n",
"# 10.4e\n",
"t_P=(t_PHL+t_PLH)/2;\n",
"print round(t_P*1e9,2),\"= t_P (ns)\"\n",
" # the answer in the textbook is slightly dirfferent due to approximation"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"3.4 = V_OH (V)\n",
"18.0 = Static current of the inverter (microA)\n",
"90.0 = Power dissipated (microW)\n",
"0.08 = The output voltage of the inverter (V) \n",
"0.24 t_PHL (ns)\n",
"0.13 = t_PHL (ns)\n",
"0.18 = t_P (ns)\n"
]
}
],
"prompt_number": 34
}
],
"metadata": {}
}
]
}