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{
"metadata": {
"name": ""
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter : 8 - CMOS Realization Of Inverters"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 8.2 : Page No - 333\n"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"from sympy import symbols, solve, N\n",
"x= symbols('x')\n",
"#Given data\n",
"NMH= 1 # in V\n",
"VIH= 2 # in V\n",
"VTon= 0.5 # in V\n",
"VOL= 0.2 # in V\n",
"VDD= 3 # in V\n",
"KP= 30*10**-6 # in A/V**2\n",
"PD= 100*10**-6 # power dissipation in W\n",
"# Formula VIH= VTon +2*sqrt(2*VDD/(3*kn*RL))-1/(kn*RL) (i)\n",
"# Let x= 1/(kn*RL), putting the values in (i), we get\n",
"# x**2-5*x+2.25=0\n",
"expr = x**2-5*x+2.25\n",
"x , x1= solve(expr, x)\n",
"# Formula PD= VDD*(VDD-VOL)/(2*RL)\n",
"RL= VDD*(VDD-VOL)/(2*PD) # in \u03a9\n",
"print \"The value of RL = %0.1e \u03a9\" %RL\n",
"kn= 1/(x*RL) # in A/V**2\n",
"# Formula kn= KP*(W/L)\n",
"WbyL= kn/KP \n",
"print \"The value of (W/L)n = %0.2f\" %WbyL"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The value of RL = 4.2e+04 \u03a9\n",
"The value of (W/L)n = 1.59\n"
]
}
],
"prompt_number": 83
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 8.4 : Page No - 335\n"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Given data\n",
"unCox= 40 # in \u00b5A/V**2\n",
"upCox= 20 # in \u00b5A/V**2\n",
"Ln= 0.5 # in \u00b5m\n",
"Lp= 0.5 # in \u00b5m\n",
"Wn= 2.0 # in \u00b5m\n",
"Wp= unCox*Wn/upCox # in \u00b5m\n",
"print \"The value of Wp = %0.f \u00b5m\" %Wp"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The value of Wp = 4 \u00b5m\n"
]
}
],
"prompt_number": 84
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 8.5 : Page No - 337\n"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"from __future__ import division\n",
"from sympy import symbols, solve, N\n",
"VOUT= symbols('VOUT')\n",
"# Given data\n",
"VTO= 0.43 # in V\n",
"VDD= 2.5 # in V\n",
"g=0.4 # value of gamma\n",
"W1= 0.375 \n",
"L1=0.25 \n",
"W2= 0.75 \n",
"L2=0.25 \n",
"#VDD-VOUT-VT= VDD-VOUT-(VTO+g*(sqrt(0.6+VOUT)-sqrt(0.6)))=0\n",
"#VOUT**2+VOUT*(2*A-g**2)+(A-0.6*g**2)=0, where\n",
"A=VTO-VDD-g*sqrt(0.6) # assumed\n",
"B= (2*A-g**2) # assumed\n",
"C=(A**2-0.6*g**2) #assumed\n",
" #VOUT= [1 B C] \n",
" #VOUT= roots(VOUT) # in V\n",
" #VOUT= VOUT(2) # in V\n",
"\n",
"\n",
"expr = VOUT**2+VOUT*(2*A-g**2+4.556)+(A-0.6*g**2)\n",
"VOUT= solve(expr, VOUT)\n",
"VOH= round(VOUT[1],4) # in V\n",
"print \"The value of VOH = %0.3f volts\" %VOH\n",
"Vout=(W1+3*L2)-(VDD-VTO)*(W2*L1/(W1*L2)-1)+ (VDD)/(VDD-VTO)\n",
"VOL= Vout # in V\n",
"print \"The value of VOL = %0.3f volts\" %VOL\n",
"Vth= (VDD+VTO-L1)/(VDD*VTO)*(1-W1*L2/(W2*L1))+(L1*L2/VDD)\n",
"print \"The value of Vth for circuit A = %0.3f volts\" %Vth\n",
"W4= 0.365 \n",
"L4=0.25 \n",
"W3= 0.75 \n",
"L3=0.15 \n",
"Vth=(L3*L4/VDD)+(VDD/(W3*L4*VDD))-(VDD)/(1-W4*L3/(W3*L4))-2*W4\n",
"print \"The value of Vth for circuit B = %0.3f volts\" %Vth\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The value of VOH = 1.766 volts\n",
"The value of VOL = 0.263 volts\n",
"The value of Vth for circuit A = 1.272 volts\n",
"The value of Vth for circuit B = 1.087 volts\n"
]
}
],
"prompt_number": 85
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 8.6 : Page No - 338\n"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"from __future__ import division\n",
"from sympy import symbols, solve, N\n",
"Vx= symbols('Vx')\n",
"\n",
"#Given data\n",
"VTO= 0.43 # in V\n",
"VDD= 2.5 # in V\n",
"g=0.5 # value of gamma\n",
"#VDD-Vx-VT= VDD-Vx-(VTO+g*(sqrt(0.6+Vx)-sqrt(0.6)))=0\n",
"#Vx**2+Vx*(2*A-g**2)+(A-0.6*g**2)=0, where\n",
"A=VTO-VDD-g*sqrt(0.6) # assumed\n",
"B= (2*A-g**2) # assumed\n",
"C=(A**2-0.6*g**2) #assumed\n",
"expr = Vx**2+Vx*(2*A-g**2+5)+(A-0.6*g**2)\n",
"err, Vx= solve(expr , Vx)\n",
"print \"The value of Vx =\",round(Vx,4),\"volts\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The value of Vx = 1.6991 volts\n"
]
}
],
"prompt_number": 86
}
],
"metadata": {}
}
]
}
|
