path: root/Electronics_Circuits_and_Systems_by_Y._N._Bapat/Ch4.ipynb

{
 "cells": [
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# Chapter 4 - Field Effect Transistors"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 4_1 Page No. 98"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 1,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "IDSS = 0.01  ampere\n",
      "VP= -4.00  volts\n",
      "VGS = -2.00  volts\n",
      "VDSmin  =VGS-VP=2.00  volts\n",
      "ID =IDSS*(1-VGS/VP)**2= 2.50e-03  ampere\n"
     ]
    }
   ],
   "source": [
    "from __future__ import division  \n",
    "IDSS=10*10**(-3)\n",
    "print \"IDSS = %0.2f\"%(IDSS),\" ampere\" #  maximum drain current\n",
    "VP=(-4)\n",
    "print \"VP= %0.2f\"%(VP),\" volts\" # pinch off voltage \n",
    "VGS=(-2)\n",
    "print \"VGS = %0.2f\"%(VGS),\" volts\" # gate to source voltage \n",
    "VDSmin=VGS-VP\n",
    "print \"VDSmin  =VGS-VP=%0.2f\"%(VDSmin),\" volts\" # Drain to source voltage \n",
    "ID=IDSS*(1-VGS/VP)**2\n",
    "print \"ID =IDSS*(1-VGS/VP)**2= %0.2e\"%(ID),\" ampere\" #   drain current"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 4_2 Page No. 99"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 6,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "IDSS = 0.01  ampere\n",
      "VP= -4.00  volts\n",
      "VGS = 0.00  volts\n",
      "RDS = 1/[(2*(IDSS/(-VP)))*(1-VGS/VP)]=200.00  ohm\n",
      "VGS = -2.00  volts\n",
      "RDS = 1/[(2*(IDSS/(-VP)))*(1-VGS/VP)]=400.00  ohm\n"
     ]
    }
   ],
   "source": [
    "from __future__ import division  \n",
    "IDSS=10*10**(-3)\n",
    "print \"IDSS = %0.2f\"%(IDSS),\" ampere\" #  maximum drain current\n",
    "VP=(-4)\n",
    "print \"VP= %0.2f\"%(VP),\" volts\" # pinch off voltage \n",
    "VGS=(0)\n",
    "print \"VGS = %0.2f\"%(VGS),\" volts\" # gate to source voltage1 \n",
    "RDS=1/((2*(IDSS/(-VP)))*(1-VGS/VP))#formula for JFET\n",
    "print \"RDS = 1/[(2*(IDSS/(-VP)))*(1-VGS/VP)]=%0.2f\"%(RDS),\" ohm\" #   drain to source resistance for VGS=0V\n",
    "VGS=(-2)\n",
    "print \"VGS = %0.2f\"%(VGS),\" volts\" # gate to source voltage2 \n",
    "RDS=1/((2*(IDSS/(-VP)))*(1-VGS/VP))\n",
    "print \"RDS = 1/[(2*(IDSS/(-VP)))*(1-VGS/VP)]=%0.2f\"%(RDS),\" ohm\" #   drain to source resistance for VGS=(-2)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 4_3 Page No. 104"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 3,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "ID = 0.01  ampere\n",
      "VDD= 24.00  volts\n",
      "VT= 5.00  volts\n",
      "VGS= 10.00  volts\n",
      "KF = ID/(VGS-VT)**2 = 4.00e-04  A/V**2\n",
      "VDS =VGS= 7.00  volts\n",
      "ID =KF*(VGS-VT)**2= 1.60e-03  ampere\n",
      "RL=(VDD-VDS)/ID= 1.06e+04 ohm\n"
     ]
    }
   ],
   "source": [
    "from __future__ import division  \n",
    "ID=10*10**(-3)\n",
    "print \"ID = %0.2f\"%(ID),\" ampere\" #  given drain current \n",
    "VDD=(24)\n",
    "print \"VDD= %0.2f\"%(VDD),\" volts\" # Drain voltage \n",
    "VT=(5)\n",
    "print \"VT= %0.2f\"%(VT),\" volts\" # Threshold voltage \n",
    "VGS=(10)\n",
    "print \"VGS= %0.2f\"%(VGS),\" volts\" # gate to source voltage1 \n",
    "KF=ID/(VGS-VT)**2\n",
    "print \"KF = ID/(VGS-VT)**2 = %0.2e\"%(KF),\" A/V**2\" # To calculate Scale factor for finding ID2\n",
    "VDS=(7)\n",
    "print \"VDS =VGS= %0.2f\"%(VDS),\" volts\" # drain to source voltage \n",
    "VGS=(VDS)\n",
    "ID=KF*(VGS-VT)**2\n",
    "print \"ID =KF*(VGS-VT)**2= %0.2e\"%(ID),\" ampere\" # New drain current for VDS=24V\n",
    "RL=(VDD-VDS)/ID\n",
    "print \"RL=(VDD-VDS)/ID= %0.2e\"%(RL)+ \" ohm\"  #calculation for load resistance at VDS=24V"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 4_4 Page No. 105"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 6,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "part(i) \n",
      "IDSS = 0.01  ampere\n",
      "VP= 2.00  volts\n",
      "IDQ = 0.00  ampere\n",
      "gm =[(2)*sqrt(IDQ*IDSS)]/VP= 4.70e-03  A/V\n",
      "part(ii) \n",
      "IDQ = 0.01  ampere\n",
      "IDSS = 0.01  ampere\n",
      "VP= 6.00  volts\n",
      "gm =[(2)*sqrt(IDQ*IDSS)]/VP= 3.17e-03  A/V\n",
      "part(iii) \n",
      "IDQ = 1.00e-03  ampere\n",
      "KF = 2.50e-04  A/V**2\n",
      "gm =sqrt(4*IDQ*KF)= 1.00e-03  A/V\n",
      "part(iv) \n",
      "IDQ = 9.10e-04  ampere\n",
      "KF = 3.75e-04  A/V**2\n",
      "gm =sqrt(4*IDQ*KF)= 1.17e-03  A/V\n"
     ]
    }
   ],
   "source": [
    "from math import sqrt\n",
    "from __future__ import division  \n",
    "print \"part(i) \"# part(i) of this question\n",
    "IDSS=5*10**(-3)\n",
    "print \"IDSS = %0.2f\"%(IDSS),\" ampere\" #  maximum drain current JFET 1\n",
    "VP=(2)\n",
    "print \"VP= %0.2f\"%(VP),\" volts\" # pinch off voltage for JFET 1\n",
    "IDQ=4.42*10**(-3)\n",
    "print \"IDQ = %0.2f\"%(IDQ),\" ampere\" #   drain current for JFET 1\n",
    "gm=((2)*sqrt(IDQ*IDSS))/VP\n",
    "print \"gm =[(2)*sqrt(IDQ*IDSS)]/VP= %0.2e\"%(gm),\" A/V\"# calculating transconductance for JFET with IDQ = 4.42 mA\n",
    "\n",
    "print \"part(ii) \"# part(ii) of this question\n",
    "IDQ=6.04*10**(-3)\n",
    "print \"IDQ = %0.2f\"%(IDQ),\" ampere\" #   drain current for JFET 1\n",
    "IDSS=15*10**(-3)\n",
    "print \"IDSS = %0.2f\"%(IDSS),\" ampere\" #  maximum drain current JFET2\n",
    "VP=(6)\n",
    "print \"VP= %0.2f\"%(VP),\" volts\" # pinch off voltage for JFET2  \n",
    "gm=((2)*sqrt(IDQ*IDSS))/VP\n",
    "print \"gm =[(2)*sqrt(IDQ*IDSS)]/VP= %0.2e\"%(gm),\" A/V\"# calculating transconductance for JFET with IDQ = 6.04 mA\n",
    "\n",
    "print \"part(iii) \"# part(iii) of this question\n",
    "IDQ=1*10**(-3)\n",
    "print \"IDQ = %0.2e\"%(IDQ),\" ampere\" #   drain current for EMOSFET 1\n",
    "KF=0.25*10**(-3)\n",
    "print \"KF = %0.2e\"%(KF),\" A/V**2\" #  Scale factor for finding EMOSFET1\n",
    "gm=sqrt(4*IDQ*KF)\n",
    "print \"gm =sqrt(4*IDQ*KF)= %0.2e\"%(gm),\" A/V\"# calculating transconductance for EMOSFET1 with IDQ = 1 mA\n",
    "\n",
    "print \"part(iv) \"# part(iv) of this question\n",
    "IDQ=0.91*10**(-3)\n",
    "print \"IDQ = %0.2e\"%(IDQ),\" ampere\" #   drain current for EMOSFET 2\n",
    "KF=0.375*10**(-3)\n",
    "print \"KF = %0.2e\"%(KF),\" A/V**2\" #  Scale factor for finding EMOSFET2\n",
    "gm=sqrt(4*IDQ*KF)\n",
    "print \"gm =sqrt(4*IDQ*KF)= %0.2e\"%(gm),\" A/V\"# calculating transconductance for EMOSFET2 with IDQ = 0.91 mA"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 4_5 Page No. 106"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 1,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "IDQmax = 0.01  ampere\n",
      "IDQmin = 0.00  ampere\n",
      "VDD= 20.00  volts\n",
      "VDSmin= 6.00  volts\n",
      "ID = 2.40e-03  ampere\n",
      "VGG= 3.00  volts\n",
      "Ri= 1.00e+05 ohm\n",
      "RF= (VGG-0)/(ID-0)= 1250.00 ohm\n",
      "R1= VDD*Ri/VGG= 6.67e+05 ohm\n",
      "R2= R1*VGG/(VDD-VGG)= 1.18e+05 ohm\n",
      "RL=[((VDD-VDSmin)/IDmax)-RF]=1550.00 ohm\n"
     ]
    }
   ],
   "source": [
    "from __future__ import division  \n",
    "IDQmax=5*10**(-3)\n",
    "print \"IDQmax = %0.2f\"%(IDQmax),\" ampere\" #  drain current for JFET for maximum transfer characteristics\n",
    "IDmax=IDQmax# maximum drain current will be given by IDQmax\n",
    "IDQmin=3*10**(-3)\n",
    "print \"IDQmin = %0.2f\"%(IDQmin),\" ampere\" #  drain current for JFET for minimum transfer characteristics \n",
    "VDD=(20)\n",
    "print \"VDD= %0.2f\"%(VDD),\" volts\" # Drain voltage supply\n",
    "VDSmin=(6)\n",
    "print \"VDSmin= %0.2f\"%(VDSmin),\" volts\" # minimum Drain to source voltage supply\n",
    "ID=2.4*10**(-3)\n",
    "print \"ID = %0.2e\"%(ID),\" ampere\" #  drain current chosen for operation within max and min limits \n",
    "VGG=3\n",
    "print \"VGG= %0.2f\"%(VGG),\" volts\" # Gate voltage from fig.\n",
    "Ri=100*10**(3)\n",
    "print \"Ri= %0.2e\"%(Ri)+ \" ohm\"  #eqivalent input resistance\n",
    "RF=(VGG-0)/(ID-0)\n",
    "print \"RF= (VGG-0)/(ID-0)= %0.2f\"%(RF)+ \" ohm\"  #calculation for  feedback resistance \n",
    "R1=VDD*Ri/VGG #using formulae VGG=VDD*Ri/R1\n",
    "print \"R1= VDD*Ri/VGG= %0.2e\"%(R1)+ \" ohm\"  #calculation for  first resistance R1 at input side\n",
    "R2=R1*VGG/(VDD-VGG)\n",
    "print \"R2= R1*VGG/(VDD-VGG)= %0.2e\"%(R2)+ \" ohm\"  #calculation for  second resistance R2 at input side\n",
    "RL=(((VDD-VDSmin)/IDmax)-RF) # using formulae VDD=IDmax(RL+RF)+VDSmin\n",
    "print \"RL=[((VDD-VDSmin)/IDmax)-RF]=%0.2f\"%(RL)+ \" ohm\"  #Load resistance calculation"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 4_6 Page No. 107"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 2,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "IDSS = 0.05  ampere\n",
      "VP= -10.00  volts\n",
      "VGSQ= -5.00  volts\n",
      "ID =IDSS*(1-VGS/VP)**2= 0.01  ampere\n",
      "RF= (VGSQ)/(ID)= 400.00 ohm\n"
     ]
    }
   ],
   "source": [
    "from __future__ import division  \n",
    "IDSS=50*10**(-3)\n",
    "print \"IDSS = %0.2f\"%(IDSS),\" ampere\" #  maximum drain current JFET \n",
    "VP=(-10)\n",
    "print \"VP= %0.2f\"%(VP),\" volts\" # pinch off voltage for JFET \n",
    "VGSQ=(-5)\n",
    "print \"VGSQ= %0.2f\"%(VGSQ),\" volts\" # Gate  operating point voltage \n",
    "ID=IDSS*(1-VGSQ/VP)**2\n",
    "print \"ID =IDSS*(1-VGS/VP)**2= %0.2f\"%(ID),\" ampere\" #   drain current JFET \n",
    "RF=abs(VGSQ/ID) \n",
    "print \"RF= (VGSQ)/(ID)= %0.2f\"%(RF)+ \" ohm\"  #calculation for  feedback resistance "
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 4_7  Page No. 108"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 3,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "IDSS = 0.01  ampere\n",
      "RL= 910.00 ohm\n",
      "RF= 2290.00 ohm\n",
      "R1= 1.20e+07 ohm\n",
      "R2= 8.57e+06 ohm\n",
      "VDD= 24.00  volts\n",
      "VP= -2.00  volts\n",
      "VGG= VDD*R2/(R1+R2)=10.00  volts\n",
      "Quadratic equation =5.244*ID**(2)-55.76*ID+144=0\n",
      "ID = 0.00  ampere\n",
      "Since ID <=IDSS, hence ID=6.214 mA cannot be chosen, so we chose ID=4.42 mA\n",
      "IDQ =0.00   A\n",
      "VGSQ = VGG-IDQ*RF = -0.12  volts\n",
      "VDSQ= VDD-IDQ*(RL+RF)= 9.86  volts\n",
      "VDGQ = VDSQ-VGSQ =9.98  volts\n",
      "VDGQ >magnitude_VP,Hence FET is in pinch off region\n"
     ]
    }
   ],
   "source": [
    "from sympy import symbols,solve\n",
    "from __future__ import division  \n",
    "IDSS=5*10**(-3)\n",
    "print \"IDSS = %0.2f\"%(IDSS),\" ampere\" #  maximum drain current JFET \n",
    "RL=910\n",
    "print \"RL= %0.2f\"%(RL)+ \" ohm\"  #Load resistance\n",
    "RF=2.29*10**(3) \n",
    "print \"RF= %0.2f\"%(RF)+ \" ohm\"  #  feedback resistance \n",
    "R1=12*10**(6)\n",
    "print \"R1= %0.2e\"%(R1)+ \" ohm\"  # first resistance R1 at input side\n",
    "R2=8.57*10**(6)\n",
    "print \"R2= %0.2e\"%(R2)+ \" ohm\"  # second resistance R2 at input side\n",
    "VDD=(24)\n",
    "print \"VDD= %0.2f\"%(VDD),\" volts\" # Drain voltage supply\n",
    "VP=(-2)\n",
    "print \"VP= %0.2f\"%(VP),\" volts\" # pinch off voltage for JFET \n",
    "VGG=(VDD*R2)/(R1+R2)\n",
    "print \"VGG= VDD*R2/(R1+R2)=%0.2f\"%(VGG),\" volts\" # Gate voltage for JFET\n",
    "print \"Quadratic equation =5.244*ID**(2)-55.76*ID+144=0\"#  Forming Quadratic equation using VGS = VGG-ID*RF and ID = IDSS(1-VGS/VP)**2 where ID in mA\n",
    "p = [5.244,  -55.76,  144]\n",
    "P=symbols('P')\n",
    "ID=solve(p[0]*P**2+p[1]*P+p[2])[0]*10**(-3)# values of ID converted into Ampere  by multiplying by 10**(-3)\n",
    "print \"ID = %0.2f\"%(ID),\" ampere\" #   drain current JFET \n",
    "print \"Since ID <=IDSS, hence ID=6.214 mA cannot be chosen, so we chose ID=4.42 mA\"\n",
    "IDQ=4.42*10**(-3) \n",
    "print \"IDQ =%0.2f\"%(IDQ),\"  A\"#Since ID <=IDSS, hence ID=6.214 mA cannot be chosen, so we chose ID=4.42 mA\n",
    "VGSQ=VGG-IDQ*RF\n",
    "print \"VGSQ = VGG-IDQ*RF = %0.2f\"%(VGSQ),\" volts\" #  Gate  operating point voltage \n",
    "VDSQ=VDD-IDQ*(RL+RF)\n",
    "print \"VDSQ= VDD-IDQ*(RL+RF)= %0.2f\"%(VDSQ),\" volts\" # Drain voltage for JFET\n",
    "VDGQ=VDSQ-VGSQ\n",
    "print \"VDGQ = VDSQ-VGSQ =%0.2f\"%(VDGQ),\" volts\" # Drain-Gate voltage for JFET\n",
    "print \"VDGQ >magnitude_VP,Hence FET is in pinch off region\""
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 4_8 Page No. 108"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 4,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "IDSS = 0.01  ampere\n",
      "RL= 910.00 ohm\n",
      "RF= 2290.00 ohm\n",
      "R1= 1.20e+07 ohm\n",
      "R2= 8.57e+06 ohm\n",
      "VDD= 24.00  volts\n",
      "VP= -6.00  volts\n",
      "VGG= VDD*R2/(R1+R2)=10.00  volts\n",
      "Quadratic equation =5.244*ID**(2)-75.68*ID+256=0\n",
      "ID = 0.01  ampere\n",
      "VDG= 9.08  volts\n",
      "since VDG < magnitude_VP for ID=9.0189 mA which is inappropriate for JFET pinch off region ,hence ID=5.4128 mA is choosen  !\n",
      "IDQ =0.01   ampere\n",
      "VGSQ = VGG-IDQ*RF = -2.39  volts\n",
      "VDSQ= VDD-IDQ*(RL+RF)= 6.69  volts\n",
      "VDGQ = VDSQ-VGSQ =9.08  volts\n",
      "VDGQ > VP,Hence FET is in pinch off region\n"
     ]
    }
   ],
   "source": [
    "from sympy import symbols,solve\n",
    "from __future__ import division  \n",
    "IDSS=15*10**(-3)\n",
    "print \"IDSS = %0.2f\"%(IDSS),\" ampere\" #  maximum drain current of JFET \n",
    "RL=910\n",
    "print \"RL= %0.2f\"%(RL)+ \" ohm\"  #Load resistance\n",
    "RF=2.29*10**(3) \n",
    "print \"RF= %0.2f\"%(RF)+ \" ohm\"  #  feedback resistance \n",
    "R1=12*10**(6)\n",
    "print \"R1= %0.2e\"%(R1)+ \" ohm\"  # first resistance R1 at input side\n",
    "R2=8.57*10**(6)\n",
    "print \"R2= %0.2e\"%(R2)+ \" ohm\"  # second resistance R2 at input side\n",
    "VDD=(24)\n",
    "print \"VDD= %0.2f\"%(VDD),\" volts\" # Drain voltage supply\n",
    "VP=(-6)\n",
    "print \"VP= %0.2f\"%(VP),\" volts\" # pinch off voltage for JFET \n",
    "VGG=(VDD*R2)/(R1+R2)\n",
    "print \"VGG= VDD*R2/(R1+R2)=%0.2f\"%(VGG),\" volts\" # Gate voltage for JFET\n",
    "print \"Quadratic equation =5.244*ID**(2)-75.68*ID+256=0\"# where ID in mA\n",
    "p = [5.244,  -75.68,  256]\n",
    "P=symbols('P')\n",
    "ID=solve(p[0]*P**2+p[1]*P+p[2])[0]*10**(-3)#values of ID converted into Ampere  by multiplying by 10**(-3)\n",
    "print \"ID = %0.2f\"%(ID),\" ampere\" #   drain current JFET \n",
    "VDG=VDD-(ID*RL)-VGG\n",
    "print \"VDG= %0.2f\"%(VDG),\" volts\" # Drain-gate voltage for JFET\n",
    "print \"since VDG < magnitude_VP for ID=9.0189 mA which is inappropriate for JFET pinch off region ,hence ID=5.4128 mA is choosen  !\" \n",
    "IDQ=5.41*10**(-3) # since VDG < magnitude_VP for ID=9.0189 mA which is inappropriate for JFET pinch off region ,hence ID=5.4128 mA is choosen  !\n",
    "print \"IDQ =%0.2f\"%(IDQ),\"  ampere\"\n",
    "VGSQ=VGG-IDQ*RF\n",
    "print \"VGSQ = VGG-IDQ*RF = %0.2f\"%(VGSQ),\" volts\" #  Gate  operating point voltage \n",
    "VDSQ=VDD-IDQ*(RL+RF)\n",
    "print \"VDSQ= VDD-IDQ*(RL+RF)= %0.2f\"%(VDSQ),\" volts\" # Drain voltage for JFET\n",
    "VDGQ=VDSQ-VGSQ\n",
    "print \"VDGQ = VDSQ-VGSQ =%0.2f\"%(VDGQ),\" volts\" # Drain-Gate voltage for JFET\n",
    "print \"VDGQ > VP,Hence FET is in pinch off region\"\n",
    "\n",
    "# NOTE :all values of book ans is wrong so give note-INCOMPLETE QUESTION\n",
    "#Roots for drain current quadratic equation are wrong in the book thus value for VGSQ,VDSQ and VDGQ are also wrong"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 4_9 Page No. 112"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 13,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "RL= 12000.00 ohm\n",
      "RF= 6000.00 ohm\n",
      "R1= 1.20e+07 ohm\n",
      "R2= 8.57e+06 ohm\n",
      "VDD= 24.00  volts\n",
      "VT= 3.00  volts\n",
      "KF= 0.00  A/V**2\n",
      "VGG= VDD*R2/(R1+R2)=10.00  volts\n",
      "Quadratic equation =9*ID**(2)-25*ID+16=0\n",
      "ID = 0.00  A\n",
      "VGS = VGG-ID*RF = 4.00  volts\n",
      "Since VGS < VT for ID=1.78 mA, hence ID = 1.78 mA cannot be chosen, so we chose ID= 1 mA as operating drain current IDQ\n",
      "IDQ =0.00 A\n",
      "VGSQ = VGG-IDQ*RF = 4.00  volts\n",
      "VDSQ= VDD-IDQ*(RL+RF)= 6.00  volts\n"
     ]
    }
   ],
   "source": [
    "from sympy import symbols,solve\n",
    "from __future__ import division  \n",
    "RL=12*10**(3)\n",
    "print \"RL= %0.2f\"%(RL)+ \" ohm\"  #Load resistance\n",
    "RF=6*10**(3) \n",
    "print \"RF= %0.2f\"%(RF)+ \" ohm\"  #  feedback resistance \n",
    "R1=12*10**(6)\n",
    "print \"R1= %0.2e\"%(R1)+ \" ohm\"  # first resistance R1 at input side\n",
    "R2=8.57*10**(6)\n",
    "print \"R2= %0.2e\"%(R2)+ \" ohm\"  # second resistance R2 at input side\n",
    "VDD=(24)\n",
    "print \"VDD= %0.2f\"%(VDD),\" volts\" # Drain voltage supply\n",
    "VT=(3)\n",
    "print \"VT= %0.2f\"%(VT),\" volts\" # Threshold voltage for n-channel EMOSFET\n",
    "KF=0.25*10**(-3)\n",
    "print \"KF= %0.2f\"%(KF),\" A/V**2\" # Constant for n-channel EMOSFET \n",
    "VGG=(VDD*R2)/(R1+R2)\n",
    "print \"VGG= VDD*R2/(R1+R2)=%0.2f\"%(VGG),\" volts\" # Gate voltage for n-channel EMOSFET \n",
    "print \"Quadratic equation =9*ID**(2)-25*ID+16=0\"# IDS=KF*(VGS-VT)**2 and VGS=VGG-ID*RD ,so Quadratic equation formed is :IDS=KF*(VGG-ID*RD-VT)**2 where ID in mA\n",
    "p = [9,  -25,  16]\n",
    "P=symbols('P')\n",
    "ID=solve(p[0]*P**2+p[1]*P+p[2])[0]*10**(-3)#values of ID converted into Ampere  by multiplying by 10**(-3)\n",
    "print \"ID = %0.2f\"%(ID),\" A\" #   drain current n-channel EMOSFET in Ampere \n",
    "VGS=VGG-ID*RF# For ID=1.78 mA and ID=1mA\n",
    "print \"VGS = VGG-ID*RF = %0.2f\"%(VGS),\" volts\" #  Gate  operating point voltage \n",
    "print \"Since VGS < VT for ID=1.78 mA, hence ID = 1.78 mA cannot be chosen, so we chose ID= 1 mA as operating drain current IDQ\"\n",
    "IDQ=1*10**(-3)\n",
    "print \"IDQ =%0.2f\"%(IDQ),\"A\"#Since VGS < VT for ID=1.78 mA, hence ID = 1.78 mA cannot be chosen, so we chose ID= 1 mA as operating drain current IDQ\n",
    "VGSQ=VGG-IDQ*RF\n",
    "print \"VGSQ = VGG-IDQ*RF = %0.2f\"%(VGSQ),\" volts\" #  Gate  operating point voltage \n",
    "VDSQ=VDD-IDQ*(RL+RF)\n",
    "print \"VDSQ= VDD-IDQ*(RL+RF)= %0.2f\"%(VDSQ),\" volts\" # Drain voltage for n-channel EMOSFET \n",
    "# NOTE:Value of VGS= -0.6676390 volts for ID=1.78 mA but in book given as -0.68 V"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 4_10 Page No. 113"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 5,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "RL= 12000.00 ohm\n",
      "RF= 6000.00 ohm\n",
      "R1= 1.20e+07 ohm\n",
      "R2= 8.57e+06 ohm\n",
      "VDD= 24.00  volts\n",
      "VT= 3.00  volts\n",
      "KF= 0.00  A/V**2\n",
      "VGG= VDD*R2/(R1+R2)=10.00  volts\n",
      "Quadratic equation =36*ID**(2)-86.67*ID+49=0\n",
      "ID = 0.00  A\n",
      "VGS = VGG-ID*RF = 4.56  volts\n",
      "Since VGS < VT for ID=1.5 mA, hence ID = 1.5 mA cannot be chosen, so we chose ID= 0.91 mA as operating drain current IDQ\n",
      "IDQ =0.00  A\n",
      "change in IDQ = 9.00  percent\n",
      "VGSQ = VGG-IDQ*RF = 4.54  volts\n",
      "VDSQ= VDD-IDQ*(RL+RF)= 7.62  volts\n"
     ]
    }
   ],
   "source": [
    "from math import sqrt\n",
    "from __future__ import division  \n",
    "RL=12*10**(3)\n",
    "print \"RL= %0.2f\"%(RL)+ \" ohm\"  #Load resistance\n",
    "RF=6*10**(3) \n",
    "print \"RF= %0.2f\"%(RF)+ \" ohm\"  #  feedback resistance \n",
    "R1=12*10**(6)\n",
    "print \"R1= %0.2e\"%(R1)+ \" ohm\"  # first resistance R1 at input side\n",
    "R2=8.57*10**(6)\n",
    "print \"R2= %0.2e\"%(R2)+ \" ohm\"  # second resistance R2 at input side\n",
    "VDD=(24)\n",
    "print \"VDD= %0.2f\"%(VDD),\" volts\" # Drain voltage supply\n",
    "VT=(3)\n",
    "print \"VT= %0.2f\"%(VT),\" volts\" # Threshold voltage for n-channel EMOSFET\n",
    "KF=0.375*10**(-3)\n",
    "print \"KF= %0.2f\"%(KF),\" A/V**2\" # Constant for  n-channel EMOSFET \n",
    "VGG=(VDD*R2)/(R1+R2)\n",
    "print \"VGG= VDD*R2/(R1+R2)=%0.2f\"%(VGG),\" volts\" # Gate voltage for  n-channel EMOSFET \n",
    "print \"Quadratic equation =36*ID**(2)-86.67*ID+49=0\"# IDS=KF*(VGS-VT)**2 and VGS=VGG-ID*RD ,so Quadratic equation formed is :IDS=KF*(VGG-ID*RD-VT)**2 ,where ID in mA\n",
    "p = [36,  -86.67,  49]\n",
    "P=symbols('P')\n",
    "ID=solve(p[0]*P**2+p[1]*P+p[2])[0]*10**(-3)#values of ID converted into Ampere  by multiplying by 10**(-3)\n",
    "print \"ID = %0.2f\"%(ID),\" A\" #   drain current n-channel EMOSFET in Ampere \n",
    "VGS=VGG-ID*RF#  Gate voltage for ID = 1.5 mA and ID = 0.91 mA\n",
    "print \"VGS = VGG-ID*RF = %0.2f\"%(VGS),\" volts\" #  Gate voltage \n",
    "print \"Since VGS < VT for ID=1.5 mA, hence ID = 1.5 mA cannot be chosen, so we chose ID= 0.91 mA as operating drain current IDQ\"\n",
    "IDQ=0.91*10**(-3)\n",
    "print \"IDQ =%0.2f\"%(IDQ),\" A\"#Since VGS < VT for ID=1.5 mA, hence ID = 1.5 mA cannot be chosen, so we chose ID= 0.91 mA as operating drain current IDQ\n",
    "change_IDQ=((1-0.91)*100)/(1)# \n",
    "print \"change in IDQ = %0.2f\"%(change_IDQ),\" percent\"# Percent change in IDQ from value 1 mA from its actual value IDQ=0.91mA\n",
    "VGSQ=VGG-IDQ*RF\n",
    "print \"VGSQ = VGG-IDQ*RF = %0.2f\"%(VGSQ),\" volts\" #  Gate  operating point voltage \n",
    "VDSQ=VDD-IDQ*(RL+RF)\n",
    "print \"VDSQ= VDD-IDQ*(RL+RF)= %0.2f\"%(VDSQ),\" volts\" # Drain voltage for n-channel EMOSFET \n",
    "\n",
    "\n",
    "# note: in the textbook author has given KF = .375 but I have work with KF = .375*10**-3A/V**2"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 4_11  Page No. 114"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 25,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "RF= 6000.00 ohm\n",
      "VDD= 20.00  volts\n",
      "part(i) \n",
      "VT= 2.00  volts\n",
      "KF= 0.00  A/V**2\n",
      "ID = 0.00  A\n",
      "RL=[VDD-VT-sqrt(ID/KF)]/ID= 16000.00 ohm\n",
      "part(ii) \n",
      "VT= 3.00  volts\n",
      "KF= 0.00  A/V**2\n",
      "Quadratic equation =(256)*ID**(2)-(546.67)*ID+289=0\n",
      "ID = 0.00  A\n",
      "VDS =VDD-ID*RL = 4.60  volts\n",
      "IDQ =0.00  A\n",
      "Percentage change= 3.80  percent\n"
     ]
    }
   ],
   "source": [
    "from math import sqrt\n",
    "from sympy import symbols,solve\n",
    "from __future__ import division  \n",
    "RF=6*10**(3) \n",
    "print \"RF= %0.2f\"%(RF)+ \" ohm\"  #  feedback resistance \n",
    "VDD=(20)\n",
    "print \"VDD= %0.2f\"%(VDD),\" volts\" # Drain voltage supply\n",
    "print \"part(i) \"# part(i) of this question\n",
    "VT=(2)\n",
    "print \"VT= %0.2f\"%(VT),\" volts\" # Threshold voltage for EMOSFET\n",
    "KF=0.25*10**(-3)\n",
    "print \"KF= %0.2f\"%(KF),\" A/V**2\" # Constant for EMOSFET \n",
    "ID=1*10**(-3)\n",
    "print \"ID = %0.2f\"%(ID),\" A\" #   drain current EMOSFET in Ampere \n",
    "RL=(VDD-VT-sqrt(ID/KF))/ID # Using formulae ID=KF*(VDD-ID*RL-VT)\n",
    "print \"RL=[VDD-VT-sqrt(ID/KF)]/ID= %0.2f\"%(RL)+ \" ohm\"  #Load resistance\n",
    "print \"part(ii) \"# part(ii) of this question\n",
    "VT=(3)\n",
    "print \"VT= %0.2f\"%(VT),\" volts\" # Threshold voltage for EMOSFET\n",
    "KF=0.375*10**(-3)\n",
    "print \"KF= %0.2f\"%(KF),\" A/V**2\" # Constant for EMOSFET \n",
    "print \"Quadratic equation =(256)*ID**(2)-(546.67)*ID+289=0\"#IDS=KF*(VGS-VT)**2 =KF*(VDS-VT)**2 and VDS=VDD-ID*RL,so Quadratic equation  is:IDS=KF*(VDD-ID*RL-VT)**2 ,where ID in mA\n",
    "p = [256,  -546.66 , 289]\n",
    "P=symbols('P')\n",
    "ID=solve(p[0]*P**2+p[1]*P+p[2])[0]*10**(-3)#values of ID converted into Ampere  by multiplying by 10**(-3)\n",
    "print \"ID = %0.2f\"%(ID),\" A\" #   drain current EMOSFET in Ampere \n",
    "VDS=VDD-ID*RL# Drain voltage for ID = 1.173 mA and ID = 0.962 mA\n",
    "print \"VDS =VDD-ID*RL = %0.2f\"%(VDS),\" volts\" #  Drain voltage \n",
    "IDQ=0.962*10**(-3)\n",
    "print \"IDQ =%0.2f\"%(IDQ),\" A\"#Since VDS < VT for ID=1.173 mA, hence ID = 1.173 mA cannot be chosen, so we chose ID= 0.962 mA as operating drain current IDQ\n",
    "Percentage_change=((1-0.962)*100)/(1)\n",
    "print \"Percentage change= %0.2f\"%(Percentage_change),\" percent\"# Percent change in IDQ  value from 1 mA(part(i)) to its  value ( of part(ii))IDQ=0.91mA\n",
    "# NOTE: part(ii):the values of ID = 1.173 mA or ID = 0.962 mA but in book given as ID= 1.197 mA and ID = 0.939 mA .Hence (correct) Percentage_change in ID= 3.8 % but in book given as 6.1 % \n",
    "# ANS is not correct check &correct"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 4_12 Page No. 115"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 15,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "VDD= 5.00  volts\n",
      "RL1= 1.25e+05 ohm\n",
      "RL2= 2.00e+05 ohm\n",
      "IDON1 =0.00  A\n",
      "IDON2 =0.00  A\n",
      "VDON1=VDD-IDON1*RL1= 0.64  volts\n",
      "VDON2=VDD-IDON2*RL2= 0.50  volts\n"
     ]
    }
   ],
   "source": [
    "from __future__ import division  \n",
    "VDD=(5)\n",
    "print \"VDD= %0.2f\"%(VDD),\" volts\" # Drain voltage supply\n",
    "RL1=125*10**(3)\n",
    "print \"RL1= %0.2e\"%(RL1)+ \" ohm\"  #Load resistance\n",
    "RL2=200*10**(3)\n",
    "print \"RL2= %0.2e\"%(RL2)+ \" ohm\"  #Load resistance\n",
    "IDON1=34.88*10**(-6)\n",
    "print \"IDON1 =%0.2f\"%(IDON1),\" A\"#Drain current for load line 1 from fig.\n",
    "IDON2=22.5*10**(-6)\n",
    "print \"IDON2 =%0.2f\"%(IDON2),\" A\"#Drain current for load line 2 from fig.\n",
    "VDON1=VDD-IDON1*RL1\n",
    "print \"VDON1=VDD-IDON1*RL1= %0.2f\"%(VDON1),\" volts\" #  output voltage  at drain terminal for IDON1\n",
    "VDON2=VDD-IDON2*RL2\n",
    "print \"VDON2=VDD-IDON2*RL2= %0.2f\"%(VDON2),\" volts\" #  output voltage  at drain terminal for IDON2"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 4_13 Page No. 120"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 28,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "IDSS = 0.01  ampere\n",
      "VP= -4.00  volts\n",
      "VGS= 0.00  volts\n",
      "VDD= 10.00  volts\n",
      "RL= 500.00 ohm\n",
      "VDS=VDD-ID*RL= 5.00  volts\n",
      "VDS>VP,so pinch off region\n",
      "RL= 750.00 ohm\n",
      "VDS=VDD-ID*RL= 2.50  volts\n",
      "VDS<VP,so ohmic region\n"
     ]
    }
   ],
   "source": [
    "from __future__ import division  \n",
    "IDSS=10*10**(-3)\n",
    "print \"IDSS = %0.2f\"%(IDSS),\" ampere\" #  maximum drain current for n-channel DEMOSFET\n",
    "ID=IDSS # since VGS=0V, so ID=maximum\n",
    "VP=(-4)\n",
    "print \"VP= %0.2f\"%(VP),\" volts\" # pinch off voltage \n",
    "VGS=(0)\n",
    "print \"VGS= %0.2f\"%(VGS),\" volts\" # Gate to source voltage \n",
    "VDD=(10)\n",
    "print \"VDD= %0.2f\"%(VDD),\" volts\" # Drain supply voltage \n",
    "RL=0.5*10**(3)\n",
    "print \"RL= %0.2f\"%(RL)+ \" ohm\"  #Load resistance\n",
    "VDS=VDD-ID*RL\n",
    "print \"VDS=VDD-ID*RL= %0.2f\"%(VDS),\" volts\" # Drain to source  voltage ,since VDS>VP DEMOSFET is in pinch off\n",
    "print \"VDS>VP,so pinch off region\"\n",
    "RL=0.75*10**(3)\n",
    "print \"RL= %0.2f\"%(RL)+ \" ohm\"  # New Load resistance value\n",
    "VDS=VDD-ID*RL\n",
    "print \"VDS=VDD-ID*RL= %0.2f\"%(VDS),\" volts\" # New Drain to source  voltage for RL=750 ohm\n",
    "print \"VDS<VP,so ohmic region\"# since VDS < VP DEMOSFET is in ohmic region for RL=750 ohm and hence will not operate as a current source"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 4_14 Page No. 122"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 2,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "KF1 = 2.50e-04  A/V**2\n",
      "KF2 = 2.50e-04  A/V**2\n",
      "IQ= 1.00e-03  ampere\n",
      "VT1 = 2.00  volts\n",
      "VT2 = 2.00  volts\n",
      "VDD= 15.00  volts\n",
      "IREF =IQ= 1.00e-03  ampere\n",
      "VGS= VT1+sqrt(2*IREF/KF1)=4.83  volts\n",
      "R= (VDD-VGS)/IREF=10171.57 ohm\n"
     ]
    }
   ],
   "source": [
    "from math import sqrt\n",
    "from __future__ import division  \n",
    "KF1=0.25*10**(-3)\n",
    "print \"KF1 = %0.2e\"%(KF1),\" A/V**2\" #  Scale factor \n",
    "KF2=KF1\n",
    "print \"KF2 = %0.2e\"%(KF2),\" A/V**2\" #  Scale factor \n",
    "IQ=1*10**(-3)\n",
    "print \"IQ= %0.2e\"%(IQ),\" ampere\" # constant current  source value\n",
    "VT1=2\n",
    "print \"VT1 = %0.2f\"%(VT1),\" volts\"# Threshold voltage\n",
    "VT2=VT1\n",
    "print \"VT2 = %0.2f\"%(VT2),\" volts\"# Threshold voltage\n",
    "VDD=(15)\n",
    "print \"VDD= %0.2f\"%(VDD),\" volts\" # Drain supply voltage \n",
    "IREF=IQ\n",
    "print \"IREF =IQ= %0.2e\"%(IREF),\" ampere\" # Reference current value\n",
    "VGS=VT1+sqrt(2*IREF/KF1) # Formulae\n",
    "print \"VGS= VT1+sqrt(2*IREF/KF1)=%0.2f\"%(VGS),\" volts\" # Gate to source voltage \n",
    "R=(VDD-VGS)/IREF\n",
    "print \"R= (VDD-VGS)/IREF=%0.2f\"%(R)+ \" ohm\"  # resistance value to obtain constant current\n",
    "#  ERROR NOTE:values of VGS and R (correct) are 4.8284271 volts and 10171.573 ohm  respectively but given in book are VGS=4V and R=11 kilo ohms"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 4_15 Page No. 123"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 17,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "RON= 100.00 ohm\n",
      "ROFF= 1.00e+10 ohm\n",
      "Vip= 1.00  volts\n",
      "Rs= 100.00 ohm\n",
      "RL= 10000.00 ohm\n",
      "part(i) \n",
      "Vo=(Vip*RL)/(RL+RON+Rs)= 0.98  volts\n",
      "ErON=[Vip*(RON+Rs)/(RL+RON+Rs)]*100= 1.96  percent\n",
      "vOFF=(Vip*RL)/ROFF= 0.00  volts\n",
      "OFF_isolation=20*log10(Vip/vOFF)= 120.00  dB\n",
      "part(ii) \n",
      "vOFF=(Vip*RON)/(Rs+RON)= 0.50  volts\n",
      "OFF_isolation=20*log10((Rs+RON)/RON)= 6.02  dB\n"
     ]
    }
   ],
   "source": [
    "from math import log10\n",
    "from __future__ import division  \n",
    "RON=100\n",
    "print \"RON= %0.2f\"%(RON)+ \" ohm\"  #ON resistance of  analog series switch\n",
    "ROFF=10**(10)\n",
    "print \"ROFF= %0.2e\"%(ROFF)+ \" ohm\"  #OFF resistance analog series switch\n",
    "Vip=1\n",
    "print \"Vip= %0.2f\"%(Vip),\" volts\"# Peak amplitude of analog voltage\n",
    "Rs=100\n",
    "print \"Rs= %0.2f\"%(Rs)+ \" ohm\"  #Voltage source resistance\n",
    "RL=10*10**(3)\n",
    "print \"RL= %0.2f\"%(RL)+ \" ohm\"  #Load resistance\n",
    "print \"part(i) \"# part(i) of this question\n",
    "Vo=(Vip*RL)/(RL+RON+Rs)\n",
    "print \"Vo=(Vip*RL)/(RL+RON+Rs)= %0.2f\"%(Vo),\" volts\"# ON voltage\n",
    "ErON=(Vip*(RON+Rs)/(RL+RON+Rs))*100\n",
    "print \"ErON=[Vip*(RON+Rs)/(RL+RON+Rs)]*100= %0.2f\"%(ErON),\" percent\"# Output voltage error \n",
    "vOFF=(Vip*RL)/ROFF\n",
    "print \"vOFF=(Vip*RL)/ROFF= %0.2f\"%(vOFF),\" volts\"# Output voltage in OFF state\n",
    "OFF_isolation=20*log10(Vip/vOFF)\n",
    "print \"OFF_isolation=20*log10(Vip/vOFF)= %0.2f\"%(OFF_isolation),\" dB\" # OFF_isolation=20*log10(Vip/vOFF) in dB# Thus ON error and OFF isolation decrease with increasing values of RL.\n",
    "print \"part(ii) \"# part(ii) of this question\n",
    "vOFF=(Vip*RON)/(Rs+RON)\n",
    "print \"vOFF=(Vip*RON)/(Rs+RON)= %0.2f\"%(vOFF),\" volts\"# Output voltage in OFF state for analog shunt switch\n",
    "OFF_isolation=20*log10((Rs+RON)/RON)# OFF_isolation of shunt switch\n",
    "print \"OFF_isolation=20*log10((Rs+RON)/RON)= %0.2f\"%(OFF_isolation),\" dB\"# Thus shunt switch is inferior to series switch in its OFF isolation property\n",
    "\n",
    "# ERROR NOTE: in question the author has given RL = 10K but in solution he has used RL = 1 k ... I have done programming using RL = 10 K."
   ]
  }
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