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| author | tslee | 2014-11-27 17:17:59 +0530 |
|---|---|---|
| committer | tslee | 2014-11-27 17:17:59 +0530 |
| commit | 7b78be04fe05bf240417e22f74b3fc22e7a77d19 (patch) | |
| tree | 1875acbe01f3225bbfcc1024266dc96e515f3ea0 /Operational_Amplifiers_and_Linear_Integrated_Circuit/C2_1.ipynb | |
| parent | 8048392490bd2efe0fdfa001945f663cba969841 (diff) | |
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diff --git a/Operational_Amplifiers_and_Linear_Integrated_Circuit/C2_1.ipynb b/Operational_Amplifiers_and_Linear_Integrated_Circuit/C2_1.ipynb new file mode 100644 index 00000000..c91c71a1 --- /dev/null +++ b/Operational_Amplifiers_and_Linear_Integrated_Circuit/C2_1.ipynb @@ -0,0 +1,212 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 2:First Experiences with Op Amp"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 2.1 Page No 19"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Vplus=15\n",
+ "Vminus=-15\n",
+ "Vsatp=13\n",
+ "Vsatm=-13 #All in Volts\n",
+ "Aol=200000.0 #gain\n",
+ "#Example 2-1(a)\n",
+ "Vam=-10*(10**-6) #voltage at minus input\n",
+ "Vap=-15*(10**-6) #voltage at plus input\n",
+ "Ed1=Vap-Vam #Differential Input Voltage\n",
+ "Vout1=Ed1*Aol #Output Voltage\n",
+ "if Vout1>15:\n",
+ " print\"Value of o/p voltage1 = 13.0000V\" #positive saturation voltage\n",
+ "elif(Vout1<-15):\n",
+ " print\"Value of o/p voltage1 = -13.0000V\" #negative saturation voltage\n",
+ "else:\n",
+ " print\" Value of o/p voltage1 = \",Vout1*Vsatp,\"V\"\n",
+ "\n",
+ "#Example 2-1(b)\n",
+ "Vbm=-10*(10**-6) #voltage at minus input\n",
+ "Vbp=+15*(10**-6) #voltage at plus input\n",
+ "Ed2=Vbp-Vbm #Differential Input Voltage\n",
+ "Vout2=Ed2*Aol #Output Voltage\n",
+ "if Vout2>15:\n",
+ " print\"Value of o/p voltage2 = 13.0000V\" #positive saturation voltage\n",
+ "elif(Vout2<-15):\n",
+ " print\"Value of o/p voltage2 = -13.0000V\" #negative saturation voltage\n",
+ "else:\n",
+ " print\" Value of o/p voltage2 = \",Vout2,\"V\"\n",
+ "\n",
+ "#Example 2-1(c)\n",
+ "Vcm=-10*(10**-6) #voltage at minus input\n",
+ "Vcp=-5*(10**-6) #voltage at plus input\n",
+ "Ed3=Vcp-Vcm #Differential Input Voltage\n",
+ "Vout3=Ed3*Aol #Output Voltage\n",
+ "if(Vout3>15):\n",
+ " print\"Value of o/p voltage3 = 13.0000V\" #positive saturation voltage\n",
+ "elif(Vout3<-15):\n",
+ " print\"Value of o/p voltage3 = -13.0000V\" #negative saturation voltage\n",
+ "else:\n",
+ " print\" Value of o/p voltage3 = \",Vout3,\"V\"\n",
+ "\n",
+ "\n",
+ "#Example 2-1(d)\n",
+ "Vdm=+1.000001 #voltage at minus input\n",
+ "Vdp=+1.000000 #voltage at plus input\n",
+ "Ed4=Vdp-Vdm #Differential Input Voltage\n",
+ "Vout4=Ed4*Aol #Output Voltage\n",
+ "if(Vout4>15):\n",
+ " print\"Value of o/p voltage4 = 13.0000V\" #positive saturation voltage\n",
+ "elif(Vout4<-15):\n",
+ " print\"Value of o/p voltage4 = -13.0000V\" #negative saturation voltage\n",
+ "else:\n",
+ " print\" Value of o/p voltage4 = \",round(Vout4,2),\"V\"\n",
+ "\n",
+ "\n",
+ "#Example 2-1(e)\n",
+ "Vem=+5*(10**-3) #voltage at minus input\n",
+ "Vep=0 #voltage at plus input\n",
+ "Ed5=Vep-Vem #Differential Input Voltage\n",
+ "Vout5=Ed5*Aol #Output Voltage\n",
+ "if(Vout5>15):\n",
+ " print\"Value of o/p voltage5 = 13.0000V\" #positive saturation voltage\n",
+ "elif(Vout5<-15):\n",
+ " print\"Value of o/p voltage5 = -13.0000V\" #negative saturation voltage\n",
+ "else:\n",
+ " print\" Value of o/p voltage5 = \",Vout5\n",
+ "\n",
+ "#Example 2-1(f)\n",
+ "Vfm=0 #voltage at minus input\n",
+ "Vfp=+5*(10**-3) #voltage at plus input\n",
+ "Ed6=Vfp-Vfm #Differential Input Voltage\n",
+ "Vout6=Ed6*Aol #Output Voltage\n",
+ "if(Vout6>15):\n",
+ " print\"Value of o/p voltage6 = 13.0000V\" #positive saturation voltage\n",
+ "elif(Vout6<-15):\n",
+ " print\"Value of o/p voltage6 = -13.0000V\" #negative saturation voltage\n",
+ "else:\n",
+ " print\" Value of o/p voltage6 = V \",Vout6\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " Value of o/p voltage1 = -13.0 V\n",
+ " Value of o/p voltage2 = 5.0 V\n",
+ " Value of o/p voltage3 = 1.0 V\n",
+ " Value of o/p voltage4 = -0.2 V\n",
+ "Value of o/p voltage5 = -13.0000V\n",
+ "Value of o/p voltage6 = 13.0000V\n"
+ ]
+ }
+ ],
+ "prompt_number": 15
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 2.2 Page No 35"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "f=50.0 #in Hz\n",
+ "Vtemp=4.0 #input signal in volts\n",
+ "Ecm=10.0 \n",
+ "\n",
+ "#Calculation\n",
+ "#Example 2-2(a)\n",
+ "T=1/f\n",
+ "Th=(Vtemp*T)/Ecm #High time in seconds\n",
+ "\n",
+ "#Example 2-2(b)\n",
+ "d=(Th/T)*100 #duty cycle in percentage\n",
+ "\n",
+ "#Result\n",
+ "print\" High Time is \",Th*1000,\"ms\"\n",
+ "print\" Duty cycle is\",d,\"percent\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " High Time is 8.0 ms\n",
+ " Duty cycle is 40.0 percent\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 2.3 Page No 37"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Vtemp=4.0 #in volts\n",
+ "Ecm=5.0 #maximum peak voltage of a sawtooth carrier wave\n",
+ "T=0.01 #in seconds\n",
+ "\n",
+ "#calculation\n",
+ "Th=T*(1-(Vtemp/Ecm)) #High Time\n",
+ "\n",
+ "#Result\n",
+ "print\" High Time is \",Th*1000,\"ms\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " High Time is 2.0 ms\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file |