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| author | Jovina Dsouza | 2014-06-18 12:43:07 +0530 |
|---|---|---|
| committer | Jovina Dsouza | 2014-06-18 12:43:07 +0530 |
| commit | 206d0358703aa05d5d7315900fe1d054c2817ddc (patch) | |
| tree | f2403e29f3aded0caf7a2434ea50dd507f6545e2 /Fundamental_of_Electronics_Devices/Ch3.ipynb | |
| parent | c6f0d6aeb95beaf41e4b679e78bb42c4ffe45a40 (diff) | |
| download | Python-Textbook-Companions-206d0358703aa05d5d7315900fe1d054c2817ddc.tar.gz Python-Textbook-Companions-206d0358703aa05d5d7315900fe1d054c2817ddc.tar.bz2 Python-Textbook-Companions-206d0358703aa05d5d7315900fe1d054c2817ddc.zip | |
adding book
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diff --git a/Fundamental_of_Electronics_Devices/Ch3.ipynb b/Fundamental_of_Electronics_Devices/Ch3.ipynb new file mode 100644 index 00000000..fcc2f572 --- /dev/null +++ b/Fundamental_of_Electronics_Devices/Ch3.ipynb @@ -0,0 +1,337 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter3 : Excess Carriers in Semiconductor"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 3.2 Page No 111"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Example 3.2\n",
+ "#What is Minimum required energy \n",
+ "\n",
+ "#given data\n",
+ "l=6000 #in Angstrum\n",
+ "h=6.6*10**(-34) #Planks constant\n",
+ "c=3*10**8 #speed of light in m/s\n",
+ "e=1.602*10**(-19) #Constant\n",
+ "\n",
+ "#calculation\n",
+ "phi=c*h/(e*l*10**(-10))\n",
+ "\n",
+ "#result\n",
+ "print\"Minimum required energy is\",round(phi,2),\"eV \"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Minimum required energy is 2.06 eV \n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 3.3 Page No 112"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Exa 3.3\n",
+ "#calculate Work function of the cathode material\n",
+ "\n",
+ "#given data\n",
+ "Emax=2.5 #maximum energy of emitted electrons in eV \n",
+ "l=2537.0 #in Angstrum\n",
+ "\n",
+ "#Calculation\n",
+ "EeV=12400.0/l #in eV\n",
+ "phi=EeV-Emax #in eV\n",
+ "\n",
+ "#result\n",
+ "print \"Work function of the cathode material is \",round(phi,2),\"eV\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Work function of the cathode material is 2.39 eV\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 3.4 Page No 115"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Example 3.4\n",
+ "#Find (i)The fraction of each photon energy unit which is converted into heat\",f\n",
+ "#(ii)Energy converted into heat in ,((2-1.43)/2)*0.009,\"J/s\"\n",
+ "#(iii)Number of photons per second given off from recombination events \",0.009/(e*2)\n",
+ "\n",
+ "#given data\n",
+ "t=0.46*10**-4 #in centi meters\n",
+ "hf1=2 #in ev\n",
+ "hf2=1.43\n",
+ "Pin=10 #in mW\n",
+ "alpha=50000 # in per cm\n",
+ "e=1.6*10**-19 #constant\n",
+ "Io=0.01 #in mW\n",
+ "\n",
+ "import math\n",
+ "\n",
+ "#Calculation\n",
+ "It=Io*math.exp(-alpha*t) #in mW\n",
+ "Iabs=Io-It\n",
+ "f=(hf1-hf2)/hf1\n",
+ "E=f*Iabs\n",
+ "N=Iabs/(e*hf1)\n",
+ "\n",
+ "#result\n",
+ "print\"(i)Thus power absorbed is \",round(Iabs,3),\"J/s\"\n",
+ "print\"(ii)Energy converted into heat is\",round(E,4),\"J/s\"\n",
+ "print\"(iii)Number of photons per second given off from recombination events \",round(N,-14)\n",
+ "#In book there is calculation mistake in Number of photons."
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i)Thus power absorbed is 0.009 J/s\n",
+ "(ii)Energy converted into heat is 0.0026 J/s\n",
+ "(iii)Number of photons per second given off from recombination events 2.81e+16\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 3.5 Page No 123"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Example 3.5\n",
+ "#What is Photoconductor gain \n",
+ "#Electron transit time.\n",
+ "\n",
+ "#given data\n",
+ "L=100 #in uM\n",
+ "A=10&-7 #in cm**2\n",
+ "th=10**-6 #in sec\n",
+ "V=12 #in Volts\n",
+ "ue=0.13 #in m**2/V-s\n",
+ "uh=0.05 #in m**2/V-s\n",
+ "\n",
+ "#Calculation\n",
+ "E=V/(L*10**-6) #in V/m\n",
+ "tn=(L*10**-6)/(ue*E)\n",
+ "Gain=(1+uh/ue)*(th/tn)\n",
+ "\n",
+ "#result\n",
+ "print\"Electron transit time in sec is \",round(tn,10),\"s\"\n",
+ "print\"Photoconductor gain is \",Gain"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Electron transit time in sec is 6.4e-09 s\n",
+ "Photoconductor gain is 216.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 29
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 3.6 Page No128"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Example3.6\n",
+ "#Calculate Current flowing through diode .\n",
+ "\n",
+ "#given datex\n",
+ "import math\n",
+ "Io=0.15 #in uA\n",
+ "V=0.12 #in mVolt\n",
+ "Vt=26 #in mVolt\n",
+ "\n",
+ "#calculation\n",
+ "I=Io*10**-6*(math.exp(V/(Vt*10**-3))-1) #in A\n",
+ "\n",
+ "#result\n",
+ "print\"Current flowing through diode is \",round(I*10**6,2),\"micra A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Current flowing through diode is 15.0 micra A\n"
+ ]
+ }
+ ],
+ "prompt_number": 30
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 3.7 Page No 128"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Exa 3.7\n",
+ "#Determine the Forward voltage \n",
+ "\n",
+ "#given data\n",
+ "import math\n",
+ "Io=2.5 #in uA\n",
+ "I=10 #in mA\n",
+ "Vt=26 #in mVolt\n",
+ "n=2 #for silicon\n",
+ "\n",
+ "#Calculation\n",
+ "V=n*Vt*10**-3*math.log((I*10**-3)/(Io*10**-6))\n",
+ "\n",
+ "#Result\n",
+ "print \"Forward voltage is \",round(V,2),\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Forward voltage is 0.43 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 31
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 3.8 Page No 128"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Example 3.8\n",
+ "#What is Reverse saturation current density \n",
+ "\n",
+ "#given data\n",
+ "ND=10**21 #in m**-3\n",
+ "NA=10**22 #in m**-3\n",
+ "De=3.4*10**-3 #in m**2-s**-1\n",
+ "Dh=1.2*10**-3 #in m**2-s**-1\n",
+ "Le=7.1*10**-4 #in meters\n",
+ "Lh=3.5*10**-4 #in meters\n",
+ "ni=1.6*10**16 #in m**-3\n",
+ "e=1.602*10**-19 #constant\n",
+ "\n",
+ "#calculation\n",
+ "IoA=e*ni**2*(Dh/(Lh*ND)+De/(Le*NA))\n",
+ "\n",
+ "#Result\n",
+ "print\"Reverse saturation current density is \",round(IoA*10**6,2),\"uA \""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Reverse saturation current density is 0.16 uA \n"
+ ]
+ }
+ ],
+ "prompt_number": 33
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [],
+ "language": "python",
+ "metadata": {},
+ "outputs": []
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
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