{
"metadata": {
"name": "",
"signature": "sha256:606a5c9e48de665a7bdbd2fe9ec84d0a776a619019e1e88b4d21b1affe9620a0"
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 4: Junction Properties"
]
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 4.1 page No. 146"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
" \n",
"#given data\n",
"import math\n",
"T=300\t\t\t #in Kelvin\n",
"ND=5*10**13\t\t #in cm**-3\n",
"NA=0\t\t\t #in cm**-3\n",
"ni=2.4*10**13\t\t#in cm**-3\n",
"\n",
"#Calculation\n",
"no=ND/2.0+math.sqrt((ND/2.0)**2+ni**2)\t#in cm**-3\n",
"po=ni**2/no\t\t#in cm**-3\n",
"\n",
"#Result\n",
"print\"Majority carrier electron concentration is \",round(no,-11),\"cm**-3\"\n",
"print\"Minority carrier hole concentration is \",round(po,-11),\" cm**-3\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Majority carrier electron concentration is 5.97e+13 cm**-3\n",
"Minority carrier hole concentration is 9.7e+12 cm**-3\n"
]
}
],
"prompt_number": 14
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 4.2 Page No.146"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
" \n",
"#given data\n",
"import math\n",
"T=300\t\t\t#in Kelvin\n",
"ND=10**16\t\t#in cm**-3\n",
"NA=0\t\t\t #in cm**-3\n",
"ni=1.5*10**10\t\t#in cm**-3\n",
"\n",
"#Calculation\n",
"no=ND/2.0+math.sqrt((ND/2.0)**2+ni**2)\t#in cm**-3\n",
"po=ni**2/no\t\t#in cm**-3\n",
"\n",
"#result\n",
"print\"Majority carrier electron concentration is \",no,\"cm**-3\"\n",
"print\"Minority carrier hole concentration is \",round(po,0),\" cm**-3\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Majority carrier electron concentration is 1e+16 cm**-3\n",
"Minority carrier hole concentration is 22500.0 cm**-3\n"
]
}
],
"prompt_number": 16
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 4.3 Page No. 147"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
" \n",
"#given data\n",
"import math\n",
"T=300\t\t\t#in Kelvin\n",
"ND=3*10**15\t\t#in cm**-3\n",
"NA=10**16\t\t#in cm**-3\n",
"ni=1.6*10**10\t\t#in cm**-3\n",
"\n",
"#Calculation\n",
"po=(NA-ND)/2+math.sqrt(((NA-ND)/2.0)**2+ni**2.0)\t#in cm**-3\n",
"no=ni**2/po\t\t#in cm**-3\n",
"\n",
"#Result\n",
"print\"Majority carrier hole concentration is\",round(po,-8),\" cm**-3\"\n",
"print\"Minority carrier electron concentration is \",round(no,0),\" cm**-3\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Majority carrier hole concentration is 7e+15 cm**-3\n",
"Minority carrier electron concentration is 36571.0 cm**-3\n"
]
}
],
"prompt_number": 19
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 4.4 Page No. 147"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
" \n",
"#Given \n",
"import math\n",
"ND=3*10**15\t\t#in cm**-3\n",
"Eg=1.12 #eV\n",
"k=8.62*10**-5 #eV/k\n",
"Nc=2.8*10**19\n",
"Nv=1.04*10**19\n",
"\n",
"#Calculation\n",
"import math\n",
"# from the equation po=(NA-ND)/2+math.sqrt(((NA-ND)/2.0)**2+ni**2.0)\t#in cm**-3\n",
"No=1.05*ND\n",
"ni=math.sqrt((No-ND/2.0)**2-0.25*ND**2)\n",
"#From ni**2=Nc*Nv*exp(-Eg/(k*t))\n",
"T=Eg/(-math.log(ni**2/(Nc*Nv))*k)\n",
"\n",
"#Result\n",
"print \"The maximum Temprature is \",round(T,1),\"K\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The maximum Temprature is 642.0 K\n"
]
}
],
"prompt_number": 45
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 4.5 Page No. 151"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
" \n",
" \n",
"#given data\n",
"import math\n",
"T=300\t\t#in Kelvin\n",
"ND=10**15\t#in cm**-3\n",
"NA=10**18\t#in cm**-3\n",
"ni=1.5*10**10\t#in cm**-3\n",
"VT=T/11600.0\t#in Volts\n",
"\n",
"#Calculation\n",
"Vbi=VT*math.log(NA*ND/ni**2)\t#in Volts\n",
"\n",
"#result\n",
"print\"Built in potential barrier is\",round(Vbi,4),\"V\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Built in potential barrier is 0.7532 V\n"
]
}
],
"prompt_number": 47
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 4.6 Page No.151"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
" \n",
"T=300\t\t #in Kelvin\n",
"ND=10**21\t #in m**-3\n",
"NA=10**21\t #in m**-3\n",
"ni=1.5*10**16 #in m**-3\n",
"VT=T/11600.0\t#in Volts\n",
"\n",
"#Calculation\n",
"import math\n",
"Vo=VT*math.log(NA*ND/ni**2)\t#in Volts\n",
"\n",
"#result\n",
"print\"Contact potential is\",round(Vo,4),\"V\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Contact potential is 0.5745 V\n"
]
}
],
"prompt_number": 52
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 4.7 Page No. 154"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
" \n",
"#given data\n",
"import math\n",
"T=300\t\t\t#in Kelvin\n",
"ND=10**15\t\t#in cm**-3\n",
"NA=10**16\t\t#in cm**-3\n",
"ni=1.5*10**10\t\t#in cm**-3\n",
"VT=T/11600.0\t\t#in Volts\n",
"e=1.6*10**-19\t #in Coulamb\n",
"\n",
"#calculation\n",
"epsilon=11.7*8.854*10**-14\t #constant\n",
"Vbi=VT*math.log(NA*ND/ni**2)\t\t#in Volts\n",
"SCW=math.sqrt((2*epsilon*Vbi/e)*(NA+ND)/(NA*ND))#in cm\n",
"SCW=SCW*10**4 #in uMeter\n",
"xn=0.864\t\t#in uM\n",
"xp=0.086\t\t#in uM\n",
"Emax=-e*ND*xn/epsilon\t#in V/cm\n",
"\n",
"#result\n",
"print\"Space charge width is\",round(SCW,2),\"micro meter\"\n",
"print\"At metallurgical junction, i.e for x=0 the electric field is \",round(Emax/10000,0),\"V\"#Note : Ans in the book is wrong"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Space charge width is 0.95 micro meter\n",
"At metallurgical junction, i.e for x=0 the electric field is -13345.0 V\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 4.8 Page No.160"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
" \n",
"#given data\n",
"import math\n",
"Ecf=0.3 #in Volts\n",
"T=27.0+273.0 #in Kelvin\n",
"delT=55 #in degree centigrade\n",
"\n",
"#calculation\n",
"#formula : Ecf=Ec-Ef=K*T*math.log(nc/ND)\n",
"#let K*math.log(nc/ND)=y\n",
"#Ecf=Ec-Ef=T*y\n",
"y=Ecf/T #assumed\n",
"Tnew=273+55 #in Kelvin\n",
"EcfNEW=y*Tnew #in Volts\n",
"\n",
"#result\n",
"print\"New position of fermi level is \",round(EcfNEW,4),\"V\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"New position of fermi level is 0.328 V\n"
]
}
],
"prompt_number": 9
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 4.9 Page No. 161"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
" \n",
" \n",
"#given data\n",
"import math\n",
"T=300\t\t\t#in Kelvin\n",
"ND=8*10**14\t\t#in cm**-3\n",
"NA=8*10**14\t\t#in cm**-3\n",
"ni=2*10**13\t\t#in cm**-3\n",
"k=8.61*10**-5\t\t#in eV/K\n",
"\n",
"#calculation\n",
"Vo=k*T*math.log(NA*ND/ni**2)\t#in Volts\n",
"\n",
"#Result\n",
"print\"Contact potential is \",round(Vo,2),\"V\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Contact potential is 0.19 V\n"
]
}
],
"prompt_number": 7
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 4.10 page No.161"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
" \n",
"#given data\n",
"ND=2*10**16 #in cm**-3\n",
"NA=5*10**15 #in cm**-3\n",
"Ao=4.83*10**21 \t#constant\n",
"T=300.0\t\t\t #in Kelvin\n",
"EG=1.1\t \t \t #in eV\n",
"kT=0.026 \t\t#in eV\n",
"\n",
"#Calculation\n",
"ni=Ao*T**(1.5)*math.exp(-EG/(2*kT))\t\t#in m**-3\n",
"p=(ni/10**6)**2/ND\t\t\t#in cm**-3\n",
"n=((ni/10**6)**2)/NA\t\t\t#in cm**-3\n",
"\n",
"#Result\n",
"\n",
"print\"Hole concentration in cm**-3 : %.1e\"%round(p,0),\"/cm**3\"\n",
"print\"electron concentration in cm**-3 :%.1e\"%round(n,0),\"/cm**3\"\n",
"print\"\\nNOTE:\\nSlight Variation in answer due to wrong value of ni in book as 1.6*10**16 instead of\",ni\n",
"if n < e:\n",
" \n",
" print\"\\n\\nthe given Si is of P-type\" \n",
"else:\n",
" print \"\\nThe given Si is of N-type\"\n",
" "
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Hole concentration in cm**-3 : 1.3e+04 /cm**3\n",
"electron concentration in cm**-3 :5.3e+04 /cm**3\n",
"\n",
"NOTE:\n",
"Slight Variation in answer due to wrong value of ni in book as 1.6*10**16 instead of 1.63166259315e+16\n",
"\n",
"The given Si is of N-type\n"
]
}
],
"prompt_number": 42
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 4.11 Page No. 168"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
" \n",
"V=5\t\t #in volts\n",
"Vo=0.7\t #in Volts\n",
"R=100\t\t#in Kohm\n",
"\n",
"#Calculation\n",
"I=(V-Vo)/R\t#in Ampere\n",
"\n",
"#result\n",
"print\"Current flowing through the circuit is\",round(I*1000,0),\"mA\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Current flowing through the circuit is 43.0 mA\n"
]
}
],
"prompt_number": 6
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 4.12 Page No. 168"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
" \n",
"V=15\t\t\t #in volts\n",
"Vo=0.7\t\t\t#in Volts\n",
"R=7\t \t \t#in Kohm\n",
"\n",
"#Calculation\n",
"I=(V-2*Vo)/R\n",
"I=(V-2*Vo)/R\t\t#in mAmpere\n",
"VA=I*R\t \t\t#in Volts\n",
"\n",
"#result\n",
"print\"Voltagee VA is \",VA,\"V\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Voltagee VA is 13.6 V\n"
]
}
],
"prompt_number": 20
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 4.13 Page No.169"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
" \n",
"V=15 #V, voltage\n",
"Vb=0.3 #V, Barrier Potential #When supply is switched on\n",
"\n",
"#Calculation\n",
"VA=V-Vb\n",
"\n",
"#Result\n",
"print\"The Voltage VA is \",VA,\"V\"\n",
"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The Voltage VA is 14.7 V\n"
]
}
],
"prompt_number": 23
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 4.14 Page No.172"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
" \n",
"#given data\n",
"Vz=5\t\t\t#in volts\n",
"to=25\t\t\t#in degree centigrade\n",
"t=100\t\t\t#in degree centigrade\n",
"Vdrop=4.8\t\t#in Volts\n",
"\n",
"#calculation\n",
"delVz=Vdrop-Vz\t\t#in Volts\n",
"delt=t-to\t\t#in degree centigrade\n",
"TempCoeff=delVz*100/(Vz*delt)\n",
"\n",
"#result\n",
"print\"Temperature coefficient f zener diode is \",round(TempCoeff,3),\"percent\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Temperature coefficient f zener diode is -0.053 percent\n"
]
}
],
"prompt_number": 22
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 4.15 Page No. 174"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
" \n",
"Vz=8.0\t\t\t#in volts\n",
"VS=12.0\t\t\t#in volts\n",
"RL=10.0\t\t\t#in Kohm\n",
"Rs=5.0\t\t\t#in Kohm\n",
"\n",
"#part (a)\n",
"Vout=Vz\t\t\t#in volts\n",
"\n",
"#part (b)\n",
"Vrs=VS-Vout\t\t#in volts\n",
"IL=Vout/RL \t\t#in mAmpere\n",
"Is=(VS-Vout)/Rs\t#in mAmpere\n",
"\n",
"#part c\n",
"Iz=Is-IL\t \t#in mAmpere\n",
"\n",
"#result\n",
"print\"(a)Output voltage will be equal to Vout=\",Vout,\" Volts\"\n",
"print\"(b)Voltage across Rs is Rs=\",Vrs,\"V\"\n",
"print\"(c)Current through zener diode is Iz=\",round(Iz,1),\"mA\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"(a)Output voltage will be equal to Vout= 8.0 Volts\n",
"(b)Voltage across Rs is Rs= 4.0 V\n",
"(c)Current through zener diode is Iz= 0.0 mA\n"
]
}
],
"prompt_number": 47
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 4.16 Page No. 175"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
" \n",
"#given data\n",
"Vz=50.\t\t\t#in volts\n",
"VSmax=120.0\t\t#in volts\n",
"VSmin=80.0\t\t#in volts\n",
"RL=10.0\t\t\t#in Kohm\n",
"Rs=5.0\t\t\t#in Kohm\n",
"\n",
"#Calculation\n",
"Vout=Vz\t\t\t#in Volts\n",
"IL=Vout/RL\t\t#in mAmpere\n",
"\n",
"ISmax=(VSmax-Vout)/Rs\t#in mAmpere\n",
"Izmax=ISmax-IL\t\t#in mA\n",
"Ismin=(VSmin-Vout)/Rs#in mAmpere\n",
"Izmin=Ismin-IL#in mA\n",
"\n",
"#Result\n",
"print\"Maximum zener diode current is \",Izmax,\"mA\"\n",
"print\"Minimum zener diode current is \",Izmin,\"mA\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Maximum zener diode current is 9.0 mA\n",
"Minimum zener diode current is 1.0 mA\n"
]
}
],
"prompt_number": 32
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 4.17 Page No. 175"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
" \n",
"Vz=15\t\t#in volts\n",
"Izk=6.0\t\t#in mA\n",
"Vout=15\t\t#in Volts\n",
"Vs=20\t\t#in Volts\n",
"ILmin=10.0\t#in mA\n",
"ILmax=20.0\t#in mA\n",
"RS=(Vs-Vz)*1000/(ILmax+Izk)\t#in ohm\n",
"\n",
"#result\n",
"print\"sereis Resistance is \",round(RS,1),\"ohm\"\n",
"print\"The zener current will be minimum i.e. Izk = 6mA when load current is maximum i.e. ILmax = 20mA\"\n",
"print\"when the load current will decrease and become 10 mA, the zener current will increase and become 6+10 i.e. 16 mA. \\nThus the current through series resistance Rs will remain unchanged at 6+20 i.e. 26 mA. \\nThus voltage drop in series resistance Rs will remain constant. Consequently, the output voltage will also remain constant. \""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"sereis Resistance is 192.3 ohm\n",
"The zener current will be minimum i.e. Izk = 6mA when load current is maximum i.e. ILmax = 20mA\n",
"when the load current will decrease and become 10 mA, the zener current will increase and become 6+10 i.e. 16 mA. \n",
"Thus the current through series resistance Rs will remain unchanged at 6+20 i.e. 26 mA. \n",
"Thus voltage drop in series resistance Rs will remain constant. Consequently, the output voltage will also remain constant. \n"
]
}
],
"prompt_number": 48
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 4.18 Page No. 175"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
" \n",
"Vs=16.0\t\t #in volts\n",
"RL=1.2\t\t\t#in Kohm\n",
"Rs=1.0\t\t\t#in Kohm\n",
"\n",
"#calculation\n",
"#If zener open circuited\n",
"VL=Vs*RL/(Rs+RL)\t#in Volts\n",
"Iz=0\t\t\t#in mA\n",
"Pz=VL*Iz\t\t#in watts\n",
"\n",
"#result\n",
"print\"When zener open circuited Voltage across load is \",round(VL,2),\"V\"\n",
"print\"Zener current is \",Iz,\"mA\"\n",
"print\"Power is\",Pz,\"watt\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"When zener open circuited Voltage across load is 8.73 V\n",
"Zener current is 0 mA\n",
"Power is 0.0 watt\n"
]
}
],
"prompt_number": 52
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 4.19 Page No. 126"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
" \n",
"Vin=20\t\t\t#in volts\n",
"Rs=220.0\t\t\t#in Kohm\n",
"Vz=10\t\t \t#in volts\n",
"RL2=50.0\t\t\t#in Kohm\n",
"RL1=200\t\t\t#in Kohm\n",
"\n",
"#calculation\n",
"# part (i) RL=50\t#in Kohm\n",
"VL1=Vin*RL1/(RL+Rs)\n",
"IR=Vin/(Rs+RL)\t#in mA\n",
"IL=IR\t\t \t#in mA\n",
"IZ=0\t\t\t #in mA\n",
"\n",
"if VL1< Vz:\n",
" \n",
" print\"Zener diode will not conduct and VL=\",round(VL1,1),\"V\" \n",
"else:\n",
" print \"Zener diode will conduct\"\n",
"\n",
" \n",
"#Result\n",
"print\"When RL=200 ohm\"\n",
"print\"IL is\",round(IL*1000,2),\"mA\"\n",
"print\"IR is\",round(IR*10**3,2),\"mA\"\n",
"print\"Iz in mA: \",round(IZ,0),\"mA\"\n",
"\n",
"# part (ii) RL=200#in Kohm\n",
"RL=200\t\t\t#in Kohm\n",
"VL2=Vin*RL2/(RL2+Rs)\n",
"IR=Vin/(Rs+RL2)\t\t#in mA\n",
"IL=IR\t\t\t#in mA\n",
"IZ=0\t\t\t#in mA\n",
"\n",
"#result\n",
"if VL2< Vz:\n",
" \n",
" print\"Zener diode will not conduct and VL=\",round(VL2,1),\"V\" \n",
"else:\n",
" print \"Zener diode will conduct\"\n",
"\n",
"print\"When RL=50 ohm\"\n",
"print\"IL is\",round(IL*1000,2),\"mA\"\n",
"print\"IR is\",round(IR*10**3,2),\"mA\"\n",
"print\"Iz in mA: \",IZ,\"mA\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Zener diode will not conduct and VL= 9.5 V\n",
"When RL=200 ohm\n",
"IL is 47.62 mA\n",
"IR is 47.62 mA\n",
"Iz in mA: 0.0 mA\n",
"Zener diode will not conduct and VL= 3.7 V\n",
"When RL=50 ohm\n",
"IL is 74.07 mA\n",
"IR is 74.07 mA\n",
"Iz in mA: 0 mA\n"
]
}
],
"prompt_number": 64
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 4.20 Page No. 176"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
" \n",
"RL=10.0\t\t\t #in Kohm\n",
"Rs=5.0 #in Kohm\n",
"Vin=100\t\t\t #in Volts\n",
"\n",
"#Calculation\n",
"V=Vin*RL/(RL+Rs)\t#in Volt\n",
"VZ=50\t\t\t#in Volts\n",
"VL=VZ\t\t\t#in volts\n",
"#Apply KVL\n",
"VR=100-50\t\t#in Volts\n",
"VR=50\t\t\t#in Volts\n",
"\n",
"if V< VZ:\n",
" \n",
" print\"Zener diode is OFF state\" \n",
"else:\n",
" print \"zener diode is ON state\"\n",
"\n",
"print\"Hence the voltage dropp across the 5 Kohm resistor in Volts is \",VR,\"V\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"zener diode is ON state\n",
"Hence the voltage dropp across the 5 Kohm resistor in Volts is 50 V\n"
]
}
],
"prompt_number": 67
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 4.21 Page No. 176"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
" \n",
"RL=120.0\t\t\t#in ohm, load resistance\n",
"Izmin=20\t\t#in mA min. diode current\n",
"Izmax=200\t\t#in mA max. diode current\n",
"VL=12\t\t\t#in Volts\n",
"VDCmin=15\t\t#in Volts\n",
"VDCmax=19.5\t\t#in Volts\n",
"Vz=12\t\t\t#in Volts\n",
"IL=VL/RL\t\t#in Ampere\n",
"IL=IL*1000\t\t#in mAmpere\n",
"\n",
"#calculation\n",
"#For VDCmin = 15 volts\n",
"VSmin=VDCmin-Vz\t\t#in Volts\n",
"#For VDCmax = 19.5 volts\n",
"VSmax=VDCmax-Vz\t\t#in Volts\n",
"ISmin=Izmin+IL\t\t#in mA\n",
"Ri=VSmin/ISmin\t\t#in Kohm\n",
"Ri=Ri*10**3\t\t#in ohm\n",
"\n",
"#result\n",
"print\"The resistance Ri is \",Ri,\"ohm\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The resistance Ri is 25.0 ohm\n"
]
}
],
"prompt_number": 72
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 4.22 Page No. 177"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
" \n",
"VRL=10\t\t\t#in Volts Diode resistance\n",
"Vi=50\t\t\t#in Volts\n",
"R=1.0\t\t\t#in Kohm Resistance\n",
"Vz=10\t\t\t#in Volts\n",
"VL=Vz\t\t\t#in Volts\n",
"Izm=32\t\t\t#in mA\n",
"IR=(Vi-VL)/R\t\t#in mA\n",
"\n",
"Izmin=0\t\t\t #in mA\n",
"ILmax=IR-Izmin\t\t#in mA\n",
"RLmin=VL/ILmax\t\t#in Ohm\n",
"Izmax=32\t\t #in mA\n",
"ILmin=IR-Izmax\t\t#in mA\n",
"VL=Vz\t\t\t #in Volts\n",
"RLmax=VL/ILmin\t\t#in Ohm\n",
"\n",
"#Result\n",
"print\"Range of RL in Kohm : From \",RLmin*1000,\"ohm to \",RLmax,\"kohm\"\n",
"print\"Range of IL in mA : From \",ILmin,\"mA to \",ILmax,\"mA\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Range of RL in Kohm : From 250.0 ohm to 1.25 kohm\n",
"Range of IL in mA : From 8.0 mA to 40.0 mA\n"
]
}
],
"prompt_number": 71
},
{
"cell_type": "code",
"collapsed": false,
"input": [],
"language": "python",
"metadata": {},
"outputs": []
}
],
"metadata": {}
}
]
}