{
"metadata": {
"name": ""
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter No.1: Special Diodes"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 1.1, Page No. 9"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Reverse Saturation current of diode\n",
"import math\n",
"\n",
"#variable declaration\n",
"\n",
"I=40*10**-3 #forword bias current in A\n",
"V=0.25 #forword bias voltage in Volt\n",
"T=20 #junction temperature in degree C\n",
"T=T+273 #junction temperature in degree K\n",
"ETA=1 #For Ge\n",
"e=1.6*10**-19 #in Coulamb(electronic charge)\n",
"k=1.38*10**-23 #in J/K(Boltzman Constant)\n",
"\n",
"#Calculation\n",
"#Formula : I=Io*(exp(e*V/(ETA*k*T))-1)\n",
"x=math.ceil((e*V/(ETA*k*T)))\n",
"Io=I/(math.exp(x)-1)\n",
"Io=math.ceil(Io*10**8)/10**8\n",
"\n",
"#Result\n",
"print(\"Reverse saturation current in micro Ampere : %.2f \"%(Io*10**6))\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Reverse saturation current in micro Ampere : 1.82 \n"
]
}
],
"prompt_number": 14
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"example 1.2, Page No. 9"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Value of forword voltage\n",
"import math\n",
"#variable declaration\n",
"Io=10*10**-6 # reverse saturation currrent in A\n",
"I=1 # forword current in Ampere\n",
"ETA=2 # For Si\n",
"T=27 # room temperature in degree C\n",
"T=T+273 # room temperature in degree K\n",
"e=1.6*10**-19 # in Coulamb(electronic charge)\n",
"k=1.38*10**-23 # in J/K(Boltzman Constant)\n",
"\n",
"#Calculation\n",
"#Formula : I=Io*(exp(%e*V/(ETA*k*T))-1)\n",
"V=(ETA*k*T/e)*math.log(I/(Io)+1)\n",
"V=math.floor(V*100)/100\n",
"#result\n",
"print(\"Forward Voltage across the diode in Volt :%.2f\"%V)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Forward Voltage across the diode in Volt :0.59\n"
]
}
],
"prompt_number": 21
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"example 1.3 , Page No. 23"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#DC Current, DC Voltage, Ripple Factor\n",
"import math\n",
"#Variable Declaration\n",
"RL=1 #load resistance in kOhm\n",
"#rf<<RL\n",
"Vrms=200 #in Volt\n",
"\n",
"#Part (i)\n",
"Vo=Vrms*math.sqrt(2) #in volt\n",
"Idc=Vo/(RL*10**3*math.pi) #in Ampere\n",
"print(\"(i)\\nDC current in load in mA :%.0f\"%(math.floor((Idc*10**3))))\n",
"\n",
"#Part (ii)\n",
"Vdc=RL*10**3*Idc #in Volt\n",
"print(\"\\n(ii)\\nDC voltage across load in volt :%.0f\"%(math.floor(Vdc)))\n",
"\n",
"#Part (iii)\n",
"#Gamma=sqrt((Irms/Idc)^2-1)=sqrt((Io/2)/(Io/%pi)-1)=sqrt((%pi/2)^2-1)\n",
"Gamma=math.sqrt((math.pi/2)**2-1) #unitless\n",
"print(\"\\n(iii)\\nRipple factor :%.2f \"%(math.floor(Gamma*100)/100))\n",
"\n",
"#Part (iv)\n",
"PIV=Vrms*math.sqrt(2) #in volt\n",
"print(\"\\n(iv)\\nPeak Inverse Voltage in volt :%.0f\"%(math.floor(PIV)))"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"(i)\n",
"DC current in load in mA :90\n",
"\n",
"(ii)\n",
"DC voltage across load in volt :90\n",
"\n",
"(iii)\n",
"Ripple factor :1.21 \n",
"\n",
"(iv)\n",
"Peak Inverse Voltage in volt :282\n"
]
}
],
"prompt_number": 31
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"example 1.4 , Page No. 23"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Average dc current rms current rectifcation efcieny PIV\n",
"import math\n",
"#variable declaration\n",
"rf=20.0 #in ohm\n",
"RL=980.0 #in Ohm\n",
"Vrms=50.0 #in Volt\n",
"Vo=Vrms*math.sqrt(2) #in Volt\n",
"Io=Vo/(RL+rf) #in Ampere\n",
"\n",
"#Part (i)\n",
"Idc=2*Io/math.pi #in Ampere\n",
"print(\"(i)\\nAverage DC current in mA :%.0f\"%(math.floor(Idc*10**3)))\n",
"\n",
"#Part (ii)\n",
"Irms=Io/math.sqrt(2) #in Ampere\n",
"print(\"\\n(ii)\\nrms value of load current in mA :%.0f\"%(math.ceil(Irms*1000)))\n",
"\n",
"#Part (iii)\n",
"Vdc=RL*Idc #in Volt\n",
"print(\"\\n(iii)\\nDC output voltage in volt :%.1f\"%(math.floor(Vdc*10)/10))\n",
"\n",
"#Part (iv)\n",
"ETA=(Idc**2*RL/(Irms**2*(RL+rf)))*100 #Rectification Efficiency in %\n",
"print(\"\\n(iv)\\nRectification Efficiency is %.1f%%\"%(math.ceil(ETA*10)/10))\n",
"\n",
"#Part (v)\n",
"PIV=2*Vo #in volt\n",
"print(\"\\n(v)\\nPeak Inverse Voltage in volt :%.1f\"%PIV)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"(i)\n",
"Average DC current in mA :45\n",
"\n",
"(ii)\n",
"rms value of load current in mA :50\n",
"\n",
"(iii)\n",
"DC output voltage in volt :44.1\n",
"\n",
"(iv)\n",
"Rectification Efficiency is 79.5%\n",
"\n",
"(v)\n",
"Peak Inverse Voltage in volt :141.4\n"
]
}
],
"prompt_number": 38
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"example 1.5, Page No.26"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Minimum value of resistance\n",
"import math\n",
"#Variable declaration\n",
"\n",
"Vin=40 #in volt\n",
"VZ=10 #in volt\n",
"Vo=10 #in volt\n",
"IZmax=50 #in mA\n",
"IL=0 #in mA\n",
"\n",
"#calculation\n",
"#Formula : I=IZ+IL=IZmax+0\n",
"I=IZmax+0 #in mA\n",
"#Formula : VZ=Vin-R*I\n",
"Rmin=(Vin-VZ)/(I*10**-3) #in Ohm\n",
"#Result\n",
"print(\"Minimum value of resistance in Ohm :%.0f \"%Rmin)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Minimum value of resistance in Ohm :600 \n"
]
}
],
"prompt_number": 40
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"example 1.6, Page No.26"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Value of series resistor\n",
"import math\n",
"#variable declaration\n",
"Vmin=15 #Minimum input voltage in volt\n",
"VZ=6.8 #Voltage across zener in volt\n",
"Vo=VZ #output voltage in volt\n",
"Vsr1=Vmin-Vo #Voltage aross series resistance in volt\n",
"print(\"If R is the series resistance, Total current in series resistance in Ampere : I=Vsr/R=8.2/R \")\n",
"ILmin=5 #in mA\n",
"print(\"current in zener diode in Ampere :IZ=I-IL=(8.2/R-IL*10-3).............eqn(1)\\n\");\n",
"Vmax=20 #mximum output voltage\n",
"Vo=VZ #output voltage in volt\n",
"Vsr2=Vmax-Vo #Voltage aross series resistance in volt\n",
"print(\"Current in series resistance circuit in Ampere : I=Vsr/R\\n\")\n",
"ILmax=15 #in mA\n",
"print(\"current in zener diode in Ampere :IZ=I-IL=(Rs/R-IL*10-3)..............eqn(2)\\n\")\n",
"print(\"For Zener diode to work as voltage regulator,(1) and (2) must be same.\");\n",
"print(\"(8.2/R-IL*10-3)=(13.2/R-IL*10-3)\")\n",
"R=(Vsr2-Vsr1)/(ILmax*10**-3-ILmin*10**-3) #in Ohm\n",
"print(\"\\nRequired value of Series Resistor in ohm : %.0f\"%R)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"If R is the series resistance, Total current in series resistance in Ampere : I=Vsr/R=8.2/R \n",
"current in zener diode in Ampere :IZ=I-IL=(8.2/R-IL*10-3).............eqn(1)\n",
"\n",
"Current in series resistance circuit in Ampere : I=Vsr/R\n",
"\n",
"current in zener diode in Ampere :IZ=I-IL=(Rs/R-IL*10-3)..............eqn(2)\n",
"\n",
"For Zener diode to work as voltage regulator,(1) and (2) must be same.\n",
"(8.2/R-IL*10-3)=(13.2/R-IL*10-3)\n",
"\n",
"Required value of Series Resistor in ohm : 500\n"
]
}
],
"prompt_number": 51
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"example 1.7 Page No.27"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Current limiting resistance and dissipated power\n",
"import math\n",
"#variable declaration\n",
"\n",
"Vin=18 #in volt\n",
"IZ=20 #in mA\n",
"ILav=(5+35)/2 #in mA\n",
"VZ=12 #in volt\n",
"Vo=12 #in volt\n",
"\n",
"#Calculation\n",
"I=IZ+ILav #in mA\n",
"R=(Vin-Vo)/(I*10**-3) #in Ohm\n",
"P=(I*10**-3)**2*R #in Watts\n",
"\n",
"#Result\n",
"print(\"Current limiting resistance in Ohm : %.0f\"%R);\n",
"print(\"Power disspation in resistance in Watt :%.2f \"%P);"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Current limiting resistance in Ohm : 150\n",
"Power disspation in resistance in Watt :0.24 \n"
]
}
],
"prompt_number": 45
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"example 1.8, Page No. 28"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Maximum and minimum input supply voltage\n",
"import math\n",
"#Variable declaration\n",
"R=1 #in kOhm\n",
"RL=5 #in kOhm\n",
"VZ=10 #in volt\n",
"Vo=10 #in volt\n",
"P=250 #in mW\n",
"\n",
"#Calculation\n",
"IL=Vo/RL #in mA\n",
"IZmin=0 #in mA\n",
"IZmax=P/VZ #in mA\n",
"Imin=IZmin+IL #in mA\n",
"Imax=IZmax+IL #in mA\n",
"Vin_min=VZ+Imin*10**-3*R*10**3 #in volt\n",
"Vin_max=VZ+Imax*10**-3*R*10**3 #in volt\n",
"\n",
"#Result\n",
"print(\"The input voltage ranges from %.0fV to %.0fV\"%(Vin_min,Vin_max));"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The input voltage ranges from 12V to 37V\n"
]
}
],
"prompt_number": 53
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"example 1.9, Page No.28"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Output voltage voltage drop and current in zener diode\n",
"import math\n",
"#Variable Declaration\n",
"R=5 #in kOhm\n",
"R=R*1000 #in Ohm\n",
"RL=10.0 #in kOhm\n",
"RL=RL*1000 #in Ohm\n",
"Vin=120.0 #in Volt\n",
"VZ=50.0 #in Volt\n",
"\n",
"#Part (i)\n",
"\n",
"#calculation\n",
"Vo=VZ #in Volt\n",
"#Result\n",
"print(\"\\n(i)\\nOutput voltage in volt :%.0f\"%Vo)\n",
"\n",
"#Part (ii)\n",
"\n",
"#calculation\n",
"VR=Vin-VZ #in Volt\n",
"#Result\n",
"print(\"\\n(ii)\\nVoltage drop across series resistance in volt :%.0f\"%VR);\n",
"\n",
"\n",
"\n",
"#Part (iii) \n",
"\n",
"#Calculation\n",
"IL=Vo/RL #in Ampere\n",
"I=VR/R #in Ampere\n",
"IZ=I-IL #in Ampere\n",
"#Result\n",
"print(\"\\n(iii)\\nLoad Current in mA :%.0f\"%(IL*1000))\n",
"print(\"Current through resistance R in mA :%.0f\"%(I*1000));\n",
"print(\"Load Current in mA :%.0f\"%(IZ*1000))"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"(i)\n",
"Output voltage in volt :50\n",
"\n",
"(ii)\n",
"Voltage drop across series resistance in volt :70\n",
"\n",
"(iii)\n",
"Load Current in mA :5\n",
"Current through resistance R in mA :14\n",
"Load Current in mA :9\n"
]
}
],
"prompt_number": 61
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"example 1.10, Page No. 30"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Maximum and Minimum LED current\n",
"import math\n",
"\n",
"#Variable declaration\n",
"VDmin=1.5 #in Volt\n",
"VDmax=2.3 #in Volt\n",
"VS=5.0 #in Volt\n",
"RS=270.0 #in Ohm\n",
"\n",
"#Calculation\n",
"Imin=(VS-VDmax)/RS #in Ampere\n",
"Imax=(VS-VDmin)/RS #in Ampere\n",
"\n",
"#Result\n",
"print(\"Minimum value of LED current in mA : %.0f\"%(Imin*1000))\n",
"print(\"Maximum value of LED current in mA : %.0f\"%(math.ceil(Imax*1000)))"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Minimum value of LED current in mA : 10\n",
"Maximum value of LED current in mA : 13\n"
]
}
],
"prompt_number": 65
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"example 1.11, Page No. 33"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Frequency range of tuning circuit\n",
"import math\n",
"\n",
"#VAriable declarion\n",
"C1min=10 #in pF\n",
"C2max=50 #in pF\n",
"L=5 #in mH\n",
"L=L*10**-3 #in H\n",
"\n",
"#Calculation\n",
"\n",
"#Formula : CT=C1*C2/(C1+C2)\n",
"#Minimum\n",
"C1=10 #in pF\n",
"C2=10 #in pF\n",
"CTmin=C1*C2/(C1+C2) #in pF\n",
"CTmin=CTmin*10**-12 #in F\n",
"#Maximum\n",
"C1=50 #in pF\n",
"C2=50 #in pF\n",
"CTmax=C1*C2/(C1+C2) #in pF\n",
"CTmax=CTmax*10**-12 #in F\n",
"\n",
"#Formula : f=1/(2*%pi*sqrt(L*C))\n",
"#maximum :\n",
"fmax=1/(2*math.pi*math.sqrt(L*CTmin));\n",
"#minimum :\n",
"fmin=1/(2*math.pi*math.sqrt(L*CTmax));\n",
"\n",
"#Result\n",
"print(\"The frequency of tuning circuit ranges from %.3fMHz to %.3fMHz.\"%(fmin/10**6,fmax/10**6));\n",
"#Note : Answer in the book is wrong."
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The frequency of tuning circuit ranges from 0.450MHz to 1.007MHz.\n"
]
}
],
"prompt_number": 68
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"example 1.12, page No.33"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Diode Capacitance\n",
"import math\n",
"\n",
"#Variable Declaration\n",
"C1=21.0 # in pF\n",
"V1=4.0 # in volt\n",
"V2=9.0 # in volt\n",
"\n",
"#Calculations\n",
"print(\"C is proportional to 1/sqrt(V)\")\n",
"print(\"So, C2/C1=sqrt(V1/V2)\")\n",
"C2=math.sqrt(V1/V2)*C1 #in pF\n",
"\n",
"#Result\n",
"print(\"At reverse bias 9V, Diode capacitance in pF : %.0f\"%math"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"C is proportional to 1/sqrt(V)\n",
"So, C2/C1=sqrt(V1/V2)\n",
"At reverse bias 9V, Diode capacitance in pF : 14\n"
]
}
],
"prompt_number": 72
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"example 1.13, Page No.39"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Photocurrent\n",
"import math\n",
"#Variable Declaration\n",
"R=0.90 #in A/W\n",
"Pop=1.0 #in mW\n",
"\n",
"#Part (i)\n",
"#calculation\n",
"IP=R*Pop #in mA\n",
"#Result\n",
"print(\"(i)\\nPower of incident light 1mW, Photocurrent in mA is :%.2f\"%IP);\n",
"\n",
"#Part (ii)\n",
"#Result\n",
"print(\"\\n(ii)\\nHere IP is not proportional to Pop(for Pop>1.5mW)\")\n",
"print(\"Hence Photourrent can not be calculated.\")"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"(i)\n",
"Power of incident light 1mW, Photocurrent in mA is :0.90\n",
"\n",
"(ii)\n",
"Here IP is not proportional to Pop(for Pop>1.5mW)\n",
"Hence Photourrent can not be calculated.\n"
]
}
],
"prompt_number": 74
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"example 1.14, Page No. 39"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Responsivity of InGaAs photodiode\n",
"import math\n",
"#variable declaration\n",
"ETA=70.0 #in %\n",
"Eg=0.75 #in eV\n",
"Eg=Eg*1.6*10**-19 #in Joule\n",
"h=6.63*10**-34 #Planks constant in J-s\n",
"c=3*10**8 #speed of light in m/s\n",
"e=1.6*10**-19 #in coulamb\n",
"\n",
"#Calcualtions\n",
"lambda1=h*c/Eg #in meter\n",
"R=(ETA/100)*e*lambda1/(h*c) #in A/W\n",
"\n",
"#Result\n",
"print(\"Wavelength in nm :%.1f\"%(lambda1*10**9))\n",
"print(\"Responsivity of InGaAs photodiode in A/W :%.3f\"%R)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Wavelength in nm :1657.5\n",
"Responsivity of InGaAs photodiode in A/W :0.933\n"
]
}
],
"prompt_number": 78
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"example 1.15, Page No. 41"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Equilibrium contact potential\n",
"import math\n",
"#variable declaration\n",
"W1=2.5 #in eV\n",
"W2=1.9 #in eV\n",
"\n",
"#Calculation\n",
"ContactPotential=W1-W2 #in Volt\n",
"\n",
"#Result\n",
"print(\"Contact potential in Volts :%.1f \"%ContactPotential)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Contact potential in Volts :0.6 \n"
]
}
],
"prompt_number": 80
}
],
"metadata": {}
}
]
}