{
"metadata": {
"name": "",
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"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter - 5 : High Frequency And High Power Devices"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example - 5.1 : Page No - 194"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"from __future__ import division\n",
"from numpy import pi, sqrt\n",
"#Given data\n",
"C1= 5 # in pF\n",
"C1= C1*10**-12 # in F\n",
"C2= 50 # in pF\n",
"C2= C2*10**-12 # in F\n",
"L= 10 # in mH\n",
"L= L*10**-3 # in H\n",
"TuningRange= 1/(2*pi*sqrt(L*C1*C2/(C1+C2))) # in Hz\n",
"print \" The tuning range for the circuit = %0.3f kHz\" %(TuningRange*10**-3)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" The tuning range for the circuit = 746.503 kHz\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example - 5.2 : Page No - 195"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Given data\n",
"C_T1= 15 # in pF\n",
"Vb1=8 # in V\n",
"Vb2= 12 # in V\n",
"# As C_T proportional to 1/sqrt(Vb), and \n",
"# C_T1/C_T2= sqrt(Vb2/Vb1), so\n",
"C_T2= C_T1*sqrt(Vb1/Vb2) # in pF\n",
"print \" The value of C_T2 = %0.2f pF\" %C_T2"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" The value of C_T2 = 12.25 pF\n"
]
}
],
"prompt_number": 4
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example - 5.3 : Page No - 195"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Given data\n",
"epsilon_Ge= 16/(36*pi*10**-11) # in f/C\n",
"A=10**-12 \n",
"d=2*10**-4 # in cm\n",
"# C_T= epsilon_0*A/d= epsilon_Ge*A/d\n",
"C_T= epsilon_Ge*A/d #in pF\n",
"print \" The space charge capacitance = %0.2f pF\" %C_T"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" The space charge capacitance = 70.74 pF\n"
]
}
],
"prompt_number": 5
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example - 5.4 : Page No - 196"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Given data\n",
"D= 0.102 # in cm\n",
"sigma_P= 0.286 # in \u03a9cm\n",
"q= 1.6*10**-19 # in C\n",
"miuP= 500 \n",
"Vb= 5+0.35 #in V\n",
"A= pi*D**2/4 # in cm**2\n",
"N_A= sigma_P/(q*miuP) # at/c\n",
"C_T= 2.92*10**-4*(N_A/Vb)**(1/2)*A # \n",
"print \" The value of transition = %0.2f pf/cm**2\" %C_T"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" The value of transition = 61.68 pf/cm**2\n"
]
}
],
"prompt_number": 6
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example - 5.5 : Page No - 196"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Given data\n",
"epsilon= 12/(36*pi*10**11) # in F/cm (value of epsilon for silicon)\n",
"q= 1.6*10**-19 # in C\n",
"# C_T= epsilon*A/d , where d= 2*epsilon*Vi/(q*NA)**(/2)\n",
"# Hence C_T/A= epsilon/d= sqrt(q*epsilon/2)*sqrt(NA/Vi)\n",
"# Let \n",
"value = sqrt(q*epsilon/2) \n",
"print \"C_T= %0.1e sqrt(NA/Vi) pF/cm**2\" %(value*10**12)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"C_T= 2.9e-04 sqrt(NA/Vi) pF/cm**2\n"
]
}
],
"prompt_number": 7
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example - 5.6 : Page No - 197"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Given data\n",
"V1= 5 # in V\n",
"IncreaseInVolt= 1.5 # in V\n",
"C_T1= 20 # in pF\n",
"# Formula C_T= lamda/sqrt(V)\n",
"lamda= C_T1*sqrt(V1) \n",
"# When\n",
"V2= V1+IncreaseInVolt # in V\n",
"C_T2= lamda/sqrt(V2) \n",
"print \" The decrease in capacitance = %0.2f pF\" %(C_T1-C_T2)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" The decrease in capacitance = 2.46 pF\n"
]
}
],
"prompt_number": 8
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example - 5.7 : Page No - 198"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Given data\n",
"Vf= 0.7 # in V\n",
"If= 10 # in mA\n",
"If= If*10**-3 # in A\n",
"toh= 70 # in ns\n",
"Cd= toh*If/Vf # in nf\n",
"print \" Diffusion capacitance for a si diode = %0.f nf\" %Cd"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" Diffusion capacitance for a si diode = 1 nf\n"
]
}
],
"prompt_number": 9
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example - 5.8 : Page No - 198"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Given data\n",
"N_A= 4*10**20 # per m**3\n",
"Vi= 0.2 # in V\n",
"q= 1.6*10**-19 \n",
"V= -1 # in V\n",
"A= 0.8*10**-6 #/ in m**2\n",
"epsilon_r= 16 \n",
"epsilon_o= 8.854*10**-12 # in F\n",
"epsilon= epsilon_o*epsilon_r \n",
"d= (2*epsilon*(Vi-V)/(q*N_A))**(1/2) \n",
"C_T= epsilon*A/d # in F\n",
"print \" The transition capacitance = %0.2f pF\" %(C_T*10**12)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" The transition capacitance = 49.17 pF\n"
]
}
],
"prompt_number": 11
}
],
"metadata": {}
}
]
}