{
"metadata": {
"name": ""
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "markdown",
"metadata": {},
"source": [
"<h1> Chapter No 3 : Excess Carriers In Semiconductors<h1>"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 3.1, Page No 145 "
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#initialisation of variables\n",
"n_o = 10**17 #in/cm^3\n",
"n_i = 1.5*10**10 #in /cm^3\n",
"\n",
"#CALCULATIONS\n",
"p_o = ((n_i)**(2))/n_o #in holes/cm^3\n",
"\n",
"#RESULTS\n",
"print('The hole concentration is =%.f in holes/cm^3' %p_o)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The hole concentration is =2250 in holes/cm^3\n"
]
}
],
"prompt_number": 20
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 3.3, Page No 146"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"#initialisation of variables\n",
"n_i = 1.5 * 10**10 #in /cm^3 for silicon\n",
"N_d = 10**17 #in atoms/cm^3\n",
"n_o = 10**17 #electrons/cm^3\n",
"KT = 0.0259\n",
"\n",
"#CALCULATIONS\n",
"# E_r - E_i = KT * log(n_o/n_i)\n",
"del_E = KT * math.log(n_o/n_i) #in eV\n",
"\n",
"#RESULTS\n",
"print('The energy band for this type material is Ei =%.2f eV' %del_E)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The energy band for this type material is Ei =0.41 eV\n"
]
}
],
"prompt_number": 21
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 3.4 Page No 147"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"#initialisation of variables\n",
"K = 1.38*10**-23 #in J/K\n",
"T = 27 #in degree C\n",
"T = T + 273 #in K\n",
"e = 1.6*10**-19 \n",
"Miu = 0.17 #in m^2/v-s\n",
"Miu1 = 0.025 #in m^2/v-s\n",
"\n",
"#CALCULATIONS\n",
"D_n = ((K*T)/e)*Miu #in m^2/s\n",
"D_p = ((K*T)/e)*Miu1 #in m^2/s\n",
"\n",
"\n",
"#RESULTS\n",
"print('The diffusion coefficient of electrons is =%.2f X 10^-4 in m^2/s ' %(D_n*(10**4)))\n",
"print('The diffusion coefficient of holes is =%.2f X 10^-4 in m^2/s ' %(D_p*(10**4)))"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The diffusion coefficient of electrons is =43.99 X 10^-4 in m^2/s \n",
"The diffusion coefficient of holes is =6.47 X 10^-4 in m^2/s \n"
]
}
],
"prompt_number": 22
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 3.5, Page No 147"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"#initialisation of variables\n",
"Miu_n = 0.15 #in m^2/v-s\n",
"K = 1.38*10**-23 \n",
"T = 300 #in K\n",
"e = 1.6*10**-19 #in C\n",
"\n",
"#CALCULATIONS\n",
"D_n = Miu_n*((K*T)/e) #in m^2/s\n",
"Torque_n = 10**-7 #in s\n",
"L_n = math.sqrt(D_n*Torque_n) #in m\n",
"del_n = 10**20 #electrons/m^3\n",
"J_n = (e*D_n*del_n)/L_n #in A/m^2\n",
"\n",
"#RESULTS\n",
"print('The diffusion length is =%.2f X 10^-4 in m' %(L_n*(10**5)))\n",
"print('The diffusion current density is =%.2f X 10^3 in A/m^2 ' %(J_n/(10**3)))"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The diffusion length is =1.97 X 10^-4 in m\n",
"The diffusion current density is =3.15 X 10^3 in A/m^2 \n"
]
}
],
"prompt_number": 23
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 3.6 Page No 147"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"#initialisation of variables\n",
"Sigma_n = 0.1 #in (ohm-cm)^-1\n",
"Miu_n = 1300\n",
"q = 1.6*10**-19 #in C\n",
"\n",
"#CALCULATIONS\n",
"n_n = Sigma_n/(Miu_n*q) #in electrons/cm^3 \n",
"n_i = 1.5*10**10\n",
"p_n = ((n_i)**2)/n_n #in holes/cm^3\n",
"p_n = p_n * 10**6 #in holes/m^3\n",
"n_n=n_n*10**6\n",
"\n",
"#RESULTS\n",
"print('Concentration of electrons is =%.2f X 10^20 per m^3 ' %(n_n/(10**20)))\n",
"print('Concentration of holes is =%.2f X 10^11 per m^3 ' %(p_n/(10**11)))"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Concentration of electrons is =4.81 X 10^20 per m^3 \n",
"Concentration of holes is =4.68 X 10^11 per m^3 \n"
]
}
],
"prompt_number": 24
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 3.7 Page No 147"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"#initialisation of variables\n",
"L = 100*10**-6 #in m\n",
"Miu_e = 0.13 #in m^2/V-s\n",
"Torque_h = 10**-6 #in s\n",
"Miu_h = 0.05 #in m^2/v-s\n",
"V = 12 #in V\n",
"\n",
"#CALCULATIONS\n",
"Torque_n = ((L)**2)/(Miu_e*V) #in s\n",
"P = (Torque_h/Torque_n)*(1+(Miu_h/Miu_e))\n",
"\n",
"#RESULTS\n",
"print('Electron transit time is =%.2f X 10^-9 in sec' %(Torque_n*(10**9)))\n",
"print('Photoconductor gain is =%.2f ' %P)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Electron transit time is =6.41 X 10^-9 in sec\n",
"Photoconductor gain is =216.00 \n"
]
}
],
"prompt_number": 25
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 3.8, Page No 147"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"#initialisation of variables\n",
"q = 1.6*10**-19 #in C\n",
"n_i = 2.5*10**13\n",
"Miu_n = 3800 #in cm^2/V-s\n",
"Miu_p = 1800 #in cm^2/V-s\n",
"\n",
"#CALCULATIONS\n",
"Sigma = n_i*(Miu_n + Miu_p)*q #in (ohm-cm)^-1\n",
"Rho = 1/Sigma #in ohm-cm\n",
"N_D =4.4*10**22/10**8 #in atoms/cm^3\n",
"Sigma_n = N_D * Miu_n*q #in (ohm-cm)^-1 \n",
"Rho1 = 1/Sigma_n #in ohm cm\n",
"\n",
"#RESULTS\n",
"print('The resistivity is =%.2f in ohm-cm ' %Rho)\n",
"print('The resistivity drops is =%.2f in ohm cm ' %Rho1)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The resistivity is =44.64 in ohm-cm \n",
"The resistivity drops is =3.74 in ohm cm \n"
]
}
],
"prompt_number": 26
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 3.9 Page No 148"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"\n",
"#initialisation of variables\n",
"n_i = 10**16 #in /m^3\n",
"N_D = 10**22 #in /m^3\n",
"\n",
"#CALCULATIONS\n",
"n = N_D #in /m^3\n",
"p = ((n_i)**2)/n #in /m^3\n",
"\n",
"\n",
"#RESULTS\n",
"print('The concentration of is =%.2f X 10^21 electrons per m^3' %(n/(10**21)))\n",
"print('The concentration of is =%.2f X 10^9 holes per m^3 ' %(p/(10**9)))"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The concentration of is =10.00 X 10^21 electrons per m^3\n",
"The concentration of is =10.00 X 10^9 holes per m^3 \n"
]
}
],
"prompt_number": 27
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 3.10 Page No 148"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"#initialisation of variables\n",
"Rho = 9.6*10**-2 #ohm-m\n",
"Sigma_n = 1/Rho #in (ohm-m)^-1\n",
"Miu_n = 1300 #in cm^2/V-s\n",
"\n",
"#CALCULATIONS\n",
"Miu_n = Miu_n * 10**-4 #in m^2/V-s\n",
"q = 1.6*10**-19 #in C\n",
"N_D = Sigma_n/(Miu_n*q) #in atoms/m^3\n",
"d = 5*10**22 #in atoms/cm^3\n",
"d = d * 10**6 #// in atoms/m^3\n",
"R_d = N_D/d #Ratio \n",
"\n",
"#RESULTS\n",
"print('Ratio of donor atom to silicon atoms per unit volume is =%.2f X 10^-8 ' %(R_d*(10**8)))"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Ratio of donor atom to silicon atoms per unit volume is =1.00 X 10^-8 \n"
]
}
],
"prompt_number": 28
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 3.11 Page No 148"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"#initialisation of variables\n",
"n_i = 1.5*10**10 #in /cm^3\n",
"n_n = 2.25*10**15 #in /cm^3\n",
"\n",
"#CALCULATIONS\n",
"p_n = ((n_i)**2)/n_n #in /cm^3\n",
"\n",
"#RESULTS\n",
"print('The concentration of is =%.2f X 10^5 holes per cm^3 ' %(p_n/(10**5)))\n",
"print('Donor impurity per cm^3 is =%.2f X 10^15 ' %(n_n/(10**15)))"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The concentration of is =1.00 X 10^5 holes per cm^3 \n",
"Donor impurity per cm^3 is =2.25 X 10^15 \n"
]
}
],
"prompt_number": 29
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example No 3.12 Page No 149"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#initialisation of variables\n",
"N_A = 2*10**16 #in /cm^3\n",
"N_D = 10**16 #in /cm^3\n",
"\n",
"#CALCULATIONS\n",
"C = N_A-N_D #in /cm^3\n",
"\n",
"#RESULTS\n",
"print('Carrier concentration in holes/cm^3 is =%.2f X 10^16 ' %(C/(10**16)))"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Carrier concentration in holes/cm^3 is =1.00 X 10^16 \n"
]
}
],
"prompt_number": 30
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example No 3.13 Page No 149"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#initialisation of variables\n",
"del_n = 10**15 #in /cm^3\n",
"Torque_p = 10*10**-6 #in sec\n",
"\n",
"#CALCULATIONS\n",
"R_G = del_n/Torque_p #in electron hole pairs/sec/cm^3\n",
"\n",
"#RESULTS\n",
"print('The rate of generation of minority carrier is : =%.2f X 10^20 electron hole pairs/sec/cm^3 ' %(R_G/(10**20)))"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The rate of generation of minority carrier is : =1.00 X 10^20 electron hole pairs/sec/cm^3 \n"
]
}
],
"prompt_number": 31
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example No 3.14 Page No 149"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#initialisation of variables\n",
"V = 1/20.0 #in cm/\u00b5sec\n",
"V=V*10**6 #in cm/sec\n",
"\n",
"#CALCULATIONS\n",
"E = 10 #in V/cm\n",
"Miu = V/E #in cm^2/V-sec\n",
"\n",
"#RESULTS\n",
"print('The mobility of minority charge carrier is =%.f cm^2/V-sec' %Miu)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The mobility of minority charge carrier is =5000 cm^2/V-sec\n"
]
}
],
"prompt_number": 32
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example No 3.15 Page No 149"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"#initialisation of variables\n",
"q = 1.6 * 10**-19 #in C\n",
"N_D = 4.5 * 10**15 #in /cm^3\n",
"del_p = 10**21 \n",
"e=10 #in cm\n",
"A = 1 #in mm^2\n",
"A = A * 10**-14 #cm^2\n",
"l = 10 #in cm\n",
"Torque_p = 1 #in microsec\n",
"\n",
"\n",
"#CALCULATIONS\n",
"Torque_p = Torque_p * 10**-6 #in sec\n",
"Torque_n = 1 #in microsec\n",
"Torque_n = Torque_n * 10**-6 #in sec\n",
"n_i = 1.5 * 10**10 #in /cm^3\n",
"D_n = 30 #cm^2/sec\n",
"D_p = 12 #in cm^2/sec\n",
"n_o = N_D #in /cm^3\n",
"p_o = ((n_i)**2)/n_o #in /cm^3\n",
"l_n = math.sqrt(D_n * Torque_n) #in cm\n",
"l_p = math.sqrt(D_p * Torque_p) #in cm\n",
"x=34.6*10*8-4 #in cm\n",
"e1 = 1.88 * 10**1 #in cm\n",
"dnBYdx = del_p * e1 #in cm^4\n",
"dpBYdx = del_p *e #in cm^4\n",
"\n",
"J_P = -(q) * D_p * dpBYdx #in A/cm^2\n",
"J_n = q * D_n * dnBYdx #in A/cm^2\n",
"\n",
"\n",
"#RESULTS\n",
"print('Hole concentration at thermal equilibrium per cm^3 is =%.1f X 10^5 ' %(p_o/(10**5)))\n",
"print('Diffusion length of holes in cm is =%.1f X 10^-3 ' %(l_p*(10**3)))\n",
"print('Concentration gradient of holes at distance in cm^4 is =%.f X 10^22 ' %(dpBYdx/(10**22)))\n",
"print('Concentration gradient of electrons in per cm^4 is =%.1f X 10^22 ' %(dnBYdx/(10**22)))\n",
"print('Current density of holes due to diffusion in A/cm^2 is =%.2f X 10^4 ' %(J_P/(10**4)))\n",
"print('Current density of electrons due to diffusion in A/cm^2 is =%.2f X 10^4 ' %(J_n/(10**4)))"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Hole concentration at thermal equilibrium per cm^3 is =0.5 X 10^5 \n",
"Diffusion length of holes in cm is =3.5 X 10^-3 \n",
"Concentration gradient of holes at distance in cm^4 is =1 X 10^22 \n",
"Concentration gradient of electrons in per cm^4 is =1.9 X 10^22 \n",
"Current density of holes due to diffusion in A/cm^2 is =-1.92 X 10^4 \n",
"Current density of electrons due to diffusion in A/cm^2 is =9.02 X 10^4 \n"
]
}
],
"prompt_number": 33
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example No 3.16 Page No 151"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#initialisation of variables\n",
"h = 6.626 * 10**-34 #in J-s\n",
"q= 1.6*10**-19 #in C\n",
"\n",
"#CALCULATIONS\n",
"h= h/q #in eV\n",
"c = 3*10**8\n",
"lembda = 5490*10**-10 #in m\n",
"E = h*c/lembda #in eV\n",
"\n",
"#RESULTS\n",
"print('The energy band gap is =%.2f eV ' %E)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The energy band gap is =2.26 eV \n"
]
}
],
"prompt_number": 34
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example No 3.17 Page No 151"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#initialisation of variables\n",
"D_n = 35 #in cm^2/sec\n",
"q = 1.6*10**-19 #in C\n",
"y2 = 6*10**16 #in /cm^3\n",
"y1 = 10**17 #in /cm^3\n",
"x2 = 2*10**-4\n",
"x1 = 0\n",
"\n",
"#CALCULATIONS\n",
"dnBYdx = (y2-y1)/(x2-x1)\n",
"J_n = q*D_n*dnBYdx #in A/cm^2\n",
"\n",
"#RESULTS\n",
"print('The current density is =%.f in A/cm^2' %J_n)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The current density is =-1120 in A/cm^2\n"
]
}
],
"prompt_number": 35
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example No 3.18 Page No 151"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#initialisation of variables\n",
"q = 1.6*10**-19 #in C\n",
"n_n = 5*10**20 #in /m^3\n",
"n_n = n_n * 10**-6 #in /cm^3\n",
"Miu_n = 0.13 #in m^2/V-sec\n",
"\n",
"\n",
"#CALCULATIONS\n",
"Miu_n = Miu_n * 10**4 #in cm^2/V-sec\n",
"Sigma_n = q*n_n*Miu_n #in ohm-cm^-1\n",
"Rho = 1/Sigma_n\n",
"A = 100 # in \u00b5m^2\n",
"A = A * 10**-8 #in cm^2\n",
"l = 0.1 #in cm\n",
"R = Rho * (l/A) #in ohm\n",
"R=R*10**-6\n",
"#RESULTS\n",
"print('The resistance of the bar is =%.2f in M ohm' %R)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The resistance of the bar is =0.96 in M ohm\n"
]
}
],
"prompt_number": 36
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example No 3.19 Page No 152"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#initialisation of variables\n",
"w = 3.0 #in \u00b5m\n",
"\n",
"#CALCULATIONS\n",
"D = w/9 #in \u00b5m\n",
"\n",
"#RESULTS\n",
"print('Depletion width on P side is =%.2f in \u00b5m ' %D)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Depletion width on P side is =0.33 in \u00b5m \n"
]
}
],
"prompt_number": 37
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example No 3.20 Page No 152"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#initialisation of variables\n",
"n_i = 1.5*10**16 #in /m^3\n",
"n_n = 5*10**20 #in /m^3\n",
"\n",
"#CALCULATIONS\n",
"p_n = ((n_i)**2)/n_n #in /m^3\n",
"\n",
"#RESULTS\n",
"print('The minority carrier density per m^3 is =%.2f X 10^11' %(p_n/(10**11)))\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The minority carrier density per m^3 is =4.50 X 10^11\n"
]
}
],
"prompt_number": 38
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example No 3.21 Page No 152"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#initialisation of variables\n",
"y2 = 10**14 #in /cm^3\n",
"y1 = 0\n",
"x1=-0.5 #in \u00b5m\n",
"\n",
"#CALCULATIONS\n",
"x1= x1*10**-4 #in cm\n",
"x2=0\n",
"dnBYdx = (y2-y1)/(x2-x1) #in /cm^4\n",
"q = 1.6*10**-19 #in C\n",
"D_n = 25 #in cm^2/sec\n",
"J_n = q*D_n*dnBYdx #in A/cm^2\n",
"\n",
"\n",
"#RESULTS\n",
"print('The collector current density is =%.f in A/cm^2' %J_n)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The collector current density is =8 in A/cm^2\n"
]
}
],
"prompt_number": 39
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example No 3.22 Page No 153"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#initialisation of variables\n",
"h = 6.64*10**-34 ##in J-s\n",
"q=1.6*10**-19 #in C\n",
"\n",
"#CALCULATIONS\n",
"h= h/q #in eV\n",
"c = 3*10**8 #in m/s\n",
"lembda = 0.87*10**-6 #in m\n",
"E_g = (h*c)/lembda #in eV\n",
"\n",
"#RESULTS\n",
"print('The band gap in eV is =%.2f ' %E_g)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The band gap in eV is =1.43 \n"
]
}
],
"prompt_number": 40
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example No 3.23 Page No 153"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"#initialisation of variables\n",
"alpha = 5*10**4 #in cm^-1\n",
"l = 0.46*10**-4 #in cm\n",
"hv = 2 #in eV\n",
"I_o = 10**-2 #in W\n",
"\n",
"#CALCULATIONS\n",
"I_t = I_o*math.exp(-alpha*l) #in W\n",
"A_p = I_o-I_t #absorbed power in W or J/s\n",
"c = 1.43\n",
"A_E = (hv-c)/hv*A_p # in J/s\n",
"e = 1.6*10**-19 #in C\n",
"P = A_p/(e*hv) #Perfect quantum efficiency in photon/s\n",
"\n",
"#RESULTS\n",
"print('Total energy absorbed is =%.f X 10^-3 in J/s' %(A_p*(10**3)))\n",
"print('Rate of excess thermal energy is =%.2f X 10^-3 in J/s' %(A_E*(10**3)))\n",
"print('Perfect quantum efficiency is =%.2f X 10^16 in photon/s' %(P/(10**16)))"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Total energy absorbed is =9 X 10^-3 in J/s\n",
"Rate of excess thermal energy is =2.56 X 10^-3 in J/s\n",
"Perfect quantum efficiency is =2.81 X 10^16 in photon/s\n"
]
}
],
"prompt_number": 41
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example No 3.24 Page No 154"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"#initialisation of variables\n",
"Mu_p = 500 #in cm^2/v-s\n",
"kT = 0.0259\n",
"Toh_p = 10**-10 #in sec\n",
"p_o = 10**17 #in cm^-3\n",
"q= 1.6*10**-19 # in C\n",
"A=0.5 #in square meter\n",
"del_p = 5 * 10**16 #in cm^-3\n",
"n_i= 1.5*10**10 #in cm^-3 \n",
"\n",
"\n",
"#CALCULATIONS\n",
"D_p = kT * Mu_p #in cm/s\n",
"L_p = math.sqrt(D_p * Toh_p) # in cm\n",
"x = 10**-5 #in cm\n",
"p = p_o+del_p*e**(x/L_p) #in cm^-3\n",
"Eip= math.log(p/n_i)*kT #in eV\n",
"Ecp= 1.1/2-Eip #value of E_c-E_p in eV\n",
"Ip= q*A*D_p/L_p*del_p*e**(x/L_p) #in A\n",
"Qp= q*A*del_p*L_p #in C\n",
"\n",
"#RESULTS\n",
"print('The hole current is : =%.2f X 10^3 in A ' %(Ip*(10**3)))\n",
"print('Approximation error')\n",
"print('The value of Qp is :=%.2f X 10^-7 in C' %(Qp*(10**7)))\n",
"\n",
"#Note: There is a calculation error to evalaute the value of hole current hence the value of hole current in the book is wrong\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The hole current is : =8.61 X 10^3 in A \n",
"Approximation error\n",
"The value of Qp is :=1.44 X 10^-7 in C\n"
]
}
],
"prompt_number": 42
}
],
"metadata": {}
}
]
}