{
"metadata": {
"name": ""
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "markdown",
"metadata": {},
"source": [
"<h1> Chapter No 4 : Junctions and Interfaces<h1>"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 4.2 Page No 184"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"#initialisation of variables\n",
"N_D = 10**17 * 10**6 #in atoms/m^3\n",
"N_A = 0.5*10**16*10**6 #in atoms/m^3\n",
"Epsilon_r = 10\n",
"\n",
"#CALCULATIONS\n",
"Epsilon_o = 8.85*10**-12\n",
"Epsilon = Epsilon_r*Epsilon_o #in F/m\n",
"e = 1.602*10**-19 #in C\n",
"V = 0\n",
"V_B = 0.7 #in V\n",
"W1 = math.sqrt( ((2*Epsilon*V_B)/e)*(1/N_A+1/N_D) ) #in m\n",
"V_o = V_B #in V\n",
"V1 = -10 #in V\n",
"V_B1 = V_o-V1 #in V\n",
"W = math.sqrt( ((2*Epsilon*V_B1)/e)*(1/N_A+1/N_D) ) #in m\n",
"\n",
"\n",
"#RESULTS\n",
"print('The junction width in meter when no external voltage is applied is =%.2f X 10^-6' %(W1*(10**6)))\n",
"print('Junction width in meter with an external voltage of -10V is =%.2f X 10^-6 m' %(W*(10**6)))\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The junction width in meter when no external voltage is applied is =0.39 X 10^-6\n",
"Junction width in meter with an external voltage of -10V is =1.54 X 10^-6 m\n"
]
}
],
"prompt_number": 34
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 4.4 Page No 208"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"#initialisation of variables\n",
"V = 5 #in V\n",
"V_Gamma = 0.6 #in V\n",
"r_F = 12 #in ohm\n",
"R = 1 #in k ohm\n",
"\n",
"#CALCULATIONS\n",
"R = R * 10**3 #in ohm\n",
"I_F = (V-V_Gamma)/(R+r_F) #in A\n",
"V_F = V_Gamma + (I_F*r_F) #in V\n",
"I_F=I_F*10**3 \n",
"#RESULTS\n",
"print('The forward diode current is =%.2f mA ' %I_F)\n",
"print('The diode voltage is =%.2f V ' %V_F)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The forward diode current is =4.35 mA \n",
"The diode voltage is =0.65 V \n"
]
}
],
"prompt_number": 35
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 4.5 Page No 214"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"#initialisation of variables\n",
"n = 4.4*10**22 #total number of Ge atoms/cm^3\n",
"n_a = 1*10**8 #number of impurity atoms\n",
"\n",
"#CALCULATIONS\n",
"N_A = n/n_a #in atoms/cm^3\n",
"N_A = N_A * 10**6 #in atoms/m^3\n",
"n_i = 2.5*10**13 #in atoms/cm^3\n",
"n_i = n_i * 10**6 #in atoms/m^3\n",
"N_D = 10**3 * N_A #in atoms/m^3\n",
"V_T = 26*10**-3 #in A\n",
"V_J = V_T*math.log( (N_A*N_D)/((n_i)**2) ) #in V\n",
"print('The contact difference of potential is =%.2f V For a silicon P-N junction ' %V_J)\n",
"n = 5*10**22\n",
"N_A = n/n_a #in atoms/cm^3\n",
"N_A = N_A * 10**6 #in atoms/m^3\n",
"N_D = 10**3 * N_A #in atoms/m^3\n",
"n_i = 1.5*10**10 #in /cm^3\n",
"V_J = V_T*math.log(N_A*N_D/n_i**2) #in V\n",
"\n",
"\n",
"#RESULTS\n",
"\n",
"print('The contact difference of potential is =%.2f V' %V_J)\n",
"\n",
"#Note: There is a calculation error to find the value of V_J in the book, so the answer in the book is wrong.\n",
"\n",
" "
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The contact difference of potential is =0.33 V For a silicon P-N junction \n",
"The contact difference of potential is =1.44 V\n"
]
}
],
"prompt_number": 36
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 4.6 Page No 214"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"#initialisation of variables\n",
"Rho_p = 2 #in ohm-cm\n",
"Rho_n = 1 #in ohm cm\n",
"q = 1.6*10**-19 #in C\n",
"n_i = 2.5*10**13 #atoms per cm^3\n",
"Miu_p = 1800\n",
"Miu_n = 3800\n",
"\n",
"#CALCULATIONS\n",
"N_A = 1/(Rho_p*q*Miu_p) #in /cm^3\n",
"N_D = 1/(Rho_n*q*Miu_n) #in /cm^3\n",
"V_T = 26 #in mV\n",
"V_T= V_T*10**-3 #in V\n",
"V_J = V_T*math.log((N_A*N_D)/((n_i)**2)) #in V\n",
"print('The height of the potential energy barrier is =%.2f in V ' %V_J)\n",
"Miu_p = 500\n",
"N_A = 1/(Rho_p*q*Miu_p) #in /cm^3\n",
"Miu_n = 1300\n",
"N_D = 1/(Rho_n*q*Miu_n) #in /cm^3\n",
"n_i = 1.5*10**10;\n",
"V_J = V_T*math.log((N_A*N_D)/((n_i)**2)) #in V\n",
"\n",
"#RESULTS\n",
"print('For silicon P-N juction The height of the potential energy barrier is =%.2f V' %V_J)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The height of the potential energy barrier is =0.22 in V \n",
"For silicon P-N juction The height of the potential energy barrier is =0.67 V\n"
]
}
],
"prompt_number": 37
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 4.7 page No 215"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"#initialisation of variables\n",
"Eta = 1\n",
"V_T = 26 #in mV\n",
"V_T= V_T*10**-3 #in V\n",
"\n",
"#CALCULATIONS\n",
"V= math.log(1-0.9)*V_T #in V\n",
"print(\"The voltage in volts is : %.2f \" %V)\n",
"V1=0.05 #in V\n",
"V2= -0.05 #in V\n",
"ratio= (math.e**(V1/(Eta*V_T))-1)/(math.e**(V2/(Eta*V_T))-1)\n",
"print(\"The ratio of the current for a forward bias to reverse bias is : %.2f \" %ratio)\n",
"\n",
"# Part (iii)\n",
"Io= 10 #in \u00b5A\n",
"Io=Io*10**-3 #in mA\n",
"#For \n",
"V=0.1 #in V\n",
"I = Io * (math.e**(V/(Eta*V_T)) - 1) #in mA\n",
"print(\"For v=0.1 V , the value of I is : %.2f mA \" %I)\n",
"\n",
"#For \n",
"V=0.2 #in V\n",
"I = Io * (math.e**(V/(Eta*V_T)) - 1) #in mA\n",
"print(\"For v=0.2 V , the value of I is : %.2f mA \" %I)\n",
"\n",
"#For \n",
"V=0.3 #in V\n",
"I = Io * (math.e**(V/(Eta*V_T)) - 1) #in mA\n",
"\n",
"#RESULTS\n",
"\n",
"I=I*10**-3\n",
"print(\"For v=0.3 V , the value of I is : %.2f mA\" %I)\n",
"print(\"From three value of I, for small rise in forward voltage, the diode current increase rapidly \")\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The voltage in volts is : -0.06 \n",
"The ratio of the current for a forward bias to reverse bias is : -6.84 \n",
"For v=0.1 V , the value of I is : 0.46 mA \n",
"For v=0.2 V , the value of I is : 21.90 mA \n",
"For v=0.3 V , the value of I is : 1.03 mA\n",
"From three value of I, for small rise in forward voltage, the diode current increase rapidly \n"
]
}
],
"prompt_number": 38
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 4.8 Page No 216"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"#initialisation of variables\n",
"# Part (i)\n",
"T1= 25 #in \u00b0C\n",
"T2= 80.0 #in \u00b0C\n",
"\n",
"#CALCULATIONS\n",
"# Formula Io2= Io1*2**((T2-T1)/10)\n",
"AntiFactor= 2**((T2-T1)/10)\n",
"print('Anticipated factor for Ge is : =%.f ' %AntiFactor)\n",
"\n",
"# Part (ii)\n",
"T1= 25.0 #in \u00b0C\n",
"T2= 150.0 #in \u00b0C\n",
"AntiFactor= 2**((T2-T1)/10)\n",
"\n",
"#RESULTS\n",
"print('Anticipated factor for Ge is : =%.f ' %AntiFactor)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Anticipated factor for Ge is : =45 \n",
"Anticipated factor for Ge is : =5793 \n"
]
}
],
"prompt_number": 39
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 4.9 Page No 216"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"#initialisation of variables\n",
"I=5.0 #in \u00b5A\n",
"V=10.0 #in V\n",
"T1= 0.11 #in \u00b0C^-1\n",
"T2= 0.07 #in \u00b0C^-1\n",
"\n",
"#CALCULATIONS\n",
"Io= T2*I/T1 #in \u00b5A\n",
"I_R= I-Io #in \u00b5A\n",
"R= V/I_R #in M\u03a9\n",
"\n",
"#RESULTS\n",
"print('The leakage resistance is : =%.1f M\u03a9 ' %R)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The leakage resistance is : =5.5 M\u03a9 \n"
]
}
],
"prompt_number": 40
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 4.10 Page No 216"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"\n",
"#initialisation of variables\n",
"Eta = 1.0\n",
"T = 125.0 #in \u00b0C\n",
"T = T + 273 #in K\n",
"V_T = 8.62 * 10**-5 * 398 #in V\n",
"I_o = 30 #in \u00b5A\n",
"\n",
"#CALCULATIONS\n",
"I_o= I_o*10**-6 #in A\n",
"v = 0.2 #in V\n",
"r_f = (Eta * V_T)/(I_o * math.e**(v/(Eta* V_T))) #in ohm\n",
"print('The dynamic resistance in the forward direction is =%.2f \u03a9 ' %r_f)\n",
"\n",
"r_r = (Eta * V_T)/(I_o * math.e**(-v/(Eta* V_T))) #in ohm\n",
"r_r=r_r*10**-3\n",
"print('The dynamic resistance in the forward direction is =%.2f K\u03a9 ' %r_r)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The dynamic resistance in the forward direction is =3.36 \u03a9 \n",
"The dynamic resistance in the forward direction is =389.08 K\u03a9 \n"
]
}
],
"prompt_number": 41
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 4.11 Page No 217"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"#initialisation of variables\n",
"epsilon = 16/(36 * math.pi * 10**11) #in F/cm\n",
"A = 1 * 10**-2\n",
"W = 2 * 10**-4\n",
"\n",
"#CALCULATIONS\n",
"C_T = (epsilon * A)/W #in F\n",
"C_T=C_T*10**12\n",
"\n",
"#RESULTS\n",
"print('The barrier capacitance is =%.2f pF ' %C_T)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The barrier capacitance is =70.74 pF \n"
]
}
],
"prompt_number": 42
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 4.12 Page No 217"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"#initialisation of variables\n",
"A = 1.0 #in mm^2\n",
"A = A * 10**-6 #in m^2\n",
"N_A = 3 * 10**20 #in atoms/m^3\n",
"q = 1.6 *10**-19 #in C\n",
"V_o = 0.2 #in V\n",
"epsilon_r=16\n",
"epsilon_o= 8.854*10**-12 #in F/m\n",
"\n",
"\n",
"#CALCULATIONS\n",
"epsilon=epsilon_r*epsilon_o\n",
"# Part (a)\n",
"V=-10 #in V\n",
"W = math.sqrt(((V_o - V) * 2 * epsilon)/(q * N_A)) #m\n",
"C_T1 = (epsilon * A)/W #in F\n",
"W=W*10**6\n",
"print('The width of the depletion layer for an applied reverse voltage of 10V is =%.2f \u00b5m ' %W)\n",
"\n",
"# Part (b)\n",
"V=-0.1 #in V\n",
"W = math.sqrt(((V_o - V) * 2 * epsilon)/(q * N_A)) #m\n",
"C_T2 = (epsilon * A)/W #in F\n",
"W=W*10**6\n",
"print('The width of the depletion layer for an applied reverse voltage of 0.1V is =%.2f \u00b5m ' %W)\n",
"\n",
"# Part (c)\n",
"V=0.1 #in V\n",
"W = math.sqrt(((V_o - V) * 2 * epsilon)/(q * N_A)) # m\n",
"W=W*10**6\n",
"print('The width of the depletion layer for an applied reverse voltage of 0.1V is =%.2f \u00b5m ' %W)\n",
"\n",
"#Part (d)\n",
"C_T1=C_T1*10**12\n",
"C_T2=C_T2*10**12\n",
"\n",
"#RESULTS\n",
"print('The space charge capacitance for an applied reverse voltage of 10V is =%.2f pF ' %C_T1)\n",
"print('The space charge capacitance for an applied reverse voltage of 0.1V is =%.2f pF' %C_T2)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The width of the depletion layer for an applied reverse voltage of 10V is =7.76 \u00b5m \n",
"The width of the depletion layer for an applied reverse voltage of 0.1V is =1.33 \u00b5m \n",
"The width of the depletion layer for an applied reverse voltage of 0.1V is =0.77 \u00b5m \n",
"The space charge capacitance for an applied reverse voltage of 10V is =18.26 pF \n",
"The space charge capacitance for an applied reverse voltage of 0.1V is =106.46 pF\n"
]
}
],
"prompt_number": 43
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example No 4.13 Page No 218"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"#initialisation of variables\n",
"I_o = 1.8 * 10**-9 #A\n",
"v = 0.6 #in V\n",
"Eta = 2\n",
"V_T = 26 #in mV\n",
"\n",
"#CALCULATIONS\n",
"V_T=V_T*10**-3 #in V\n",
"I = I_o *(math.e**(v/(Eta * V_T))) #in A\n",
"I=I*10**3\n",
"\n",
"#RESULTS\n",
"print('The current in the junction is =%.2f mA ' %I)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The current in the junction is =0.18 mA \n"
]
}
],
"prompt_number": 44
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example No 4.14 Page No 218"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"#initialisation of variables\n",
"I_o = 2.4 * 10**-14\n",
"I = 1.5 #in mA\n",
"I=I*10**-3 #in A\n",
"Eta = 1\n",
"V_T = 26 #in mV\n",
"\n",
"#CALCULATIONS\n",
"V_T= V_T*10**-3 #in V\n",
"v =math.log((I + I_o)/I_o) * V_T #in V\n",
"\n",
"#RESULTS\n",
"print('The forward biasing voltage across the junction is =%.2f V' %v)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The forward biasing voltage across the junction is =0.65 V\n"
]
}
],
"prompt_number": 45
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example No 4.15 Page No 218"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#initialisation of variables\n",
"I_o = 10 #in nA\n",
"\n",
"#CALCULATIONS\n",
"I = I_o * (-1) #in nA\n",
"\n",
"#RESULTS\n",
"print('The Diode current is = %.f nA ' %I)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The Diode current is = -10 nA \n"
]
}
],
"prompt_number": 46
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example No 4.16 Page No 218"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"#initialisation of variables\n",
"R = 4.5 #in ohm\n",
"I = 44.4 #in mA\n",
"\n",
"#CALCULATIONS\n",
"I=I*10**-3 #in A\n",
"V = R * I #in V\n",
"Eta = 1\n",
"V_T = 26 #in mV\n",
"V_T=V_T*10**-3 #in V\n",
"I_o = I/((math.e**(V/(Eta * V_T))) -1) #in A\n",
"V = 0.1 #in V\n",
"r_f = (Eta * V_T)/(I_o * ((math.e**(V/(Eta * V_T)))-1)) #in ohm\n",
"\n",
"#RESULTS\n",
"print('The diode dynamic resistance is =%.2f \u03a9' %r_f)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The diode dynamic resistance is =27.78 \u03a9\n"
]
}
],
"prompt_number": 47
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example No 4.18 Page No 219"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"#initialisation of variables\n",
"V = 0.25 #in V\n",
"I_o = 1.2 #in \u00b5A\n",
"\n",
"#CALCULATIONS\n",
"I_o = I_o * 10**-6 #in A\n",
"V_T = 26 #in mV\n",
"V_T = V_T * 10**-3 #in V\n",
"Eta = 1\n",
"r = (Eta * V_T)/(I_o * (math.e**(V/(Eta * V_T)))) #in ohm\n",
"\n",
"#RESULTS\n",
"print('The ac resistance of the diode is =%.2f \u03a9' %r)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The ac resistance of the diode is =1.45 \u03a9\n"
]
}
],
"prompt_number": 48
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
" Example No 4.19 Page No 219"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"#initialisation of variables\n",
"q = 1.6 * 10**-19 #in C\n",
"N_A = 3 * 10**20 #in /m^3\n",
"A = 1 #in \u00b5m^2\n",
"\n",
"#CALCULATIONS\n",
"A = A * 10**-6 #in m^2\n",
"V = -10 #in V\n",
"V_J = 0.25 #in V\n",
"V_B = V_J - V #in V\n",
"epsilon_o = 8.854 #in pF/m\n",
"epsilon_o = epsilon_o * 10**-12 #in F/m\n",
"epsilon_r = 16\n",
"epsilon = epsilon_o * epsilon_r\n",
"W = math.sqrt((V_B * 2 * epsilon)/(q * N_A)) #in m \n",
"\n",
"#RESULTS\n",
"C_T = (epsilon * A)/W #in pF\n",
"W=W*10**6\n",
"C_T=C_T*10**12\n",
"print('The width of depletion layer is =%.2f \u00b5m ' %W)\n",
"print('The space charge capacitance is =%.2f pF ' %C_T)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The width of depletion layer is =7.78 \u00b5m \n",
"The space charge capacitance is =18.21 pF \n"
]
}
],
"prompt_number": 49
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
" Example No 4.20 Page No 220"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"#initialisation of variables\n",
"C_T = 100.0 #in pF\n",
"C_T=C_T*10**-12 #in F\n",
"epsilon_r = 12\n",
"epsilon_o = 8.854 * 10**-12 #in F/m\n",
"\n",
"#CALCULATIONS\n",
"epsilon = epsilon_r * epsilon_o\n",
"Rho_p = 5 #in ohm-cm\n",
"Rho_p = Rho_p * 10**-2 #in ohm-m\n",
"V_j = 0.5 #in V\n",
"V = -4.5 #in V\n",
"Mu_p = 500 #in cm^2\n",
"Mu_p = Mu_p * 10**-4 #in m^2\n",
"Sigma_p = 1/Rho_p #in per ohm-m\n",
"qN_A = Sigma_p/ Mu_p\n",
"V_B = V_j - V\n",
"W = math.sqrt((V_B * 2 * epsilon)/qN_A)\n",
"A = (C_T * W)/ epsilon #in m\n",
"D = math.sqrt(A * (4/math.pi)) #in m\n",
"D = D * 10**3 #in mm\n",
"\n",
"#RESULTS\n",
"print('The diameter is =%.2f mm ' %D)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The diameter is =1.40 mm \n"
]
}
],
"prompt_number": 50
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
" Example No 4.21 Page No 221"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#initialisation of variables\n",
"q = 1.6 * 10**-19 #in C\n",
"Mu_p = 500 #in cm^2/V-sec\n",
"Rho_p = 3.5 #in ohm-cm\n",
"Mu_n = 1500 #in cm^2/V-sec\n",
"Rho_n = 10 #in ohm-cm\n",
"\n",
"#CALCULATIONS\n",
"N_A = 1/(Rho_p * Mu_p * q) # in /cm^3\n",
"N_D = 1/(Rho_n * Mu_n * q) # in /cm^3\n",
"V_J = 0.56 # in V\n",
"n_i = 1.5 * 10**10 #in /cm^3\n",
"V_T = V_J/math.log((N_A * N_D)/(n_i)**2) #in V\n",
"T = V_T * 11600 #in K\n",
"T = T - 273 #in \u00b0C\n",
"\n",
"#RESULTS\n",
"print('The Temperature of junction is =%.2f \u00b0C ' %T)\n",
"print('Approximation error')"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The Temperature of junction is =14.28 \u00b0C \n",
"Approximation error\n"
]
}
],
"prompt_number": 51
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
" Example No 4.22 Page No 221"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"#initialisation of variables\n",
"R = 5 #in ohm\n",
"I = 50 #in mA\n",
"I=I*10**-3 #in A\n",
"V = R * I #in V\n",
"Eta = 1\n",
"V_T = 26 #in mV\n",
"\n",
"#CALCULATIONS\n",
"V_T=V_T*10**-3 #in V\n",
"I_o = I/((math.e**(V/(Eta * V_T))) - 1) #in A\n",
"v1 = 0.2 #in V\n",
"r = (Eta * V_T)/(I_o * (math.e**(v1/(Eta * V_T)))) #in ohm\n",
"\n",
"#RESULTS\n",
"I_o=I_o*10**6\n",
"print('Reverse saturation current is =%.2f \u00b5A ' %I_o)\n",
"print('Dynamic resistance of the diode is =%.2f \u03a9 ' %r)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Reverse saturation current is =3.33 \u00b5A \n",
"Dynamic resistance of the diode is =3.56 \u03a9 \n"
]
}
],
"prompt_number": 52
}
],
"metadata": {}
}
]
}