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{
"metadata": {
"name": ""
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"CHAPTER 3 DIODE THEORY"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 3-2, Page 63"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Example 3.2.py\n",
"#A diode has a power rating of 5W. If diode voltage is 1.2V and diode current is 1.75A,\n",
"#What is the power dissipation? Will the diode be destroyed?\n",
"\n",
"#Variable declaration\n",
"Pr=5 #Power rating(W)\n",
"Vd=1.2 #diode voltage(V)\n",
"Id=1.75 #diode current(A)\n",
"\n",
"#Calculation\n",
"PD=Vd*Id #Power dissipaion(W)\n",
"\n",
"#Result\n",
"print 'Power Dissipation =',PD,'W'\n",
"print 'PD(',PD,'W) < ''Pr(',Pr,'W), So diode will not be destroyed.'"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Power Dissipation = 2.1 W\n",
"PD( 2.1 W) < Pr( 5 W), So diode will not be destroyed.\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 3-3, Page 65"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Example 3.3.py\n",
"#Use the ideal diode to calculate the VL & IL in Fig. 3-6a.\n",
"\n",
"\n",
"#Variable declaration\n",
"Vs=10 #Source voltage(V)\n",
"RL=1 #Load resistance(KOhm)\n",
"VL=Vs #LOad voltage(V)\n",
"\n",
"#Calculation\n",
"IL=VL/RL #Load current(mA)\n",
"\n",
"#Result\n",
"print 'Vs will be appearing across RL'\n",
"print 'Load voltage VL =',Vs,'V'\n",
"print 'Load current IL =',IL,'mA'"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Vs will be appearing across RL\n",
"Load voltage VL = 10 V\n",
"Load current IL = 10 mA\n"
]
}
],
"prompt_number": 3
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 3-4, Page 65"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Example 3.4.py\n",
"#Calculate the VL & IL in Fig. 3-6b using an ideal diode.\n",
"\n",
"#Variable Declaration\n",
"print 'As per figure 3-6b, Thevenize the circuit to the left of the diode'\n",
"R1=6 #Resistance(KOhm)\n",
"R2=3 #Resistance(KOhm)\n",
"RL=1 #Load Resistance(KOhm)\n",
"Vs=36 #Supply voltage(V)\n",
"\n",
"#Calculation\n",
"Vth=R2*Vs/(R1+R2) #ThCevenin voltage(V)\n",
"Rth=(R1*R2)/(R1+R2) #Thevenin resistance(KOhm)\n",
"Rt=Rth+RL #total resistance(KOhm)\n",
"IL=Vth/Rt #Load current(mA)\n",
"VL=IL*RL #Load voltage(V)\n",
"\n",
"#Result\n",
"print 'Vth =',Vth,'V & Rth =',Rth,'KOhm'\n",
"print 'Visualize diode as closed switch,'\n",
"print 'IL =',IL,'mA'\n",
"print 'VL =',VL,'V'"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"As per figure 3-6b, Thevenize the circuit to the left of the diode\n",
"Vth = 12 V & Rth = 2 KOhm\n",
"Visualize diode as closed switch,\n",
"IL = 4 mA\n",
"VL = 4 V\n"
]
}
],
"prompt_number": 4
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 3-5, Page 67"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Example 3.5.py\n",
"#Use the second approximation to calculate the VL, IL & PD in Fig 3-8. \n",
"\n",
"#Variable Declaration\n",
"print 'As per Second approximation in Fig.3-8,'\n",
"Vd=0.7 #diode voltgage(V)\n",
"Vs=10 #supply voltage(V)\n",
"RL=1 #Load resistance(KOhm)\n",
"\n",
"#Calculation\n",
"VL=Vs-Vd #Load voltage(v)\n",
"IL=VL/RL #Load current(mA)\n",
"PD=Vd*IL #diode power(mW)\n",
"\n",
"#Result\n",
"print 'IL =',IL,'mA & VL =',VL,'V'\n",
"print 'Diode power PD =',PD,'mW'"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"As per Second approximation in Fig.3-8,\n",
"IL = 9.3 mA & VL = 9.3 V\n",
"Diode power PD = 6.51 mW\n"
]
}
],
"prompt_number": 5
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 3-6, Page 67"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Example 3.6.py\n",
"#Calculate the VL, IL & PD in fig. 3-9a using the second approximation.\n",
"\n",
"#Variable Declaration\n",
"print 'As per Second approximation in Fig.3-9a,'\n",
"Vd=0.7 #diode voltgage(V)\n",
"Vs=36 #supply voltage(V)\n",
"R1=6 #Resistance(KOhm)\n",
"R2=3 #Resistance(KOhm)\n",
"RL=1 #Load resistance(KOhm)\n",
"\n",
"#Calculation\n",
"Vth=R2*Vs/(R1+R2)#Thevenin Voltage(V)\n",
"Rth=(R1*R2)/(R1+R2)#Thevenin resistance(KOhm)\n",
"IL=(Vth-Vd)/R2#Load current(mA)\n",
"VL=IL*RL#Load voltage(V)\n",
"PD=Vd*IL#diode power(mW)\n",
"\n",
"#Result\n",
"print 'Thevenize the circuit to the left of the diode'\n",
"print 'Vth =',Vth,'V & Rth =',Rth,'KOhm'\n",
"print 'VL =',round(VL,2),'V & IL =',round(IL,2),'mA'\n",
"print 'Diode power PD =',round(PD,2),'mW'"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"As per Second approximation in Fig.3-9a,\n",
"Thevenize the circuit to the left of the diode\n",
"Vth = 12 V & Rth = 2 KOhm\n",
"VL = 3.77 V & IL = 3.77 mA\n",
"Diode power PD = 2.64 mW\n"
]
}
],
"prompt_number": 7
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 3-7, Page 68"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Example 3.7.py\n",
"#The IN4001 has a bulk resistance of 0.23 Ohm. What is VL, IL & PD?\n",
"\n",
"#Variable Declaration\n",
"print 'In Fig.3-11a,'\n",
"Vd=0.7 #diode voltgage(V)\n",
"Vs=10 #supply voltage(V)\n",
"RL=1000L #Load resistance(Ohm)\n",
"Rb=0.23 #bulk resistance\n",
"\n",
"#Calculation\n",
"print 'As per third approximation, we get fig.3-11b'\n",
"if Rb<(RL/100):\n",
" print'If Rb < 0.01RL than ignore Rb & use second approximation.'\n",
" VL=Vs-Vd #Load voltage(V)\n",
" IL=(VL/RL)*1000 #Load current(mA)\n",
" PD=Vd*IL #diode power(mW)\n",
"\n",
"#Result \n",
" print 'IL =',IL,'mA & VL =',VL,'V'\n",
" print 'Diode power PD =',PD,'mW'"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"In Fig.3-11a,\n",
"As per third approximation, we get fig.3-11b\n",
"If Rb < 0.01RL than ignore Rb & use second approximation.\n",
"IL = 9.3 mA & VL = 9.3 V\n",
"Diode power PD = 6.51 mW\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 3-8, Page 69"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Example 3.8.py\n",
"#Repeat the preceding example for RL of 10 Ohm.\n",
"\n",
"#Variable Declaration\n",
"print 'In Fig.3-11a, take RL = 10 Ohm'\n",
"Vd=0.7 #diode voltgage(V)\n",
"Vs=10 #supply voltage(V)\n",
"RL=10 #Load resistance(Ohm)\n",
"Rb=0.23 #bulk resistance\n",
"RT=Rb+RL #Total reistance(Ohm)\n",
"VT=Vs-Vd #total voltage(V)\n",
"\n",
"#Calculation\n",
"print 'RT =',RT,'Ohm & VT =',VT,'V'\n",
"IL=VT/RT #Load current(mA)\n",
"VL=IL*RL #Load voltage(V)\n",
"VD=Vd+(IL*Rb) \n",
"PD=VD*IL #diode power(W)\n",
"\n",
"#Result\n",
"print 'IL =',round(IL,2),'mA & VL =',round(VL,2),'V'\n",
"print 'VD =',round(VD,2),'V'\n",
"print 'Diode power PD =',round(PD,2),'W'"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"In Fig.3-11a, take RL = 10 Ohm\n",
"RT = 10.23 Ohm & VT = 9.3 V\n",
"IL = 0.91 mA & VL = 9.09 V\n",
"VD = 0.91 V\n",
"Diode power PD = 0.83 W\n"
]
}
],
"prompt_number": 9
},
{
"cell_type": "code",
"collapsed": false,
"input": [],
"language": "python",
"metadata": {},
"outputs": []
}
],
"metadata": {}
}
]
}
|
