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-rw-r--r--Electronic_Principles_/Chapter_5_New.ipynb62
1 files changed, 0 insertions, 62 deletions
diff --git a/Electronic_Principles_/Chapter_5_New.ipynb b/Electronic_Principles_/Chapter_5_New.ipynb
index 44f83157..cfea03f3 100644
--- a/Electronic_Principles_/Chapter_5_New.ipynb
+++ b/Electronic_Principles_/Chapter_5_New.ipynb
@@ -27,23 +27,17 @@
"cell_type": "code",
"collapsed": false,
"input": [
- "#Example 5.1.py\n",
- "#Zener diode of figure 5.4a has breakdown voltage of 10V. \n",
- "#What are the minimum & maximum zener currents? \n",
"\n",
- "#Variable declaration\n",
"Vsmin=20 #Source voltage minimum(V)\n",
"Vsmax=40 #Source voltage maximum(V)\n",
"Vbd=10 #Breakdown voltage(V)\n",
"R=0.82 #Resistance(KOhm)\n",
"\n",
- "#Calculation\n",
"Vr1=Vsmin-Vbd #voltage across resistor(V)\n",
"Is1=Vr1/R #Minimum current(mA)\n",
"Vr2=Vsmax-Vbd #voltage across resistor(V)\n",
"Is2=Vr2/R #Maximum current(mA)\n",
"\n",
- "#Result\n",
"print 'Ideally, zener diode acts as a battery(of breakdown voltage = 10V) shown in figure 5-4b'\n",
"print 'Minimum current Is1=',round(Is1,2),'mA'\n",
"print 'Maximum current Is1=',round(Is2,2),'mA'"
@@ -75,19 +69,14 @@
"cell_type": "code",
"collapsed": false,
"input": [
- "#Example 5.2.py\n",
- "#Is the zener diode of figure 5-6a operating in the breakdown region?\n",
"\n",
- "#Variable declaration\n",
"Vs=18 #supply voltage(V)\n",
"Rs=0.27 #source resistance(KOhm)\n",
"RL=1 #Load resistance(KOhm)\n",
"Vz=10 #Zener voltage(V)\n",
"\n",
- "#Calculation\n",
"VTH=(RL/(Rs+RL))*Vs #Thevenin voltage(V)\n",
"\n",
- "#Result\n",
"print 'Thevenin voltage VTH = ',round(VTH,2),'V'\n",
"print 'Thevenin voltage is greater than zener voltage, zener diode is operating in breakdown region.'"
],
@@ -117,22 +106,17 @@
"cell_type": "code",
"collapsed": false,
"input": [
- "#Example 5.3.py\n",
- "#What does the zener current equal in Figure 5-6b?\n",
"\n",
- "#Variable declaration\n",
"Vs=18 #supply voltage(V)\n",
"Rs=0.27 #source resistance(KOhm)\n",
"RL=1 #Load resistance(KOhm)\n",
"Vbd=10 #Zener voltage(V)\n",
"\n",
- "#Calculation\n",
"Vr=Vs-Vbd #voltage across resistor(V)\n",
"Is=Vr/Rs #Current(mA)\n",
"IL=Vbd/RL #Current(mA)\n",
"Iz=Is-IL #Zener current(mA)\n",
"\n",
- "#Result\n",
"print 'Load current IL = ',IL,'mA'\n",
"print 'Zener current Iz = ',round(Iz,2),'mA'"
],
@@ -162,20 +146,14 @@
"cell_type": "code",
"collapsed": false,
"input": [
- "#Example 5.7.py\n",
- "#The zener diode of figure 5-12 has a breakdown voltage of 10V and a zener resistance of 8.5 Ohm.\n",
- "#Use second approx. to calculate load voltage when zener current is 20 mA.\n",
"\n",
- "#Variable declaration\n",
"Iz=20 #zener current(mA)\n",
"Rz=8.5 #zener resistance(Ohm)\n",
"Vbd=10 #Zener voltage(V)\n",
"\n",
- "#Calculation\n",
"DVL=Iz*Rz/1000 #change in load voltage(V)\n",
"VL=Vbd+DVL #Load voltage(V)\n",
"\n",
- "#Result\n",
"print 'Change in load voltage DVL =',DVL,'V'\n",
"print 'Load voltage with second approx., VL =',VL,'V'"
],
@@ -205,19 +183,13 @@
"cell_type": "code",
"collapsed": false,
"input": [
- "#Example 5.8.py\n",
- "#In figure 5-12, Rs = 270 Ohm, Rz=8.5 Ohm and VR(in)=2V. \n",
- "#What is the approximate ripple voltage across the load?\n",
"\n",
- "#Variable declaration\n",
"Rs=270 #Source resistance (Ohm)\n",
"Rz=8.5 #zener resistance(Ohm)\n",
"VRin=2 #Zener voltage(V)\n",
"\n",
- "#Calculation\n",
"VRout=(Rz/Rs)*VRin*1000 #Load ripple voltage(V)\n",
"\n",
- "#Result\n",
"print 'Load ripple voltage VRout=',round(VRout,2),'mV'"
],
"language": "python",
@@ -245,21 +217,15 @@
"cell_type": "code",
"collapsed": false,
"input": [
- "#Example 5.10.py\n",
- "#A zener regulator has an i/p voltage varies from 22 to 30 V. \n",
- "#If the regulated output voltage is 12 V & load resistance varies from 140 Ohm to 10 KOhm, Find maximum allowable series resistance.\n",
"\n",
- "#Variable declaration\n",
"Vil=22 #input voltage range low(V)\n",
"Vih=30 #input voltage range high(V)\n",
"Vz=12 #regulated output voltage(V)\n",
"Rl=140 #Load resistance low(KOhm)\n",
"Rh=10 #Load resistance high(KOhm)\n",
"\n",
- "#Calculation\n",
"RSmax=Rl*(float(Vil)/float(Vz)-1) #Maximum series resistance\n",
"\n",
- "#Result\n",
"print 'Maximum series resistance RSmax =',round(RSmax,2),'V'"
],
"language": "python",
@@ -287,22 +253,16 @@
"cell_type": "code",
"collapsed": false,
"input": [
- "#Example 5.11.py\n",
"\n",
- "#A zener regulator has an input voltage varies from 15 to 20 V. \n",
- "#If zenerge is 6.8V and load current varies from 5 mA to 20 mA,find maximum allowable series resistance.\n",
"\n",
- "#Variable declaration\n",
"Vil=15 #input voltage range low(V)\n",
"Vih=20 #input voltage range high(V)\n",
"Vz=6.8 #regulated output voltage(V)\n",
"Il=5 #Load current low(mA)\n",
"Ih=20 #Load current high(mA)\n",
"\n",
- "#Calculation\n",
"RSmax=(Vil-float(Vz))/Ih*1000 #Maximum series resistance\n",
"\n",
- "#Result\n",
"print 'Maximum series resistance RSmax =',RSmax,'V'"
],
"language": "python",
@@ -330,18 +290,13 @@
"cell_type": "code",
"collapsed": false,
"input": [
- "#Example 5.12.py\n",
- "#Figure 5-23a shows voltage-polarity tester. Find approx. LED current if the dc input voltage is 50V & series resistance is 2.2 KOhm.\n",
"\n",
- "#Variable declaration\n",
"Vi=50 #voatage supply(V)\n",
"Rs=2.2 #series resistance(KOhm)\n",
"Vf=2 #forward approx. voltage\n",
" \n",
- "#Calculation\n",
"Is=(Vi-Vf)/Rs\n",
"\n",
- "#Result\n",
"print 'LED current Is =',round(Is,2),'mA'"
],
"language": "python",
@@ -369,18 +324,13 @@
"cell_type": "code",
"collapsed": false,
"input": [
- "#Example 5.13.py\n",
- "#Figure 5-23b shows continuity tester. find approx. LED current if the series resistance is 470 Ohm.\n",
"\n",
- "#Variable declaration\n",
"Vs=9 #voatage supply(V)\n",
"Rs=470.0 #series resistance(Ohm)\n",
"Vf=2 #forward approx. voltage\n",
" \n",
- "#Calculation\n",
"Is=(Vs-Vf)/Rs\n",
"\n",
- "#Result\n",
"print 'LED current Is =',round((Is*1000),2),'mA'"
],
"language": "python",
@@ -408,23 +358,17 @@
"cell_type": "code",
"collapsed": false,
"input": [
- "#Example 5.14.py\n",
- "#In figure 5-24, find avg. LED current if the ac source voltage is 20V rms and series resistance is 680 Ohm.\n",
- "#Also calculate approx. power dissipation in series resistor.\n",
"\n",
"import math\n",
"\n",
- "#Variable declaration\n",
"Vac=20 #AC voatage supply(V)\n",
"Rs=680.0 #series resistance(KOhm)\n",
" \n",
- "#Calculation\n",
"Vacp=1.414*Vac #peak source voltage(V)\n",
"Is1=(Vacp/Rs)*1000 #approx. peak current(mA)\n",
"Is2=Is1/math.pi #average of half-wave current through LED(mA)\n",
"P=(Vac)**2/Rs #Power dissipation(W)\n",
"\n",
- "#Result\n",
"print 'approx. peak LED current Is1 =',round(Is1,2),'mA'\n",
"print 'average of half-wave current through LED Is2 =',round(Is2,2),'mA'\n",
"print 'Power dissipation P =',round(P,2),'W'"
@@ -456,24 +400,18 @@
"cell_type": "code",
"collapsed": false,
"input": [
- "#Example 5.15.py\n",
- "#Figure 5-25, find capacitor is used instead of resistor. \n",
- "#what is average LED current if the capacitance is 0.68 uf?\n",
"\n",
"import math\n",
"\n",
- "#Variable declaration\n",
"Vs=120 #AC voatage supply(V)\n",
"f=60 #frequency(Hz)\n",
"C=0.68 #series resistance(KOhm)\n",
" \n",
- "#Calculation\n",
"Xc=1/(2*math.pi*f*C)*1000 #capacitive reactance(KOhm)\n",
"Vacp=Vs*1.414\n",
"Is1=(Vacp/Xc) #approx. peak current(mA)\n",
"Is2=Is1/math.pi #average current through LED(mA)\n",
"\n",
- "#Result\n",
"print 'Capacitance reactance Xc = ',round(Xc,2),'KOhm'\n",
"print 'approx. peak LED current Is1 =',round(Is1,2),'mA'\n",
"print 'average current through LED Is2 =',round(Is2,2),'mA'"
generated by cgit (git 2.34.1) at 2025-01-11 00:55:53 +0000