{
"cells": [
 {
		   "cell_type": "markdown",
	   "metadata": {},
	   "source": [
       "# Chapter 3: Diode Circuits and Rectifiers"
	   ]
	},
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 3.10: to_determine_the_effect_of_reverse_recovery_time_on_the_avg_output_voltage.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clear;\n",
"clc;\n",
"V_s=230;//V;\n",
"t_rr=40*10^-6;//s reverde recovery time\n",
"V_o=2*sqrt(2)*V_s/%pi;\n",
"V_m=sqrt(2)*V_s;\n",
"disp('when f=50Hz');\n",
"f=50;\n",
"V_r=(V_m/%pi)*(1-cosd(2*%pi*f*t_rr*180/%pi));\n",
"v_avg=V_r*100/V_o*10^3;    printf('percentage reduction in avg o/p voltage=%.3fx10^-3',v_avg);\n",
"\n",
"disp('when f=2500Hz');\n",
"f=2500;\n",
"V_r=(V_m/%pi)*(1-cosd(2*%pi*f*t_rr*180/%pi));\n",
"v_avg=V_r*100/V_o;    printf('percentage reduction in avg o/p voltage=%.3f',v_avg);"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 3.11: EX3_11.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clear;\n",
"clc;\n",
"V_s=230;//V\n",
"R=10;//ohm\n",
"V_m=sqrt(2)*V_s;\n",
"V_o=2*V_m/%pi;    printf('avg value of o/p voltage=%.2f V',V_o);\n",
"I_o=V_o/R;    printf('\navg value of o/p current=%.3f A',I_o);\n",
"I_DA=I_o/2;    printf('\navg value of diode current=%.3f A',I_DA);\n",
"I_Dr=I_o/sqrt(2);    printf('\nrms value of diode current=%.3f A',I_Dr);\n",
"\n",
"printf('\nrms value of o/p current=%.3f A',I_o);\n",
"printf('\nrms value of i/p current=%.3f A',I_o);\n",
"pf=(V_o/V_s);    printf('\nsupply pf=%.2f',pf);"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 3.12: EX3_12.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clear;\n",
"clc;\n",
"V_s=230;//V\n",
"R=1000;//ohm\n",
"R_D=20;//ohm\n",
"V_m=sqrt(2)*V_s;\n",
"I_om=V_m/(R+R_D);    printf('peak load current=%.4f A',I_om);\n",
"I_o=I_om/%pi;    printf('\ndc load current=%.5f A',I_o);\n",
"V_D=I_o*R_D-V_m/%pi;    printf('\ndc diode voltage=%.1f V',V_D);\n",
"V_on=V_m/%pi;    printf('\nat no load, load voltage=%.3f V',V_on);\n",
"V_o1=I_o*R;    printf('\nat given load, load voltage=%.3f V',V_o1);\n",
"vr=(V_on-V_o1)*100/V_on;    printf('\nvoltage regulation(in percent)=%.3f',vr);"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 3.13: to_design_a_zener_voltage_regulator_to_meet_given_specifications.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clear;\n",
"clc;\n",
"V_L=6.8;//V\n",
"V_smax=20*1.2;//V\n",
"V_smin=20*.8;//V\n",
"I_Lmax=30*1.5;//mA\n",
"I_Lmin=30*.5;//mA\n",
"I_z=1;//mA\n",
"\n",
"R_smax=(V_smax-V_L)/((I_Lmin+I_z)*10^-3);    printf('max source resistance=%.0f ohm',R_smax);\n",
"R_smin=(V_smin-V_L)/((I_Lmax+I_z)*10^-3);    printf('\nmin source resistance=%.0f ohm',R_smin);//in book solution, error is committed in putting the values in formulea(printing error) but solution is correct\n",
"R_Lmax=V_L*1000/I_Lmin;    printf('\nmax load resistance=%.1f ohm',R_Lmax);\n",
"R_Lmin=V_L*1000/I_Lmax;    printf('\nmin load resistance=%.1f ohm',R_Lmin);\n",
"\n",
"V_d=.6;//V\n",
"V_r=V_L-V_d;    printf('\nvoltage rating of zener diode=%.1f V',V_r);"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 3.14: to_find_R1_and_R2.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clear;\n",
"clc;\n",
"I2=200*10^-6;//A\n",
"V_z=20;//V\n",
"R_G=500;//ohm\n",
"R2=(V_z/I2)-R_G;    printf('R2=%.1f kilo-ohm',R2/1000);\n",
"\n",
"V_v=25;//V\n",
"I1=I2;\n",
"R1=(V_v-V_z)/I1;    printf('\nR1=%.0f kilo-ohm',R1/1000);"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 3.15: to_find_the_VA_rating_of_the_transformer.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clear;\n",
"clc;\n",
"V_s=2*230;//V\n",
"V_o=(sqrt(2)*V_s)/%pi;\n",
"R=60;//ohm\n",
"P_dc=(V_o)^2/R;\n",
"TUF=0.2865;\n",
"VA=P_dc/TUF;    printf('kVA rating of the transformer=%.1f kVA',VA/1000);"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 3.16: EX3_16.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clear;\n",
"clc;\n",
"tr=0.5;    //turns ratio\n",
"I_o=10;\n",
"V=230;\n",
"V_s=V/tr;\n",
"V_m=sqrt(2)*V_s;\n",
"V_o=2*V_m/%pi;    printf('o/p voltage=%.2f V',V_o);\n",
"phi1=0;    //displacemnt angle=0 as fundamnetal component of i/p source current in phase with source voltage\n",
"DF=cosd(phi1);    printf('\ndistortion factor=%.0f',DF);\n",
"I_s1=4*I_o/(sqrt(2)*%pi);\n",
"I_s=sqrt(I_o^2*%pi/%pi);\n",
"CDF=I_s1/I_o;    printf('\ncurrent displacent factor=%.1f',CDF);\n",
"pf=CDF*DF;    printf('\ni/p pf=%.2f',pf);\n",
"HF=sqrt((I_s/I_s1)^2-1);    printf('\nharmonic factor=%.2f',HF);\n",
"CF=I_o/I_s;    printf('\ncreast factor=%.2f',CF);"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 3.17: to_determine_diode_rating_and_transformer_rating.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clear;\n",
"clc;\n",
"V_o=230;\n",
"R=10;\n",
"V_s=V_o*%pi/(2*sqrt(2));\n",
"I_o=V_o/R;\n",
"I_m=sqrt(2)*V_s/R;    printf('peak diode current=%.2f A',I_m);\n",
"I_DAV=I_m/%pi;    printf('\nI_DAV=%.2f A',I_DAV);    //avg value of diode current\n",
"I_Dr=I_m/2;    printf('\nI_Dr=%.2f A',I_Dr);    //rms value of diode current\n",
"PIV=sqrt(2)*V_s;    printf('\nPIV=%.1f V',PIV);\n",
"I_s=I_m/sqrt(2);\n",
"TF=V_s*I_s;    printf('\ntransformer rating=%.2f kVA',TF/1000);"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 3.18: EX3_18.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clear;\n",
"clc;\n",
"tr=5;\n",
"V=1100;\n",
"R=10;\n",
"disp('in case of 3ph-3pulse type');\n",
"V_ph=V/tr;\n",
"V_mp=sqrt(2)*V_ph;\n",
"V_o=3*sqrt(3)*V_mp/(2*%pi);    printf('avg o/p voltage=%.1f V',V_o);\n",
"I_mp=V_mp/R;\n",
"I_D=(I_mp/%pi)*sin(%pi/3);    printf('\navg value of diode current=%.3f A',I_D);\n",
"I_Dr=I_mp*sqrt((1/(2*%pi))*(%pi/3+.5*sin(2*%pi/3)));    printf('\nrms value of diode current=%.2f A',I_Dr);\n",
"V_or=V_mp*sqrt((3/(2*%pi))*(%pi/3+.5*sin(2*%pi/3)));\n",
"P=(V_or^2)/R;    printf('\npower delivered=%.1f W',P);\n",
"\n",
"disp('in case of 3ph-M6 type');\n",
"V_ph=V_ph/2;\n",
"V_mp=sqrt(2)*V_ph;\n",
"V_o=3*V_mp/(%pi);    printf('avg o/p voltage=%.2f V',V_o);\n",
"I_mp=V_mp/R;\n",
"I_D=(I_mp/%pi)*sin(%pi/6);    printf('\navg value of diode current=%.4f A',I_D);\n",
"I_Dr=I_mp*sqrt((1/(2*%pi))*(%pi/6+.5*sin(2*%pi/6)));    printf('\nrms value of diode current=%.3f A',I_Dr);\n",
"V_or=V_mp*sqrt((6/(2*%pi))*(%pi/6+.5*sin(2*%pi/6)));\n",
"P=(V_or^2)/R;    printf('\npower delivered=%.0f W',P);\n",
""
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 3.19: determine_the_ratings_of_diodes_and_of_3ph_deltastar_transformer.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clear;\n",
"clc;\n",
"V_o=400;\n",
"R=10;\n",
"V_ml=V_o*%pi/3;\n",
"V_s=V_ml/(sqrt(2)*sqrt(3));\n",
"I_m=V_ml/R;\n",
"I_s=.7804*I_m;\n",
"tr=3*V_s*I_s;    printf('transformer rating=%.1f VA',tr);\n",
"\n",
"I_Dr=.5518*I_m;    printf('\nrms value of diode current=%.3f A',I_Dr);\n",
"I_D=I_m/%pi;    printf('\navg value of diode current=%.3f A',I_D);\n",
"printf('\npeak diode current=%.2f A',I_m);\n",
"PIV=V_ml;    printf('\nPIV=%.2f V',PIV);"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 3.20: EX3_20.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clear;\n",
"clc;\n",
"V_l=230;\n",
"E=240;\n",
"R=8;\n",
"V_ml=sqrt(2)*V_l;\n",
"V_o=3*V_ml/%pi;\n",
"I_o=(V_o-E)/R;\n",
"P_b=E*I_o;    printf('power delivered to battery=%.1f W',P_b);\n",
"P_d=E*I_o+I_o^2*R;    printf('\npower delivered to load=%.2f W',P_d);\n",
"\n",
"phi1=0;\n",
"DF=cosd(phi1);\n",
"printf('\ndisplacement factor=%.0f',DF);\n",
"I_s1=2*sqrt(3)*I_o/(sqrt(2)*%pi);\n",
"I_s=sqrt(I_o^2*2*%pi/(3*%pi));\n",
"CDF=I_s1/I_s;    printf('\ncurrent distortion factor=%.3f',CDF);\n",
"pf=DF*CDF;    printf('\ni/p pf=%.3f',pf);\n",
"HF=sqrt(CDF^-2-1);    printf('\nharmonic factor=%.4f',HF);\n",
"tr=sqrt(3)*V_l*I_o*sqrt(2/3);    printf('\ntranformer rating=%.2f VA',tr);\n",
"//answers have small variations from the book due to difference in rounding off of digits"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 3.21: design_capaitor_filter_and_avg_value_of_output_voltage_with_and_without_filter.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clear;\n",
"clc;\n",
"f=50;    //Hz\n",
"V=230;\n",
"V_m=sqrt(2)*V;\n",
"R=400;\n",
"RF=0.05;\n",
"C=(1/(4*f*R))*(1+(1/(sqrt(2)*RF)));    printf('capacitor value=%.1f uF',C/10^-6);\n",
"V_o=V_m*(1-1/(4*f*R*C));    printf('\no/p voltage with filter=%.3f V',V_o);\n",
"V_o=2*V_m/%pi;    printf('\no/p voltage without filter=%.2f V',V_o);"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 3.22: to_find_the_value_of_L_with_different_R_and_CRF_without_L.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clear;\n",
"clc;\n",
"f=50;\n",
"CRF=0.05;\n",
"R=300;\n",
"L=sqrt((CRF/(.4715*R))^-2-R^2)/(2*2*%pi*f);    printf('L=%.4f H',L);\n",
"R=30;\n",
"L=sqrt((CRF/(.4715*R))^-2-R^2)/(2*2*%pi*f);    printf('\nL=%.4f H',L);\n",
"L=0;\n",
"CRF=.4715*R/sqrt(R^2+(2*2*%pi*f*L)^2);    printf('\nCRF=%.4f',CRF);"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 3.23: to_design_LC_filter.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clear;\n",
"clc;\n",
"R=50;\n",
"L_L=10*10^-3;\n",
"f=50;\n",
"w=2*%pi*f;\n",
"C=10/(2*w*sqrt(R^2+(2*w*L_L)^2));    printf('C=%.2f uF',C*10^6);\n",
"VRF=0.1;\n",
"L=(1/(4*w^2*C))*((sqrt(2)/(3*VRF))+1);    printf('\nL=%.3f mH',L*10^3);"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 3.2: EX3_2.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clear;\n",
"clc;\n",
"V_s=400;//V\n",
"V_o=100;//V\n",
"L=100;//uH\n",
"C=30;//uF\n",
"t_o=%pi*sqrt(L*C);    printf('conduction time of diode=%.2f us',t_o);\n",
"//in book solution is t_o=54.77 us. The ans is incorrect as %pi is not muliplied in ans. Formulae mentioned in correct.\n",
"I_p=(V_s-V_o)*sqrt(C/L);    printf('\npeak current through diode=%.2f A',I_p);\n",
"v_D=-V_s+V_o;    printf('\nvoltage across diode=%.0f V',v_D);"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 3.6: to_determine_conduction_time_of_diode_and_rate_of_change_of_current.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clc\n",
"clear\n",
"R=10;//ohm\n",
"L=.001;//H\n",
"C=5*10^-6;//F\n",
"V_s=230;//V\n",
"xi=R/(2*L);\n",
"w_o=1/sqrt(L*C);\n",
"w_r=sqrt((1/(L*C))-(R/(2*L))^2);\n",
"t=%pi/w_r;    printf('conduction time of diode=%.3f us',t*10^6);\n",
"t=0;\n",
"//di=di/dt\n",
"di=V_s/L;    printf('\nrate of change of current at t=0 is %.0f A/s',di);"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 3.7: to_find_the_time_required_to_deliver_a_charge_of_200Ah.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clear;\n",
"clc;\n",
"I_or=100;//A\n",
"R=1;//assumption\n",
"V_m=I_or*2*R;\n",
"I_o=V_m/(%pi*R);\n",
"q=200;//Ah\n",
"t=q/I_o;    printf('time required to deliver charge=%.04f hrs',t);"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 3.8: EX3_8.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clear;\n",
"clc;\n",
"V_s=230;//V\n",
"P=1000;//W\n",
"R=V_s^2/P;\n",
"V_or=sqrt(2)*V_s/2;\n",
"P_h=V_or^2/R;    printf('power delivered to the heater=%.0f W',P_h);\n",
"V_m=sqrt(2)*230;\n",
"I_m=V_m/R;    printf('\npeak value of diode current=%.4f A',I_m);\n",
"pf=V_or/V_s;    printf('\ninput power factor=%.3f',pf);"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 3.9: EX3_9.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clear;\n",
"clc;\n",
"V_s=230;//V\n",
"V_m=V_s*sqrt(2);\n",
"E=150;//V\n",
"theta1=asind(E/(sqrt(2)*V_s));\n",
"R=8;//ohm\n",
"f=50;//Hz\n",
"I_o=(1/(2*%pi*R))*((2*sqrt(2)*V_s*cosd(theta1))-E*(%pi-2*theta1*%pi/180));\n",
"printf('avg value of charging current=%.4f A',I_o);\n",
"P_d=E*I_o;    printf('\npower delivered to battery=%.2f W',P_d);\n",
"I_or=sqrt((1/(2*%pi*R^2))*((V_s^2+E^2)*(%pi-2*theta1*%pi/180)+V_s^2*sind(2*theta1)-4*V_m*E*cosd(theta1)));    printf('\nrms value of the load current=%.4f A',I_or);\n",
"pf=(E*I_o+I_or^2*R)/(V_s*I_or);    printf('\nsupply pf=%.3f',pf);\n",
"P_dd=I_or^2*R;    printf('\npower dissipated in the resistor=%.2f W',P_dd);\n",
"q=1000.00;//Wh\n",
"t=q/P_d;    printf('\ncharging time=%.3f hr',t);\n",
"n=P_d*100/(P_d+P_dd);    disp(n,'rectifier efficiency(%)');\n",
"PIV=sqrt(2)*V_s+E;    printf('PIV of diode=%.2f V',PIV);\n",
"//solutions have small variations due to difference in rounding off of digits"
   ]
   }
],
"metadata": {
		  "kernelspec": {
		   "display_name": "Scilab",
		   "language": "scilab",
		   "name": "scilab"
		  },
		  "language_info": {
		   "file_extension": ".sce",
		   "help_links": [
			{
			 "text": "MetaKernel Magics",
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