{
"cells": [
 {
		   "cell_type": "markdown",
	   "metadata": {},
	   "source": [
       "# Chapter 2: Diode Applications"
	   ]
	},
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 2.10: Average_load_current_and_rectification_efficiency.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"// Exa 2.10\n",
"clc;\n",
"clear;\n",
"close;\n",
"format('v',6)\n",
"// Given data\n",
"R_L = 1;// in k ohm\n",
"R_L = R_L * 10^3;// in ohm\n",
"Rf = 1;// in ohm\n",
"R2 = 2;// in ohm\n",
"N1 = 4;\n",
"N2 = 1;\n",
"V1rms = 240;// in V\n",
"V2rms = (N2/N1)*V1rms;// in V\n",
"Vm = sqrt(2)*V2rms;// in V\n",
"// The average load current \n",
"I_LDC = (2*Vm)/(%pi*(R2+Rf+R_L));// in A\n",
"I_LDC= I_LDC *10^3;// in mA\n",
"disp(I_LDC,'The average load current in mA is');\n",
"I_LDC= I_LDC *10^-3;// in A\n",
"// The average load voltage at no load \n",
"V_NL = (2*Vm)/%pi;// in V\n",
"disp(V_NL,'The average load voltage at no load in V is');\n",
"// The average load voltage at full load \n",
"V_LDC = I_LDC*R_L;// in V\n",
"disp(V_LDC,'The average load voltage at full load in V is');\n",
"// The percentage load regulation \n",
"Per_loadReg= (V_NL-V_LDC)/V_LDC*100;// in %\n",
"disp(Per_loadReg,'The percentage load regulation in % is : ')\n",
"Im = Vm/(R_L+R2+Rf);// in A\n",
"Irms = Im/2;// in A\n",
"P_iAC = (Vm^2)/(2*(R2+Rf+R_L));// in W \n",
"P_ODC = (I_LDC^2)*R_L;// in W\n",
"// The rectification efficiency \n",
"Eta = (P_ODC/P_iAC)*100;// in %\n",
"disp(Eta,'The rectification efficiency in % is');"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 2.11: Output_waveform.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"// Exa 2.11\n",
"clc;\n",
"clear;\n",
"close;\n",
"format('v',6)\n",
"// Given data\n",
"R1 = 2;// in k ohm\n",
"R2 = 2;// in k ohm\n",
"V_AB = 20;// in V\n",
"Vo = V_AB*(R1/(R1+R2));// in V\n",
"// The required PIV \n",
"V_AC = Vo;// in V\n",
"disp(V_AC,'The required PIV in V is');"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 2.12: Idc_and_Irms.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"// Exa 2.12\n",
"clc;\n",
"clear;\n",
"close;\n",
"format('v',6)\n",
"// Given data\n",
"Vrms = 10;// in V\n",
"r_f = 0.3;// in ohm\n",
"R_L = 2;// in ohm\n",
"Vm = sqrt(2)*Vrms;// in V\n",
"Im = Vm/(R_L+r_f);// in A\n",
"// The value of Idc \n",
"Idc = Im/%pi;// in A\n",
"disp(Idc,'The value of Idc in A is');\n",
"// The RMS value of output current \n",
"Irms = Im/2;// in A\n",
"disp(Irms,'The RMS value of output current in A is');"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 2.13: Ripple_factor.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"// Exa 2.13\n",
"clc;\n",
"clear;\n",
"close;\n",
"format('v',6)\n",
"// Given data\n",
"Vdc = 50;// in V\n",
"Vrms = 5;// in V\n",
"// The ripple factor,\n",
"Gamma = Vrms/Vdc;\n",
"disp(Gamma,'The ripple factor is');"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 2.14: Mean_and_rms_load_current_and_output_efficiency.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"// Exa 2.14\n",
"clc;\n",
"clear;\n",
"close;\n",
"format('v',6)\n",
"// Given data\n",
"Vrms = 50;// in V\n",
"r_f = 20;// in ohm\n",
"R_L = 980;// in ohm\n",
"Vm = sqrt(2)*Vrms;// in V\n",
"Im = (Vm)/(R_L+r_f);\n",
"// The mean load current \n",
"Idc = (2*Im)/%pi;// in A\n",
"Idc = round(Idc * 10^3);// in mA\n",
"disp(Idc,'The mean load current in mA is');\n",
"// The RMS load current \n",
"Irms = Im/sqrt(2);// in A\n",
"Irms = Irms*10^3;// in mA\n",
"disp(Irms,'The RMS load current in mA is');\n",
"a = 0.812;// assumed\n",
"// The output efficiency \n",
"Eta = a/(1+(r_f/R_L));\n",
"Eta = Eta * 100;// in %\n",
"disp(Eta,'The output efficiency in % is');"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 2.15: Output_voltage_for_a_complete_cycle.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"// Exa 2.15\n",
"clc;\n",
"clear;\n",
"close;\n",
"format('v',6)\n",
"// Given data\n",
"V = 10;// in V\n",
"V1 = 2.5;// in V\n",
"R = 1;// in Mohm\n",
"R = R * 10^6;// in ohm\n",
"i = (V-V1)/R;// in A\n",
"i = i * 10^6;// in µA\n",
"// The output voltage for a complete cycle \n",
"Vo1 = (i*10^-6*R)+V1;// in V\n",
"disp(Vo1,'The output voltage for a complete cycle in V is');\n",
"// The output voltage for half neagtive cycle \n",
"Vo2 = V1;// in V\n",
"disp(Vo2,'The output voltage for half negative cycle in V is');"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 2.17: DC_output_voltage_PIV_and_rectification_efficiency.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"// Exa 2.17\n",
"clc;\n",
"clear;\n",
"close;\n",
"format('v',5)\n",
"// Given data\n",
"V1 = 230;// iin V\n",
"N2 = 1;\n",
"N1 = 5;\n",
"R_L = 100;// in ohm\n",
"V2 = V1*N2/N1;// in V\n",
"Vrms = V2;// in V\n",
"Vs = V2/2;// in V\n",
"Vm = sqrt(2)*Vs;// in V\n",
"// The dc output voltage \n",
"Vdc = (2*Vm)/%pi;// in V\n",
"disp(Vdc,'The dc output voltage in V is');\n",
"// The PIV value \n",
"PIV = round(2*Vm);// in V\n",
"disp(PIV,'The PIV value in V is');\n",
"// The rectification efficiency \n",
"Eta = 0.812;\n",
"Eta = Eta*100;// in %\n",
"disp(Eta,'The rectification efficiency in % is');"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 2.18: DC_output_voltage_and_output_frequency.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"// Exa 2.18\n",
"clc;\n",
"clear;\n",
"close;\n",
"format('v',6)\n",
"// Given data\n",
"V1 = 220;// in V\n",
"N2 = 1;\n",
"N1 = 10;\n",
"R_L = 250;// in ohm\n",
"V2 = V1 * (N2/N1);// in V\n",
"Vm = sqrt(2)*V2;// in V\n",
"Im =Vm/R_L;// in A\n",
"Iav = (2*Im)/%pi;// in A\n",
"Idc = Iav;// in A\n",
"// The dc output volatge \n",
"Vdc = Idc* R_L;// in V\n",
"disp(Vdc,'The dc output volatge in V is');\n",
"Pdc = (Idc^2)*R_L;// in W\n",
"Irms = (Im)/sqrt(2);// in A\n",
"Pac = (Irms^2)*R_L;// in W\n",
"// The rectification efficiency \n",
"Eta = (Pdc/Pac)*100;// in %\n",
"disp(Eta,'The rectification efficiency in % is');\n",
"// The peak inverse volatge \n",
"PIV = Vm;// in V\n",
"disp(PIV,'The peak inverse volatge in V is');\n",
"f_in = 50;// in Hz\n",
"// The output frequency \n",
"f_out = 2*f_in;// in Hz\n",
"disp(f_out,'The output frequency in Hz is');\n",
"\n",
"// Note: The answer of rectification efficiency in the book is not accurate."
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 2.1: Average_current_and_load_voltage.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"// Exa 2.1\n",
"clc;\n",
"clear;\n",
"close;\n",
"format('v',6)\n",
"// Given data\n",
"N2 = 4;\n",
"N1 = 1;\n",
"R_L = 1*10^3;// in ohm\n",
" Vm = 40;// in V\n",
"V_Lav = Vm/%pi;// in V\n",
"// The average load voltage \n",
"V_LDC = V_Lav;// in V\n",
"disp(V_LDC,'The average load voltage in V is');\n",
"Im = Vm/R_L;// in A\n",
"I_DC = Im/%pi;// in A\n",
"I_DC= I_DC*10^3;// in mA\n",
"disp(I_DC,'The average load current in mA is');\n",
"I_DC= I_DC*10^-3;// in A\n",
"// The RMS voltage\n",
"V_Lrms = Vm/2;// in V\n",
"disp(V_Lrms,'The RMS voltage in V is');\n",
"// The RMS current\n",
"Irms = V_Lrms/R_L;// in A\n",
"Irms= Irms*10^3;// in mA\n",
"disp(Irms,'The RMS current in mA is');\n",
"Irms= Irms*10^-3;// in A\n",
"//Eta = (P_ODC/P_iAC)*100;\n",
"I_Loc = I_DC;// in A\n",
"P_ODC = (I_Loc^2)*R_L;// in W\n",
"P_iAC = (Irms^2)*R_L;// in W\n",
"// The efficiency of rectification \n",
"Eta = (P_ODC/P_iAC)*100;// in %\n",
"disp(Eta,'The efficiency of rectification in % is');\n",
"V2rms = Vm/sqrt(2);\n",
"I2rms = Irms;// in A\n",
"// The value of TUF \n",
"TUF = ((P_ODC)/(V2rms*I2rms))*100;// in %\n",
"disp(TUF,'The value of TUF in % is');\n",
"// The ripple factor \n",
"Gamma = (sqrt((V_Lrms^2)-(V_LDC^2)))/V_LDC;\n",
"Gamma = round(Gamma * 100);// in % done by own\n",
"disp(Gamma,'The ripple factor in % is');"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 2.21: Filter_capacitor.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"// Exa 2.21\n",
"clc;\n",
"clear;\n",
"close;\n",
"format('v',6)\n",
"// Given data\n",
"R_F = 0.01;\n",
"Vdc = 30;// in V\n",
"R_L = 1;// in  k ohm\n",
"R_L = R_L * 10^3;// in ohm\n",
"Idc = Vdc/R_L;// in A\n",
"Idc = Idc * 10^3;// in mA\n",
"// Vdc = Vm-( (5000*Idc)/C );\n",
"Gamma = 0.01;// ripple factor\n",
"//Gamma = 2900/(C*R_L);\n",
"C = 2900/(Gamma*R_L);// in F\n",
"Vm = Vdc + ((5000*Idc*10^-3)/C);// in V\n",
"// The input voltage required \n",
"V2 = (2*Vm)/sqrt(2);// in V \n",
"disp(V2,'The input voltage required in V is');\n",
"\n",
"//Note: The value of Vm in the book is not accurate, So the answer in the book is wrong."
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 2.22: Designing_of_a_filter_for_full_wave_circuit.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"// Exa 2.22\n",
"clc;\n",
"clear;\n",
"close;\n",
"format('v',7)\n",
"// Given data\n",
"V_L = 25;// in V\n",
"I_L = 200;// in mA\n",
"I_L = I_L * 10^-3;// in A\n",
"R_L = V_L/I_L;// in ohm\n",
"Gamma = 3/100;\n",
"//Gamma = 1/(6*sqrt(2)*(omega^2)*L*C);\n",
"f = 50;// in Hz\n",
"omega = 2*%pi*f;// in rad/sec\n",
"//LC = 1/( 6*sqrt(2)*(omega^2)*Gamma )\n",
"L = R_L/(3*omega);// in H\n",
"disp(L,'The value of L in H is');\n",
"C = 1/( 6*sqrt(2)*(omega^2)*Gamma*L );// in F\n",
"C = C * 10^6;// in µF\n",
"disp(C,'The value of C in µF is');"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 2.23: Output_voltage_current_and_ripple.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"// Exa 2.23\n",
"clc;\n",
"clear;\n",
"close;\n",
"format('v',6)\n",
"// Given data\n",
"Vm  = 15;// in V\n",
"// The output voltage \n",
"Vdc = (2*sqrt(2)*Vm)/%pi;// in V\n",
"disp(Vdc,'The output voltage in V is');\n",
"R_L = 5;// in ohm\n",
"Idc = Vdc/R_L;// in A\n",
"disp(Idc,'The current in A is');\n",
"L = 50;// in mH\n",
"L = L * 10^-3;// in H\n",
"C = 1000;// in µF\n",
"C = C * 10^-6;// in F\n",
"f = 50;// in Hz\n",
"omega = 2*%pi*f;// in rad/sec\n",
"// The ripple factor \n",
"Gamma = 1/( 6*sqrt(2)*(omega^2)*L*C );\n",
"disp(Gamma,'The ripple factor is');\n",
"// Im =Vm/X_L = (Vm*sqrt(2))/(2*%pi*f*L);\n",
"Im = (Vm*sqrt(2))/(2*%pi*f*L);// in A\n",
"I_Lmin = Im;// in A\n",
"// The maximum value of R_L \n",
"R_Lmax = Vdc/I_Lmin;// in ohm\n",
"disp(R_Lmax,'The maximum value of R_L in ohm is');"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 2.24: Percentage_ripple.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"// Exa 2.24\n",
"clc;\n",
"clear;\n",
"close;\n",
"format('v',6)\n",
"// Given data\n",
"f = 50;// in Hz\n",
"Vrms = 280;// in V\n",
"Vm = sqrt(2)*Vrms;// in V\n",
"V_Lmax = Vm;// in V\n",
"Idc = 100;// in mA\n",
"Idc =Idc * 10^-3;// in A\n",
"C2 = 10;// in µF\n",
"C2 = C2 * 10^-6;// in F\n",
"C1 = C2 ;// in F\n",
"R_L =V_Lmax/Idc;// in ohm\n",
"L = 104;// in H\n",
"omega = 2*%pi*f;// in rad/sec\n",
"// The percentage ripple \n",
"Gamma = sqrt(2)/( 8*omega^3*C1*C2*L*R_L)*100;// in %\n",
"disp(Gamma,'The percentage ripple in % is');\n",
"\n",
"// Note: There is calculation error to find the value of gamma, So the answer in the book is wrong."
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 2.25: Peak_output_voltage.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"// Exa 2.25\n",
"clc;\n",
"clear;\n",
"close;\n",
"format('v',6)\n",
"// Given data\n",
"Vin = 15;// in V\n",
"// Peak output voltage,\n",
"Vout = Vin;// in V\n",
"disp(Vout,'Peak output voltage in V is');"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 2.26: Peak_output_voltage.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"// Exa 2.26\n",
"clc;\n",
"clear;\n",
"close;\n",
"format('v',6)\n",
"// Given data\n",
"R_L = 2;// in k ohm\n",
"R_L = R_L * 10^3;// in ohm\n",
"R = 2;// in k ohm\n",
"R = R * 10^3;// in ohm\n",
"Vin = 5;// in V\n",
"// The peak output voltage \n",
"Vout = (R_L/(R+R_L))*Vin;// in V\n",
"disp(Vout,'The peak output voltage in V is');"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 2.27: IR_and_IRmax.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"// Exa 2.27\n",
"clc;\n",
"clear;\n",
"close;\n",
"format('v',6)\n",
"// Given data\n",
"Vi = 10;// in V\n",
"V1 = 6;// in V\n",
"R = 10;// in k ohm\n",
"R = R * 10^3;// in ohm\n",
"// The value of i_Rmax\n",
"i_Rmax = (Vi-V1)/R;// in A\n",
"i_Rmax = i_Rmax * 10^3;// in mA\n",
"Vi = -10;// in V\n",
"V1 = 8;// in V\n",
"// The value of i_R \n",
"i_R =(Vi+V1)/R;// in A\n",
"i_R  =i_R * 10^3;// in mA\n",
"disp(i_Rmax,'The value of i_Rmax in mA is : ')\n",
"disp(i_R,'The value of i_R in mA is : ')"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 2.28: Output_voltage_and_voltage_across_R.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"// Exa 2.28\n",
"clc;\n",
"clear;\n",
"close;\n",
"format('v',6)\n",
"// Given data\n",
"Vout = -0.7;// in V\n",
"V = -12;// in V\n",
"// The output voltage \n",
"V_R =V-Vout;// in V\n",
"disp(V_R,'The output voltage in V is');"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 2.29: Output_waveform.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"// Exa 2.29\n",
"clc;\n",
"clear;\n",
"close;\n",
"format('v',6)\n",
"// Given data\n",
"Vin = 10;// in V\n",
"V1 = 2;// in V\n",
"// Vin -V_C - V1 = 0;\n",
"V_C =Vin-V1;// in V\n",
"// During positive half cycle the output voltage \n",
"Vout = Vin-V_C;// in V\n",
"disp(Vout,'During positive half cycle the output voltage in V is : ')\n",
"Vin = -10;// in V\n",
"V1 = 8;// in V\n",
"// Vin-V1-Vout = 0;\n",
"// During negative half cycle the output voltage \n",
"Vout = Vin-V1;// in V\n",
"disp(Vout,'During negative half cycle the output voltage in V is : ')\n",
""
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 2.2: Input_AC_power.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"// Exa 2.2\n",
"clc;\n",
"clear;\n",
"close;\n",
"format('v',6)\n",
"// Given data\n",
"Rf = 10;// in ohm\n",
"R_L = 1;// in k ohm\n",
"R_L = R_L * 10^3;// in ohm\n",
"Vi = 230;// in V\n",
"Vm = Vi*sqrt(2);\n",
"//I_DC = Im/%pi;\n",
"I_DC = Vm/((R_L+Rf)*%pi);// in A\n",
"Irms = Vm/((R_L+Rf)*2);// in A\n",
"// The input ac power \n",
"P_iAC = (Irms^2)*(Rf+R_L);// in W \n",
"disp(P_iAC,'The input ac power in W is');\n",
"// The output ac power \n",
"P_ODC = (I_DC^2)*R_L;// in W\n",
"disp(P_ODC,'The output ac power in W is');\n",
"// The efficiency \n",
"Eta = (P_ODC/P_iAC)*100;// in %\n",
"disp(Eta,'The efficiency in % is');\n",
"// The percentage regulation \n",
"R = (Rf/R_L)*100;// in %\n",
"disp(R,'The percentage regulation in % is');\n",
"\n",
"// Note: The calculated value of input a.c. power in the book is wrong."
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 2.30: Output_voltage.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"// Exa 2.30\n",
"clc;\n",
"clear;\n",
"close;\n",
"format('v',6)\n",
"// Given data\n",
"Vin = 10;// in V\n",
"V1 = 2;// in V\n",
"// Vin-V_C+V1 = 0;\n",
"V_C = Vin+V1;// in V\n",
"//During positive half cycle the output voltage \n",
"Vout = Vin-V_C;// in V\n",
"disp(Vout,'During positive half cycle the output voltage in V is');\n",
"Vin = -10;// iin V\n",
"V1 = 12;// in V\n",
"// Vin-V1-Vout = 0;\n",
"//During negative half cycle the output voltage \n",
"Vout = Vin-V1;// in V\n",
"disp(Vout,'During negative half cycle the output voltage in V is');\n",
""
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 2.31: Output_voltage.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"// Exa 2.31\n",
"clc;\n",
"clear;\n",
"close;\n",
"format('v',6)\n",
"// Given data\n",
"Vi = 20;// in V\n",
"V1 = 5;// in V\n",
"Vc = Vi-V1;// in V\n",
"Vo = -5;// in V\n",
"// The value of Vo,\n",
"Vo = Vi+Vc;// in V\n",
"disp(Vo,'The value of Vo in V is');"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 2.3: Output_DC_voltage_and_current.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"// Exa 2.3\n",
"clc;\n",
"clear;\n",
"close;\n",
"format('v',6)\n",
"// Given data\n",
"V1 = 230;// in V\n",
"N2= 1;\n",
"N1 = 4;\n",
"R_L = 1;// in k ohm\n",
"R_L = R_L * 10^3;// in ohm\n",
"Vd = 0.7;// in V\n",
"// V_LDC = (Vm-Vd)/%pi;// in V\n",
"V2 = V1*(N2/N1);// in V\n",
"// Vm = sqrt(2)*Vrms;\n",
"Vm = sqrt(2)*V2;// in V\n",
"// The output dc voltage \n",
"V_LDC = (Vm-Vd)/%pi;// in V\n",
"disp(V_LDC,'The output dc voltage in V is');\n",
"// The current for a load resistance \n",
"I_LDC = V_LDC/R_L;// in A\n",
"I_LDC = I_LDC * 10^3;// in mA\n",
"disp(I_LDC,'The current for a load resistance in mA is');"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 2.7: PIV_rating.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"// Exa 2.7\n",
"clc;\n",
"clear;\n",
"close;\n",
"format('v',6)\n",
"// Given data\n",
"R1 = 2.2;// in k ohm\n",
"R2 = 2.2;// in  kohm\n",
"R3 = 4.7;// in k ohm\n",
"R = (R2*R3)/(R2+R3);// in k ohm\n",
"Vin = 200;\n",
"// Vo = (R/(R1+R))*Vin;\n",
"// The PIV rating for first diode \n",
"Vomax= round(R/(R1+R)*Vin);// in V\n",
"disp(Vomax,'The PIV rating for first diode in V is : ')\n",
"Rdas = (R1*R3)/(R1+R3);// in k ohm\n",
"// Vo = (Rdas/(R1+Rdas))*(-Vin);\n",
"// The PIV rating for second diode \n",
"Vomax=round(Rdas/(R1+Rdas)*Vin);// in V\n",
"disp(Vomax,'The PIV rating for second diode in V is : ')"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 2.9: PIV_of_diode.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"// Exa 2.9\n",
"clc;\n",
"clear;\n",
"close;\n",
"format('v',5)\n",
"// Given data\n",
"// Vi = 15*sind(314*t);\n",
"Vm = 15;// in V\n",
"R_L = 1;// in k ohm\n",
"R_L = R_L * 10^3;// in ohm\n",
"Im = Vm/R_L;// in A\n",
"Im = Im * 10^3;// in mA\n",
"Idc = Im/%pi;// in mA\n",
"// The average current \n",
"I_Dav = Idc;// in mA\n",
"disp(I_Dav,'The average current in mA is');\n",
"// The RMS current \n",
"I_Drms = Im/2;// in mA\n",
"disp(I_Drms,'The RMS current in mA is');\n",
"// The peak diode current \n",
"I_Dpeak = Im;// in mA\n",
"disp(I_Dpeak,'The peak diode current in mA is');\n",
"// The PIV of diode \n",
"PIV = 2*Vm;// in V\n",
"disp(PIV,'The PIV of diode in V is');"
   ]
   }
],
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