Merge pull request #1 from prashantsinalkar/masterHEADmaster

Initial commit

14 files changed, 4359 insertions, 0 deletions

diff --git a/Electronic_Devices_and_Circuits_by_D_C_Kulshreshtha/1-Introduction_to_Electronics.ipynb b/Electronic_Devices_and_Circuits_by_D_C_Kulshreshtha/1-Introduction_to_Electronics.ipynb
new file mode 100644
index 0000000..00c8a26
--- /dev/null
+++ b/Electronic_Devices_and_Circuits_by_D_C_Kulshreshtha/1-Introduction_to_Electronics.ipynb
@@ -0,0 +1,210 @@
+{
+"cells": [
+ {
+		   "cell_type": "markdown",
+	   "metadata": {},
+	   "source": [
+       "# Chapter 1: Introduction to Electronics"
+	   ]
+	},
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 1.1: Find_the_range_of_tolerance.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Find the range of tolerance\n",
+"clear;\n",
+"clc;\n",
+"//soltion\n",
+"//given\n",
+"\n",
+"//color coding\n",
+"orange=3;\n",
+"gold=5;\n",
+"yellow=4;\n",
+"violet=7;\n",
+"//band pattern\n",
+"band1=yellow;\n",
+"band2=violet;\n",
+"band3=orange;\n",
+"band4=gold;\n",
+"//resistor color coding\n",
+"r=(band1*10+band2)*10^(band3);\n",
+"tol=r*(band4/100)//tolerance\n",
+"ulr=r+tol;//upper limit of resistance\n",
+"llr=r-tol;//lower limit of resistance\n",
+"printf('The Range of resistance is %.2f kΩ to %.2f kΩ',llr/1000,ulr/1000);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 1.2: Find_the_range_of_tolerance.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Find the range of tolerance\n",
+"clear;\n",
+"clc;\n",
+"//soltion\n",
+"//given\n",
+"\n",
+"//color coding\n",
+"blue=6;\n",
+"gold=-1;\n",
+"gray=8;\n",
+"silver=10;\n",
+"//band pattern\n",
+"band1=gray;\n",
+"band2=blue;\n",
+"band3=gold;\n",
+"band4=silver;\n",
+"//resistor color coding\n",
+"r=(band1*10+band2)*10^(band3);\n",
+"tol=r*(band4/100)//tolerance\n",
+"ulr=r+tol;//upper limit of resistance\n",
+"llr=r-tol;//lower limit of resistance\n",
+"printf('The Range of resistance is %.2f Ω to %.2f Ω',llr,ulr);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 1.3: Find_the_equivalent_current_source.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"\n",
+"//Find the equivalent current source\n",
+"clear;\n",
+"clc;\n",
+"//soltion\n",
+"//given\n",
+"Vs=2;//Volts             //dc voltage source\n",
+"Rs=1;//ohm               //internal resistance\n",
+"Rl=1;//ohm              //load resistance\n",
+"Ise=Vs/Rs;//ampere      //equivalent current source\n",
+"\n",
+"// In accordance to figure 1.23a\n",
+"Il1=Ise*(Rs/(Rs+Rl));//using current divider concept\n",
+"Vl1=Il1*Rl;\n",
+"printf('\nIn accordance to figure 1.23a \n');\n",
+"printf('The Load current (current source) Il= %dA\n',Il1);\n",
+"printf('The Load voltage (current source) Vl= %dV\n\n',Vl1);\n",
+"\n",
+"// In accordance to figure 1.23b\n",
+"Vl2=Vs*(Rs/(Rs+Rl));//using voltage divider concept\n",
+"Il2=Vl2/Rl;\n",
+"printf('\nIn accordance to figure 1.23b \n');\n",
+"printf('The Load voltage (voltage source) Vl= %dV\n',Vl2);\n",
+"printf('The Load current (voltage source) Il= %dA\n\n',Il2);\n",
+"printf('Therefore they both provide same voltage and current to load');"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 1.4: Find_percentage_variation_in_load_current_and_load_voltage.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"\n",
+"//Find percentage variation in load current and load voltage\n",
+"clear;\n",
+"clc;\n",
+"//soltion\n",
+"//given\n",
+"Vs=10;//volt//Supply voltage\n",
+"Rs=100;//ohm//internal resistance\n",
+"\n",
+"// In accordance to figure 1.24a\n",
+"//For 1Ω - 10 Ω\n",
+"Rl11=1;//ohm//min extreme value of Rl\n",
+"Rl12=10;//ohm//max extreme value of Rl\n",
+"Il11=Vs/(Rs+Rl11);\n",
+"Il12=Vs/(Rs+Rl12);\n",
+"Pi1=(Il11-Il12)*100/Il11;//Percentage variation in current\n",
+"Vl11=Il11*Rl11;\n",
+"Vl12=Il12*Rl12;\n",
+"Pv1=(Vl12-Vl11)*100/Vl12;//Percentage variation in voltage\n",
+"printf('\nIn accordance to figure 1.24a \n');\n",
+"printf('Percentage variation in Current(1-10 Ω) %.2f percent\n',Pi1);\n",
+"printf('Percentage variation in Voltage(1-10 Ω) %.1f percent\n\n',Pv1);\n",
+"\n",
+"// In accordance to figure 1.24b\n",
+"//For 1kΩ - 10kΩ\n",
+"Rl21=1000;//ohm//min extreme value of Rl\n",
+"Rl22=10000;//ohm//max extreme value of Rl\n",
+"Il21=Vs/(Rs+Rl21);\n",
+"Il22=Vs/(Rs+Rl22);\n",
+"Pi2=(Il21-Il22)*100/Il21;//Percentage variation in current\n",
+"Vl21=Il21*Rl21;\n",
+"Vl22=Il22*Rl22;\n",
+"Pv2=(Vl22-Vl21)*100/Vl22;//Percentage variation in voltage\n",
+"printf('\nIn accordance to figure 1.24b \n');\n",
+"printf('Percentage variation in Current(1-10 Ω) %d percent \n',Pi2);\n",
+"printf('Percentage variation in Voltage(1-10 Ω) %.1f percent \n\n',Pv2);\n",
+"// In book the percentage variation in voltage(1kΩ-10kΩ) is 9 percent due to \n",
+"// the incorrect value of Il22 i.e. 0.000999 Amp correct value is 0.0009901 Amp"
+   ]
+   }
+],
+"metadata": {
+		  "kernelspec": {
+		   "display_name": "Scilab",
+		   "language": "scilab",
+		   "name": "scilab"
+		  },
+		  "language_info": {
+		   "file_extension": ".sce",
+		   "help_links": [
+			{
+			 "text": "MetaKernel Magics",
+			 "url": "https://github.com/calysto/metakernel/blob/master/metakernel/magics/README.md"
+			}
+		   ],
+		   "mimetype": "text/x-octave",
+		   "name": "scilab",
+		   "version": "0.7.1"
+		  }
+		 },
+		 "nbformat": 4,
+		 "nbformat_minor": 0
+}
diff --git a/Electronic_Devices_and_Circuits_by_D_C_Kulshreshtha/10-Feedback_in_Amplifiers.ipynb b/Electronic_Devices_and_Circuits_by_D_C_Kulshreshtha/10-Feedback_in_Amplifiers.ipynb
new file mode 100644
index 0000000..f939871
--- /dev/null
+++ b/Electronic_Devices_and_Circuits_by_D_C_Kulshreshtha/10-Feedback_in_Amplifiers.ipynb
@@ -0,0 +1,333 @@
+{
+"cells": [
+ {
+		   "cell_type": "markdown",
+	   "metadata": {},
+	   "source": [
+       "# Chapter 10: Feedback in Amplifiers"
+	   ]
+	},
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 10.10: Calculate_the_percentage_of_negative_feedback_to_input.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Calculate the percentage of negative feedback to input\n",
+"clear;\n",
+"clc;\n",
+"//soltion\n",
+"//given\n",
+"\n",
+"Adb=60;//dB      //internal gain in dB\n",
+"A=10^(Adb/(20));      //taking antilog\n",
+"Ro=12*10^3;//ohm      //output resistance\n",
+"Rof=600;//ohm\n",
+"B=(Ro/Rof-1)/A;\n",
+"printf('The percentage of negative feedback to input= %.1f percent',B*100);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 10.1: Determine_the_gain_of_feedback_amplifier.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Determine the gain of feedback amplifier\n",
+"clear;\n",
+"clc;\n",
+"//soltion\n",
+"//given\n",
+"\n",
+"A=100;      //internal gain\n",
+"B=0.1;//feedback factor\n",
+"Af=A/(1+A*B);\n",
+"printf('The gain of feedback amplifier %.2f',Af);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 10.2: Determine_the_gain_of_feedback_amplifier_in_dB.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Determine the gain of feedback amplifier in dB\n",
+"clear;\n",
+"clc;\n",
+"//soltion\n",
+"//given\n",
+"\n",
+"Ad=60;//dB      //internal gain in dB\n",
+"A=10^(Ad/20);    //internal gain\n",
+"B=1/20;//feedback factor\n",
+"Af=A/(1+A*B);\n",
+"Afd=20*log10(Af);\n",
+"printf('The gain of feedback amplifier %.2f dB',Afd);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 10.3: Calculate_the_percentage_of_output_fed_back_to_input.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Calculate the percentage of output fed back to input\n",
+"clear;\n",
+"clc;\n",
+"//soltion\n",
+"//given\n",
+"\n",
+"A=600;      //internal gain\n",
+"Af=50;      //gain of feedback amplifier\n",
+"B=(A/Af-1)/A;\n",
+"printf('The percentage of output fed back to input= %.3f percent',B*100);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 10.4: Calculate_the_internal_gain_and_percentage_of_output_fed_back_to_input.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Calculate the internal gain and percentage of output fed back to input\n",
+"clear;\n",
+"clc;\n",
+"//soltion\n",
+"//given\n",
+"\n",
+"Af=80;      //gain of feedback amplifier\n",
+"Vi=0.05;//V    //input with feedback\n",
+"Vi_=4*10^-3;//V   //input without feedback\n",
+"Vo_=Af*Vi;\n",
+"A=Vo_/Vi_;\n",
+"printf('The internal gain is %.0f\n',A);\n",
+"B=(A/Af-1)/A;\n",
+"printf('The percentage of output fed back to input= %.2f percent',B*100);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 10.5: Calculate_the_gain_with_and_without_feedback_and_feedback_factor.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Calculate the gain with and without feedback and feedback factor\n",
+"clear;\n",
+"clc;\n",
+"//soltion\n",
+"//given\n",
+"\n",
+"Vo_=5;//V         //output voltage\n",
+"Vi=0.2;//V    //input with feedback\n",
+"Vi_=0.05;//V   //input without feedback\n",
+"A=Vo_/Vi_;\n",
+"Af=Vo_/Vi;\n",
+"printf('The gain without feedback is %.0f\n',A);\n",
+"printf('The gain with feedback is %.0f\n',Af);\n",
+"B=(A/Af-1)/A;\n",
+"printf('The feedback factor= %.0f percent',B*100);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 10.6: Calculate_the_gain_of_feedback_amplifier_and_feedback_factor.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Calculate the gain of feedback amplifier and feedback factor\n",
+"clear;\n",
+"clc;\n",
+"//soltion\n",
+"//given\n",
+"\n",
+"A=100;      //internal gain\n",
+"N=20;//dB   //negative feedback\n",
+"B=(10^(-N/(-20))-1)/A;      //taking antilog\n",
+"Af=A/(1+A*B);\n",
+"printf('The feedback factor= %.0f percent\n',B*100);\n",
+"printf('The gain of the feedback amplifier is %.0f\n',Af);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 10.7: Calculate_percentage_change_in_the_overall_gain.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Calculate percentage change in the overall gain\n",
+"clear;\n",
+"clc;\n",
+"//soltion\n",
+"//given\n",
+"\n",
+"A=1000;      //internal gain\n",
+"N=40;//dB   //negative feedback\n",
+"D=10^((-N)/-20);     //D=(1+AB)   desensitivity\n",
+"dA_A=10;//percent    //dA/A\n",
+"dAf_Af=dA_A/D;      //dAf/Af\n",
+"printf('The percentage change in the overall gain= %.1f percent',dAf_Af);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 10.8: Calculate_percentage_change_in_the_overall_gain.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Calculate percentage change in the overall gain\n",
+"clear;\n",
+"clc;\n",
+"//soltion\n",
+"//given\n",
+"\n",
+"Adb=60;//dB      //internal gain in dB\n",
+"B=0.005;        //feedback factor\n",
+"A=10^(Adb/(20));      //taking antilog\n",
+"dA_A=-12;//percent    //dA/A\n",
+"D=(1+A*B);     //D=(1+AB)   desensitivity\n",
+"dAf_Af=dA_A/D;      //dAf/Af\n",
+"printf('The percentage change in the overall gain reduces by %.1f percent',-dAf_Af);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 10.9: Determine_the_input_resistance_of_feedback_amplifier.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Determine the input resistance of feedback amplifier\n",
+"clear;\n",
+"clc;\n",
+"//soltion\n",
+"//given\n",
+"\n",
+"A=250;      //internal gain\n",
+"B=0.1;//feedback factor\n",
+"Ri=1.1*10^3;//ohm    //input resistance\n",
+"Rif=Ri*(1+A*B);\n",
+"printf('The input resistance of feedback amplifier %.1f kΩ',Rif/1000);\n",
+"//The ans in book is incorrect due to use of (2+A*B) instead of (1+A*B) the ans in book is 29.7 kΩ"
+   ]
+   }
+],
+"metadata": {
+		  "kernelspec": {
+		   "display_name": "Scilab",
+		   "language": "scilab",
+		   "name": "scilab"
+		  },
+		  "language_info": {
+		   "file_extension": ".sce",
+		   "help_links": [
+			{
+			 "text": "MetaKernel Magics",
+			 "url": "https://github.com/calysto/metakernel/blob/master/metakernel/magics/README.md"
+			}
+		   ],
+		   "mimetype": "text/x-octave",
+		   "name": "scilab",
+		   "version": "0.7.1"
+		  }
+		 },
+		 "nbformat": 4,
+		 "nbformat_minor": 0
+}
diff --git a/Electronic_Devices_and_Circuits_by_D_C_Kulshreshtha/11-Tuned_Voltage_Amplifiers.ipynb b/Electronic_Devices_and_Circuits_by_D_C_Kulshreshtha/11-Tuned_Voltage_Amplifiers.ipynb
new file mode 100644
index 0000000..b1914b5
--- /dev/null
+++ b/Electronic_Devices_and_Circuits_by_D_C_Kulshreshtha/11-Tuned_Voltage_Amplifiers.ipynb
@@ -0,0 +1,154 @@
+{
+"cells": [
+ {
+		   "cell_type": "markdown",
+	   "metadata": {},
+	   "source": [
+       "# Chapter 11: Tuned Voltage Amplifiers"
+	   ]
+	},
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 11.1: EX11_1.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Calculate frequency and impedance and current and voltage across each element at resonance\n",
+"clear;\n",
+"clc;\n",
+"//soltion\n",
+"//given\n",
+"\n",
+"R=12;//ohm\n",
+"L=200*10^-6;//H\n",
+"C=300*10^-12;//F\n",
+"Vs=9;//V\n",
+"fo=1/(2*%pi*sqrt(L*C));\n",
+"Z=R;     //impedance\n",
+"printf('The Resonant frequency= %.1f kHz\n',fo/1000);\n",
+"printf('The impedance Z= %.0f Ω\n',Z);\n",
+"\n",
+"Io=Vs/R;\n",
+"printf('The Source current= %.2f A\n',Io);\n",
+"\n",
+"Vl=Io*(2*%pi*fo*L);\n",
+"Vc=Io/(2*%pi*fo*C);\n",
+"Vr=Io*R;\n",
+"printf('The voltage across the inductor =%.1f V\n',Vl);\n",
+"printf('The voltage across the capacitor =%.1f V\n',Vc);\n",
+"printf('The voltage across the resistor =%.0f V\n',Vr);\n",
+"//There is a slight variation in voltage across capacitor due to the approaximation"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 11.2: EX11_2.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Calculate frequency and impedance and current at resonance and current in coil and capacitor\n",
+"clear;\n",
+"clc;\n",
+"//soltion\n",
+"//given\n",
+"\n",
+"R=10;//ohm\n",
+"L=100*10^-6;//H\n",
+"C=100*10^-12;//F\n",
+"Vs=10;//V\n",
+"fo=1/(2*%pi*sqrt(L*C));\n",
+"Zp=L/(C*R);     //impedance\n",
+"printf('The Resonant frequency= %.3f MHz\n',fo/10^6);\n",
+"printf('The impedance Z= %.0f kΩ\n',Zp/1000);\n",
+"\n",
+"Io=Vs/Zp;\n",
+"printf('The Source current= %.0f uA\n',Io*10^6);\n",
+"\n",
+"Xl=(2*%pi*fo*L);\n",
+"Xc=1/(2*%pi*fo*C);\n",
+"Z1=sqrt(Xl^2+R^2);\n",
+"Z2=Xc;\n",
+"Ic=Vs/Z2;\n",
+"Il=Ic;\n",
+"printf('The current in the coil = %.0f mA\n',Il*1000);\n",
+"printf('The current in the capacitor = %.0f mA\n',Ic*1000);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 11.3: Calculate_impedance_and_quality_factor_and_bandwidth.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Calculate impedance and quality factor and bandwidth\n",
+"clear;\n",
+"clc;\n",
+"//soltion\n",
+"//given\n",
+"\n",
+"R=10;//ohm\n",
+"L=150*10^-6;//H\n",
+"C=100*10^-12;//F\n",
+"fo=1/(2*%pi*sqrt(L*C));\n",
+"Zp=L/(C*R);     //impedance\n",
+"printf('The impedance Z= %.0f kΩ\n',Zp/1000);\n",
+"\n",
+"Xl=(2*%pi*fo*L);\n",
+"Q=Xl/R;\n",
+"BW=fo/Q;\n",
+"printf('The Quality factor of the circuit =%.1f \n',Q);\n",
+"printf('The Band width of the circuit =%.1f kHz\n',BW/1000);"
+   ]
+   }
+],
+"metadata": {
+		  "kernelspec": {
+		   "display_name": "Scilab",
+		   "language": "scilab",
+		   "name": "scilab"
+		  },
+		  "language_info": {
+		   "file_extension": ".sce",
+		   "help_links": [
+			{
+			 "text": "MetaKernel Magics",
+			 "url": "https://github.com/calysto/metakernel/blob/master/metakernel/magics/README.md"
+			}
+		   ],
+		   "mimetype": "text/x-octave",
+		   "name": "scilab",
+		   "version": "0.7.1"
+		  }
+		 },
+		 "nbformat": 4,
+		 "nbformat_minor": 0
+}
diff --git a/Electronic_Devices_and_Circuits_by_D_C_Kulshreshtha/12-Sinusoidal_Oscillators.ipynb b/Electronic_Devices_and_Circuits_by_D_C_Kulshreshtha/12-Sinusoidal_Oscillators.ipynb
new file mode 100644
index 0000000..a0efb53
--- /dev/null
+++ b/Electronic_Devices_and_Circuits_by_D_C_Kulshreshtha/12-Sinusoidal_Oscillators.ipynb
@@ -0,0 +1,191 @@
+{
+"cells": [
+ {
+		   "cell_type": "markdown",
+	   "metadata": {},
+	   "source": [
+       "# Chapter 12: Sinusoidal Oscillators"
+	   ]
+	},
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 12.1: Calculate_frequency_of_oscillations.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Calculate frequency of oscillations\n",
+"clear;\n",
+"clc;\n",
+"//soltion\n",
+"//given\n",
+"\n",
+"L=55*10^-6;//H\n",
+"C=300*10^-12;//F\n",
+"fo=1/(2*%pi*sqrt(L*C));\n",
+"printf('The frequency of oscillations= %.0f kHz\n',fo/1000);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 12.2: Calculate_frequency_of_oscillations_and_feedback_factor_and_voltage_gain.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Calculate frequency of oscillations and feedback factor and voltage gain\n",
+"clear;\n",
+"clc;\n",
+"//soltion\n",
+"\n",
+"function [z]=prll(r1,r2)//Function for the parallel combination of resistor\n",
+"    z=r1*r2/(r1+r2);\n",
+"endfunction\n",
+"\n",
+"//given\n",
+"C1=0.001*10^-6;//F\n",
+"C2=0.01*10^-6;//F\n",
+"L=15*10^-6;//H\n",
+"C=prll(C1,C2);\n",
+"fo=1/(2*%pi*sqrt(L*C));\n",
+"printf('The frequency of oscillations= %.2f MHz\n',fo/10^6);\n",
+"B=C1/C2;\n",
+"Amin=1/B;\n",
+"printf('The feedback factor of the circuit =%.1f \n',B);\n",
+"printf('The circuit needs a minimum voltage gain of %.0f',Amin);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 12.3: Calculate_frequency_of_oscillations.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Calculate frequency of oscillations\n",
+"clear;\n",
+"clc;\n",
+"//soltion\n",
+"//given\n",
+"\n",
+"R=10*10^3;//ohm\n",
+"C=0.01*10^-6;//F\n",
+"fo=1/(2*%pi*R*C*sqrt(6));\n",
+"printf('The frequency of oscillations= %.1f Hz\n',fo);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 12.4: Calculate_frequency_of_oscillations.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Calculate frequency of oscillations\n",
+"clear;\n",
+"clc;\n",
+"//soltion\n",
+"//given\n",
+"\n",
+"R=22*10^3;//ohm\n",
+"C=100*10^-12;//F\n",
+"fo=1/(2*%pi*R*C);\n",
+"printf('The frequency of oscillations= %.2f KHz\n',fo/1000);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 12.5: Determine_the_series_and_parallel_resonant_frequencies.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Determine the series and parallel resonant frequencies\n",
+"clear;\n",
+"clc;\n",
+"//soltion\n",
+"\n",
+"function [z]=prll(r1,r2)//Function for the parallel combination of resistor\n",
+"    z=r1*r2/(r1+r2);\n",
+"endfunction\n",
+"\n",
+"//given\n",
+"\n",
+"L=3;//H\n",
+"Cm=10*10^-12;//F\n",
+"Cs=0.05*10^-12;//F\n",
+"fs=1/(2*%pi*sqrt(L*Cs));\n",
+"printf('The series resonant frequency =%.0f kHz\n',fs/1000);\n",
+"\n",
+"Cp=prll(Cm,Cs);\n",
+"fp=1/(2*%pi*sqrt(L*Cp));\n",
+"printf('The parallel resonant frequency =%.0f kHz',fp/1000);"
+   ]
+   }
+],
+"metadata": {
+		  "kernelspec": {
+		   "display_name": "Scilab",
+		   "language": "scilab",
+		   "name": "scilab"
+		  },
+		  "language_info": {
+		   "file_extension": ".sce",
+		   "help_links": [
+			{
+			 "text": "MetaKernel Magics",
+			 "url": "https://github.com/calysto/metakernel/blob/master/metakernel/magics/README.md"
+			}
+		   ],
+		   "mimetype": "text/x-octave",
+		   "name": "scilab",
+		   "version": "0.7.1"
+		  }
+		 },
+		 "nbformat": 4,
+		 "nbformat_minor": 0
+}
diff --git a/Electronic_Devices_and_Circuits_by_D_C_Kulshreshtha/14-Operational_Amplifiers.ipynb b/Electronic_Devices_and_Circuits_by_D_C_Kulshreshtha/14-Operational_Amplifiers.ipynb
new file mode 100644
index 0000000..7a735fa
--- /dev/null
+++ b/Electronic_Devices_and_Circuits_by_D_C_Kulshreshtha/14-Operational_Amplifiers.ipynb
@@ -0,0 +1,188 @@
+{
+"cells": [
+ {
+		   "cell_type": "markdown",
+	   "metadata": {},
+	   "source": [
+       "# Chapter 14: Operational Amplifiers"
+	   ]
+	},
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 14.1: Calculate_voltage_gain_and_input_and_output_resistance.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Calculate voltage gain and input and output resistance\n",
+"clear;\n",
+"clc;\n",
+"//soltion\n",
+"//given\n",
+"\n",
+"R1=20*10^3;//ohm\n",
+"Rf=2000*10^3;//ohm\n",
+"Acl=-Rf/R1;\n",
+"Ricl=R1;\n",
+"Ro=0;\n",
+"printf('The voltage gain= %.0f\n',Acl);\n",
+"printf('The input resistance =%.0f kΩ\n',R1/1000);\n",
+"printf('The output resistance =%.0f Ω\n',Ro);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 14.2: Find_the_output_voltage.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Find the output voltage\n",
+"clear;\n",
+"clc;\n",
+"//soltion\n",
+"//given\n",
+"\n",
+"R1=20*10^3;//ohm\n",
+"Rf=2000*10^3;//ohm\n",
+"v1=4;//V\n",
+"v2=3.8;//V\n",
+"vo=v2*(1+Rf/R1)-(Rf/R1)*v1;\n",
+"printf('The output voltage= %.1f V',vo);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 14.4: Design_an_adder_circuit_using_an_op_amp.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Design an adder circuit using an op amp\n",
+"clear;\n",
+"clc;\n",
+"//soltion\n",
+"//given\n",
+"\n",
+"//Vo=-(V1+10*V2+100*V3)\n",
+"Rf=100*10^3;//ohm\n",
+"C1=1;         //coefficient of V1\n",
+"C2=10;         //coefficient of V2\n",
+"C3=100;         //coefficient of V3\n",
+"R1=Rf/C1;\n",
+"R2=Rf/C2;\n",
+"R3=Rf/C3;\n",
+"printf('R1 = %.0f kΩ\n',R1/1000);\n",
+"printf('R2 = %.0f kΩ\n',R2/1000);\n",
+"printf('R3 = %.0f kΩ\n',R3/1000);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 14.5: Calculate_CMRR_in_dB.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Calculate CMRR in dB\n",
+"clear;\n",
+"clc;\n",
+"//soltion\n",
+"//given\n",
+"\n",
+"Ad=100;          //differential mode gain\n",
+"Ac=0.01;         //common mode gain\n",
+"CMRR=20*log10(Ad/Ac);\n",
+"printf('The CMRR in dB %.0f dB',CMRR);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 14.6: Calculate_the_output_voltage.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Calculate the output voltage\n",
+"clear;\n",
+"clc;\n",
+"//soltion\n",
+"//given\n",
+"\n",
+"Ad=2000;          //differential mode gain\n",
+"CMRR=10000;\n",
+"V1=10^-3;//V\n",
+"V2=0.9*10^-3;//V\n",
+"Vd=V1-V2;\n",
+"Vc=(V1+V2)/2;\n",
+"Vo=Ad*Vd*(1+Vc/(CMRR*Vd));\n",
+"printf('The output voltage is %.2f mV',Vo*1000);"
+   ]
+   }
+],
+"metadata": {
+		  "kernelspec": {
+		   "display_name": "Scilab",
+		   "language": "scilab",
+		   "name": "scilab"
+		  },
+		  "language_info": {
+		   "file_extension": ".sce",
+		   "help_links": [
+			{
+			 "text": "MetaKernel Magics",
+			 "url": "https://github.com/calysto/metakernel/blob/master/metakernel/magics/README.md"
+			}
+		   ],
+		   "mimetype": "text/x-octave",
+		   "name": "scilab",
+		   "version": "0.7.1"
+		  }
+		 },
+		 "nbformat": 4,
+		 "nbformat_minor": 0
+}
diff --git a/Electronic_Devices_and_Circuits_by_D_C_Kulshreshtha/15-Electronic_Instruments.ipynb b/Electronic_Devices_and_Circuits_by_D_C_Kulshreshtha/15-Electronic_Instruments.ipynb
new file mode 100644
index 0000000..ffd29fa
--- /dev/null
+++ b/Electronic_Devices_and_Circuits_by_D_C_Kulshreshtha/15-Electronic_Instruments.ipynb
@@ -0,0 +1,281 @@
+{
+"cells": [
+ {
+		   "cell_type": "markdown",
+	   "metadata": {},
+	   "source": [
+       "# Chapter 15: Electronic Instruments"
+	   ]
+	},
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 15.1: Calculate_shunt_resistance_and_multiplying_factor.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Calculate shunt resistance and multiplying factor\n",
+"clear;\n",
+"clc;\n",
+"//soltion\n",
+"//given\n",
+"\n",
+"Im=5*10^-3;//A\n",
+"Rm=20;//ohm\n",
+"I=5;//A\n",
+"Rsh=Rm*Im/(I-Im);\n",
+"n=I/Im;\n",
+"printf('Shunt resistance= %.5f Ω\n',Rsh);\n",
+"printf('Multiplying factor= %.0f',n);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 15.2: Calculate_shunt_resistance.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Calculate shunt resistance\n",
+"clear;\n",
+"clc;\n",
+"//soltion\n",
+"//given\n",
+"\n",
+"//At I= 1 mA\n",
+"I1=1*10^-3;//A\n",
+"Im=0.1*10^-3;//A\n",
+"Rm=500;//ohm\n",
+"Rsh=Rm*Im/(I1-Im);\n",
+"printf('Shunt resistance= %.4f Ω\n',Rsh);\n",
+"\n",
+"\n",
+"//At I= 1 mA\n",
+"I2=10*10^-3;//A\n",
+"Rsh=Rm*Im/(I2-Im);\n",
+"printf('Shunt resistance= %.4f Ω\n',Rsh);\n",
+"\n",
+"\n",
+"//At I= 1 mA\n",
+"I3=100*10^-3;//A\n",
+"Rsh=Rm*Im/(I3-Im);\n",
+"printf('Shunt resistance= %.4f Ω\n',Rsh);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 15.3: Caluclate_the_series_resistance_to_convert_it_into_voltmeter.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Caluclate the series resistance to convert it into voltmeter\n",
+"clear;\n",
+"clc;\n",
+"//soltion\n",
+"//given\n",
+"\n",
+"Im=100*10^-6;//A\n",
+"Rm=100;//ohm\n",
+"V=100;//V\n",
+"Rs=V/Im-Rm;\n",
+"printf('The value of series resistance is %.1f kΩ',Rs/1000);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 15.4: Calculate_multiplier_resistance_and_voltage_multiplying_factor.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Calculate multiplier resistance and voltage multiplying factor\n",
+"clear;\n",
+"clc;\n",
+"//soltion\n",
+"//given\n",
+"\n",
+"Im=50*10^-6;//A\n",
+"Rm=1000;//ohm\n",
+"V=50;//V\n",
+"Rs=V/Im-Rm;\n",
+"printf('The value of multiplier resistance is %.0f kΩ\n',Rs/1000);\n",
+"Vm=Im*Rm;\n",
+"n=V/Vm;\n",
+"printf('Voltage multiplying factor =%.0f',n);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 15.5: Calculate_reading_and_error_of_each_voltmeter.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Calculate reading and error of each voltmeter\n",
+"clear;\n",
+"clc;\n",
+"//soltion\n",
+"\n",
+"function [z]=prll(r1,r2)//Function for the parallel combination of resistor\n",
+"    z=r1*r2/(r1+r2);\n",
+"endfunction\n",
+"\n",
+"//given\n",
+"\n",
+"S_A=1000;// Ω/V        //sensitivity\n",
+"S_B=20000;// Ω/V        //sensitivity\n",
+"R=50;//V           //range of voltmeter\n",
+"Vs=150;//V          //Supply\n",
+"R1=100*10^3;//ohm\n",
+"R2=50*10^3;//ohm\n",
+"Vt=Vs*(R2/(R1+R2));\n",
+"\n",
+"//Voltmeter A\n",
+"Ri1=S_A*R;\n",
+"Rxy_A=prll(Ri1,R2);      //total resistance at X and Y\n",
+"V1=Vs*(Rxy_A/(Rxy_A+R1));\n",
+"printf('The voltmeter indicates %.0f V\n',V1);\n",
+"\n",
+"//Voltmeter B\n",
+"Ri2=S_B*R;\n",
+"Rxy_B=prll(Ri2,R2);      //total resistance at X and Y\n",
+"V2=Vs*(Rxy_B/(Rxy_B+R1));\n",
+"printf('The voltmeter indicates %.2f V\n',V2);\n",
+"\n",
+"e1=(Vt-V1)*100/Vt;\n",
+"e2=(Vt-V2)*100/Vt;\n",
+"printf('The error in the reading of voltmeter A = %.0f percent\n',e1);\n",
+"printf('The error in the reading of voltmeter A = %.1f percent',e2);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 15.6: Determine_rms_value_of_the_ac_voltage.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Determine rms value of the ac voltage\n",
+"clear;\n",
+"clc;\n",
+"//soltion\n",
+"//given\n",
+"\n",
+"l=8.3;//cm         //length of the trace\n",
+"D=5;// V/cm        //deflection sensitivity\n",
+"Vpp=l*D;\n",
+"Vrms=Vpp/(2*sqrt(2));\n",
+"printf('The rms value of the ac voltage %.2f V',Vrms);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 15.7: Determine_rms_value_and_frequency_of_the_sine_voltage.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Determine rms value and frequency of the sine voltage\n",
+"clear;\n",
+"clc;\n",
+"//soltion\n",
+"//given\n",
+"\n",
+"l=3.5;//cm         //length of the trace\n",
+"D=2;// V/cm        //deflection sensitivity\n",
+"Vpp=l*D;\n",
+"Vrms=Vpp/sqrt(2);\n",
+"printf('The rms value of the sine voltage = %.2f V\n',Vrms);\n",
+"x=4;// cm         //one cycle length on x axis\n",
+"t=0.5*10^-3;// s/cm    //timebase setting\n",
+"T=x*t;\n",
+"f=1/T;\n",
+"printf('The frequency of the sine voltage = %.1f kHz',f/1000);"
+   ]
+   }
+],
+"metadata": {
+		  "kernelspec": {
+		   "display_name": "Scilab",
+		   "language": "scilab",
+		   "name": "scilab"
+		  },
+		  "language_info": {
+		   "file_extension": ".sce",
+		   "help_links": [
+			{
+			 "text": "MetaKernel Magics",
+			 "url": "https://github.com/calysto/metakernel/blob/master/metakernel/magics/README.md"
+			}
+		   ],
+		   "mimetype": "text/x-octave",
+		   "name": "scilab",
+		   "version": "0.7.1"
+		  }
+		 },
+		 "nbformat": 4,
+		 "nbformat_minor": 0
+}
diff --git a/Electronic_Devices_and_Circuits_by_D_C_Kulshreshtha/2-Semiconductor_Physics.ipynb b/Electronic_Devices_and_Circuits_by_D_C_Kulshreshtha/2-Semiconductor_Physics.ipynb
new file mode 100644
index 0000000..34aa4a2
--- /dev/null
+++ b/Electronic_Devices_and_Circuits_by_D_C_Kulshreshtha/2-Semiconductor_Physics.ipynb
@@ -0,0 +1,96 @@
+{
+"cells": [
+ {
+		   "cell_type": "markdown",
+	   "metadata": {},
+	   "source": [
+       "# Chapter 2: Semiconductor Physics"
+	   ]
+	},
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 2.1: Calculate_the_conductivity_and_resistivity_of_germanium.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Calculate the conductivity and resistivity of germanium\n",
+"clear;\n",
+"clc;\n",
+"//soltion\n",
+"//given\n",
+"q=1.6*10^-19;//Coulomb        //charge of an electron\n",
+"ni=2.5*10^19;//per m^3        //concentration\n",
+"un=0.36;//m^2/Vs              //mobility of electron\n",
+"up=0.17;//m^2/Vs             //mobility of holes\n",
+"con=q*ni*(un+up);           //conductivity\n",
+"res=(1/con);                //resistivity\n",
+"printf('The conductivty is %.2f S/m \n',con);\n",
+"printf('The resistivity is %.2f Ωm',res);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 2.2: Determine_the_conductivity_of_extrinsic_semiconductor.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Determine the conductivity of extrinsic semiconductor\n",
+"clear;\n",
+"clc;\n",
+"//soltion\n",
+"//given\n",
+"e=1.6*10^-19;//Coulomb        //charge of an electron\n",
+"ni=1.5*10^16;//per m^3        //concentration\n",
+"un=0.13;//m^2/Vs             //mobility of electron\n",
+"up=0.05;//m^2/Vs            //mobility of holes\n",
+"Si=5*10^28;//per m^3        //atomic density in silicon\n",
+"dop=(1/(2*10^8));          //concentration of an antimony per silicon atoms\n",
+"Nd=dop*Si;//per m^3       //donor concentraion\n",
+"n=Nd;//per m^3           //free electron concentration\n",
+"p=(ni^2/Nd);//per m ^3   // hole concentration\n",
+"con=e*(n*un+p*up);\n",
+"printf('The conductivty is %.1f S/m \n',con);"
+   ]
+   }
+],
+"metadata": {
+		  "kernelspec": {
+		   "display_name": "Scilab",
+		   "language": "scilab",
+		   "name": "scilab"
+		  },
+		  "language_info": {
+		   "file_extension": ".sce",
+		   "help_links": [
+			{
+			 "text": "MetaKernel Magics",
+			 "url": "https://github.com/calysto/metakernel/blob/master/metakernel/magics/README.md"
+			}
+		   ],
+		   "mimetype": "text/x-octave",
+		   "name": "scilab",
+		   "version": "0.7.1"
+		  }
+		 },
+		 "nbformat": 4,
+		 "nbformat_minor": 0
+}
diff --git a/Electronic_Devices_and_Circuits_by_D_C_Kulshreshtha/3-Semiconductor_Diode.ipynb b/Electronic_Devices_and_Circuits_by_D_C_Kulshreshtha/3-Semiconductor_Diode.ipynb
new file mode 100644
index 0000000..f765fd6
--- /dev/null
+++ b/Electronic_Devices_and_Circuits_by_D_C_Kulshreshtha/3-Semiconductor_Diode.ipynb
@@ -0,0 +1,546 @@
+{
+"cells": [
+ {
+		   "cell_type": "markdown",
+	   "metadata": {},
+	   "source": [
+       "# Chapter 3: Semiconductor Diode"
+	   ]
+	},
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 3.10: find_dc_power_supplied_to_load_and_efficiency_and_PIV_rating_of_the_diode.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//find dc power supplied to load and efficiency and PIV rating of the diode\n",
+"clear;\n",
+"clc;\n",
+"//soltion\n",
+"//given\n",
+"rf=2;//Ω//Dynamic forward resistance\n",
+"Rs=5;//Ω//resistaqnce of secondary\n",
+"Rl=25;//Ω//Load resistance\n",
+"Idc=0.1;//A//dc current to a load\n",
+"Pdc=Idc^2*Rl;      //dc power\n",
+"n=(81.2*Rl)/(Rl+rf+Rs);        //efficiency\n",
+"Im=(%pi*Idc)/2;              //peak value current\n",
+"Vm=Im*(Rl+rf+Rs);              //peak voltage\n",
+"Vlm=Vm-Im*(rf+Rs);              //peak voltage across load\n",
+"PIV=Vm+Vlm;\n",
+"printf('The dc power supplied to the load is %.2f W\n',Pdc);\n",
+"printf('Efficiency = %.2f percent\n',n);\n",
+"printf('The peak inverse voltage is %.2f V',PIV);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 3.11: EX3_11.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Calculate output voltage and current through load and voltage across series resistor and current and power dissipated in zener diode\n",
+"clear;\n",
+"clc;\n",
+"//soltion\n",
+"//given\n",
+"Vi=110;//V       //input voltage\n",
+"Rl=6*10^3;// ohm    //load resistance\n",
+"Rs=2*10^3;//ohm       //series resistance\n",
+"Vz=60;//V        //Zener voltage\n",
+"V=Vi*Rl/(Rs+Rl);\n",
+"\n",
+"//This V>Vz therefore Zener diode is ON\n",
+"\n",
+"Vo=Vz;          //output voltage\n",
+"Il=Vo/Rl;       //Current through load resistance\n",
+"Vs=Vi-Vo;       //Voltage drop across the series resistor\n",
+"Is=Vs/Rs         //current through the series resistor\n",
+"Iz=Is-Il        ///By applying kirchhoff's law\n",
+"Pz=Vz*Iz        //Power dissipated accross zener diode\n",
+"\n",
+"printf('The output voltage is %.0f V\n',Vo);\n",
+"printf('The current through load resistance is %.0f mA\n',Il*1000);\n",
+"printf('Voltage across series resistor is %.0f V\n',Vs)\n",
+"printf('Current in zener diode is %.0f mA\n',Iz*1000)\n",
+"printf('Power dissipated by zener diode %.0f mW',Pz*1000);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 3.12: Calculate_max_and_min_values_of_zener_diode_current.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"\n",
+"//Calculate max and min values of zener diode current\n",
+"clear;\n",
+"clc;\n",
+"//soltion\n",
+"//given\n",
+"Vimin=80;//V       //minimum input voltage\n",
+"Vimax=120;//V       //maximum input voltage\n",
+"Rl=10*10^3;// ohm    //load resistance\n",
+"Rs=5*10^3;//ohm       //series resistance\n",
+"Vz=50;//V        //Zener voltage\n",
+"V=Vimin*Rl/(Rs+Rl);\n",
+"\n",
+"//This V>Vz therefore Zener diode is ON\n",
+"\n",
+"//For minimum value of zener diode\n",
+"\n",
+"Vo=Vz;          //output voltage\n",
+"Vs=Vimin-Vo;       //Voltage drop across the series resistor\n",
+"Is=Vs/Rs         //current through the series resistor\n",
+"Il=Vo/Rl;       //Current through load resistance\n",
+"Izmin=Is-Il;\n",
+"printf('\nMinimum values of zener diode current is %.0f mA\n',Izmin*1000);\n",
+"\n",
+"//For maximum value of zener diode\n",
+"\n",
+"Vo=Vz;          //output voltage\n",
+"Vs=Vimax-Vo;       //Voltage drop across the series resistor\n",
+"Is=Vs/Rs         //current through the series resistor\n",
+"Il=Vo/Rl;       //Current through load resistance\n",
+"Izmax=Is-Il;\n",
+"printf('Maximum values of zener diode current is %.0f mA',Izmax*1000);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 3.13: determine_value_of_the_series_resistor_and_wattage_rating.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//determine value of the series resistor and wattage rating\n",
+"clear;\n",
+"clc;\n",
+"//soltion\n",
+"//given\n",
+"Vi=12;//V       //input voltage\n",
+"Vz=7.2;//V        //Zener voltage\n",
+"Izmin=10*10^-3;//A             //min current through zener diode\n",
+"Ilmax=100*10^-3;//A            //max current through load\n",
+"Ilmin=12*10^-3;//A            //min current through load\n",
+"\n",
+"Vs=Vi-Vz;       //Voltage drop across the series resistor\n",
+"Is=Izmin+Ilmax;     //Current through the series resistor\n",
+"Rs=Vs/Is;\n",
+"printf('The series resistor so that 10mA current flow through zener diode is %.1f Ω\n',Rs);\n",
+"\n",
+"Izmax=Is-Ilmin;      //max zener through zener diode\n",
+"Pmax=Izmax*Vz;\n",
+"printf('The maximum wattage rating is %.1f mW',Pmax*1000);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 3.14: Find_the_capacitance_of_a_varactor_diode.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Find the capacitance of a varactor diode\n",
+"clear;\n",
+"clc;\n",
+"//soltion\n",
+"//given\n",
+"C=5;//pf//capcitance of varactor diode at V=4V\n",
+"V=4;//V\n",
+"K=C*sqrt(4);\n",
+"\n",
+"//When bias voltage is increased upto 6 V\n",
+"Vn=6;//V//new bias voltage\n",
+"Cn=K/(sqrt(Vn));\n",
+"printf('Capacitance (At 6 V ) = %.3f pf',Cn);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 3.1: find_the_value_of_threshold_voltage.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//find the value of threshold voltage\n",
+"clear;\n",
+"clc;\n",
+"//soltion\n",
+"//given\n",
+"t1=25;//°C//initial temperature\n",
+"t2=100;//°C//final temperature\n",
+"V=2*10^-3;//V per celsius degree//decrease in barrier potential per degree\n",
+"V0=0.7//V//Potential at normal temperature\n",
+"Vd=(t2-t1)*V;//decrease in barrier potential\n",
+"Vt=V0-Vd;//threshold volatge at 100°C\n",
+"printf('Threshold volatge at 100°C = %.2f V',Vt);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 3.2: detrenmine_dc_resistance_of_silicon_diode.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//detrenmine dc resistance of silicon diode\n",
+"clear;\n",
+"clc;\n",
+"//soltion\n",
+"//given\n",
+"\n",
+"//At Id = 2 mA\n",
+"Id=2*10^-3;//Ampere//diode current\n",
+"Vd=0.5;//V//voltage(from given curve)\n",
+"Rf=(Vd/Id);\n",
+"printf('The dc resistance is %d Ω\n',Rf);\n",
+"\n",
+"//At Id = 20 mA\n",
+"Id=20*10^-3;//Ampere//diode current\n",
+"Vd=0.75;//V//voltage(from given curve)\n",
+"Rf=(Vd/Id);\n",
+"printf('The dc resistance is %.1f Ω\n',Rf);\n",
+"\n",
+"//At Vd = - 10 V \n",
+"Id=-2*10^-6;//Ampere//diode current(from given curve)\n",
+"Vd=-10;//V//voltage\n",
+"Rf=(Vd/Id);\n",
+"printf('The dc resistance is %d MΩ\n',Rf/10^6);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 3.3: determine_dc_and_ac_resistance_of_silicon_diode.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//determine dc & ac resistance of silicon diode\n",
+"clear;\n",
+"clc;\n",
+"//soltion\n",
+"//given\n",
+"Id=20*10^-3;//A//diode current\n",
+"Vd=0.75;//V// as given in the V-I graph\n",
+"Rf=Vd/Id;\n",
+"printf('The dc resistance of diode is %.1f Ω\n',Rf);\n",
+"\n",
+"//From Graph the values of dynamic voltage and current are\n",
+"//which is equal to MN and NL repectively (in graph)\n",
+"del_Vd=(0.8-0.68);//V\n",
+"del_Id=(40-0)*10^-3;//A\n",
+"rf=del_Vd/del_Id;\n",
+"printf('The ac resistance of the diode is %d Ω',rf)"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 3.4: determine_ac_resistance_of_silicon_diode.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//determine ac resistance of silicon diode\n",
+"clear;\n",
+"clc;\n",
+"//soltion\n",
+"//given\n",
+"\n",
+"//At Id =10mA\n",
+"Id=10;//mA\n",
+"rf=25/Id;\n",
+"printf('The ac resistance of the diode is(At Id= 10mA) %.1f Ω\n',rf)\n",
+"\n",
+"//At Id =20mA\n",
+"Id=20;//mA\n",
+"rf=25/Id;\n",
+"printf('The ac resistance of the diode is(At Id= 20mA) %.2f Ω',rf)"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 3.5: Find_current_through_diode.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Find current through diode\n",
+"clear;\n",
+"clc;\n",
+"//soltion\n",
+"//given\n",
+"Vt=0.3;//V//Threshold voltage\n",
+"rf=25;//ohm// average resistance\n",
+"\n",
+"//assuming it to be ideal\n",
+"//from fig 3.19\n",
+"Vaa=10;//V//supply\n",
+"R1=45;//ohm\n",
+"R2=5;//ohm\n",
+"Vab=Vaa*R2/(R1+R2);\n",
+"//Vab>Vt therefore diode is forward bias and no current flow through R2\n",
+"Idi=Vaa/R1;       //for ideal\n",
+"printf('The diode current (for ideal) is %.0f mA\n',Idi*1000);\n",
+"\n",
+"//assuming it to be real\n",
+"//Thevenin's equivalent circuit parameters of fig 3.19\n",
+"Vth=Vaa*R2/(R1+R2);\n",
+"Rth=R1*R2/(R1+R2);\n",
+"Idr=(Vth-Vt)/(Rth+rf);          //for real\n",
+"printf('The diode current (for real) is %.1f mA',Idr*1000);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 3.6: Find_current_through_resistance_in_given_figure.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Find current through resistance in given figure\n",
+"clear;\n",
+"clc;\n",
+"//soltion\n",
+"\n",
+"//From fig\n",
+"Vaa=20;//V//supply\n",
+"Vt=0.7;//V//threshold voltage of diode\n",
+"rf=5;//ohm //forward resistance\n",
+"R=90;//ohm//given resistor\n",
+"\n",
+"//Diode D1 and D4 are forward bias and D2 and D3 are reverse biased\n",
+"\n",
+"Vnet=Vaa-Vt-Vt;\n",
+"Rt=R+rf+rf;\n",
+"I=Vnet/Rt;\n",
+"printf('Current through 90 ohm resistor is %.0f mA',I*1000);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 3.7: Find_current_drawn_by_the_battery.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Find current drawn by the battery\n",
+"clear;\n",
+"clc;\n",
+"//soltion\n",
+"\n",
+"//From fig\n",
+"Vaa=10;//V//supply\n",
+"R1=100;//ohm\n",
+"R2=100;//ohm\n",
+"\n",
+"//Forward Bias\n",
+"Id=Vaa/R1;\n",
+"printf('Current drawn from battery (forward bias) %.1f A\n',Id);\n",
+"\n",
+"//Reverse Bias\n",
+"Rnet=R1+R2;\n",
+"Id=Vaa/Rnet;\n",
+"printf('Current drawn from battery (reverse bias) %.2f A',Id);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 3.8: EX3_8.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//determine dc current through load and rectification efficiency and peak inverse voltage\n",
+"clear;\n",
+"clc;\n",
+"//soltion\n",
+"//given\n",
+"TR=31/2;//Turn ratio of the transformer\n",
+"rf=20;//Ω//Dynamic forward resistance\n",
+"Rl=1000;//Ω//Load resistance\n",
+"Vt=0.66;//V//Threshold voltage of diode\n",
+"V=220;//V//input voltage of transformer\n",
+"Vp=sqrt(2)*220//V//peak value of primary voltage\n",
+"Vm=(1/TR)*Vp;\n",
+"Im=(Vm-Vt)/(rf+Rl);\n",
+"Idc=Im/%pi;\n",
+"n=40.6/(1+rf/Rl);\n",
+"printf('The dc current through load is %d mA\n',Idc*1000);\n",
+"printf('The rectification efficiency is %.1f percent\n',n);\n",
+"printf('Peak inverse voltage =Vm = %.2f V\n',Vm)"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 3.9: determine_dc_voltage_across_load_and_peak_inverse_voltage_across_each_diode.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//determine dc voltage across load and peak inverse voltage across each diode\n",
+"clear;\n",
+"clc;\n",
+"//soltion\n",
+"//given\n",
+"TR=12/1;//Turn ratio of the transformer\n",
+"V=220;//V//input voltage of transformer\n",
+"Vp=sqrt(2)*220//V//peak value of primary voltage\n",
+"Vm=(1/TR)*Vp;\n",
+"Vdc=(2*Vm)/%pi;\n",
+"printf('The dc voltage across load %.1f V\n',Vdc);\n",
+"printf('Peak inverse voltage (for bridge rectifier) = %.1f V\n',Vm);\n",
+"printf('Peak inverse voltage (for centre tap rectifier) = %.2f V\n',2*Vm);"
+   ]
+   }
+],
+"metadata": {
+		  "kernelspec": {
+		   "display_name": "Scilab",
+		   "language": "scilab",
+		   "name": "scilab"
+		  },
+		  "language_info": {
+		   "file_extension": ".sce",
+		   "help_links": [
+			{
+			 "text": "MetaKernel Magics",
+			 "url": "https://github.com/calysto/metakernel/blob/master/metakernel/magics/README.md"
+			}
+		   ],
+		   "mimetype": "text/x-octave",
+		   "name": "scilab",
+		   "version": "0.7.1"
+		  }
+		 },
+		 "nbformat": 4,
+		 "nbformat_minor": 0
+}
diff --git a/Electronic_Devices_and_Circuits_by_D_C_Kulshreshtha/4-Bipolar_Junction_Transistors.ipynb b/Electronic_Devices_and_Circuits_by_D_C_Kulshreshtha/4-Bipolar_Junction_Transistors.ipynb
new file mode 100644
index 0000000..25c3508
--- /dev/null
+++ b/Electronic_Devices_and_Circuits_by_D_C_Kulshreshtha/4-Bipolar_Junction_Transistors.ipynb
@@ -0,0 +1,355 @@
+{
+"cells": [
+ {
+		   "cell_type": "markdown",
+	   "metadata": {},
+	   "source": [
+       "# Chapter 4: Bipolar Junction Transistors"
+	   ]
+	},
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 4.10: calculate_ac_current_gain_in_CE_and_CC_configuration.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//calculate ac current gain in CE and CC configuration\n",
+"clear;\n",
+"clc;\n",
+"//soltion\n",
+"//given\n",
+"a=0.99;\n",
+"B=a/(1-a);\n",
+"printf('The ac current gain in CE configuration is %.0f\n',B);\n",
+"y=1+B;\n",
+"printf('The ac current gain in CC configuration is %.0f',y);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 4.1: determine_the_collector_and_base_current.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//determine the collector and base current\n",
+"clear;\n",
+"clc;\n",
+"//soltion\n",
+"//given\n",
+"a=0.98;//dc alpha\n",
+"Ie=5*10^-3;//A//emitter current\n",
+"Ico=2*10^-6;//A//collector reverse leakage current\n",
+"Ic=a*Ie+Ico;\n",
+"Ib=Ie-Ic;\n",
+"printf('The collector current is %.3f mA\n',Ic*1000);\n",
+"printf('The base current is %.0f uA',Ib*10^6);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 4.2: determine_the_base_and_collector_current_and_exact_and_approax_dc_alpha.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//determine the base and collector current and exact and approax dc alpha \n",
+"clear;\n",
+"clc;\n",
+"//soltion\n",
+"//given\n",
+"Ie=8.4*10^-3//A//emitter current\n",
+"Icbo=0.1*10^-6;//A//reverse leakage current\n",
+"Ib=0.008*Ie;//A//base current\n",
+"Ic=Ie-Ib;\n",
+"Icinj=Ic-Icbo;\n",
+"a0=Icinj/Ie;\n",
+"a=Ic/Ie;\n",
+"printf('Base current is %.1f uA\n',Ib*10^6);\n",
+"printf('Collector current %.4f mA\n',Ic*1000);\n",
+"printf('Exact value of alphha = %.7f\n',a0);\n",
+"printf('Approax value of alpha = %.3f',a);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 4.3: Determine_the_base_current.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Determine the base current\n",
+"clear;\n",
+"clc;\n",
+"//soltion\n",
+"//given\n",
+"a=0.96;            //dc alpha\n",
+"Rc=2*10^3;//ohm      //resistor across collector\n",
+"Vc=4;//V            //Voltage drop across the collector resistor\n",
+"Ic=Vc/Rc;         //Colletor current\n",
+"Ie=Ic/a;          //Emmiter current\n",
+"Ib=Ie-Ic;         //Base current\n",
+"printf('The base current is %.0f uA',Ib*10^6)"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 4.4: determine_dynamic_input_resistance.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//determine dynamic input resistance\n",
+"clear;\n",
+"clc;\n",
+"//soltion\n",
+"//given\n",
+"Ie=2;//mA\n",
+"Vcb=10;//V\n",
+"\n",
+"//Taking points around Ie & Vcb from graph\n",
+"del_Ie=(2.5-1.5)*10^-3;//A\n",
+"\n",
+"//corresponding change in Veb\n",
+"del_Veb=(0.9-0.8);//V\n",
+"rib=del_Veb/del_Ie;\n",
+"printf('The dynamic input resistance of transistor is %.0f Ω',rib);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 4.5: find_dc_current_gain_in_common_emitter_configuration.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//find dc current gain in common emitter configuration\n",
+"clear;\n",
+"clc;\n",
+"//soltion\n",
+"//given\n",
+"a=0.98;//dc current gain in common base configuration\n",
+"B=a/(1-a);\n",
+"printf('The dc current gain in common emitter configuration is %.0f',B);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 4.6: calculate_ac_alpha_and_beta.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//calculate ac alpha and beta\n",
+"clear;\n",
+"clc;\n",
+"//soltion\n",
+"//given\n",
+"ic=0.995//mA//Emitter current change\n",
+"ie=1//mA//collector current change\n",
+"a=ic/ie;\n",
+"B=a/(1-a);\n",
+"printf('The ac alpha is %.3f\n',a)\n",
+"printf('The common emitter ac current gain is %.0f',B);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 4.7: Calculate_beta_and_Iceo_and_exact_and_approax_collector_current.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Calculate beta and Iceo and exact and approax collector current\n",
+"clear;\n",
+"clc;\n",
+"//soltion\n",
+"//given\n",
+"a0=0.992;//dc current gain in common base configuration\n",
+"Icbo=48*10^-9;//A\n",
+"Ib=30*10^-6;//A//base current\n",
+"B=a0/(1-a0);\n",
+"Iceo=Icbo/(1-a0);\n",
+"printf('Beta= %.0f\n',B);\n",
+"printf('Iceo= %0.f uA\n',Iceo*10^6);\n",
+"Ic=B*Ib+Iceo;\n",
+"Ica=B*Ib;//approax\n",
+"printf('Exact collector current %.3f mA\n',Ic*1000);\n",
+"printf('Approax collector current %.2f mA',Ica*1000);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 4.8: determine_dynamic_input_resistance.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//determine dynamic input resistance\n",
+"clear;\n",
+"clc;\n",
+"//soltion\n",
+"//given\n",
+"Vbe=0.75;//V\n",
+"Vce=2;//V\n",
+"\n",
+"//Taking points around Vbe=0.75V from graph\n",
+"del_Vbe=(0.98-0.9);//V\n",
+"\n",
+"//corresponding change in ib\n",
+"del_ib=(68-48)*10^-6;//A\n",
+"\n",
+"rie=del_Vbe/del_ib;\n",
+"printf('The dynamic input resistance of transistor is %.0f kΩ',rie/1000);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 4.9: determine_dynamic_input_resistance_and_dc_and_ac_current_gain.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//determine dynamic input resistance and dc and ac current gain\n",
+"clear;\n",
+"clc;\n",
+"//soltion\n",
+"//given\n",
+"Ib=30*10^-6;//A\n",
+"Vce=10;//V\n",
+"Ic=3.6*10^-3;//A        //from graph\n",
+"\n",
+"//Taking points around Vce = 10V from graph\n",
+"del_Vce=(12.5-7.5);//V\n",
+"\n",
+"//corresponding change in ic\n",
+"del_ic=(3.7-3.5)*10^-3;//A\n",
+"\n",
+"roe=del_Vce/del_ic;\n",
+"printf('The dynamic output resistance of transistor is %.0f kΩ\n',roe/1000);\n",
+"\n",
+"//dc current gain\n",
+"Bo=Ic/Ib;\n",
+"printf('The dc current gain is %.0f\n',Bo);\n",
+"\n",
+"//ac current gain\n",
+"\n",
+"del_ic=(4.7-2.5)*10^-3;    //the collector current change is from 3.5mA to 4.7mA as we can see from graph when we change ib from 40mA to 20mA\n",
+"del_ib=(40-20)*10^-6;\n",
+"B=del_ic/del_ib;\n",
+"printf('The ac current gain is %.0f\n',B);"
+   ]
+   }
+],
+"metadata": {
+		  "kernelspec": {
+		   "display_name": "Scilab",
+		   "language": "scilab",
+		   "name": "scilab"
+		  },
+		  "language_info": {
+		   "file_extension": ".sce",
+		   "help_links": [
+			{
+			 "text": "MetaKernel Magics",
+			 "url": "https://github.com/calysto/metakernel/blob/master/metakernel/magics/README.md"
+			}
+		   ],
+		   "mimetype": "text/x-octave",
+		   "name": "scilab",
+		   "version": "0.7.1"
+		  }
+		 },
+		 "nbformat": 4,
+		 "nbformat_minor": 0
+}
diff --git a/Electronic_Devices_and_Circuits_by_D_C_Kulshreshtha/5-Field_Effect_Transistors.ipynb b/Electronic_Devices_and_Circuits_by_D_C_Kulshreshtha/5-Field_Effect_Transistors.ipynb
new file mode 100644
index 0000000..7b747d7
--- /dev/null
+++ b/Electronic_Devices_and_Circuits_by_D_C_Kulshreshtha/5-Field_Effect_Transistors.ipynb
@@ -0,0 +1,301 @@
+{
+"cells": [
+ {
+		   "cell_type": "markdown",
+	   "metadata": {},
+	   "source": [
+       "# Chapter 5: Field Effect Transistors"
+	   ]
+	},
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 5.1: Calculate_saturation_voltage_and_saturation_current.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Calculate saturation voltage and saturation current\n",
+"clear;\n",
+"clc;\n",
+"//soltion\n",
+"//given\n",
+"Vp=-4//V          //pinch off voltage\n",
+"Idss=12*10^-3;//A      //drain to source current with gate shorted\n",
+"Vgs=-2;//V             //gate to source voltage\n",
+"Vds=Vgs-Vp;\n",
+"Id=Idss*(Vds/Vp)^2;\n",
+"printf('Saturation Voltage is %.0f V\n',Vds);\n",
+"printf('Saturation current is %.0f mA',Id*10^3);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 5.2: Find_the_value_of_drain_current.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Find the value of drain current\n",
+"clear;\n",
+"clc;\n",
+"//soltion\n",
+"//given\n",
+"Vgso=-5;//V          //gate to source cut off voltage\n",
+"Idss=20*10^-3;//A     //drain to source current with gate shorted\n",
+"\n",
+"//At vgs = -2 V\n",
+"vgs=-2;//V        input voltage\n",
+"id=Idss*(1-(vgs/Vgso))^2;          //Schockley's equation\n",
+"printf('Drain current is (At vgs = -2 V) %.1f mA\n',id*10^3);\n",
+"\n",
+"//At vgs = -4 V\n",
+"vgs=-4;//V        input voltage\n",
+"id=Idss*(1-(vgs/Vgso))^2;          //Schockley's equation\n",
+"printf('Drain current is (At vgs = -4 V) %.1f mA\n',id*10^3);\n",
+"\n",
+"//At vgs = -8 V\n",
+"printf('Drain current is 0 A (At vgs = -8 V) because gate is biased beyond cut off ');"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 5.3: Calculate_Vgs_and_Vds_saturation.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Calculate Vgs and Vds saturation\n",
+"clear;\n",
+"clc;\n",
+"//soltion\n",
+"//given\n",
+"Vp=5//V          //pinch off voltage\n",
+"Idss=-15*10^-3;//A      //drain to source current with gate shorted\n",
+"Id=-3*10^-3;//A         //saturation current\n",
+"Vgs=Vp*(1-sqrt(Id/Idss));\n",
+"Vds=Vgs-Vp;\n",
+"printf('The gate to source voltage (Vgs) is %.3f V\n',Vgs);\n",
+"printf('The saturation voltage is Vds(sat)= %.3f V',Vds);\n",
+"\n",
+"// THe value of Vgs = 2.115V and Vds= -2.885V in book because of the calculation error"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 5.4: Calculate_drain_current_Id_for_N_channel.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Calculate drain current Id for N channel\n",
+"clear;\n",
+"clc;\n",
+"//soltion\n",
+"//given\n",
+"Vp=5//V          //pinch off voltage\n",
+"Idss=18*10^-3;//A     //drain to source current with gate shorted\n",
+"\n",
+"//For Vgs= - 3 V\n",
+"Vgs=-3;//V\n",
+"Id=Idss*(1-(Vgs/(-Vp)))^2;\n",
+"printf('The drain current Id(For Vgs= -3V) = %.2f mA\n',Id*10^3);\n",
+"\n",
+"//For Vgs= 2.5 V\n",
+"Vgs=2.5;//V\n",
+"Id=Idss*(1-(Vgs/(-Vp)))^2;\n",
+"printf('The drain current Id(For Vgs= 2.5V) = %.1f mA',Id*10^3);\n",
+""
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 5.5: Calculate_drain_current_Id_for_P_channel.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Calculate drain current Id for P channel\n",
+"clear;\n",
+"clc;\n",
+"//soltion\n",
+"//given\n",
+"Vp=-5//V          //pinch off voltage\n",
+"Idss=18*10^-3;//A     //drain to source current with gate shorted\n",
+"\n",
+"//For Vgs= -3V\n",
+"Vgs=-3;//V\n",
+"Id=Idss*(1-(Vgs/(-Vp)))^2;\n",
+"printf('The drain current Id (For Vgs= -3V) = %.2f mA\n',Id*10^3);\n",
+"\n",
+"//For Vgs= 2.5V\n",
+"Vgs=2.5;//V\n",
+"Id=Idss*(1-(Vgs/(-Vp)))^2;\n",
+"printf('The drain current Id (For Vgs= 2.5V) = %.1f mA',Id*10^3);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 5.6: Find_the_value_of_drain_current.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Find the value of drain current\n",
+"clear;\n",
+"clc;\n",
+"//soltion\n",
+"//given\n",
+"Vt=2;//V      //threshold voltage\n",
+"K=0.25*10^-3;// A/V^2      //conductivity parameter\n",
+"Vgs=3;//V    //gate supply\n",
+"Vds=2;//V      //saturation voltage\n",
+"Vdsm=Vgs-Vt;        //minimum voltage required to pinch off\n",
+"\n",
+"// Vds > Vdsm therefore the device is in saturation region\n",
+"\n",
+"Id=K*(Vgs-Vt)^2;\n",
+"printf('The drain current is %.2f mA',Id*1000);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 5.7: Find_the_value_of_Id.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Find the value of Id\n",
+"clear;\n",
+"clc;\n",
+"//soltion\n",
+"//given\n",
+"Vt=1.5;//V      //threshold voltage\n",
+"Id=2*10^-3;//A\n",
+"Vgs=3;//V    //gate supply\n",
+"Vds=5;//V      //saturation voltage\n",
+"Vdsm=Vgs-Vt;        //minimum voltage required to pinch off\n",
+"\n",
+"// Vds > Vdsm therefore the device is in saturation region\n",
+"\n",
+"// Calculating K\n",
+"K=Id/((Vgs-Vt)^2);       // A/V^2      //conductivity parameter\n",
+"\n",
+"//Calculating Id for Vgs= 5 V and Vds= 6 V\n",
+"Vgs=5;//V    //gate supply\n",
+"Vds=6;//V      //saturation voltage\n",
+"Id=K*((Vgs-Vt)^2);\n",
+"printf('The drain current is %.2f mA',Id*1000);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 5.8: Calculate_the_dynamic_drain_resistance.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Calculate the dynamic drain resistance\n",
+"clear;\n",
+"clc;\n",
+"//soltion\n",
+"//given\n",
+"gm=200*10^-6;//S          trans conductance\n",
+"u=80;//         amplification factor\n",
+"rd=u/gm;\n",
+"printf('The dynamic drain resistance is %.0f kΩ',rd/1000);"
+   ]
+   }
+],
+"metadata": {
+		  "kernelspec": {
+		   "display_name": "Scilab",
+		   "language": "scilab",
+		   "name": "scilab"
+		  },
+		  "language_info": {
+		   "file_extension": ".sce",
+		   "help_links": [
+			{
+			 "text": "MetaKernel Magics",
+			 "url": "https://github.com/calysto/metakernel/blob/master/metakernel/magics/README.md"
+			}
+		   ],
+		   "mimetype": "text/x-octave",
+		   "name": "scilab",
+		   "version": "0.7.1"
+		  }
+		 },
+		 "nbformat": 4,
+		 "nbformat_minor": 0
+}
diff --git a/Electronic_Devices_and_Circuits_by_D_C_Kulshreshtha/6-Transistor_Biasing_and_Stabilization.ipynb b/Electronic_Devices_and_Circuits_by_D_C_Kulshreshtha/6-Transistor_Biasing_and_Stabilization.ipynb
new file mode 100644
index 0000000..0521c2c
--- /dev/null
+++ b/Electronic_Devices_and_Circuits_by_D_C_Kulshreshtha/6-Transistor_Biasing_and_Stabilization.ipynb
@@ -0,0 +1,603 @@
+{
+"cells": [
+ {
+		   "cell_type": "markdown",
+	   "metadata": {},
+	   "source": [
+       "# Chapter 6: Transistor Biasing and Stabilization"
+	   ]
+	},
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 6.10: Determine_the_collector_current_at_two_different_B.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Determine the collector current at two different B\n",
+"clear;\n",
+"clc;\n",
+"//soltion\n",
+"//given\n",
+"\n",
+"//At B=50\n",
+"\n",
+"B=50;            //dc beta\n",
+"Rc=2;//ohm      //resistor connected to collector\n",
+"Re=1000;//ohm      //resistor connected to emitter\n",
+"Rb=300*10^3;//ohm      //resistor connected to base\n",
+"Vcc=9;//V            //Voltage supply across the collector resistor\n",
+"Ib=Vcc/(Rb+B*Re);         //Base current\n",
+"Ic=B*Ib;         //Colletor current\n",
+"printf('the collector current at (B=50)= %.3f mA\n',Ic*1000);\n",
+"\n",
+"//At B=150\n",
+"\n",
+"B1=150;            //dc beta\n",
+"Ib1=Vcc/(Rb+B1*Re);         //Base current\n",
+"Ic1=B1*Ib1;         //Colletor current\n",
+"printf('the collector current at (B=150)= %.0f mA\n',Ic1*1000);\n",
+"printf('The factor at which collector current increases %.2f',Ic1/Ic);\n",
+"\n",
+"//IN BOOK Ic(AT B=50)= 1.25 mA and Ic1/Ic=2.4 DUE TO APPROAXIMATION"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 6.11: Calculate_Q_point_in_voltage_divider.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Calculate Q point in voltage divider\n",
+"clear;\n",
+"clc;\n",
+"//soltion\n",
+"//given\n",
+"B=100;            //dc beta\n",
+"Rc=2*10^3;//ohm      //resistor connected to collector\n",
+"R1=10*10^3;//ohm      //voltage divider resistor 1\n",
+"R2=1*10^3;//ohm      //voltage divider resistor 2\n",
+"Re=200;//ohm      //resistor connected to emitter\n",
+"Vcc=10;//V            //Voltage supply across the collector resistor\n",
+"Vbe=0.3;//V          //base to emitter voltage\n",
+"\n",
+"I=Vcc/(R1+R2);       //current through voltage divider\n",
+"Vb=I*R2;        //voltage at base\n",
+"Ve=Vb-Vbe;\n",
+"Ie=Ve/Re;\n",
+"Ic=Ie        //approaximating Ib is nearly equal to 0\n",
+"Vc=Vcc-Ic*Rc;\n",
+"Vce=ceil(Vc)-Ve; \n",
+"printf('The Q point is (%.1f V, %.0f mA)',Vce,Ic*1000);\n",
+"\n",
+"Ibc=I/20;     //critical value of base current\n",
+"Ib=Ic/B;      //actual base current\n",
+"\n",
+"//Since Ib < Ibc, hence assumption is alright"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 6.12: Solve_the_voltage_divider_accurately_by_applying_thevenins_theorem.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"\n",
+"//Solve the voltage divider accurately by applying thevenin's theorem\n",
+"clear;\n",
+"clc;\n",
+"//soltion\n",
+"//given\n",
+"B=100;            //dc beta\n",
+"Rc=2*10^3;//ohm      //resistor connected to collector\n",
+"R1=10*10^3;//ohm      //voltage divider resistor 1\n",
+"R2=1*10^3;//ohm      //voltage divider resistor 2\n",
+"Re=200;//ohm      //resistor connected to emitter\n",
+"Vcc=10;//V            //Voltage supply across the collector resistor\n",
+"Vbe=0.3;//V          //base to emitter voltage\n",
+"\n",
+"Vth=Vcc*R2/(R1+R2);\n",
+"Rth=R1*R2/(R1+R2);\n",
+"\n",
+"printf('\nThevenin equivalent voltage Vth = %.5f V',Vth);\n",
+"printf('\nThevenin equivalent resistance Rth = %.2f ohm',Rth);\n",
+"\n",
+"Ib=(Vth-Vbe)/(Rth+(1+B)*Re);\n",
+"Ic=B*Ib;\n",
+"Ie=Ic+Ib;\n",
+"Vce=Vcc-Ic*Rc-Ie*Re; \n",
+"printf('\nThe accurate value of Ic = %.5f mA',Ic*10^3);\n",
+"printf('\nThe accurate value of Vce = %.6f V',Vce);\n",
+"Icp=3*10^-3; // Current calculated by voltage divider in previous example\n",
+"Vcep=3.4; // Voltage calculated by voltage divider in previous example\n",
+"Err_Ic=(Ic-Icp)*100/Ic;\n",
+"Err_Vce=(Vce-Vcep)*100/Vce;\n",
+"printf('\nError in Ic= %.1f percent\n',Err_Ic);\n",
+"printf('Error in Vce= %.0f percent',Err_Vce);\n",
+"\n",
+"// The errors and The accurate values are different \n",
+"// because of the approaximation in Vth and Rth in book\n",
+"\n",
+"// In Book Ic = 2.8436 mA and Vce = 3.73839 V\n",
+"// Error in Ic = -5.5% \n",
+"// Error in Vce = +9% \n",
+""
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 6.13: determine_the_Q_point_for_the_emitter_bias_circuit.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//determine the Q point for the emitter bias circuit\n",
+"clear;\n",
+"clc;\n",
+"//soltion\n",
+"//given\n",
+"B=100;               //dc beta\n",
+"Rc=5*10^3;//ohm      //resistor connected to collector\n",
+"Rb=10*10^3;//ohm      //resistor connected to base\n",
+"Re=10*10^3;//ohm      //resistor connected to emitter\n",
+"Vcc=12;//V            //Voltage supply across the collector resistor\n",
+"Vee=15;//V          //supply at emitter\n",
+"Ie=Vee/Re;\n",
+"Ic=Ie;\n",
+"Vce=Vcc-Ic*Rc;\n",
+"printf('The Q point is (%.1f V, %.1f mA)',Vce,Ic*1000);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 6.14: Calculate_Vgs_and_Rs.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Calculate Vgs and Rs\n",
+"clear;\n",
+"clc;\n",
+"//soltion\n",
+"//given\n",
+"\n",
+"Vp=2;//V\n",
+"Idss=1.75*10^-3;//A      //drain current at Vgs=0\n",
+"Vdd=24;//V             //drain to supply source\n",
+"Id=1*10^-3;//A       //drain current\n",
+"Vgs=(-Vp)*(1-sqrt(Id/Idss));\n",
+"Rs=abs(Vgs)/Id;\n",
+"printf('Vgs = %.3f V\n',Vgs);\n",
+"printf('Rs = %.0f Ω',Rs);\n",
+" "
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 6.1: Determine_the_Q_point.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Determine the Q point\n",
+"clear;\n",
+"clc;\n",
+"//soltion\n",
+"//given\n",
+"B=50;            //dc beta\n",
+"Rc=2.2*10^3;//ohm      //resistor connected to collector\n",
+"Rb=270*10^3;//ohm      //resistor connected to base\n",
+"Vcc=9;//V            //Voltage supply across the collector resistor\n",
+"Vbe=0.7;//V          //base to emitter voltage\n",
+"Ib=(Vcc-Vbe)/Rb;         //Base current\n",
+"Ic=B*Ib;         //Colletor current\n",
+"Ics=Vcc/Rc;       //Colletor saturation current\n",
+"\n",
+"//Actual Ic is the smaller of the above two values\n",
+"\n",
+"Vce=Vcc-Ic*Rc;\n",
+"printf('The Q point is (%.2f V, %.1f mA)',Vce,Ic*1000);\n",
+"//In book Vce = 5.7 V because of approaximation"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 6.2: Determine_the_Q_point.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Determine the Q point\n",
+"clear;\n",
+"clc;\n",
+"//soltion\n",
+"//given\n",
+"B=150;            //dc beta\n",
+"Rc=1*10^3;//ohm      //resistor connected to collector\n",
+"Rb=100*10^3;//ohm      //resistor connected to base\n",
+"Vcc=10;//V            //Voltage supply across the collector resistor\n",
+"Vbe=0.7;//V          //base to emitter voltage\n",
+"Ib=(Vcc-Vbe)/Rb;         //Base current\n",
+"Ic=B*Ib;         //Colletor current\n",
+"Ics=Vcc/Rc;       //Colletor saturation current\n",
+"\n",
+"//Actual Ic is the smaller of the above two values i.e. Ic(sat) and since the transistor is in saturation mode therefore Vce will become 0\n",
+"\n",
+"Vce=0;\n",
+"printf('The Q point is (%d V, %.0f mA)',Vce,Ics*1000);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 6.3: Determine_Rb_and_percentage_change_in_collector_current_due_to_temperature_rise.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Determine Rb and percentage change in collector current due to temperature rise\n",
+"clear;\n",
+"clc;\n",
+"//soltion\n",
+"//given\n",
+"\n",
+"//Calculating the base resistance\n",
+"B=20;            //dc beta\n",
+"Rc=1*10^3;//ohm      //resistor connected to collector\n",
+"Ic=1*10^-3;//A       //collector current\n",
+"Vcc=6;//V            //Voltage supply across the collector resistor\n",
+"Vbe=0.3;//V       //for germanium\n",
+"Icbo=2*10^-6;//A       //collector  to base leakage current\n",
+"\n",
+"Ib=(Ic-(1+B)*Icbo)/B;\n",
+"Rb=(Vcc-Vbe)/Ib;\n",
+"\n",
+"printf('The value of resistor Ib is %.4f kΩ = 120 kΩ \n',Rb/1000);\n",
+"\n",
+"Rb=120*10^3;//ohm        approax\n",
+"\n",
+"//Now when temperature rise\n",
+"Icbo=10*10^-6;//A       //collector  to base leakage current\n",
+"B=25;            //dc beta\n",
+"Ic1=B*Ib+(B+1)*Icbo;          //changed collector current\n",
+"perc=(Ic1-Ic)*100/Ic;        //percentage increase\n",
+"printf('The percentage change in collector current is %.0f percent',perc);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 6.4: Determine_the_Q_point_at_two_different_B.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Determine the Q point at two different B\n",
+"clear;\n",
+"clc;\n",
+"//soltion\n",
+"//given\n",
+"\n",
+"//At B=50\n",
+"\n",
+"B=50;            //dc beta\n",
+"Rc=2*10^3;//ohm      //resistor connected to collector\n",
+"Rb=300*10^3;//ohm      //resistor connected to base\n",
+"Vcc=9;//V            //Voltage supply across the collector resistor\n",
+"Ib=Vcc/Rb;         //Base current\n",
+"Ic=B*Ib;         //Colletor current\n",
+"Ics=Vcc/Rc;       //Colletor saturation current\n",
+"\n",
+"//Actual Ic is the smaller of the above two values\n",
+"\n",
+"Vce=Vcc-Ic*Rc;\n",
+"printf('The Q point (At B=50) is (%.0f V, %.1f mA)\n',Vce,Ic*1000);\n",
+"\n",
+"//At B=150\n",
+"\n",
+"B1=150;            //dc beta\n",
+"Ic1=B*Ib;         //Colletor current\n",
+"Ics1=Vcc/Rc;       //Colletor saturation current\n",
+"\n",
+"//Actual Ic is the smaller of the above two values i.e. Ic(sat) and since the transistor is in saturation mode therefore Vce will become 0\n",
+"\n",
+"Vce=0;\n",
+"printf('The Q point (At B=150) is (%d V, %.1f mA)\n',Vce,Ics*1000);\n",
+"\n",
+"printf('The factor at which collector current increases %.0f',Ics1/Ic);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 6.5: determine_Q_point_in_collector_to_base_bias_circuit.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//determine Q point in collector to base bias circuit\n",
+"clear;\n",
+"clc;\n",
+"//soltion\n",
+"//given\n",
+"B=100;            //dc beta\n",
+"Rc=500;//ohm      //resistor connected to collector\n",
+"Rb=500*10^3;//ohm      //resistor connected to base\n",
+"Vcc=10;//V            //Voltage supply across the collector resistor\n",
+"Ib=Vcc/(Rb+B*Rc);         //Base current\n",
+"Ic=B*Ib;         //Colletor current\n",
+"Ics=Vcc/Rc;       //Colletor saturation current\n",
+"\n",
+"//Actual Ic is the smaller of the above two values\n",
+"\n",
+"Vce=Vcc-(Ic+Ib)*Rc;\n",
+"printf('The Q point is (%.1f V, %.1f mA)',Vce,Ic*1000);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 6.6: EX6_6.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Calculate the collector current and change in it if B is changed by three times of previous B\n",
+"clear;\n",
+"clc;\n",
+"//soltion\n",
+"//given\n",
+"B=50;            //dc beta\n",
+"Rc=2*10^3;//ohm      //resistor connected to collector\n",
+"Rb=300*10^3;//ohm      //resistor connected to base\n",
+"Vcc=9;//V            //Voltage supply across the collector as it is PNP so taking positive\n",
+"Ib=Vcc/(Rb+B*Rc);         //Base current\n",
+"Ic=B*Ib;         //Colletor current\n",
+"printf('Collector current (B=50)= %.3f mA\n',Ic*1000);\n",
+"//Now B=150\n",
+"B=3*B;      //three times of previous B\n",
+"Ib1=Vcc/(Rb+B*Rc);         //Base current\n",
+"Ic1=B*Ib1;         //Colletor current\n",
+"printf('Collector current (B=150)= %.2f mA\n',Ic1*1000);\n",
+"printf('The factor at which collector current increases %.0f',Ic1/Ic);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 6.7: Calculate_the_value_of_all_three_current_Ie_and_Ic_and_Ib.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Calculate the value of all three current Ie and Ic and Ib\n",
+"clear;\n",
+"clc;\n",
+"//soltion\n",
+"//given\n",
+"B=90;            //dc beta\n",
+"Rc=1*10^3;//ohm      //resistor connected to collector\n",
+"Rb=500*10^3;//ohm      //resistor connected to base\n",
+"Re=500;//ohm      //resistor connected to emitter\n",
+"Vcc=9;//V            //Voltage supply across the collector resistor\n",
+"Ib=Vcc/(Rb+B*Re);         //Base current\n",
+"Ic=B*Ib;         //Colletor current\n",
+"Ie=Ic+Ib;        //Emitter current\n",
+"printf('Base current = %.1f uA \nCollector current = %.3f mA \nEmitter current = %.4f mA',Ib*10^6,Ic*10^3,Ie*10^3);\n",
+" "
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 6.8: Calculate_max_and_min_value_of_emitter_current.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Calculate max and min value of emitter current\n",
+"clear;\n",
+"clc;\n",
+"//soltion\n",
+"//given\n",
+"\n",
+"//At B=50\n",
+"\n",
+"B=50;            //dc beta\n",
+"Rc=75;//ohm      //resistor connected to collector\n",
+"Re=100;//ohm      //resistor connected to emitter\n",
+"Rb=10*10^3;//ohm      //resistor connected to base\n",
+"Vcc=6;//V            //Voltage supply across the collector resistor\n",
+"Vbe=0.3;//V       //for germanium\n",
+"Ib=(Vcc-Vbe)/(Rb+(1+B)*Re);         //Base current\n",
+"Ie=(1+B)*Ib;\n",
+"Vce=Vcc-(Rc+Re)*Ie\n",
+"printf('Minimum emitter current %.2f mA\n',Ie*10^3);\n",
+"printf('The collector to emitter volatge is %.2f V\n',Vce);\n",
+"\n",
+"//At B=300 \n",
+"\n",
+"B1=300;            //dc beta\n",
+"Ib1=(Vcc-Vbe)/(Rb+(1+B1)*Re);         //Base current\n",
+"Ie1=(1+B1)*Ib1;\n",
+"Vce1=Vcc-(Rc+Re)*Ie1\n",
+"//Here Vce1= -1.4874 V but can never have negative voltage because Ie1 is wrong as it can't be more than saturation value therefore\n",
+"Ie1=Vcc/(Rc+Re);\n",
+"\n",
+"//And Vce=0 V\n",
+"\n",
+"Vce1=0;//V\n",
+"printf('Maximum emitter current %.2f mA\n',Ie1*10^3);\n",
+"printf('The collector to emitter volatge(saturation) is %.0f V\n',Vce1);\n",
+""
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 6.9: Determine_the_value_of_base_resistance.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Determine the value of base resistance\n",
+"clear;\n",
+"clc;\n",
+"//soltion\n",
+"//given\n",
+"\n",
+"B=100;            //dc beta\n",
+"Rc=200;//ohm      //resistor connected to collector\n",
+"Re=500;//ohm      //resistor connected to emitter\n",
+"Vcc=9;//V         //Voltage supply across the collector as it is PNP so taking positive\n",
+"Vce=4.5;//V       //Collector to emitter voltage\n",
+"Ic=(Vcc-Vce)/(Rc+Re);\n",
+"Ib=Ic/B;\n",
+"Rb=(Vcc-B*Re*Ib)/Ib;\n",
+"printf('The value of base resistance is %.0f kΩ',Rb/1000);\n",
+" "
+   ]
+   }
+],
+"metadata": {
+		  "kernelspec": {
+		   "display_name": "Scilab",
+		   "language": "scilab",
+		   "name": "scilab"
+		  },
+		  "language_info": {
+		   "file_extension": ".sce",
+		   "help_links": [
+			{
+			 "text": "MetaKernel Magics",
+			 "url": "https://github.com/calysto/metakernel/blob/master/metakernel/magics/README.md"
+			}
+		   ],
+		   "mimetype": "text/x-octave",
+		   "name": "scilab",
+		   "version": "0.7.1"
+		  }
+		 },
+		 "nbformat": 4,
+		 "nbformat_minor": 0
+}
diff --git a/Electronic_Devices_and_Circuits_by_D_C_Kulshreshtha/7-Small_Signal_Single_Stage_Amplifier.ipynb b/Electronic_Devices_and_Circuits_by_D_C_Kulshreshtha/7-Small_Signal_Single_Stage_Amplifier.ipynb
new file mode 100644
index 0000000..6e45c5a
--- /dev/null
+++ b/Electronic_Devices_and_Circuits_by_D_C_Kulshreshtha/7-Small_Signal_Single_Stage_Amplifier.ipynb
@@ -0,0 +1,519 @@
+{
+"cells": [
+ {
+		   "cell_type": "markdown",
+	   "metadata": {},
+	   "source": [
+       "# Chapter 7: Small Signal Single Stage Amplifier"
+	   ]
+	},
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 7.10: Determine_the_small_signal_voltage_gain_and_input_and_output_resistance.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Determine the small signal voltage gain and input and output resistance\n",
+"clear;\n",
+"clc;\n",
+"//soltion\n",
+"//given\n",
+"\n",
+"Rd=2*10^3;//ohm\n",
+"rd=100*10^3;//ohm\n",
+"Rg=1*10^6;//ohm\n",
+"gm=2*10^-3;//S\n",
+"Av=-gm*(rd*Rd/(rd+Rd));\n",
+"Ri=Rg;\n",
+"Ro=rd*Rd/(rd+Rd);\n",
+"printf('The small signal voltage gain = %.0f\ninput resistance= %.0f MΩ\noutput resistance = %.0f kΩ',Av,Ri/10^6,Ro/1000);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 7.11: Determine_the_small_signal_voltage_gain_and_input_and_output_resistance.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Determine the small signal voltage gain and input and output resistance\n",
+"clear;\n",
+"clc;\n",
+"//soltion\n",
+"//given\n",
+"\n",
+"R1=500*10^3;//ohm\n",
+"R2=50*10^3;//ohm\n",
+"Rd=5*10^3;//ohm\n",
+"Rs=100;//ohm\n",
+"Rl=5*10^3;//ohm\n",
+"gm=1.5*10^-3;//S\n",
+"rd=200*10^3;//ohm\n",
+"Rg=R1*R2/(R1+R2);\n",
+"Rac=Rd*Rl/(Rd+Rl);\n",
+"Av=-gm*Rac;\n",
+"Ri=Rg;\n",
+"Ro=(rd*Rac/(rd+Rac));\n",
+"printf('The small signal voltage gain = %.2f\ninput resistance= %.2f kΩ\noutput resistance = %.1f kΩ',Av,Ri/1000,Ro/1000);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 7.12: Calculate_the_voltage_gain_of_the_FET.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Calculate the voltage gain of the FET\n",
+"clear;\n",
+"clc;\n",
+"//soltion\n",
+"//given\n",
+"\n",
+"Idss=8*10^-3;//A\n",
+"Vp=4;//V\n",
+"rd=25*10^3;//ohm\n",
+"Rd=2.2*10^3;//ohm     //by the help of figure\n",
+"Vgs=-1.8;//V\n",
+"gmo=2*Idss/(abs(Vp));\n",
+"gm=gmo*(1-(Vgs/(-Vp)));\n",
+"Av=-gm*(rd*Rd/(rd+Rd));\n",
+"printf('The voltage gain of the FET %.2f',Av);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 7.1: EX7_1.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Calculate max current and check will the capacitor act as short for given frequency\n",
+"clear;\n",
+"clc;\n",
+"//soltion\n",
+"//given\n",
+"\n",
+"C=100*10^-6;//Farad\n",
+"Rs=1*10^3;//ohm\n",
+"Rl=4*10^3;//ohm\n",
+"Vs=1;//V\n",
+"Imax=Vs/(Rs+Rl);\n",
+"fc=1/(2*%pi*(Rs+Rl)*C)     //critical frequency\n",
+"fh=10*fc;                  //Border frequency\n",
+"printf('Maximum current is %.0f uA\n',Imax*10^6);\n",
+"printf('fh = %.2f Hz\n',fh);\n",
+"printf('As long as source frequency is greater than %.2f Hz, the coupling capacitor acts like an ac short for 20Hz to 20kHz.',fh)\n",
+"\n",
+"//In book Imax is 200mA but there is misprinting of 'm' in mA it should be uA"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 7.2: EX7_2.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Check whether the capacitor is an effective bypass for the signal currents of lowest frequency 20 Hz\n",
+"clear;\n",
+"clc;\n",
+"//soltion\n",
+"//given\n",
+"\n",
+"C=100*10^-6;//Farad\n",
+"Rs=1*10^3;//ohm\n",
+"Rl=4*10^3;//ohm\n",
+"f=20;//Hz          //lowest frequency\n",
+"Xc=1/(2*%pi*f*C)         //reactance of capacitor at 20Hz\n",
+"Rth=Rs*Rl/(Rs+Rl);      //Thevenin's equivalent resistance\n",
+"printf('Xc < Rth/10 = %.1f Ω < %.1f Ω is satisfied\n',Xc,Rth/10);\n",
+"printf('The capacitor of 100uF will work as a good bypass for frequencies greater than 20 Hz ')"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 7.3: Calculate_the_value_of_capacitor_required.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Calculate the value of capacitor required\n",
+"clear;\n",
+"clc;\n",
+"//soltion\n",
+"//given\n",
+"\n",
+"Rs1=20*10^3;//ohm\n",
+"Rs2=30*10^3;//ohm\n",
+"Rl1=40*10^3;//ohm\n",
+"Rl2=80*10^3;//ohm\n",
+"Rl3=80*10^3;//ohm\n",
+"Rth=Rs1*Rs2/(Rs1+Rs2);          //Thevenin's equivalent resistance\n",
+"Rl_=Rl2*Rl3/(Rl2+Rl3);\n",
+"Rl=Rl1*Rl_/(Rl1+Rl_);    //Equivalent load\n",
+"f=50;//Hz          //lowest frequency\n",
+"R=Rth+Rl;\n",
+"C=10/(2*%pi*f*R)\n",
+"printf('The required value of coupling capacitor is %.0f uF',C*10^6);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 7.4: Calculate_voltage_and_current_gain_and_input_and_output_resistance.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Calculate voltage and current gain and input and output resistance\n",
+"clear;\n",
+"clc;\n",
+"//soltion\n",
+"\n",
+"function [z]=prll(r1,r2)//Function for the parallel combination of resistor\n",
+"    z=r1*r2/(r1+r2);\n",
+"endfunction\n",
+"\n",
+"//given\n",
+"\n",
+"//DC analysis\n",
+"Vcc=12;//V\n",
+"Rb=200*10^3;//ohm\n",
+"Rc=1*10^3;//ohm\n",
+"B=100;// beta\n",
+"Ib=Vcc/Rb;\n",
+"Ic=B*Ib;\n",
+"Icsat=Vcc/Rc;\n",
+"Vce=Vcc-Ic*Rc;\n",
+"printf('The Q point of DC analysis= (%.0f V, %.0f mA)\n',Vce,Ic*1000);\n",
+"\n",
+"//AC analysis\n",
+"Rl=1*10^3;//ohm\n",
+"hfe=B;\n",
+"hie=2*10^3;//ohm\n",
+"hoe_1=40*10^3;//ohm      // 1/hoe\n",
+"Rac=prll(Rc,Rl);\n",
+"Av=-hfe*Rac/hie;\n",
+"printf('The voltage gain = %.0f\n',Av);\n",
+"\n",
+"//Siince (1/hoe) > Rac therefore entire current will flows through Rac\n",
+"Io=-100*Ib;\n",
+"Ac=Io/Ib;\n",
+"printf('The current gain = %.0f\n',Ac);\n",
+"\n",
+"Ri=prll(Rb,hie);\n",
+"Ro=prll(Rl,prll(Rc,hoe_1));\n",
+"printf('The input resistance= %.0f kΩ\n',Ri/1000);\n",
+"printf('The output resistance= %.1f kΩ',Ro/1000);\n",
+"\n",
+"//In book the voltage gain is 25 due to skipping of '-' in printing"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 7.5: Solve_previous_example_using_hybrid_pie_model.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Solve previous example using hybrid pie model\n",
+"clear;\n",
+"clc;\n",
+"//soltion\n",
+"function [z]=prll(r1,r2)//Function for the parallel combination of resistor\n",
+"    z=r1*r2/(r1+r2);\n",
+"endfunction\n",
+"\n",
+"//given\n",
+"\n",
+"Vcc=12;//V\n",
+"Rb=200*10^3;//ohm\n",
+"Rc=1*10^3;//ohm\n",
+"Rl=1*10^3;//ohm\n",
+"B=100;// beta\n",
+"hie=2*10^3;//ohm\n",
+"hoe_1=40*10^3;//ohm      // 1/hoe\n",
+"\n",
+"Ib=Vcc/Rb;\n",
+"Ic=B*Ib;\n",
+"Rac=prll(Rc,Rl);\n",
+"gm=Ic/(25*10^-3);\n",
+"rpi=B/gm;\n",
+"ri=hie;\n",
+"rb=ri-rpi;\n",
+"ro=hoe_1;\n",
+"Vi=poly(0,'Vi');      //let the input be Vi\n",
+"Vpi=Vi*rpi/(rpi+rb);\n",
+"Vo=-gm*Vpi*Rac;    //output voltage\n",
+"Av=Vo/Vi;\n",
+"printf('The voltage gain ');\n",
+"disp(Av);\n",
+"//In book voltage gain is -24.96 due to appraoximation"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 7.6: Determine_the_value_of_output_voltage.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Determine the value of output voltage\n",
+"clear;\n",
+"clc;\n",
+"//soltion\n",
+"//given\n",
+"\n",
+"Vcc=12;//V\n",
+"Rb=150*10^3;//ohm\n",
+"Rc=5*10^3;//ohm\n",
+"B=200;// beta\n",
+"hie=2*10^3;//ohm\n",
+"ro=60*10^3;//ohm      // 1/hoe\n",
+"Vi=1*10^-3;//V\n",
+"Ib=Vcc/Rb;\n",
+"Ic=B*Ib;\n",
+"Icsat=Vcc/Rc;\n",
+"// Icsat < Ic therefore transistor is in saturation mode and outpuut voltage wil be zero\n",
+"Vo=0;\n",
+"printf('The output voltage= %.0f V',Vo);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 7.7: Calculate_voltage_gain_and_input_resistance.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Calculate voltage gain and input resistance\n",
+"clear;\n",
+"clc;\n",
+"//soltion\n",
+"\n",
+"function [z]=prll(r1,r2)//Function for the parallel combination of resistor\n",
+"    z=r1*r2/(r1+r2);\n",
+"endfunction\n",
+"\n",
+"//given\n",
+"R1=75*10^3;//ohm\n",
+"R2=7.5*10^3;//ohm\n",
+"Rc=4.7*10^3;//ohm\n",
+"Re=1.2*10^3;//ohm\n",
+"Rl=12*10^3;//ohm\n",
+"B=150;\n",
+"ri=2*10^3;//ohm\n",
+"Vcc=15;//V\n",
+"Vb=Vcc*R2/(R1+R2);\n",
+"Ve=Vb;        //since Vbe=0\n",
+"Ie=Ve/Re;\n",
+"Ic=Ie;\n",
+"Icsat=Vcc/(Rc+Re);\n",
+"// Ic < Icsat therefore transistor is in active mode\n",
+"Vce=Vcc-Ic*(Rc+Re);\n",
+"printf('The Q point of DC analysis= (%.1f V, %.3f mA)\n',Vce,Ic*1000);\n",
+"\n",
+"Rac=prll(Rc,Rl);\n",
+"Av=-B*Rac/ri;\n",
+"printf('The voltage gain = %.1f\n',Av);\n",
+"Ri_=prll(ri,R2);\n",
+"printf('The input resistance= %.2f kΩ\n',Ri_/1000);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 7.8: Calculate_the_value_of_gm_at_different_values_of_Vgs.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"\n",
+"//Calculate the value of gm at different values of Vgs\n",
+"clear;\n",
+"clc;\n",
+"//soltion\n",
+"//given\n",
+"\n",
+"Idss=8*10^-3;//A\n",
+"Vp=4;//V\n",
+"//At Vgs= -0.5 V\n",
+"Vgs= -0.5;//V\n",
+"gmo=2*Idss/(abs(Vp));\n",
+"gm=gmo*(1-(Vgs/(-Vp)));\n",
+"printf('gmo = %.0f mS\n',gmo*1000);\n",
+"printf('gm (At Vgs = -0.5V) =%.1f mS\n',gm*1000);\n",
+"\n",
+"//At Vgs= -1.5 V\n",
+"Vgs= -1.5;//V\n",
+"gmo=2*Idss/(abs(Vp));\n",
+"gm=gmo*(1-(Vgs/(-Vp)));\n",
+"printf('gm (At Vgs = -1.5V) =%.1f mS\n',gm*1000);\n",
+"\n",
+"//At Vgs= -2.5 V\n",
+"Vgs= -2.5;//V\n",
+"gmo=2*Idss/(abs(Vp));\n",
+"gm=gmo*(1-(Vgs/(-Vp)));\n",
+"printf('gm (At Vgs = -2.5V) =%.1f mS\n',gm*1000);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 7.9: Find_the_output_signal_voltage_of_the_amplifier.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Find the output signal voltage of the amplifier\n",
+"clear;\n",
+"clc;\n",
+"//soltion\n",
+"//given\n",
+"\n",
+"Rd=12*10^3;//ohm\n",
+"Rg=1*10^6;//ohm\n",
+"Rs=1*10^3;//ohm\n",
+"Cs=25*10^-6;//F\n",
+"u=80;      //amplification factor\n",
+"rd=200*10^3;//ohm\n",
+"Vi=0.1;//V\n",
+"f=1*10^3;//Hz      //input frequency\n",
+"Xcs=1/(2*%pi*f*Cs);\n",
+"//This is much smaller than Rs therefore it is bypassed\n",
+"\n",
+"gm=u/rd;\n",
+"Av=gm*(rd*Rd/(rd+Rd));\n",
+"Vo=Av*Vi;\n",
+"printf('The output voltage is %.3f V',Vo);"
+   ]
+   }
+],
+"metadata": {
+		  "kernelspec": {
+		   "display_name": "Scilab",
+		   "language": "scilab",
+		   "name": "scilab"
+		  },
+		  "language_info": {
+		   "file_extension": ".sce",
+		   "help_links": [
+			{
+			 "text": "MetaKernel Magics",
+			 "url": "https://github.com/calysto/metakernel/blob/master/metakernel/magics/README.md"
+			}
+		   ],
+		   "mimetype": "text/x-octave",
+		   "name": "scilab",
+		   "version": "0.7.1"
+		  }
+		 },
+		 "nbformat": 4,
+		 "nbformat_minor": 0
+}
diff --git a/Electronic_Devices_and_Circuits_by_D_C_Kulshreshtha/8-Multistage_Amplifiers.ipynb b/Electronic_Devices_and_Circuits_by_D_C_Kulshreshtha/8-Multistage_Amplifiers.ipynb
new file mode 100644
index 0000000..5d5517d
--- /dev/null
+++ b/Electronic_Devices_and_Circuits_by_D_C_Kulshreshtha/8-Multistage_Amplifiers.ipynb
@@ -0,0 +1,340 @@
+{
+"cells": [
+ {
+		   "cell_type": "markdown",
+	   "metadata": {},
+	   "source": [
+       "# Chapter 8: Multistage Amplifiers"
+	   ]
+	},
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 8.1: Express_the_gain_in_decibel.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Express the gain in decibel\n",
+"clear;\n",
+"clc;\n",
+"//soltion\n",
+"//given\n",
+"\n",
+"//Powere gain of 1000\n",
+"Pg1=1000;\n",
+"Pgd1=10*log10(Pg1);\n",
+"printf('Power gain (in dB)= %.0f dB\n',Pgd1);\n",
+"\n",
+"//Voltage gain of 1000\n",
+"Vg1=1000;\n",
+"Vgd1=20*log10(Vg1);\n",
+"printf('Voltage gain (in dB)= %.0f dB\n',Vgd1);\n",
+"\n",
+"//Powere gain of 1/100\n",
+"Pg2=1/100;\n",
+"Pgd2=10*log10(Pg2);\n",
+"printf('Power gain (in dB)= %.0f dB\n',Pgd2);\n",
+"\n",
+"//Voltage gain of 1/100\n",
+"Vg2=1/100;\n",
+"Vgd2=20*log10(Vg2);\n",
+"printf('Voltage gain (in dB)= %.0f dB\n',Vgd2);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 8.2: Determine_power_and_voltage_gain.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"\n",
+"//Determine power and voltage gain\n",
+"clear;\n",
+"clc;\n",
+"//soltion\n",
+"//given\n",
+"\n",
+"//For  Gain = 10 dB\n",
+"G=10;//dB\n",
+"Pg1=10^(G/10);       //taking antilog\n",
+"Vg1=10^(G/20);       //taking antilog\n",
+"printf('\nFor Gain = %.0f dB',G)\n",
+"printf('\nPower gain ratio = %.0f \n',Pg1);\n",
+"printf('Voltage gain ratio = %.2f \n',Vg1);\n",
+"\n",
+"//For Gain 3 dB\n",
+"G=3;//dB\n",
+"Pg2=10^(G/10);       //taking antilog\n",
+"Vg2=10^(G/20);       //taking antilog\n",
+"printf('\nFor Gain = %.0f dB\n',G)\n",
+"printf('Power gain ratio = %.0f \n',Pg2);\n",
+"printf('Voltage gain ratio = %.3f \n',Vg2);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 8.3: Calculate_the_overall_voltage_gai.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Calculate the overall voltage gain\n",
+"clear;\n",
+"clc;\n",
+"//soltion\n",
+"//given\n",
+"\n",
+"A1=80\n",
+"A2=50\n",
+"A3=30\n",
+"Ad=20*log10(A1)+20*log10(A2)+20*log10(A3);;\n",
+"\n",
+"//Alternatively\n",
+"A=A1*A2*A3;\n",
+"Ad=20*log10(A);\n",
+"printf('The Voltage gain is %.2f dB',Ad);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 8.4: Calculate_quiescent_output_voltage_and_small_signal_voltage_gain.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Calculate quiescent output voltage and small signal voltage gain\n",
+"clear;\n",
+"clc;\n",
+"//soltion\n",
+"//given\n",
+"\n",
+"//At input Voltage =3V\n",
+"Vi1=3;//V      //input voltage\n",
+"Vbe=0.7;//V\n",
+"B=250;\n",
+"Vcc=10;//V     //Supply\n",
+"Re1=1*10^3;//ohm\n",
+"Rc1=3*10^3;//ohm\n",
+"Re2=2*10^3;//ohm\n",
+"Rc2=4*10^3;//ohm\n",
+"Vb1=Vi1;        //Voltage at the base of transistor T1\n",
+"Ve1=Vb1-Vbe;      //Voltage at the emitter of transistor T1\n",
+"Ie1=Ve1/Re1;\n",
+"Ic1=Ie1;\n",
+"Vc1=Vcc-Ic1*Rc1;\n",
+"Vb2=Vc1;\n",
+"Ve2=Vb2-Vbe;\n",
+"Ie2=Ve2/Re2;\n",
+"Ic2=Ie2;\n",
+"Vo1=Vcc-Ic2*Rc2;\n",
+"printf('The quiescent output voltage(At input Voltage =3 V ) is %.1f V\n',Vo1);\n",
+"\n",
+"//At input Voltage =3.2 V\n",
+"Vi2=3.2;//V      //input voltage\n",
+"Vb1=Vi2;        //Voltage at the base of transistor T1\n",
+"Ve1=Vb1-Vbe;      //Voltage at the emitter of transistor T1\n",
+"Ie1=Ve1/Re1;\n",
+"Ic1=Ie1;\n",
+"Vc1=Vcc-Ic1*Rc1;\n",
+"Vb2=Vc1;\n",
+"Ve2=Vb2-Vbe;\n",
+"Ie2=Ve2/Re2;\n",
+"Ic2=Ie2;\n",
+"Vo2=Vcc-Ic2*Rc2;\n",
+"printf('The quiescent output voltage (At input Voltage =3.2 V) is %.1f V\n',Vo2);\n",
+"\n",
+"//Small Signal input and output voltage\n",
+"vi=Vi2-Vi1;\n",
+"vo=Vo2-Vo1;\n",
+"Av=vo/vi;\n",
+"printf('The small signal voltage gain is %.0f ',Av)"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 8.5: Calculate_the_maximum_voltage_gain_and_bandwidth_of_multistage_amplifier.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Calculate the maximum voltage gain and bandwidth of multistage amplifier\n",
+"clear;\n",
+"clc;\n",
+"//soltion\n",
+"//FUNCTIONS\n",
+"\n",
+"function [z]=prll(r1,r2)//Function for the parallel combination of resistor\n",
+"    z=r1*r2/(r1+r2);\n",
+"endfunction\n",
+"\n",
+"//given\n",
+"rin=10*10^6;//ohm         //input resistance of JFET\n",
+"Rd=10*10^3;//ohm\n",
+"Rs=500;//ohm\n",
+"Rg=470*10^3;//ohm\n",
+"Rl=470*10^3;//ohm\n",
+"Cc=0.01*10^-6;//Farad\n",
+"Csh=100*10^-12;//Farad\n",
+"Cs=50*10^-6;//Farad\n",
+"rd=100*10^3;//ohm\n",
+"gm=2*10^-3;//S\n",
+"Rac2=prll(Rd,Rl);\n",
+"Rac1=prll(Rd,Rg);\n",
+"Req=prll(rd,prll(Rd,Rl));\n",
+"Am=ceil(gm*Req);\n",
+"Am2=Am*Am;       //Voltage gain of two stage amplifier\n",
+"printf('Voltage gain of two stage amplifier= %.0f\n',Am2);\n",
+"R_=prll(rd,Rd)+prll(Rg,rin);\n",
+"f1=1/(2*%pi*Cc*R_);      //lower cutoff frequency\n",
+"f1_=f1/(sqrt(sqrt(2)-1));\n",
+"f2=1/(2*%pi*Csh*Req);      //upper cutoff frequency\n",
+"f2_=f2*(sqrt(sqrt(2)-1));\n",
+"BW=f2_-f1_;\n",
+"printf('Bandwidth= %.1f kHz',BW/1000);\n",
+"\n",
+"//There is a slight error in f1 due to use of R'(here R_)=479 kΩ and in f2 due to approaximation of Req there is a slight variation"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 8.6: Calculate_the_midband_voltage_gain_and_bandwidth_of_cascade_amplifier.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Calculate the midband voltage gain and bandwidth of cascade amplifier\n",
+"clear;\n",
+"clc;\n",
+"//soltion\n",
+"//given\n",
+"Am=8;         //midband voltage gain of individual MOSFET\n",
+"BW=500*10^3//Hz\n",
+"f2=BW;\n",
+"n=4;\n",
+"A2m=Am^n;\n",
+"f2_=f2*(sqrt((2^(1/n))-1));\n",
+"printf('Midband voltage gain = %.0f\n',A2m);\n",
+"printf('Overall Bandwidth= %.1f kHz',f2_/1000);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 8.7: Calculate_the_input_and_output_impedance_and_voltage_gain.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Calculate the input and output impedance and voltage gain\n",
+"clear;\n",
+"clc;\n",
+"//soltion\n",
+"//FUNCTIONS\n",
+"\n",
+"function [z]=prll(r1,r2)//Function for the parallel combination of resistor\n",
+"    z=r1*r2/(r1+r2);\n",
+"endfunction\n",
+"\n",
+"hie=1.1*10^3;//ohm      = rin\n",
+"hfe=120;//    = B\n",
+"\n",
+"//the values of Rac2, Zi, Zo are as per diagram\n",
+"Rac2=prll(3.3*10^3,2.2*10^3);\n",
+"Rac1=prll(6.8*10^3,prll(56*10^3,prll(5.6*10^3,1.1*10^3)));\n",
+"Zi=prll(5.6*10^3,prll(56*10^3,1.1*10^3));\n",
+"Zo=prll(3.3*10^3,2.2*10^3);\n",
+"\n",
+"printf('Input Resistance = %.3f kΩ\nOutput Resistance = %.2f kΩ\n',Zi/1000,Zo/1000);\n",
+"\n",
+"Am2=-hfe*Rac2/(hie);\n",
+"Am1=-hfe*Rac1/(hie);\n",
+"Am=Am1*Am2;\n",
+"Am=20*log10(Am);\n",
+"printf('The Overall Voltage gain is %.2f dB',Am);"
+   ]
+   }
+],
+"metadata": {
+		  "kernelspec": {
+		   "display_name": "Scilab",
+		   "language": "scilab",
+		   "name": "scilab"
+		  },
+		  "language_info": {
+		   "file_extension": ".sce",
+		   "help_links": [
+			{
+			 "text": "MetaKernel Magics",
+			 "url": "https://github.com/calysto/metakernel/blob/master/metakernel/magics/README.md"
+			}
+		   ],
+		   "mimetype": "text/x-octave",
+		   "name": "scilab",
+		   "version": "0.7.1"
+		  }
+		 },
+		 "nbformat": 4,
+		 "nbformat_minor": 0
+}
diff --git a/Electronic_Devices_and_Circuits_by_D_C_Kulshreshtha/9-Power_Amplifiers.ipynb b/Electronic_Devices_and_Circuits_by_D_C_Kulshreshtha/9-Power_Amplifiers.ipynb
new file mode 100644
index 0000000..0e42087
--- /dev/null
+++ b/Electronic_Devices_and_Circuits_by_D_C_Kulshreshtha/9-Power_Amplifiers.ipynb
@@ -0,0 +1,242 @@
+{
+"cells": [
+ {
+		   "cell_type": "markdown",
+	   "metadata": {},
+	   "source": [
+       "# Chapter 9: Power Amplifiers"
+	   ]
+	},
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 9.1: Determine_the_turns_ratio_of_the_transformer.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Determine the turns ratio of the transformer\n",
+"clear;\n",
+"clc;\n",
+"//soltion\n",
+"//given\n",
+"\n",
+"Rl=8;//ohm\n",
+"Rl_=5*10^3;//ohm\n",
+"TR=sqrt(Rl_/Rl);       //Turns ratio\n",
+"printf('Turns Ratio= %.0f : 1',TR);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 9.2: Determine_the_output_impedance_of_the_transistor.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Determine the output impedance of the transistor\n",
+"clear;\n",
+"clc;\n",
+"//soltion\n",
+"//given\n",
+"\n",
+"TR=16/1;    //turn ratio\n",
+"Rl=4;//ohm   //loudspeaker impedance\n",
+"ro=(TR^2)*Rl;\n",
+"printf('The output impedance of the transistor %.0f Ω',ro);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 9.3: Determine_the_efficiency_of_a_single_ended_transformer.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"\n",
+"//Determine the efficiency of a single ended transformer\n",
+"clear;\n",
+"clc;\n",
+"//soltion\n",
+"//given\n",
+"\n",
+"Vceq=10;//V     //supply voltage\n",
+"\n",
+"//At Vp=10V\n",
+"Vp=10;//V\n",
+"Vce_max1=Vceq+Vp;\n",
+"Vce_min1=Vceq-Vp;\n",
+"n1=50*((Vce_max1-Vce_min1)/(Vce_max1+Vce_min1))^2;\n",
+"printf('Efficiency (At Vp = 10V)= %.0f percent\n',n1);\n",
+"\n",
+"//At Vp=5V\n",
+"Vp=5;//V\n",
+"Vce_max2=Vceq+Vp;\n",
+"Vce_min2=Vceq-Vp;\n",
+"n2=50*((Vce_max2-Vce_min2)/(Vce_max2+Vce_min2))^2;\n",
+"printf('Efficiency (At Vp = 5V)= %.1f percent\n',n2);\n",
+"\n",
+"//At Vp=1V\n",
+"Vp=1;//V\n",
+"Vce_max3=Vceq+Vp;\n",
+"Vce_min3=Vceq-Vp;\n",
+"n3=50*((Vce_max3-Vce_min3)/(Vce_max3+Vce_min3))^2;\n",
+"printf('Efficiency (At Vp = 1V)= %.1f percent\n',n3);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 9.4: Determine_input_and_output_power_and_efficiency.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Determine input and output power and efficiency\n",
+"clear;\n",
+"clc;\n",
+"//soltion\n",
+"//given\n",
+"\n",
+"Vcc=20;//V     //supply voltage\n",
+"Rl=4;//Ω\n",
+"Vp=15;//V\n",
+"Ip=Vp/Rl;\n",
+"Idc=Ip/%pi;\n",
+"Pi=Vcc*Idc;\n",
+"Po=((Vp/2)^2)/Rl;\n",
+"n=100*Po/Pi;\n",
+"printf('Input power %.2f W\n',Pi);\n",
+"printf('Output power %.2f W\n',Po);\n",
+"printf('Efficiency = %.0f percent\n',n);\n",
+""
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 9.5: Calculate_the_percentage_increase_in_output_power.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Calculate the percentage increase in output power\n",
+"clear;\n",
+"clc;\n",
+"//soltion\n",
+"//given\n",
+"\n",
+"D=0.2;//harmonic distortion\n",
+"P=(1+D^2);//Total power increase\n",
+"\n",
+"//percent increase= (Pi*(1+D^2)-Pi)*100/Pi;\n",
+"//taking out and cancelling Pi\n",
+"PI=(P-1)*100;\n",
+"printf('The percentage increase in output power= %.0f percent',PI);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 9.6: EX9_6.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Calculate harmonic distortion and percentage increase in output voltage due to this\n",
+"clear;\n",
+"clc;\n",
+"//soltion\n",
+"//given\n",
+"\n",
+"I1=60;//A\n",
+"I2=6;//A\n",
+"I3=1.2;//A\n",
+"I4=0.6;//A\n",
+"D2=I2/I1;\n",
+"D3=I3/I1;\n",
+"D4=I4/I1;\n",
+"printf('The Harmonic distortion of each component\nD2= %.0f percent\nD3= %.0f percent\nD4= %.0f percent\n',D2*100,D3*100,D4*100);\n",
+"\n",
+"D=sqrt((D2)^2+(D3)^2+(D4)^2);\n",
+"printf('The Total Harmonic distortion = %.0f percent\n',D*100);\n",
+"P=(1+D^2);//Total power increase\n",
+"\n",
+"//percent increase= (Pi*(1+D^2)-Pi)*100/Pi;\n",
+"//taking out and cancelling Pi\n",
+"PI=(P-1)*100;\n",
+"printf('The percentage increase in output power= %.0f percent',PI);"
+   ]
+   }
+],
+"metadata": {
+		  "kernelspec": {
+		   "display_name": "Scilab",
+		   "language": "scilab",
+		   "name": "scilab"
+		  },
+		  "language_info": {
+		   "file_extension": ".sce",
+		   "help_links": [
+			{
+			 "text": "MetaKernel Magics",
+			 "url": "https://github.com/calysto/metakernel/blob/master/metakernel/magics/README.md"
+			}
+		   ],
+		   "mimetype": "text/x-octave",
+		   "name": "scilab",
+		   "version": "0.7.1"
+		  }
+		 },
+		 "nbformat": 4,
+		 "nbformat_minor": 0
+}