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+{
+"cells": [
+ {
+		   "cell_type": "markdown",
+	   "metadata": {},
+	   "source": [
+       "# Chapter 6: Bipolar junction transistor"
+	   ]
+	},
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 6.10: determine_IC_and_IB.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Example 6.10\n",
+"clc\n",
+"format(7)\n",
+"IE=12\n",
+"beta=100\n",
+"IB=IE/(1+beta)\n",
+"disp(IB,'We know that base current,  IB(mA) = IE / (1 + beta) = ')\n",
+"format(8)\n",
+"IC=IE-IB\n",
+"disp(IC,'and collector current,  IC(mA) = IE - IB = ')"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 6.11: beta_and_alpha_and_IE.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Example 6.11\n",
+"clc\n",
+"format(6)\n",
+"IB=100*10^-6\n",
+"IC=2*10^-3\n",
+"beta=IC/IB\n",
+"disp('(a) To find beta of the transistor ')\n",
+"disp(beta,'beta = IC / IB =')\n",
+"alpha=beta/(beta+1)\n",
+"disp('(b) To find alpha of the transistor')\n",
+"disp(alpha,'alpha = beta / (1+beta) =')\n",
+"IE=IB+IC\n",
+"IE1=IE*10^3\n",
+"disp('(c) To find emitter current, IE')\n",
+"disp(IE1,'IE(mA) = IB + IC =')  // answer in the textbook is wrong\n",
+"disp('(d) To find the new value of beta when delta_IB = 25uA and delta_IC = 0.6mA')\n",
+"delta_IB=25*10^-6\n",
+"delta_IC=0.6*10^-3\n",
+"IB1=IB+delta_IB\n",
+"IB11=IB1*10^6\n",
+"IC1=IC+delta_IC\n",
+"IC11=IC1*10^3\n",
+"disp(IB11,'Therefore,  IB(uA) = ')\n",
+"disp(IC11,'            IC(mA) = ')\n",
+"beta1=IC1/IB1\n",
+"disp('New value of beta of the transistor,')\n",
+"disp(beta1,'beta = IC / IB = ')"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 6.12: find_IC_and_IE.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Example 6.12.\n",
+"clc\n",
+"format(6)\n",
+"alpha=0.98\n",
+"ICO=5*10^-6\n",
+"ICBO=ICO\n",
+"IB=100*10^-6\n",
+"IC=((alpha*IB)/(1-alpha))+(ICO/(1-alpha))\n",
+"IC1=IC*10^3\n",
+"disp(IC1,'The collector current is, IC(mA) = ((alpha*IB)/(1-alpha))+(ICO/(1-alpha))')\n",
+"IE=IB+IC\n",
+"IE1=IE*10^3\n",
+"disp(IE1,'The emitter current is, IE(mA) = IB + IC = ')"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 6.13: IC_and_new_collector_current.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Example 6.13.\n",
+"clc\n",
+"format(6)\n",
+"ICBO=10*10^-6\n",
+"hFE=50\n",
+"beta=hFE\n",
+"IB=0.25*10^-3\n",
+"IC=(beta*IB)+((1+beta)*ICBO)\n",
+"IC1=IC*10^3\n",
+"disp('(a) To find the value of collector current when IB = 0.25mA')\n",
+"disp(IC1,'IC(mA) = (beta*IB) + ((1+beta)*ICBO)')\n",
+"T1=27\n",
+"T2=50\n",
+"format(5)\n",
+"I_CBO = ICBO * (2^((T2-T1)/10))\n",
+"I_CBO1=I_CBO*10^6\n",
+"disp('(b) To find the value of new collector current if temperature rises to 50 C')\n",
+"disp(I_CBO1,'I''CBO(beta=50)(in uA) = ICBO*(2^((T2-T1)/10)) = ')\n",
+"format(6)\n",
+"IC2=(beta*IB)+((1+beta)*I_CBO)\n",
+"IC3=IC2*10^3\n",
+"disp('Therefore, the collector current at 50 C is')\n",
+"disp(IC3,'IC(mA) = (beta*IB) + ((1+beta)*I''CBO) = ')"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 6.14: find_the_current_gain.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Example 6.14.\n",
+"clc\n",
+"format(6)\n",
+"delta_IC=0.99*10^-3\n",
+"delta_IE=1*10^-3\n",
+"alpha=delta_IC/delta_IE\n",
+"disp(alpha,'The current gain of the transistor is alpha = delta_IC/delta_IE = ')"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 6.15: dc_current_gain_in_CB_mode.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Example 6.15\n",
+"clc\n",
+"format(5)\n",
+"beta_dc=100\n",
+"alpha_dc=beta_dc/(1+beta_dc)\n",
+"disp(alpha_dc,'The d.c. current gain of the transistor in CB mode is, alpha_dc = beta_dc/(1+beta_dc) = ')"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 6.16: current_gain_alpha_and_beta.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Example 6.16.\n",
+"clc\n",
+"format(6)\n",
+"delta_IC=0.995*10^-3\n",
+"delta_IE=1*10^-3\n",
+"alpha=delta_IC/delta_IE\n",
+"disp(alpha,'Common base current gain is, alpha = delta_IC/delta_IE = ')\n",
+"beta=alpha/(1-alpha)\n",
+"disp(beta,'Common-emitter current gain is beta = alpha / (1-alpha) = ')"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 6.17: current_gain_and_base_current.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Example 6.17.\n",
+"clc\n",
+"format(6)\n",
+"beta=49\n",
+"alpha=beta/(1+beta)\n",
+"disp('We know that,  alpha = beta/(1+beta)')\n",
+"disp(alpha,'Therefore, the common base current gain is, alpha = ')\n",
+"disp('We also know that, alpha = IC / IE')\n",
+"IE=3*10^-3\n",
+"IC=alpha*IE\n",
+"IC1=IC*10^3\n",
+"disp(IC1,'Therefore,  IC(mA) = alpha * IE = ')"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 6.18: determine_IC_and_IE_and_alpha.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Example 6.18.\n",
+"clc\n",
+"format(6)\n",
+"IB=15*10^-6\n",
+"beta=150\n",
+"IC=beta*IB\n",
+"IC1=IC*10^3\n",
+"disp(IC1,'The collector current, IC(mA) = beta * IB = ')\n",
+"IE=IC+IB\n",
+"IE1=IE*10^3\n",
+"disp(IE1,'The emitter current, IE(mA) = IC + IB = ')\n",
+"format(7)\n",
+"alpha=beta/(1+beta)\n",
+"disp(alpha,'Common-base current gain, alpha = beta/(1+beta) = ')"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 6.19: IB_IC_IE_and_VCE.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Example 6.19.\n",
+"clc\n",
+"format(6)\n",
+"disp('Referring to fig.6.18, the base current is,')\n",
+"VBB=4\n",
+"VBE=0.7\n",
+"RB=200*10^3\n",
+"IB=(VBB-VBE)/RB\n",
+"IB1=IB*10^6\n",
+"disp(IB1,'IB(uA) = (VBB - VBE) / RB = ')\n",
+"beta=200\n",
+"IC=beta*IB\n",
+"IC1=IC*10^3\n",
+"disp(IC1,'The collector current is, IC(mA) = beta*IB = ')\n",
+"format(7)\n",
+"IE=IC+IB\n",
+"IE1=IE*10^3\n",
+"disp(IE1,'The emitter current is, IE(mA) = IC + IB = ')\n",
+"format(6)\n",
+"VCC=10\n",
+"RC=2*10^3\n",
+"VCE=VCC-(IC*RC)\n",
+"disp(VCE,'Therefore,  VCE(V) = VCC - IC*RC = ')"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 6.1: find_value_of_the_base_current_IB.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Example 6.1.\n",
+"clc\n",
+"format(5)\n",
+"IE=10\n",
+"IC=9.8\n",
+"disp('The emitter current is,')\n",
+"disp('IE = IB + IC')\n",
+"disp('10 = IB + 9.8')\n",
+"IB=IE-IC\n",
+"disp(IB,'Therefore,    IB(mA) = ')"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 6.20: calculate_IC_and_IE.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Example 6.20.\n",
+"clc\n",
+"format(6)\n",
+"alpha_dc=0.99\n",
+"ICBO=5*10^-6\n",
+"IB=20*10^-6\n",
+"IC=((alpha_dc*IB)/(1-alpha_dc))+(ICBO/(1-alpha_dc))\n",
+"IC1=IC*10^3\n",
+"disp(IC1,'IC(mA) = ((alpha_dc*IB)/(1-alpha_dc)) + (ICBO/(1-alpha_dc)) = ')\n",
+"IE=IB+IC\n",
+"IE1=IE*10^3\n",
+"disp(IE1,'Therefore,  IE(mA)= IB + IC = ')"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 6.21: alpha_dc_and_beta_dc.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Example 6.21.\n",
+"clc\n",
+"format(6)\n",
+"ICBO=0.2*10^-6\n",
+"ICEO=18*10^-6\n",
+"IB=30*10^-3\n",
+"disp('The leakage current  ICBO = 0.2 uA')\n",
+"disp('                     ICEO = 18 uA')\n",
+"disp('Assume that          IB = 30 mA')\n",
+"disp('IE = IB + IC')\n",
+"disp('IC = IE - IB = (beta*IB)+((1+beta)*ICBO)')\n",
+"disp('We know that,  ICEO = ICBO/(1-alpha) = (1+beta)*ICBO')\n",
+"beta=(ICEO/ICBO)-1\n",
+"disp(beta,'beta = (ICEO / ICBO)-1 = ')\n",
+"IC=(beta*IB)+((1+beta)*ICBO)\n",
+"disp(IC,'IC(A) = (beta*IB) + ((1+beta)*ICBO) = ')\n",
+"alpha_dc=1-(ICBO/ICEO)\n",
+"disp(alpha_dc,'alpha_dc = 1 - (ICBO / ICEO) = ')\n",
+"format(4)\n",
+"beta_dc=(IC-ICBO)/(IB-ICEO)\n",
+"disp(beta_dc,'beta_dc = (IC-ICBO) / (IB-ICEO) = ')"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 6.22: find_emitter_current.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Example 6.22.\n",
+"clc\n",
+"format(6)\n",
+"alpha_dc=0.99\n",
+"ICBO=50*10^-6\n",
+"IB=1*10^-3\n",
+"IC=((alpha_dc*IB)/(1-alpha_dc))+(ICBO/(1-alpha_dc))\n",
+"IC1=IC*10^3\n",
+"disp('Assume that,  IB = 1 mA')\n",
+"disp(IC1,'IC(mA) = ((alpha_dc*IB) / (1-alpha_dc)) + (ICBO/(1-alpha_dc)) = ')\n",
+"IE=IC+IB\n",
+"IE1=IE*10^3\n",
+"disp(IE1,'IE(mA) = IC + IB = ')"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 6.23: dc_and_ac_load_line_and_operating_point.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Example 6.23.refer fig.6.22(a)\n",
+"clc\n",
+"format(6)\n",
+"disp('(i) DC load line:')\n",
+"disp('Refer fig.6.22(a), we have VCC = VCE + IC*RC')\n",
+"disp('To draw the d.c. load line,we need two end points, viz. maximum VCE point(at IC = 0) and maximum IC point(at VCE = 0)')\n",
+"disp('Maximum VCE = VCC = 24V')\n",
+"IC=24/(8*10^3)  //in Ampere\n",
+"x1=IC*10^3  //in mA\n",
+"disp(x1,'Maximum  IC(mA) = VCC / RC =')\n",
+"disp('Therefore, the d.c. load line AB is drawn with the point B(OB = 24V) on the VCE axis and the point A(OA = 3mA) on the IC axis, as shown in fig.6.22(b)')\n",
+"disp('')\n",
+"disp('(ii) For fixing the optimum operating point Q, mark the middle of the d.c. load line AB and the corresponding VCE and IC values can be found')\n",
+"VCEQ=24/2\n",
+"disp(VCEQ,'Here,    VCEQ(V) = VCC / 2 =') //in volts\n",
+"disp('         ICQ = 1.5 mA')\n",
+"disp('')\n",
+"disp('(iii) AC load line')\n",
+"disp('To draw the a.c. load line, we need two end points, viz. maximum VCE and maximum IC when signal is applied')\n",
+"Rac=(8*24)/(8+24) //in k-ohm\n",
+"disp(Rac,'AC load,   R_a.c.(k-ohm) = RC || RL =')\n",
+"VCE=12+((1.5*10^-3)*(6*10^3)) //in Volts\n",
+"disp(VCE,'Therefore, maximum VCE(V) = VCEQ + ICQ*R_a.c. =')\n",
+"disp('This locates the point D(OD = 21V) on the VCE axis')\n",
+"IC=(1.5*10^-3)+(12/(6*10^3))  //in Ampere\n",
+"x3=IC*10^3  //in mA\n",
+"disp(x3,'Maximum  IC(mA) = ICQ + VCEQ/R_a.c. =')\n",
+"disp('This locates the point C(OC = 3.5mA) on the IC axis. By joining points C and D a.c. load line CD is constructed. ')\n",
+"x=[24,0]\n",
+"y=[0,3]\n",
+"plot2d(x,y,style=2)\n",
+"x1=[21,0]\n",
+"y1=[0,3.5]\n",
+"plot2d(x1,y1,style=1)\n",
+"legend('d.c. load line AB','a.c. load line CD')\n",
+"title('Fig.6.22(b)')\n",
+"xlabel('VCE(V)')\n",
+"ylabel('IC(mA)')"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 6.24: ac_and_dc_load_line_and_operating_point.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Example 6.24. refer fig.6.23(a).\n",
+"clc\n",
+"format(6)\n",
+"disp('(i) DC load line:')\n",
+"disp('Refer fig.6.23(a), we have VCC = VCE + IC*(RC+RE)')\n",
+"disp('To draw the d.c. load line,we need two end points, viz. maximum VCE point(at IC = 0) and maximum IC point(at VCE = 0)')\n",
+"disp('Maximum VCE = VCC = 12V, which locates the point B(OB = 12V) of the d.c. load line')\n",
+"IC=12/(2*10^3)  //in Ampere\n",
+"x1=IC*10^3  //in mA\n",
+"disp(x1,'Maximum  IC(mA) = VCC / (RC+RE) =')\n",
+"disp('This locates the point A(OA = 6mA) of the d.c. load line. Fig.6.23(b) shows the d.c. load line AB, with (12V,6mA)')\n",
+"disp('')\n",
+"disp('(ii) Operating point Q')\n",
+"disp('The voltage across  R2 is V2 = (R2/R1+R2)*VCC')\n",
+"V2=((4*10^3)/(12*10^3))*12  //in V\n",
+"disp(V2,'Therefore,    V2(V) =')\n",
+"disp('       V2 = VBE + IE*RE')\n",
+"IE=(4-0.7)/(1*10^3)  //in Ampere\n",
+"x2=IE*10^3  //in mA\n",
+"disp(x2,'Therefore,  IE(mA) = V2-VBE / RE =')\n",
+"IC=x2 //in mA\n",
+"disp(IC,'        IC(mA) = IE(mA) = ')\n",
+"VCE=12-((3.3*10^-3)*(2*10^3)) //in volts\n",
+"disp(VCE,'VCE(V) = VCC - IC(RC+RE) =')\n",
+"disp('Therefore, the operating point Q is at 5.4V and 3.3mA, which is shown on the d.c. load line')\n",
+"disp('')\n",
+"disp('(iii) AC load line')\n",
+"disp('To draw the a.c. load line, we need two end points, viz. maximum VCE and maximum IC when signal is applied')\n",
+"Rac=1.5/2.5 //in k-ohm\n",
+"disp(Rac,'AC load,   Ra.c.(k-ohm) = RC || RL =')\n",
+"VCE=5.4+((3.3*10^-3)*(0.6*10^3)) //in Volts\n",
+"disp(VCE,'Therefore, maximum VCE(V) = VCEQ + ICQ*Ra.c. =')\n",
+"disp('This locates the point C(OC = 6.24V) on the VCE axis')\n",
+"IC=(3.3*10^-3)+(5.4/(0.6*10^3))  //in Ampere\n",
+"x3=IC*10^3  //in mA\n",
+"disp(x3,'Maximum  IC(mA) = ICQ + VCEQ/Ra.c. =')\n",
+"disp('This locates the point D(OD = 12.3mA) on the IC axis. By joining points C and D a.c. load line CD is constructed. ')\n",
+"x=[7.38,0]\n",
+"y=[0,12.3]\n",
+"plot2d(x,y,style=2)\n",
+"x1=[12,0]\n",
+"y1=[0,6]\n",
+"plot2d(x1,y1,style=1)\n",
+"legend('a.c. load line CD','d.c. load line AB')\n",
+"title('Fig.6.23(b)')\n",
+"xlabel('VCE(V) -->')\n",
+"ylabel('IC(mA) -->')"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 6.25: Design_circuit_in_fig_6_24.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Example 6.25.\n",
+"clc\n",
+"format(6)\n",
+"ICQ=1*10^-3\n",
+"VCEQ=6\n",
+"VCC=10\n",
+"beta=100\n",
+"VBE=0.7\n",
+"RC=(VCC-VCEQ)/ICQ\n",
+"RC1=RC*10^-3\n",
+"RC2=round(RC1)\n",
+"disp(RC2,'The collector resistance is, RC(k-ohm) = (VCC - VCEQ) / ICQ = ')\n",
+"IBQ=ICQ/beta\n",
+"IBQ1=IBQ*10^6\n",
+"disp(IBQ1,'The base current is, IBQ(uA) = ICQ / beta = ')\n",
+"RB=(VCC-VBE)/IBQ\n",
+"RB1=RB*10^-6\n",
+"disp(RB1,'The base resistance is, RB(M-ohm) = (VCC - VBE(on)) / IBQ = ')"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 6.26: characteristics_circuit_in_fig_6_25.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Example 6.26.\n",
+"clc\n",
+"format(6)\n",
+"beta=100\n",
+"VBE=0.7\n",
+"VCC=10\n",
+"RB=20*10^3\n",
+"RC=0.4*10^3\n",
+"RE=0.6*10^3\n",
+"VBB=5\n",
+"disp('Referring to fig.6.25, Kirchhoff voltage law equation is,')\n",
+"disp('VBB = IB*RB + VBE(on) + IE*RE')\n",
+"disp('Also,  IE = IB + IC = IB + beta*IB = (1 + beta)*IB')\n",
+"IB=(VBB-VBE)/(RB+((1+beta)*RE))\n",
+"IB1=IB*10^6\n",
+"disp(IB1,'The base current,  IB(uA) = (VBB - VBE(on)) / (RB + ((1+beta)*RE)) = ')\n",
+"IC=beta*IB\n",
+"IC1=IC*10^3\n",
+"disp(IC1,'Therefore,  IC(mA) = beta*IB = ')\n",
+"IE=IC+IB\n",
+"IE1=IE*10^3\n",
+"disp(IE1,'IE(mA) = IC + IB')\n",
+"VCE=VCC-(IC*RC)-(IE*RE)\n",
+"disp(VCE,'VCE(V) = VCC - (IC*RC) - (IE*RE) = ')\n",
+"disp('The Q point is at')\n",
+"disp(VCE,'VCEQ(V) = ')\n",
+"disp(IC1,'and ICQ(mA) = ')"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 6.27: dc_load_line_and_operating_point_and_S.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Example 6.27. refer fig.6.26.\n",
+"clc\n",
+"format(6)\n",
+"disp('(i) DC load line:')\n",
+"disp('      VCE = VCC - IC*RC')\n",
+"disp('When  IC = 0, VCE = VCC = 6V')\n",
+"IC=6/(3*10^3) //in Ampere\n",
+"x1=IC*10^3  //in mA\n",
+"disp(x1,'When  VCE = 0, IC(mA) = VCC/RC =')\n",
+"disp('')\n",
+"disp('(ii) Operating point Q:')\n",
+"disp('     For silicon transistor,  VBE = 0.7V')\n",
+"disp('                              VCC = IB*RB + VBE')\n",
+"IB=(6-0.7)/(530*10^3)\n",
+"x2=IB*10^6\n",
+"disp(x2,'Therefore,              IB(uA) = VCC-VBE / RB =')\n",
+"IC=100*10*10^-6 // in Ampere\n",
+"x3=IC*10^3 // in mA\n",
+"disp(x3,'Therefore,              IC(mA) = beta*IB =')\n",
+"VCE=6-((1*10^-3)*(3*10^3)) // in volts\n",
+"disp(VCE,'                           VCE(V) = VCC - IC*RC =')\n",
+"disp('Therefore operating point is VCEQ = 3 V and ICQ = 1 mA')\n",
+"disp('')\n",
+"disp('(iii) Stability factor:    S = 1 + beta = 1 + 100 = 101')\n",
+"x=[6,0]\n",
+"y=[0,2]\n",
+"plot2d(x,y,style=1)\n",
+"xtitle('DC load line','VCE (V) --->','IC (mA) --->')"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 6.28: RB_and_S_and_operating_point.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Example 6.28.\n",
+"clc\n",
+"format(6)\n",
+"VCC=12\n",
+"beta=100\n",
+"VBE=0.7\n",
+"disp('Refer fig.6.26. We know that for a silicon transistor, VBE = 0.7 V')\n",
+"disp('(a) To determine RB :')\n",
+"VCE=7\n",
+"IC=1*10^-3\n",
+"RC=(VCC-VCE)/IC\n",
+"RC1=RC*10^-3\n",
+"disp(RC1,'RC(k-ohm) = (VCC - VCE) / IC = ')\n",
+"IB=IC/beta\n",
+"IB1=IB*10^6\n",
+"disp(IB1,'IB(uA) = IC / beta = ')\n",
+"RB=(VCC-VBE-(IC*RC))/IB\n",
+"RB1=RB*10^-3\n",
+"disp(RB1,'RB(k-ohm) = (VCC - VBE - (IC*RC)) / IB = ')\n",
+"S=(1+beta)/(1+(beta*(RC/(RC+RB))))\n",
+"format(5)\n",
+"disp(S,'(b) Stability factor, S =(1 + beta) / (1 + (beta*(RC / (RC+RB)))) = ')\n",
+"beta1=50\n",
+"format(6)\n",
+"disp('(c) VCC = (beta*IB*RC) + (IB*RB) + VBE')\n",
+"disp('        = IB * ((beta*RC) + RB) + VBE')\n",
+"IB=(VCC-VBE)/((beta1*RC)+RB)\n",
+"IB1=IB*10^6\n",
+"disp(IB1,'IB(uA) = (VCC-VBE) / ((beta*RC)+RB) = ')\n",
+"IC=beta1*IB\n",
+"IC1=IC*10^3\n",
+"disp(IC1,'IC(mA) = beta*IB = ')\n",
+"VCE=VCC-(IC*RC)\n",
+"disp(VCE,'VCE = VCC - (IC*RC) = ')\n",
+"disp('Therefore the coordinates of new operating point are :')\n",
+"disp(VCE,'VCEQ(V) = ')\n",
+"disp(IC1,'ICQ(mA) = ')"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 6.29: calculate_RB_and_stability_factor.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Example 6.29.\n",
+"clc\n",
+"format(6)\n",
+"VCC=12\n",
+"RC=250\n",
+"IB=0.25*10^-3\n",
+"beta=100\n",
+"VCEQ=8\n",
+"RB=VCEQ/IB\n",
+"RB1=RB*10^-3\n",
+"disp(RB1,'RB(k-ohm) = VCEQ / IB = ')\n",
+"S=(1+beta)/(1+(beta*(RC/(RC+RB))))\n",
+"disp(S,'Stability factor, S = (1+beta) / 1 + (beta*(RC/RC+RB)) = ')"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 6.2: common_base_dc_current_gain.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Example 6.2.\n",
+"clc\n",
+"format(6)\n",
+"IE=6.28\n",
+"IC=6.20\n",
+"disp('The common-base d.c. current gain,')\n",
+"alpha=IC/IE\n",
+"disp(alpha,'alpha = IC/IE =')"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 6.30: operating_point_coordinates_and_stability_factor.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Example 6.30. Refer fig. 6.27.\n",
+"clc\n",
+"format(5)\n",
+"VCC=16\n",
+"RC=3*10^3\n",
+"RE=2*10^3\n",
+"R1=56*10^3\n",
+"R2=20*10^3\n",
+"alpha=0.985\n",
+"VBE=0.3\n",
+"disp('For a germanium transistor, VBE=0.3V. As alpha=0.985')\n",
+"beta=alpha/(1-alpha)\n",
+"beta1=round(beta)\n",
+"disp(beta1,'beta = alpha / ( 1 - alpha) = ')\n",
+"disp('(a) To find the coordinates of the operating point')\n",
+"disp('Referring to fig. 6.29,')\n",
+"VT=(R2/(R1+R2))*VCC\n",
+"disp(VT,'Thevenin voltage,     VT(V) = (R2 / (R1+R2)) * VCC = ')\n",
+"format(7)\n",
+"RB=(R1*R2)/(R1+R2)\n",
+"RB1=RB*10^-3\n",
+"disp(RB1,'Thevenin resistance,  RB(k-ohm) = (R1 * R2) / (R1 + R2) =')\n",
+"disp('The loop equation around the base circuit is,')\n",
+"disp('VT = ( IB * RB) + VBE + ((IB + IC)*RE)')\n",
+"disp('VT = ((IC / beta) * RB) + VBE + (((IC / beta) + IC)*RE)')\n",
+"format(5)\n",
+"IC=(VT-VBE)/((RB/beta)+(RE/beta)+RE)\n",
+"IC1=IC*10^3\n",
+"disp(IC1,'Therefore,   IC(mA) = ')\n",
+"disp('Since IB is very small, IC ~ IE = 1.73 mA')\n",
+"IE=IC\n",
+"VCE=VCC-(IC*RC)-(IE*RE)\n",
+"disp(VCE,'Therefore,    VCE(V) = VCC - (IC*RC) - (IE*RE) = ')\n",
+"disp('Therefore, the coordinates of the operating point are :')\n",
+"disp(IC1,'IC(mA) = ')\n",
+"disp(VCE,'VCE(V) = ')\n",
+"disp('(b) To find the stability factor S')\n",
+"disp('S = (1+beta)*((1+(RB/RE))/(1+beta+(RB/RE)))')\n",
+"format(6)\n",
+"S = (1+beta)*((1+(RB/RE))/(1+beta+(RB/RE)))\n",
+"disp(S,'S = ')"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 6.31: resistors_RE_and_R1_and_R2.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Example 6.31.\n",
+"clc\n",
+"format(4)\n",
+"VCE=12\n",
+"IC=2*10^-3\n",
+"VCC=24\n",
+"VBE=0.7\n",
+"beta=50\n",
+"RC=4.7*10^3\n",
+"S=5.1\n",
+"disp('(a) To determine RE,')\n",
+"disp('VCE = VCC - (IC*RC) - (IE*RE)')\n",
+"RE = (VCC - (IC*RC) - VCE)/IC  //IC=IE\n",
+"RE1=RE*10^-3\n",
+"disp(RE1,'Therefore,  RE(k-ohm) = ')\n",
+"disp('')\n",
+"disp('(b) To determine R1 and R2,')\n",
+"disp('Stability factor,  S = (1+beta)/(1+beta(RE+(RE+RB))),    where RB = (R1*R2)/(R1+R2)')\n",
+"RB=((RE*beta)/(((1+beta)/S)-1))-RE\n",
+"RB1=(RB*10^-3)\n",
+"disp(RB1,'Therefore,  RB(k-ohm) = ((RE*beta) / (((1+beta)/S)-1)) - RE =')\n",
+"disp('Also, for a good voltage divider, the value of resistor R2 = 0.1*beta*RE')\n",
+"R2=0.1*beta*RE\n",
+"R2_1=R2*10^-3\n",
+"disp(R2_1,'Therefore,    R2(k-ohm) = ')\n",
+"disp('RB = (R1*R2) / (R1+R2)')\n",
+"R1=(5.9*10^3*R2)/(R2-(5.9*10^3)) //RB=5.9\n",
+"R1_1=round(R1*10^-3)\n",
+"disp(R1_1,'Therefore, R1(k-ohm) = R2 / ((R2/RB)-1)')"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 6.32: determine_the_Q_point.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Example 6.32. refer fig.6.30.\n",
+"clc\n",
+"format(5)\n",
+"R1=56*10^3\n",
+"R2=12.2*10^3\n",
+"RC=2*10^3\n",
+"RE=400\n",
+"VCC=10\n",
+"VBE=0.7\n",
+"beta=150\n",
+"disp('From the Thevenin equivalent circuit shown in fig.6.30(b),')\n",
+"RTH=(R1*R2)/(R1+R2)\n",
+"RTH1=round(RTH*10^-3)\n",
+"disp(RTH1,'RTH(k-ohm) = R1 || R2 =')\n",
+"VTH=(R2/(R1+R2))*VCC\n",
+"disp(VTH,'VTH(V) = (R2 / (R1+R2)) * VCC =')\n",
+"disp('By kirchhoff voltage law equation,')\n",
+"IBQ=(VTH-VBE)/(RTH+((1+beta)*RE))\n",
+"IBQ1=IBQ*10^6\n",
+"disp(IBQ1,'IBQ(uA) = (VTH-VBE(on)) / (RTH + ((1+beta)*RE)) = ')\n",
+"ICQ=beta*IBQ\n",
+"ICQ1=ICQ*10^3\n",
+"disp(ICQ1,'Therefore,    ICQ(mA) = beta * IBQ = ')\n",
+"format(6)\n",
+"IEQ=IBQ+ICQ\n",
+"IEQ1=IEQ*10^3\n",
+"disp(IEQ1,'IEQ(mA) = IBQ + ICQ')\n",
+"VCEQ=VCC-(ICQ*RC)-(IEQ*RE)\n",
+"disp(VCEQ,'VCEQ(V) = VCC - (ICQ*RC) - (IEQ*RE)')\n",
+"disp('The Q point is at :')\n",
+"disp(VCEQ,'VCEQ(V) = ')\n",
+"format(5)\n",
+"disp(ICQ1,'ICQ(mA) = ')"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 6.33: IB_IC_and_VCE_and_S.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Example 6.33. refer from fig.6.31.\n",
+"clc\n",
+"VCC=22\n",
+"RC=2*10^3\n",
+"beta=60\n",
+"VBE=0.6\n",
+"R1=100*10^3\n",
+"R2=5*10^3\n",
+"RE=100\n",
+"disp('For the given circuit')\n",
+"disp('          VCC = R1*(I1+IB) + I1*R2')\n",
+"disp('           I1 = (VCC - IB*R1) / (R1 + R2)                  Eq.1')\n",
+"disp('Further,  VCC = R1*[I1+IB] + VBE + IE*RE')\n",
+"disp('As,        IE = IC + IB')\n",
+"disp('              = beta*IB + IB = (1 + beta)*IB')\n",
+"disp('Hence,    VCC = R1*[I1 + IB] + VBE + (1 + beta)*IB*RE')\n",
+"disp('Substituting for I1 from Eq.1,')\n",
+"disp('          VCC = R1*[[(VCC - IB*R1)/R1+R2] - IB] + VBE + (1 + beta)*IB*RE')\n",
+"disp('          VCC = R1*[(VCC + IB*R2)/R1+R2] + VBE + (1 + beta)*IB*RE')\n",
+"format(6)\n",
+"a=VCC-VBE-((R1*VCC)/(R1+R2))\n",
+"c=(((R1*R2)/(R1+R2))+((1+beta)*RE))\n",
+"IB=a/c\n",
+"IB1=IB*10^6\n",
+"disp('Substituting for VCC, R1, R2, VBE, beta and RE, ')\n",
+"disp(IB1,'       IB(uA) =')\n",
+"format(5)\n",
+"IC=beta*IB\n",
+"IC1=IC*10^3\n",
+"disp(IC1,'       IC(mA) =')\n",
+"disp('Applying KVL to collector circuit,')\n",
+"disp('         VCC = IC*RC + VCE + IE*RE = IC*RC + VCE + (1+beta)*IB*RE')\n",
+"disp('Hence,   VCE = VCC - IC*RC - (1+beta)*IB*RE')\n",
+"format(7)\n",
+"VCE = VCC - (IC*RC) - ((1+beta)*IB*RE)\n",
+"disp(VCE,'     VCE(V) = ')\n",
+"disp('To find stability factor, (S):')\n",
+"disp('Stability factor for voltage divider bias is')\n",
+"format(5)\n",
+"RB=(R1*R2)/(R1+R2)\n",
+"S=(1+beta)/(1+(beta*(RE/(RE+RB))))\n",
+"disp(S,'          S =(1+beta)/(1+(beta*(RE/(RE+RB)))) =        where RB = R1 || R2')"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 6.34: Q_point_and_stability_factor.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Example 6.34.\n",
+"clc\n",
+"format(6)\n",
+"VCC=10\n",
+"RC=2*10^3\n",
+"beta=50\n",
+"RB=100*10^3\n",
+"VBE=0.7  //collector to base resistor\n",
+"disp('To determine quiescent point')\n",
+"disp('the collector to base transistor circuit')\n",
+"disp('              VCC = (beta*IB*RC) + IB*RB + VBE')\n",
+"disp('Therefore,    IB  = (VCC - VBE) / (RB + (beta*RC))')\n",
+"IB=(VCC-VBE)/(RB+(beta*RC))\n",
+"IB1=IB*10^6\n",
+"disp(IB1,'        IB(uA) =')\n",
+"IC=beta*IB\n",
+"IC1=IC*10^3\n",
+"disp(IC1,'Hence, IC(mA) = beta * IB = ')\n",
+"VCE=VCC-(IC*RC)\n",
+"disp(VCE,'       VCE(V) = VCC - IC*RC =')\n",
+"disp('Therefore,the co-ordinates of the new operating point are:')\n",
+"disp(VCE,'VCEQ(V) = ')\n",
+"disp(IC1,'ICQ(mA) = ')\n",
+"disp('To find the stability factor S')\n",
+"S=(1+beta)/(1+(beta*[RC/(RC+RB)]))\n",
+"disp(S,'S = (1+beta) / (1 + (beta*[RC/(RC+RB)])) = ')"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 6.3: find_value_of_base_current.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Example 6.3.\n",
+"clc\n",
+"format(6)\n",
+"alpha=0.967\n",
+"IE=10\n",
+"disp('The common-base d.c. current gain (alpha) is,')\n",
+"disp('alpha = 0.967 = IC/IE = IC/10')\n",
+"IC=alpha*IE\n",
+"disp(IC,'Therefore,    IC(mA) = ')\n",
+"disp('The emitter current,    IE = IB + IC')\n",
+"IB=IE-IC\n",
+"disp(IB,'Therefore,    IB(mA) = ')"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 6.4: find_values_of_IC_and_IB.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Example 6.4.\n",
+"clc\n",
+"format(6)\n",
+"IE=10\n",
+"alpha=0.98\n",
+"disp('The common-base d.c. current gain, alpha = IC/IE')\n",
+"IC=alpha*IE\n",
+"disp(IC,'Therefore,  IC(mA) = ')\n",
+"disp('The emitter current, IE = IB + IC')\n",
+"IB=IE-IC\n",
+"disp(IB,'Therefore,  IB(mA) = ')"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 6.5: find_value_of_beta_and_alpha.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Example 6.5.\n",
+"clc\n",
+"format(6)\n",
+"alpha=0.97\n",
+"disp('If alpha=0.97, beta = alpha/(1 - alpha)')\n",
+"beta=alpha/(1-alpha)\n",
+"disp(beta,'beta = ')\n",
+"beta1=200\n",
+"disp('If beta=200, alpha = beta/(beta + 1)')\n",
+"alpha1 =beta1/(beta1+1)\n",
+"disp(alpha1,'alpha = ')"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 6.6: find_value_of_emitter_current.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Example 6.6.\n",
+"clc\n",
+"format(6)\n",
+"beta=100\n",
+"IC=40\n",
+"disp('beta = 100 = IC / IB')\n",
+"IB=IC/beta\n",
+"disp(IB,'Therefore,    IB(mA) = ')\n",
+"disp('IE = IB + IC')\n",
+"IE=IB+IC\n",
+"disp(IE,'IE(mA) = ')"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 6.7: collector_and_base_currents.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Example 6.7.\n",
+"clc\n",
+"format(6)\n",
+"beta=150\n",
+"IE=10\n",
+"alpha=beta/(beta+1)\n",
+"disp(alpha,'The common-base current gain, alpha = beta / (beta + 1) = ')\n",
+"disp('Also,    alpha = IC / IE')\n",
+"format(5)\n",
+"IC=alpha*IE\n",
+"disp(IC,'Therefore,    IC(mA) = ')\n",
+"disp('the emitter current,    IE = IB + IC')\n",
+"IB=IE-IC\n",
+"disp(IB,'Therefore,    IB(mA) = ')"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 6.8: calculate_IB_and_IE.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Example 6.8.\n",
+"clc\n",
+"format(5)\n",
+"beta=170\n",
+"IC=80\n",
+"disp('We know that (beta),  beta = 170 = IC / IB')\n",
+"IB=IC/beta\n",
+"disp(IB,'Therefore,    IB(mA) = ')\n",
+"format(6)\n",
+"IE=IB+IC\n",
+"disp(IE,'and    IE(mA) = IB + IC = ')"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 6.9: determine_IC_and_IE.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Example 6.9.\n",
+"clc\n",
+"format(7)\n",
+"IB=0.125\n",
+"beta=200\n",
+"disp('beta = 200 = IC / IB')\n",
+"IC=beta*IB\n",
+"disp(IC,'Therefore,    IC(mA) = ')\n",
+"IE=IB+IC\n",
+"disp(IE,'and    IE(mA) = IB + IC = ')"
+   ]
+   }
+],
+"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
+}