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| author | hardythe1 | 2015-06-03 15:27:17 +0530 |
|---|---|---|
| committer | hardythe1 | 2015-06-03 15:27:17 +0530 |
| commit | df60071cf1d1c18822d34f943ab8f412a8946b69 (patch) | |
| tree | ab059cf19bad4a1233a464ccf5d72cf8b3fb323c /Integrated_Electronics__Analog_And_Digital_Circuits_and_Systems/Chapter16.ipynb | |
| parent | fba055ce5aa0955e22bac2413c33493b10ae6532 (diff) | |
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diff --git a/Integrated_Electronics__Analog_And_Digital_Circuits_and_Systems/Chapter16.ipynb b/Integrated_Electronics__Analog_And_Digital_Circuits_and_Systems/Chapter16.ipynb new file mode 100755 index 00000000..d113c07e --- /dev/null +++ b/Integrated_Electronics__Analog_And_Digital_Circuits_and_Systems/Chapter16.ipynb @@ -0,0 +1,626 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 16 : Integrated Circuits as Analog System Building blocks"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 16.1, Page No 621"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "\n",
+ "#Caption:Fourth Order Butterworth Filter\n",
+ "#Given Data\n",
+ "fo=1.0\t\t#Cutoff Frequency in Hz\n",
+ "#For n = 4\n",
+ "k1=0.765\n",
+ "k2=1.848\n",
+ "Av1 = 3-k1\n",
+ "Av2 = 3-k2\n",
+ "\n",
+ "#Calculations\n",
+ "print('For a fourth order Buttworth filter we cacade 2 second order Buttworth filter with parameters R1 R2 R1d R2d R C')\n",
+ "#we arbitrarily choose\n",
+ "R1=10.0 \t#in K\n",
+ "print(\"The value of R1= %.2f K \" %R1)\n",
+ "#Av1=(R1+R1d)/R1\n",
+ "R1d=(Av1*R1)-R1\n",
+ "print(\"The value of R1d= %.2f K \" %R1d)\n",
+ "\n",
+ "R2 = 10.0 #in K\n",
+ "print(\"The value of R2= %.2f K \" %R2)\n",
+ "R2d=(Av2*R2)-R2\n",
+ "print(\"The value of R2d= %.2f K \" %R2d)\n",
+ "\n",
+ "\n",
+ "#Results\n",
+ "#To satisfy fo = 1/(2*math.pi*r*c) = 1kHz\n",
+ "R=1#in K\n",
+ "C = 1/(2*math.pi*R*fo)\n",
+ "print(\"The value of R = %.2f K \" %R)\n",
+ "print(\"The value of C = %.2f microF \" %C)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "For a fourth order Buttworth filter we cacade 2 second order Buttworth filter with parameters R1 R2 R1d R2d R C\n",
+ "The value of R1= 10.00 K \n",
+ "The value of R1d= 12.35 K \n",
+ "The value of R2= 10.00 K \n",
+ "The value of R2d= 1.52 K \n",
+ "The value of R = 1.00 K \n",
+ "The value of C = 0.16 microF \n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 16.2, Page No 626"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#initialisation of variables\n",
+ "Ao=50.0 #Gain\n",
+ "fo=160.0 #center frequency\n",
+ "B=16.0 #Bandwidth in Hz\n",
+ "C1=0.1 #in microF\n",
+ "C2=0.1 #in microF\n",
+ "\n",
+ "#Required Formulae\n",
+ "\n",
+ "#Calculations\n",
+ "Q=fo/B\n",
+ "R1=(1000*Q)/(Ao*2*math.pi*fo*C1)\n",
+ "R3=(1000*Q)/((2*math.pi*fo)*(C1*C2/(C1+C2)))\n",
+ "#As C is in microFarad to compensate for it 1000 is multiplied\n",
+ "#Let r = R'\n",
+ "r=(10**6)/((2*math.pi*fo)**2*R3*C1*C2)\n",
+ "R2=(R1*r)/(R1-r)\n",
+ "\n",
+ "#Results\n",
+ "print(\"The value of R1= %.2f K \" %R1)\n",
+ "print(\"The value of R3= %.2f K \" %R3)\n",
+ "print(\"The value of r= %.2f ohm \" %(r*10**3))\n",
+ "print(\"The value of R2= %.2f ohm \" %(R2*10**3))"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of R1= 1.99 K \n",
+ "The value of R3= 198.94 K \n",
+ "The value of r= 497.36 ohm \n",
+ "The value of R2= 663.15 ohm \n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 16.3 Page No 638"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#initialisation of variables\n",
+ "\n",
+ "Avo=-25.0\n",
+ "Vagc=20.0 #in V\n",
+ "Vcc=6.0 #in V\n",
+ "hfe=50.0 \n",
+ "rbb=50.0 #in ohm\n",
+ "Cs=5.0 #in pF\n",
+ "Cl=5.0 #in pF\n",
+ "Ie1=1.0 #in mA\n",
+ "ft=900.0 #in MHz\n",
+ "Vt=26.0 #in V\n",
+ "n=2.0 #eeta\n",
+ "#re2 = infinity\n",
+ "\n",
+ "#Calculations\n",
+ "#Since Vagc=0 , transistor Q2 is in cut off region and collector current of Q1 flows through Q3....So\n",
+ "Ie2=0\n",
+ "Ie3=1.0 #in mA\n",
+ "re3 = (n*Vt)/Ie3 #in ohm\n",
+ "print(\"The value of re3 = %.2f ohm \" %re3)\n",
+ "gm = (Ie1)/Vt #in ohm^-1\n",
+ "print(\"The value of gm = %.2f ohm^-1 \" %gm)\n",
+ "rbe=hfe/gm\n",
+ "print(\"The value of rbe = %.2f ohm \" %rbe)\n",
+ "Ce=gm/(2*math.pi*ft*10**-6)\n",
+ "print(\"The value of Ce = %.2f pF \" %Ce)\n",
+ "a3=1.0 #we make an assumption that alpha is one\n",
+ "s=0\n",
+ "#Av0 = -((a3*gm)/(re3*rbb))*(1/(((1/rbb)+(1/rbe)+(s*Ce))*((1/re3)+(s*Cs))*((1/Rl)+(s*(Cs+Cl)))))\n",
+ "#From here we can find Rl\n",
+ "k = -((a3*gm)/(re3*rbb))*(1/(((1/rbb)+(1/rbe)+(s*Ce))*((1/re3)+(s*Cs))))\n",
+ "Rl=Avo/k\n",
+ "print(\"The value of Rl = %.2f ohm \" %Rl)\n",
+ "#C is in picoFarad so to compensate the whole equation some constants are multiplied\n",
+ "f1 = 1.0/(2*math.pi*Rl*(Cs+Cl)*10**-6)\n",
+ "\n",
+ "#Results\n",
+ "print(\"The value of f1 = %.2f MHz \" %f1)\n",
+ "f2 = 1.0/(2*math.pi*Ce*10**-6*((rbe*rbb)/(rbe+rbb)))\n",
+ "print(\"The value of f2 = %.2f MHz \" %f2)\n",
+ "f3 = 1.0/(2*math.pi*Cs*re3*10**-6)\n",
+ "print(\"The value of f3 = %.2f MHz \" %f3)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of re3 = 52.00 ohm \n",
+ "The value of gm = 0.04 ohm^-1 \n",
+ "The value of rbe = 1300.00 ohm \n",
+ "The value of Ce = 6.80 pF \n",
+ "The value of Rl = 675.00 ohm \n",
+ "The value of f1 = 23.58 MHz \n",
+ "The value of f2 = 486.00 MHz \n",
+ "The value of f3 = 612.13 MHz \n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 16.4a, Page No 656"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "\n",
+ "Vbb = 1.15 #in V\n",
+ "Vee=5.20 #in V\n",
+ "Vbe5=0.7 #in V\n",
+ "R=1.18 #in K\n",
+ "r=300.0 #in ohm\n",
+ "Vbecutin=0.5 #in V\n",
+ "\n",
+ "#Calculations\n",
+ "#If all inputs are low then we assume that Q1,Q2 and Q3 are cutoff and Q4 is conducting\n",
+ "Ve=-Vbb-Vbe5#Voltage at Common Emitter in V\n",
+ "#Current I in 1.18K Resistor\n",
+ "I = (Ve+Vee)/R#in mA\n",
+ "I1=I\n",
+ "print(\"Current in 300 ohm resistance I= %.2f mA \" %I)\n",
+ "#Output Voltage at Y\n",
+ "vy = -(r*I/1000)-Vbe5 #I is in mA so 1000 is multiplied\n",
+ "Vbe = vy-Ve\n",
+ "print(\"The value of Vbe is %.2f v \" %Vbe)\n",
+ "if Vbe<Vbecutin :\n",
+ " print('Input transistors are non conducting as was assumed')\n",
+ " print('If atleast one input is high then it is assumed that curent in 1.18K resistance is switched to R and Q4 is cutoff')\n",
+ " print('Drop in 300 ohm resistance is zero.Since the base aand collector are tied together Q5 now behaves as a diode')\n",
+ " print('Across Q5')\n",
+ " v=0.7#voltage across Q5 in V\n",
+ " rQ5 = 1.5#in K\n",
+ " i = (Vee-v)/rQ5\n",
+ " v = 0.75#from the graph in V\n",
+ " print(\"The value of i is %.2f mA \" %i)\n",
+ " print(\"The value of V is %.2f v \" %v)\n",
+ " Ve = -v-Vbe5\n",
+ " Vbe4=-Vbb-Ve\n",
+ " print(\"The value of Vbe4 is %.2f v \" %Vbe4)\n",
+ "\n",
+ "#Results \n",
+ "print('The total output swing between two logic gates')\n",
+ "vo = -vy-v\n",
+ "print(\"The value of vo is %.2f v \" %vo)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Current in 300 ohm resistance I= 2.84 mA \n",
+ "The value of Vbe is 0.30 v \n",
+ "Input transistors are non conducting as was assumed\n",
+ "If atleast one input is high then it is assumed that curent in 1.18K resistance is switched to R and Q4 is cutoff\n",
+ "Drop in 300 ohm resistance is zero.Since the base aand collector are tied together Q5 now behaves as a diode\n",
+ "Across Q5\n",
+ "The value of i is 3.00 mA \n",
+ "The value of V is 0.75 v \n",
+ "The value of Vbe4 is 0.30 v \n",
+ "The total output swing between two logic gates\n",
+ "The value of vo is 0.80 v \n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 16.4b Page No 656"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "Vbb = 1.15 #in V\n",
+ "Vee=5.20 #in V\n",
+ "Vbe5=0.7 #in V\n",
+ "R=1.18 #in K\n",
+ "r=300 #in ohm\n",
+ "Vbecutin=0.5 #in V\n",
+ "\n",
+ "#Calculations\n",
+ "#If all inputs are low then we assume that Q1,Q2 and Q3 are cutoff and Q4 is conducting\n",
+ "Ve=-Vbb-Vbe5 #Voltage at Common Emitter in V\n",
+ "#Current I in 1.18K Resistor\n",
+ "I = (Ve+Vee)/R#in mA\n",
+ "I1=I\n",
+ "#Output Voltage at Y\n",
+ "vy = -(r*I/1000)-Vbe5 #I is in mA so 1000 is multiplied\n",
+ "Vbe = vy-Ve\n",
+ "if Vbe<Vbecutin :\n",
+ " v=0.7 #voltage across Q5 in V\n",
+ " rQ5 = 1.5#in K\n",
+ " i = (Vee-v)/rQ5\n",
+ " v = 0.75 #from the graph in V\n",
+ " Ve = -v-Vbe5\n",
+ " Vbe4=-Vbb-Ve\n",
+ "\n",
+ "vo = -vy-v\n",
+ "\n",
+ "#Results\n",
+ "#Calculation of noise margin\n",
+ "vn = Vbecutin-Vbe4\n",
+ "print('Positive noise spike which will cause the gate to malfunction')\n",
+ "print(\"The value of vn is %.2f v \" %vn)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Positive noise spike which will cause the gate to malfunction\n",
+ "The value of vn is 0.20 v \n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 16.4c Page No 656"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#initialisation of variables\n",
+ "\n",
+ "#Verify that conducting transistor is in active region\n",
+ "#Given Data\n",
+ "Vbb = 1.15 #in V\n",
+ "Vee=5.20 #in V\n",
+ "Vbe5=0.7 #in V\n",
+ "R=1.18 #in K\n",
+ "r=300.0 #in ohm\n",
+ "Vbecutin=0.5 #in V\n",
+ "\n",
+ "#Calculations\n",
+ "#If all inputs are low then we assume that Q1,Q2 and Q3 are cutoff and Q4 is conducting\n",
+ "Ve=-Vbb-Vbe5#Voltage at Common Emitter in V\n",
+ "#Current I in 1.18K Resistor\n",
+ "I = (Ve+Vee)/R #in mA\n",
+ "I1=I\n",
+ "#Output Voltage at Y\n",
+ "vy = -(r*I/1000)-Vbe5 #I is in mA so 1000 is multiplied\n",
+ "Vbe = vy-Ve\n",
+ "if Vbe<Vbecutin :\n",
+ " v=0.7 #voltage across Q5 in V\n",
+ " rQ5 = 1.5 #in K\n",
+ " i = (Vee-v)/rQ5\n",
+ " v = 0.75 #from the graph in V\n",
+ " Ve = -v-Vbe5\n",
+ " Vbe4=-Vbb-Ve\n",
+ " \n",
+ "vo = -vy-v\n",
+ "\n",
+ "Vb4 = Vbb\n",
+ "Vc4 = -(I*r)/1000 #in V\n",
+ "Vcb4 = Vc4+Vb4\n",
+ "print(\"The value of Vcb4 is %.2f v \" %Vcb4)\n",
+ "if Vcb4>0 :\n",
+ " print('For on npn transistor this represents a reverse bias and Q4 must be in active region')\n",
+ "\n",
+ "Vb1 = v\n",
+ "Vc1 = vy+Vbe5\n",
+ "Vcb1 = Vc1 + Vb1\n",
+ "\n",
+ "#Results\n",
+ "print(\"The value of Vc1 is %.2f v \" %Vc1)\n",
+ "print(\"The value of Vcb1 is %.2f v \" %Vcb1)\n",
+ "if Vcb1<0 :\n",
+ " print('For an npn transistor this represents a forward bias.... therefore Q1 is in saturation region')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of Vcb4 is 0.30 v \n",
+ "For on npn transistor this represents a reverse bias and Q4 must be in active region\n",
+ "The value of Vc1 is -0.85 v \n",
+ "The value of Vcb1 is -0.10 v \n",
+ "For an npn transistor this represents a forward bias.... therefore Q1 is in saturation region\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 16.4d, Page No 656"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "Vbb = 1.15 #in V\n",
+ "Vee=5.20 #in V\n",
+ "Vbe5=0.7 #in V\n",
+ "R=1.18 #in K\n",
+ "r=300.0 #in ohm\n",
+ "Vbecutin=0.5 #in V\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#If all inputs are low then we assume that Q1,Q2 and Q3 are cutoff and Q4 is conducting\n",
+ "Ve=-Vbb-Vbe5#Voltage at Common Emitter in V\n",
+ "#Current I in 1.18K Resistor\n",
+ "I = (Ve+Vee)/R#in mA\n",
+ "I1=I\n",
+ "#Output Voltage at Y\n",
+ "vy = -(r*I/1000.0)-Vbe5#I is in mA so 1000 is multiplied\n",
+ "Vbe = vy-Ve\n",
+ "if Vbe<Vbecutin :\n",
+ " v=0.7#voltage across Q5 in V\n",
+ " rQ5 = 1.5#in K\n",
+ " i = (Vee-v)/rQ5\n",
+ " v = 0.75#from the graph in V\n",
+ " Ve = -v-Vbe5\n",
+ " Vbe4=-Vbb-Ve\n",
+ "\n",
+ "vo = -vy-v\n",
+ "\n",
+ "#Verify that conducting transistor is in active region\n",
+ "Vb4 = Vbb\n",
+ "Vc4 = -(I*r)/1000#in V\n",
+ "Vcb4 = Vc4+Vb4\n",
+ "Vb1 = v\n",
+ "Vc1 = vy+Vbe5\n",
+ "Vcb1 = Vc1 + Vb1\n",
+ "\n",
+ "Vbe1 = Vbe5\n",
+ "Ve = -(Vb1+Vbe1)\n",
+ "\n",
+ "#Results\n",
+ "\n",
+ "print(\"The value of Ve is %.2f v \" %Ve)\n",
+ "I = (Ve + Vee)/R\n",
+ "I2=I\n",
+ "R = -Vc1/I\n",
+ "print(\"The value of R is %.2f ohm \" %R)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of Ve is -1.45 v \n",
+ "The value of R is 0.27 ohm \n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 16.4e Page No 656"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math \n",
+ "\n",
+ "#initialisation of variables\n",
+ "\n",
+ "Vbb = 1.15#in V\n",
+ "Vee=5.20#in V\n",
+ "Vbe5=0.7#in V\n",
+ "R=1.18#in K\n",
+ "r=300.0 #in ohm\n",
+ "Vbecutin=0.5#in V\n",
+ "\n",
+ "#Calculations\n",
+ "#If all inputs are low then we assume that Q1,Q2 and Q3 are cutoff and Q4 is conducting\n",
+ "Ve=-Vbb-Vbe5#Voltage at Common Emitter in V\n",
+ "#Current I in 1.18K Resistor\n",
+ "I = (Ve+Vee)/R#in mA\n",
+ "I1=I\n",
+ "#Output Voltage at Y\n",
+ "vy = -(r*I/1000.0)-Vbe5#I is in mA so 1000 is multiplied\n",
+ "Vbe = vy-Ve\n",
+ "if Vbe<Vbecutin :\n",
+ " v=0.7#voltage across Q5 in V\n",
+ " rQ5 = 1.5#in K\n",
+ " i = (Vee-v)/rQ5\n",
+ " v = 0.75#from the graph in V\n",
+ " Ve = -v-Vbe5\n",
+ " Vbe4=-Vbb-Ve\n",
+ "\n",
+ "vo = -vy-v\n",
+ "\n",
+ "Vb4 = Vbb\n",
+ "Vc4 = -(I*r)/1000.0#in V\n",
+ "Vcb4 = Vc4+Vb4\n",
+ "Vb1 = v\n",
+ "Vc1 = vy+Vbe5\n",
+ "Vcb1 = Vc1 + Vb1\n",
+ "\n",
+ "Vbe1 = Vbe5\n",
+ "Ve = -(Vb1+Vbe1)\n",
+ "I = (Ve + Vee)/R\n",
+ "I2=I\n",
+ "\n",
+ "I =(I1+I2)/2.0\n",
+ "\n",
+ "#Results\n",
+ "print(\"The value of I is %.2f mA \" %I)\n",
+ "I2 = (Vee-v)/rQ5\n",
+ "I3 = (Vee+vy)/rQ5\n",
+ "I = I + I2 + I3\n",
+ "P = Vee*I\n",
+ "print(\"Power dissipated is %.2f mW \" %P)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of I is 3.01 mA \n",
+ "Power dissipated is 43.72 mW \n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 16.5 Page No 668"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math \n",
+ "#initialisation of variables\n",
+ "C=50.0 #nF\n",
+ "Ra=10.0 #kOhm\n",
+ "Vcc=5.0 #V\n",
+ "beta=20.0 \n",
+ "\n",
+ "#Calculations\n",
+ "T=1.1*Ra*C\n",
+ "Ib=(Vcc-0.7)/100\n",
+ "Ic=beta*Ib\n",
+ "dt=C*(Vcc/Ic)\n",
+ "\n",
+ "#Results\n",
+ "print(\"Another pulse can be generated at hte output after time period of = %.2f ns \" %dt)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Another pulse can be generated at hte output after time period of = 290.70 ns \n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file |