{
"metadata": {
"name": ""
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 19 : Power amplifiers"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 19.1, Page No 810"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"Rpy=40.0\n",
"N1=74.0\n",
"N2=14.0\n",
"R2=3.7*10**3\n",
"R1=4.7*10**3\n",
"Vbe=0.7\n",
"Re=1.0*10**3\n",
"Vcc=13.0\n",
"Rl=56.0\n",
"\n",
"#Calculations\n",
"print(\"Q-point\")\n",
"Vb=Vcc*(R2/(R1+R2))\n",
"Ic=(Vb-Vbe)/Re\n",
"Ie=Ic\n",
"Vce=Vcc-Ic*(Rpy+Re)\n",
"rl=(N1/N2)**2 *Rl\n",
"rl=rl+Rpy\n",
"Ic=5*10**-3\n",
"Vce=Ic*rl\n",
"\n",
"#Results\n",
"print(\"The value of Vce is %.2f v \" %Vce)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Q-point\n",
"The value of Vce is 8.02 v \n"
]
}
],
"prompt_number": 44
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 19.2, Page No 814"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"#initialisation of variables\n",
"\n",
"Vcc=13.0\n",
"Icq=5.0*10**-3\n",
"Vceq=8.0\n",
"Vp=Vceq\n",
"Ip=Icq\n",
"nt=0.8\n",
"\n",
"#Calculations\n",
"Pi=Vcc*Icq\n",
"Po=.5*Vp*Ip\n",
"P0=nt*Po\n",
"n=(P0/Pi)*100.0\n",
"\n",
"#Results\n",
"print(\" maximum efficiency is %3.2f percentage \" %n)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" maximum efficiency is 24.62 percentage \n"
]
}
],
"prompt_number": 45
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 19.4 Page No 821"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"#initialisation of variables\n",
"N1=60.0\n",
"N2=10.0\n",
"Rl=16.0\n",
"Rpy=0\n",
"R6=56.0\n",
"Vcc=27.0\n",
"Vce=0.5\n",
"n=0.79\n",
"\n",
"#Calculations\n",
"print(\" Referred laod\")\n",
"rl=(N1/N2)**2 *Rl\n",
"print(\" tatol ac load line in series with each of Q2 and Q3\")\n",
"Rl=rl+R6+Rpy\n",
"print(\" peak primary current\")\n",
"Ip=(Vcc-Vce)/Rl\n",
"print(\"peak primary voltage\")\n",
"Vp=Vcc-Vce-(Ip*R6)\n",
"print(\"power delivered to primary\")\n",
"Po=.5*Vp*Ip\n",
"\n",
"#Calculations\n",
"Po=Po*n#n is power efficiency\n",
"print(\"power delivered to the load %.2f W \" %Po)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" Referred laod\n",
" tatol ac load line in series with each of Q2 and Q3\n",
" peak primary current\n",
"peak primary voltage\n",
"power delivered to primary\n",
"power delivered to the load 0.40 W \n"
]
}
],
"prompt_number": 46
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 19.5, Page No 824"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"#initialisation of variables\n",
"\n",
"Po=4.0\n",
"nt=0.8\n",
"Vcc=30.0\n",
"Vp=Vcc\n",
"Rl=16.0\n",
"\n",
"#Calculations\n",
"P0=Po/nt\n",
"rl=(Vp)**2 /(2*P0)\n",
"rl=4*rl\n",
"print(\"transformer specification Po=4 %Rl=16 rl=360\")\n",
"Vce=2.0*Vcc\n",
"Ip=(2.0*P0)/Vp\n",
"Pi=Vcc*.636*Ip\n",
"Pt=0.5*(Pi-P0)\n",
"\n",
"#Results\n",
"print(\" transistor specification is Py=.68W Vce=60 Ip=333mA\")\n",
"print(\"power delivered to the load Pi = %.2f W \" %Pi)\n",
"print(\"power delivered to the load Pt = %.2f W \" %Pt)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"transformer specification Po=4 %Rl=16 rl=360\n",
" transistor specification is Py=.68W Vce=60 Ip=333mA\n",
"power delivered to the load Pi = 6.36 W \n",
"power delivered to the load Pt = 0.68 W \n"
]
}
],
"prompt_number": 47
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 19.6 Page No 830"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"#initialisation of variables\n",
"Rl=50.0\n",
"Po=1.0\n",
"hFE=50.0\n",
"Vbe=0.7\n",
"Vrc=4.0\n",
"Vre=1.0\n",
"Vd1=0.7\n",
"\n",
"#Calculations\n",
"Vd2=Vd1\n",
"Vp=math.sqrt(2*Rl*Po)\n",
"Ip=Vp/Rl\n",
"Re3=.1*Rl\n",
"Re2=4.7#use stabdard value\n",
"Re2=Re3\n",
"Icq=.1*Ip\n",
"Vb=Vbe+Icq*(Re2+Re3)+Vbe\n",
"Vc1=Vrc\n",
"Ib2=Ip/hFE\n",
"Irc=Ib2+1*10**-3\n",
"Rc=Vrc/Irc\n",
"Rc=680.0 #use standard value\n",
"Vcc=2.0*(Vp+Vre+Vbe+Vrc)\n",
"Vcc=32#use standard value\n",
"Vrcdc=.5*(Vcc-Vb)\n",
"Ic1=Vrcdc/Rc\n",
"Rb=(Vb-Vd1-Vd2)/Ic1\n",
"\n",
"#Results\n",
"print(\"The value of Rb is %.2f kOhm \" %Rb)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The value of Rb is 8.95 kOhm \n"
]
}
],
"prompt_number": 48
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 19.7 Page No 832"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"#initialisation of variables\n",
"\n",
"Vcc=32.0\n",
"Vce=32.0\n",
"Ip=200.0*10**-3\n",
"Po=1.0\n",
"\n",
"#Calculations\n",
"Ic=1.1*Ip\n",
"Pi=0.35*Vcc*Ip\n",
"Pt=0.5*(Pi-Po)\n",
"\n",
"#Results\n",
"print(\"power delivered to the load Pi = %.2f \" %Pi)\n",
"print(\"power delivered to the load Pt = %.2f \" %Pt)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"power delivered to the load Pi = 2.24 \n",
"power delivered to the load Pt = 0.62 \n"
]
}
],
"prompt_number": 49
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 19.8, Page No 832"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"#initialisation of variables\n",
"f=50.0\n",
"hib=2.0\n",
"Rl=50.0\n",
"\n",
"#Calculations\n",
"Ce=1.0/(2*3.14*f*hib)\n",
"Co=1.0/(2*3.14*50*.1*Rl)\n",
"\n",
"#Results\n",
"print(\"The value of Ce is %.2f pF \" %(Ce*10**3))\n",
"print(\"The value of Co is %.2f pF \" %(Co*10**3))"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The value of Ce is 1.59 pF \n",
"The value of Co is 0.64 pF \n"
]
}
],
"prompt_number": 50
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 19.9 Page No 834"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"\n",
"#initialisation of variables\n",
"hFE=2000.0\n",
"Vbe=1.4\n",
"Vp=10.0\n",
"Ip=200.0*10**-3\n",
"Icq2=20.0*10**-3\n",
"Re3=4.7\n",
"Re2=4.7\n",
"Vd=0.7\n",
"Ve1=3.0\n",
"Vc1=15.2\n",
"\n",
"#Calculations\n",
"Vrc=Vc1\n",
"Vb=Vbe+Icq*(Re2+Re3)+Vbe\n",
"Vcc=Vrc+Vc1+Vb\n",
"Ib2=Ip/hFE\n",
"Irc=1.0*10**-3\n",
"Vrcac=4.0\n",
"Rc=Vrcac/Irc\n",
"Ic1=Vrc/Rc\n",
"Rb=(Vb-(4*Vd))/Ic1\n",
"\n",
"#Results\n",
"print(\"The value of Rb is %.2f kohm \" %Rb)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The value of Rb is 49.47 kohm \n"
]
}
],
"prompt_number": 51
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 19.10 Page No 838"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"#initialisation of variables\n",
"Vb=3.2\n",
"Ic1=5*10**-3\n",
"Vce=3.2\n",
"Vbe=0.7\n",
"\n",
"#Calculations\n",
"Vbmin=Vb-0.5\n",
"Vbmax=Vb+0.5\n",
"I10=.1*Ic1\n",
"R10=(Vce-Vbe)/I10\n",
"R10=4.7*10**3#use standard value\n",
"print(\" for Vce=3.7\")\n",
"Vce=3.7\n",
"I10max=(Vce-Vbe)/R10\n",
"print(\"Vce=2.7V\")\n",
"Vce=2.7\n",
"I10min=(Vce-Vbe)/R10\n",
"R=Vbe/I10min\n",
"R11=Vbe/I10max\n",
"R12=R-R11\n",
"\n",
"#Results\n",
"print(\"The value of R12 is %.2f kohm \" %R12)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" for Vce=3.7\n",
"Vce=2.7V\n",
"The value of R12 is 548.33 kohm \n"
]
}
],
"prompt_number": 52
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 19.11 Page No 843"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"#initialisation of variables\n",
"\n",
"Rl=16.0\n",
"Po=6.0\n",
"Vbe=0.7\n",
"\n",
"#Calculations\n",
"Vp=math.sqrt(2.0*Rl*Po)\n",
"Vr14=0.1*Vp\n",
"Vr15=Vr14\n",
"R14=0.1*Rl\n",
"R15=R14\n",
"Vce3=1.0\n",
"Vce4=Vce3\n",
"Vr9=3.0\n",
"Vr11=Vr9\n",
"Vcc=(Vp+Vr14+Vbe+Vce3+Vr9)\n",
"Vee=-Vcc\n",
"Ip=Vp/Rl\n",
"print(\" DC power inpit from supply line\")\n",
"Pi=(Vcc-Vee)*.35*Ip\n",
"Pt=.5*(Pi-Po)\n",
"Vce=2*Vcc\n",
"Ic=1.1*Ip\n",
"\n",
"#Results\n",
"print(\" output transistor specification %.2f mA\" %Ic)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" DC power inpit from supply line\n",
" output transistor specification 0.95 mA\n"
]
}
],
"prompt_number": 53
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 19.12, Page No 844"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"#initialisation of variables\n",
"\n",
"hFE7=20.0\n",
"Icbo=50.0*10**-6\n",
"hFE5=70.0\n",
"Vr9=3.0\n",
"Ip=869.0*10**-3\n",
"R15=1.5\n",
"R8=15.0*10**3\n",
"Vbe=0.7\n",
"Vr11=3.0\n",
"Vee=20.0\n",
"\n",
"#Calculations\n",
"R12=0.01/Icbo\n",
"R12=220#use standard value\n",
"R13=R12\n",
"Ib5=Ip/(hFE7*hFE5)\n",
"Ic3=2.0*10**-3\n",
"R9=Vr9/Ic3\n",
"R11=R9\n",
"Iq78=0.1*Ip\n",
"Vr14=Iq78*R15\n",
"Vr15=Vr14\n",
"Vr10=(Vr14+Vr15)+(Vr14+Vr15)/2\n",
"R10=Vr10/Ic3\n",
"Ir8=(Vr11+Vbe)/R8\n",
"R7=(Vee-(Vr11+Vbe))/Ir8\n",
"\n",
"#Results\n",
"print(\"The value of R7 is %.2f kohm \" %(R7/1000))"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The value of R7 is 66.08 kohm \n"
]
}
],
"prompt_number": 54
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 19.13, Page No 848"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"#initialisation of variables\n",
"\n",
"Rl=20.0\n",
"Po=2.5\n",
"Rd=4.0\n",
"Vr6=1.0\n",
"Vr9=Vr6\n",
"Vth=1.0\n",
"gFS=250.0*10**-3\n",
"Vbe=0.7\n",
"\n",
"#Calculations\n",
"Vp=math.sqrt(2*Rl*Po)\n",
"Ip=Vp/Rl\n",
"Vcc=(Vp+Ip*Rd)\n",
"vr6=Ip/gFS\n",
"Vr2=vr6+1\n",
"Vce=Vr2\n",
"Vce3=1.0\n",
"Vr2=Vcc-Vce\n",
"Vee=Vcc\n",
"Vr3=Vee-Vbe\n",
"Vr7=Vr2-Vr6\n",
"Vr8=Vcc-(-Vee)-Vr6-Vr7-Vr9\n",
"\n",
"#Results\n",
"print(\"The value of Vr8 is %.2f V \" %Vr8)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The value of Vr8 is 14.00 V \n"
]
}
],
"prompt_number": 55
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 19.14, Page No 849"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"#initialisation of variables\n",
"\n",
"R6=100.0*10**3\n",
"R9=R6\n",
"Vth=1.0\n",
"Vr7=8.0\n",
"Vr8=14.0\n",
"Vr3=11.3\n",
"Vpout=10.0\n",
"Vpin=800.0*10**-3\n",
"\n",
"#Calculations\n",
"I6=Vth/R6\n",
"R7=Vr7/I6\n",
"R8=Vr8/I6\n",
"Ic1=1*10**-4\n",
"Ic2=Ic1\n",
"Vr2=9\n",
"R2=Vr2/Ic1\n",
"R3=Vr3/(Ic1+Ic2)\n",
"R5=4.7*10**3\n",
"Acl=Vpout/Vpin\n",
"R4=R5/(Acl-1.0)\n",
"\n",
"#Results\n",
"print(\"The value of R4 is %.2f kohm \" %(R4/1000))"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The value of R4 is 0.41 kohm \n"
]
}
],
"prompt_number": 56
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 19.15, Page No 854"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"#initialisation of variables\n",
"\n",
"Vce=1.5\n",
"Vcc=17.0\n",
"Vd1=0.7\n",
"R8=1.5*10**3\n",
"R9=R8\n",
"Rl=100.0\n",
"R6=8.2\n",
"\n",
"#Calculations\n",
"I4=(Vcc-Vd1)/(R8+R9)\n",
"Vc3=Vcc-(I4*R8)\n",
"print(\" bootstrap capacitance terminal voltage is %3.1fV \" %Vc3)\n",
"V=Vcc-Vce#V=Vp+Vr6\n",
"Ip=V/(Rl+R6)\n",
"Vp=Ip*Rl\n",
"print(\" peak output voltage is %3.1fV \" %Vp)\n",
"Po=(Vp)**2.0/(2.0*Rl)\n",
"\n",
"#Results\n",
"print(\" peak output power is %dW \" %Po)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" bootstrap capacitance terminal voltage is 8.8V \n",
" peak output voltage is 14.3V \n",
" peak output power is 1W \n"
]
}
],
"prompt_number": 57
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 19.16, Page No 856"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"#initialisation of variables\n",
"\n",
"Rl=8.0\n",
"Po=6.0\n",
"vs=0.1\n",
"hFE=1000.0\n",
"Vce=2.0\n",
"f=50.0*10**3\n",
"Vd1=0.7\n",
"\n",
"#Calculations\n",
"Vp=math.sqrt(2*Rl*Po)\n",
"Ip=Vp/Rl\n",
"R6=.1*Rl\n",
"R7=R6\n",
"Vcc=Vp+Ip*R6+Vce\n",
"Ib=Ip/hFE\n",
"I4=2*10**-3\n",
"R4=(Vcc-Vd1-Vd1)/I4\n",
"R8=.5*R4\n",
"Acl=Vp/vs\n",
"R3=100*10**3\n",
"R2=R3/(Acl-1)\n",
"SR=(2*3.14*f*Vp)*10**-6\n",
"\n",
"#Results\n",
"print(\" slew rate is %.2f V/us \" %SR)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" slew rate is 3.08 V/us \n"
]
}
],
"prompt_number": 58
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 19.17, Page No 856"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"#initialisation of variables\n",
"\n",
"f=50.0\n",
"R1=100.0*10**3\n",
"R2=1.0*10**3\n",
"R8=2.7*10**3\n",
"\n",
"#Calculations\n",
"R9=R8\n",
"C1=1/(2*3.14*f*.1*R1)\n",
"C2=1/(2*3.14*f*R2)\n",
"Xc3=.1*((R8*R9)/(R8+R9))\n",
"C3=1/(2*3.14*f*Xc3)\n",
"C4=C3\n",
"\n",
"#Results\n",
"print(\"The value of C4 is %.2f pF \" %(C4*10**6))"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The value of C4 is 23.59 pF \n"
]
}
],
"prompt_number": 59
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 19.18, Page No 860"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"#initialisation of variables\n",
"\n",
"Ismin=1.8*10**-3\n",
"Ismax=3.4*10**-3\n",
"R7=820.0\n",
"R5=390.0\n",
"R6=18.0*10**3\n",
"Vi=100.0*10**-3\n",
"Rl=10.0\n",
"\n",
"#Calculations\n",
"Vgsmin=Ismin*R7\n",
"Vgsmax=Ismax*R7\n",
"Acl=(R5+R6)/R5\n",
"Vp=Acl*Vi\n",
"Ip=Vp/Rl\n",
"print(\"peak output current is %3.3fA \" %Ip)\n",
"Po=(Vp*Ip)/2.0\n",
"\n",
"#Results\n",
"print(\"peak output power is %3.2fW \" %Po)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"peak output current is 0.472A \n",
"peak output power is 1.11W \n"
]
}
],
"prompt_number": 60
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 19.19, Page No 862"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"#initialisation of variables\n",
"\n",
"Vbe=0.7\n",
"R2=560.0\n",
"R3min=0\n",
"R3max=1.0*10**3\n",
"Is=2.0*10**-3\n",
"\n",
"#Calculations\n",
"Ic2max=Vbe/(R2+R3min)\n",
"Ic2min=Vbe/(R2+R3max)\n",
"Vgsmin=(Is+Ic2min)*820.0\n",
"Vgsmax=(Is+Ic2max)*820.0\n",
"\n",
"#Results\n",
"print(\"The value of Vgsmax is %.2f v \" %Vgsmax)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The value of Vgsmax is 2.67 v \n"
]
}
],
"prompt_number": 61
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 19.20, Page No 865"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"#initialisation of variables\n",
"\n",
"Vcc=12.0\n",
"Rl=10.0\n",
"Rd=0.5\n",
"gfs=2.5\n",
"R7=820.0\n",
"V9=1.0*10**3\n",
"\n",
"#Calculations\n",
"R10=R9\n",
"Vp=(Vcc*Rl)/(Rd+Rl)\n",
"Ip=Vp/Rl\n",
"Vgs=Ip/gfs\n",
"Vr7=Is*R7\n",
"Vs=Vcc-Vr7-Vgs\n",
"Vr9=(Vp*R9)/(R9+R10)\n",
"\n",
"#Results\n",
"print(\"op-amp peak output voltage is %.2f v \" %Vr9)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"op-amp peak output voltage is 5.71 v \n"
]
}
],
"prompt_number": 62
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 19.21, Page No 867"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"#initialisation of variables\n",
"Vbe=0.7\n",
"R2=470.0\n",
"R3=1.0*10**3\n",
"Is=0.5*10**-3\n",
"R7=1.5*10**3\n",
"Vcc=15\n",
"\n",
"#Calculations\n",
"Ic2max=Vbe/R2\n",
"Ic2min=Vbe/(R2+R3)\n",
"Vgs=(Is+Ic2max)*R7\n",
"print(\" MOSFET maximum gate source voltage is %.1fV \" %Vgs)\n",
"Vs=Vcc-Vgs\n",
"\n",
"#Results\n",
"print(\" op-amp minimum suppy is %.2fV \" %Vs)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" MOSFET maximum gate source voltage is 3.0V \n",
" op-amp minimum suppy is 12.02V \n"
]
}
],
"prompt_number": 63
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 19.22, Page No 868"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"#initialisation of variables\n",
"Vcc=15.0\n",
"Rl=15.0\n",
"Rd=0.3\n",
"R5=2.2*10**3\n",
"R6=33.0*10**3\n",
"C2=3.9*10**-6\n",
"C4=100.0*10**-12\n",
"\n",
"#Calculations\n",
"print(\" power output\")\n",
"Vp=(Vcc*Rl)/(Rd+Rl)\n",
"Ip=Vp/Rl\n",
"Po=(Vp*Ip)/2.0\n",
"print(\" voltage gain\")\n",
"Av=(R5+R6)/R5\n",
"print(\"cutoff frequency\")\n",
"f1=1.0/(2*3.14*C2*R5)\n",
"f2=1.0/(2*3.14*C4*R6)\n",
"\n",
"#Results\n",
"print(\" cutoff frequency f1 %.2f \" %f1)\n",
"print(\" cutoff frequency f2 %.2f \" %f2)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" power output\n",
" voltage gain\n",
"cutoff frequency\n",
" cutoff frequency f1 18.56 \n",
" cutoff frequency f2 48253.23 \n"
]
}
],
"prompt_number": 64
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 19.23, Page No 871"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"#initialisation of variables\n",
"Vcc=23.0\n",
"Rl=8.0\n",
"Rf2=100.0*10**3\n",
"Rf1=5.6*10**3\n",
"Cf=1.0*10**-6\n",
"Vp=Vcc-5\n",
"\n",
"#Calculations\n",
"Po=(Vp)**2/(2*Rl)\n",
"print(\"maximum output power is %3.2fW \" %Po)\n",
"Acl=(Rf1+Rf2)/Rf1\n",
"print(\" voltage gain %3.1f \" %Acl)\n",
"f=1/(2*3.14*Cf*Rf1)\n",
"\n",
"#Results\n",
"print(\"lower cutoff frequency is %dHz \" %f)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"maximum output power is 20.25W \n",
" voltage gain 18.9 \n",
"lower cutoff frequency is 28Hz \n"
]
}
],
"prompt_number": 65
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 19.24, Page No 875"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"#initialisation of variables\n",
"Rf=15.0*10**3\n",
"R1=5.6*10**3\n",
"vs=0.5\n",
"Vp=2.7\n",
"\n",
"#Calculations\n",
"Acl=(2.0*Rf)/R1\n",
"Vo=Acl*vs\n",
"Po=(Vp)**2.0/(2.0*Rl)\n",
"\n",
"#Results\n",
"print(\"load power dissipation is %.2fW \" %Po)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"load power dissipation is 0.46W \n"
]
}
],
"prompt_number": 66
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 19.24, Page No 875"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"#initialisation of variables\n",
"Vcc=10.0\n",
"Rl=1.0*10**3\n",
"f=3.0*10**6\n",
"Ip=25.0*10**-3\n",
"Vce=0.3\n",
"\n",
"#Calculations\n",
"Vp=Vcc-Vce\n",
"Po=(Vp)**2 /(2*Rl)\n",
"T=1.0/f\n",
"t=(Po*T)/(Ip*Vp)\n",
"angle=(t/T)*360\n",
"print(\" conduction angle is %3.1fdegree \" %angle)\n",
"Idc=Po/Vp\n",
"Pi=Vcc*Idc\n",
"print( \"dc input power is %3.4fW \" %Pi)\n",
"n=(Po/Pi)*100#efficiency\n",
"\n",
"#Results\n",
"print(\" maximum efficiency is %3.2f percentage \" %n)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" conduction angle is 69.8degree \n",
"dc input power is 0.0485W \n",
" maximum efficiency is 97.00 percentage \n"
]
}
],
"prompt_number": 67
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 19.26, Page No 882"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"#initialisation of variables\n",
"f=1.0*10**6\n",
"Xc=120.0\n",
"Vce=0.5\n",
"Vcc=30.0\n",
"Rl=1.2*10**3\n",
"O=100.0\n",
"\n",
"#Calculations\n",
"Cp=1.0/(2*3.14*f*Xc)\n",
"Cp=1300*10**-12#use standard value\n",
"Lp=1/(((2*3.14*f)**2)*Cp)\n",
"Vp=Vcc-Vce\n",
"Po=((Vp)**2) /(2*Rl)\n",
"Idc=Po/Vp\n",
"T=1.0/f\n",
"t=(O*T)/360.0\n",
"Ip=(Idc*T)/t\n",
"\n",
"#Results\n",
"print(\"The value of Ip is %.2f mA \" %(Ip*10**3))"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The value of Ip is 44.25 mA \n"
]
}
],
"prompt_number": 68
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 19.27, Page No 883"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"#initialisation of variables\n",
"Rw=0.1\n",
"f=1.0*10**6\n",
"Lp=19.5*10**-6\n",
"Rl=1.2*10**3\n",
"Vcc=30.0\n",
"\n",
"#Calculations\n",
"Idc=12.3*10**-3\n",
"QL=(2*3.14*f*Lp)/Rw\n",
"Qp=Rl/(2*3.14*f*Lp)\n",
"B=f/Qp\n",
"Il=(.707*Vp)/(2*3.14*f*Lp)\n",
"Pl=(Il)**2 *Rw\n",
"Pi=(Vcc*Idc)+Pl\n",
"n=(Po/Pi)*100.0\n",
"\n",
"#Results\n",
"print(\" maximum efficiency is %3.2f percentage \" %n)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" maximum efficiency is 97.50 percentage \n"
]
}
],
"prompt_number": 69
}
],
"metadata": {}
}
]
}