{
"metadata": {
"name": "",
"signature": "sha256:7b8ef11e11019a3b1cc284475151e0d8bb5d6632f9c468e0f90289ce47134c38"
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
" Chapter 2: TRANSISTORS AND OTHER DEVICES"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 2.1,Page number 89"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Variable declaration\n",
"Rb=200 #base resistance(ohm)\n",
"Vbe=0.7 #base emitter voltage drop(V) in active region \n",
"Vbb=5 #base voltage of bipolar transistor(V) \n",
"beeta=100 #current gain\n",
"Rc=3 #collector resistance(k ohms)\n",
"Vcc=10 #voltage given to the collector(V)\n",
"\n",
"#Calculations\n",
"Ib=(Vbb-Vbe)/Rb #base current(mA) \n",
"Ic=beeta*Ib #collector current(mA) \n",
"Vcb=-Vbe-(Rc*Ic)+Vcc #collector base voltage drop(V)\n",
"\n",
"#Results\n",
"print\"base current\",Ib,\"mA\" \n",
"print\"collector current\",Ic,\"mA\"\n",
"print\"reverse bias collector junction is\",Vcb,\"V\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"base current 0.0215 mA\n",
"collector current 2.15 mA\n",
"reverse bias collector junction is 2.85 V\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 2.2,Page number 90"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Variable declaration\n",
"Vbb=5 #base voltage of bipolar transistor(V)\n",
"Vbe=0.7 #base emitter voltage drop(V) in active region \n",
"Rb=150 #base resistance(ohm)\n",
"beeta=125 #curret gain \n",
"Rc=3 #collector resistance(k ohms) \n",
"Vcc=10 #supply voltage(V)\n",
"Vce=0.2 #collector to emitter voltage(V)\n",
"\n",
"#Calculations\n",
"#Part a\n",
"Ib=(Vbb-Vbe)/Rb #base current(mA)\n",
"Ic=beeta*Ib #collector current(mA)\n",
"Vcb=-Vbe-(Rc*Ic)+Vcc #collector base voltage drop(V)\n",
"\n",
"#Part b -for npn transistor\n",
"Vbe=0.8 #base emitter voltage drop(V) in saturation\n",
"Ic=(Vcc-Vce)/Rc #collector current(mA)\n",
"Ib=(Vbb-Vbe)/Rb #base current(mA)\n",
"Ibmin=Ic/beeta #minimum base current(mA) to go into saturation(mA)\n",
"\n",
"#Results\n",
"print\"In active region, base current is\",round(Ib/1E-3),\"*10**-3 mA and collector current is\",round(Ic,2),\"mA\" \n",
"print\"base current and collector current in npn are\",round((Ib/1E-3),2),\"*10**-3 mA\",\"and\",round(Ic,2),\"mA resp.\"\n",
"print\"base current minimum is\",round(Ibmin,3),\"mA\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"In active region, base current is 28.0 *10**-3 mA and collector current is 3.27 mA\n",
"base current and collector current in npn are 28.0 *10**-3 mA and 3.27 mA resp.\n",
"base current minimum is 0.026 mA\n"
]
}
],
"prompt_number": 3
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 2.3,Page number 91"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Variable declaration\n",
"Vbb=5 #base voltage of bipolar transistor(V)\n",
"Vbe=0.7 #base emitter voltage drop(V) in active region \n",
"Rb=50 #base resistance(ohm)\n",
"beeta=50 #current gain\n",
"Re=1.8 #emitter resistance(k ohms) \n",
"Vcc=10 #supply voltage(V)\n",
"Vce=0.2 #collector to emitter voltage(V)\n",
"\n",
"#Calculations\n",
"Ib=(Vbb-Vbe)/(Rb+Re*(beeta+1)) #base current(mA)\n",
"Ic=beeta*Ib #collector current(mA)\n",
"Ie=Ib+Ic #emitter current(mA)\n",
"\n",
"#Results\n",
"print\"values are Ib:\",round(Ib,2),\"mA,Ic:\",round(Ic,2),\"mA and Ie:\",round(Ie,2),\"mA\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"values are Ib: 0.03 mA,Ic: 1.52 mA and Ie: 1.55 mA\n"
]
}
],
"prompt_number": 4
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 2.4,Page number 91"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Variable declaration\n",
"Vbe=0.7 #base to emitter voltage(V)\n",
"Rb=250 #base resistance(k ohms)\n",
"Vcc=10 #supply voltage(V)\n",
"Rl=0.5 #load resistance(k ohms)\n",
"\n",
"#Calculations\n",
"Ic=Vcc/Rl #collector current(mA)\n",
"IbQ=(Vcc-Vbe)/Rb #Ib at operating point(uA)\n",
"IcQ=8 #Ic at operating point(mA)\n",
"VceQ=6 #Vce at operating point(V)\n",
"\n",
"#Results\n",
"print\"values are IbQ:\",IbQ,\"uA,IcQ:\",IcQ,\"mA and Vcc:\",Vcc,\"V\"\n",
"print\"collector current Ic is\",Ic,\"mA and output voltage,vL=6-2 sinwt V\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"values are IbQ: 0.0372 uA,IcQ: 8 mA and Vcc: 10 V\n",
"collector current Ic is 20.0 mA and output voltage,vL=6-2 sinwt V\n"
]
}
],
"prompt_number": 6
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 2.5,Page number 104"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Variable declaration\n",
"Vgs=12 #gate to source voltage(V)\n",
"Vt=4 #threshold voltage(V)\n",
"Id=12.8 #drain current(mA)\n",
"K=0.0002 #device parameter \n",
"Vdd=24 #drain voltage(V)\n",
"Vds=Vgs=8 #drain to source voltage(V) \n",
"\n",
"#Calculations\n",
"Id=K*((Vds-Vt)**2) #drain current at Vds=8V\n",
"Rd=(Vdd-Vds)/Id #drain resistance(k ohms)\n",
"\n",
"#Result\n",
"print\"diode resistance is\",Rd,\"ohms\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"diode resistance is 5000.0 ohms\n"
]
}
],
"prompt_number": 4
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 2.7,Page number 106"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Variable declaration\n",
"Vds=7.5 #drain to source voltage(V)\n",
"Id=5 #drain current(mA) \n",
"\n",
"#Calculations\n",
"Vgs=-1.5 #gate to source voltage(V)\n",
"Vgg=-Vgs #gate voltage=gate to source voltage(V) \n",
"\n",
"#Result\n",
"print\"gate voltage is\",Vgg,\"V\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"gate voltage is 1.5 V\n"
]
}
],
"prompt_number": 5
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 2.8,Page number 107"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Variable declaration\n",
"Vds=7.5 #drain to source voltage(V)\n",
"Idss=8. #drain current for Vgs(V)\n",
"Vgs=2. #gate to source voltage(V)\n",
"Vp=4. #peak voltage(V)\n",
"\n",
"#Calculations\n",
"Id=Idss*((Vp-Vgs)/Vp)**2 #drain current(mA)\n",
"\n",
"#Result\n",
"print\"diode current is\",Id,\"mA\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"diode current is 2.0 mA\n"
]
}
],
"prompt_number": 13
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 2.10,Page number 115"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Variable declaration\n",
"beeta=160 #current gain\n",
"Vee=10 #emitter voltage(V)\n",
"Rb=400 #base resistance(k ohms)\n",
"Veb=0.8 #emitter to base voltage(V)\n",
"Re=2.5 #emitter resistance(k ohms)\n",
"Rc=1.5 #collector resistance(k ohms) \n",
"\n",
"#Calculations\n",
"#Part a \n",
"Ib=(Vee-Veb)/((Re*(1+beeta))+Rb) #base current(uA)\n",
"Ic=beeta*Ib #collector current(mA)\n",
"Ie=(beeta+1)*Ib #emitter current(mA) \n",
"Vce=Vee-(Re*Ie)-(Rc*Ic) #emitter to collector voltage(V) \n",
"Vce=-Vce #collector to emitter voltage(V)\n",
"\n",
"#Part b\n",
"beeta=80 #current gain \n",
"Ib1=(Vee-Veb)/((Re*(1+beeta))+Rb) #base current(uA)\n",
"Ic1=beeta*Ib1 #collector current(mA)\n",
"Ie1=(beeta+1)*Ib1 #emitter current(mA) \n",
"Vce1=-(Vee-(Ie1*Re)-(Rc*Ic1)) #collector to emitter voltage(V) \n",
" \n",
"#Result\n",
"print\"collector current and Vce for beeta=160 are\",round(Ic,2),\"mA\",\"and\",round(Vce,2),\"V\" \n",
"print\"Ic and Vce for beeta=80 are\",round(Ic,2),\"mA\",\"and\",round(Vce1,2),\"V\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"collector current and Vce for beeta=160 are 1.83 mA and -2.63 V\n",
"Ic and Vce for beeta=80 are 1.83 mA and -5.08 V\n"
]
}
],
"prompt_number": 7
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 2.13,Page number 120"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"from sympy import*\n",
"import math \n",
"\n",
"#Variable declaration\n",
"K=2 #device parameter\n",
"Rd=Rs=2.5*10**3 #drain resistance(k ohms)\n",
"R1=100*10**3 #resistance(ohms)\n",
"R2=200*10**3 #resistance(ohms)\n",
"Vdd=12 #drain voltage(V)\n",
"Vt=4 #threshold voltage(V)\n",
"\n",
"#Calculations\n",
"Vgg=(R2*Vdd)/(R1+R2)\n",
"Id=symbols('Id') #gate voltage(V)\n",
"expr=solve(Id**2-3.28*Id+2.56,Id)\n",
"print expr\n",
"Id=1.28\n",
"Vds=Vdd-5*Id\n",
"\n",
"#Result\n",
"print\"Id is\",Id,\"mA and Vds is\",Vds,\"V\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"[1.28000000000000, 2.00000000000000]\n",
"Id is 1.28 mA and Vds is 5.6 V\n"
]
}
],
"prompt_number": 10
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 2.14,Page number 121"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Variable declaration\n",
"k=2. #device parameter\n",
"Vt=-1. #threshold voltage(V)\n",
"Vdd=-12. #drain voltage(V)\n",
"R1=300. #resistance(kohms)\n",
"R2=100. #resistance(kohms)\n",
"\n",
"#Calculations\n",
"#Part a\n",
"Vgs=-2 #gate to source voltage(V)\n",
"Vgg=(R2*Vdd)/(R1+R2) #gate voltage(V)\n",
"Id=k*((Vgs-Vt)**2) #drain current(mA)\n",
"Rs=(Vgs-Vgg)/Id #source resistance(k ohms) as Id=Is,Kvl in GS loop\n",
"Is=Id\n",
"\n",
"#Part b\n",
"Vds=-4 #drain to source voltage(V)\n",
"Rd=(-Vdd+Vds-(Is*Rs))/Id #applying kvl in DS loop\n",
"\n",
"#Part c \n",
"Vt=-1.5 #threshold voltage(V) \n",
"Vgg=-1.5 #gate voltage using Id formula \n",
"R2new=(Vgg*R1)/(Vdd-Vgg) #new resistance(k ohms)\n",
"\n",
"#Results\n",
"print\"a)source resistance is\",Rs,\"kohm\" \n",
"print\"b)drain resistance is\",Rd,\"kohm\"\n",
"print\"c)R2new is\",round(R2new,2),\"kohm\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"a)source resistance is 0.5 kohm\n",
"b)drain resistance is 3.5 kohm\n",
"c)R2new is 42.86 kohm\n"
]
}
],
"prompt_number": 6
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 2.15,Page number 122"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"#Variable declaration\n",
"Vp=-4 #peak voltage(V)\n",
"Idss=10 #drain current for Vgs(V)\n",
"Vdd=18 #drain voltage(V) \n",
"Rs=2 #source resistance(ohms) \n",
"Rd=2 #drain resistance(ohms)\n",
"R1=450*10**3 #resistance(ohms)\n",
"R2=90*10**3 #resistance(ohms)\n",
"\n",
"#Calculations\n",
"Vgg=(R2*Vdd)/(R1+R2)\n",
"from sympy import*\n",
"Id=symbols('Id')\n",
"expr=solve(20*Id**2-148*Id+245,Id)\n",
"print expr\n",
"Id1=2.5\n",
"Vds=Vdd-((Rs+Rd)*Id1)\n",
"\n",
"#Result\n",
"print\"Id is\",Id1,\"mA and Vds is\",Vds,\"V\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"[5/2, 49/10]\n",
"Id is 2.5 mA and Vds is 8.0 V\n"
]
}
],
"prompt_number": 20
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 2.16,Page number 123"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"#Variable declaration\n",
"Vp=4 #peak voltage(V)\n",
"Idss=12. #drain current for Vgs(V) \n",
"Vdd=12 #drain voltage(V)\n",
"Id=4. #drain current(mA)\n",
"Vds=6 #drain to source voltage(V)\n",
"\n",
"#Calculations\n",
"Rs=(Vp/4)*(1-(math.sqrt(Id/Idss))) #by Id=Idss(1-(Vgs/Vp))^2 and putting Vgs=4Rs in it and solving\n",
"Rd=((Vdd+Vds)/Id)-Rs #solving equation -Vdd-Vds+(Id*(Rd+Rs))=0 \n",
" \n",
"#Result\n",
"print\"source resistance is\",round(Rs,2),\"kohm\" \n",
"print\"drain resistance\",round(Rd,2),\"kohms\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"source resistance is 0.42 kohm\n",
"drain resistance 4.08 kohms\n"
]
}
],
"prompt_number": 3
}
],
"metadata": {}
}
]
}