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{
"metadata": {
"name": "",
"signature": "sha256:6ddc78aaecb14f72535c7221afdce9fdaaec875cc64d4211459abad2c8c70077"
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter05:Field Effect Transistors (FETs)"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example E1 - Pg 161"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Calculate saturation voltage and saturation current\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",
"print '%s %.f %s' %(\"Saturation Voltage is =\",Vds,\"V\\n\");\n",
"print '%s %.f %s' %(\"Saturation current is =\",Id*10**3,\"mA\");\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Saturation Voltage is = 2 V\n",
"\n",
"Saturation current is = 3 mA\n"
]
}
],
"prompt_number": 10
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example E2 - Pg 162"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Find the value of drain current\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.; #Schockleys equation\n",
"print '%s %.1f %s' %(\"Drain current is (At vgs = -2 V) =\",Id*10**3,\"mA\\n\");\n",
"\n",
"#At vgs = -4 V\n",
"vgs=-4.;#V input voltage\n",
"Id=Idss*(1.-(vgs/Vgso))**2.; #Schockleys equation\n",
"print '%s %.1f %s' %(\"Drain current is (At vgs = -4 V) =\",Id*10**3,\"mA\\n\");\n",
"\n",
"#At vgs = -8 V\n",
"print '%s' %(\"Drain current is 0 A (At vgs = -8 V) because gate is biased beyond cut off \");\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Drain current is (At vgs = -2 V) = 7.2 mA\n",
"\n",
"Drain current is (At vgs = -4 V) = 0.8 mA\n",
"\n",
"Drain current is 0 A (At vgs = -8 V) because gate is biased beyond cut off \n"
]
}
],
"prompt_number": 3
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example E3 - Pg 163"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Calculate Vgs and Vds saturation\n",
"#given\n",
"import math\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.-math.sqrt(Id/Idss));\n",
"Vds=Vgs-Vp;\n",
"print '%s %.3f %s' %(\"The gate to source voltage (Vgs) is =\",Vgs,\"V\\n\");\n",
"print '%s %.3f %s' %(\"The saturation voltage is Vds(sat) =\",Vds,\"V\");\n",
"\n",
"print '\\nThe value of Vgs = 2.115V and Vds= -2.885V in book because of the calculation error'\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The gate to source voltage (Vgs) is = 2.764 V\n",
"\n",
"The saturation voltage is Vds(sat) = -2.236 V\n",
"\n",
"The value of Vgs = 2.115V and Vds= -2.885V in book because of the calculation error\n"
]
}
],
"prompt_number": 4
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example E4 - Pg 167"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Calculate drain current Id for N channel\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",
"print '%s %.2f %s' %(\"The drain current Id(For Vgs= -3V) =\",Id*10**3,\"mA\\n\");\n",
"\n",
"#For Vgs= 2.5 V\n",
"Vgs=2.5;#V\n",
"Id=Idss*(1.-(Vgs/(-Vp)))**2.;\n",
"print '%s %.1f %s' %(\"The drain current Id(For Vgs= 2.5V) =\",Id*10**3,\"mA\");\n",
"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The drain current Id(For Vgs= -3V) = 2.88 mA\n",
"\n",
"The drain current Id(For Vgs= 2.5V) = 40.5 mA\n"
]
}
],
"prompt_number": 5
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example E5 - Pg 167"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Calculate drain current Id for P channel\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",
"print '%s %.2f %s' %(\"The drain current Id (For Vgs= -3V) =\",Id*10**3,\"mA\\n\");\n",
"\n",
"#For Vgs= 2.5V\n",
"Vgs=2.5;#V\n",
"Id=Idss*(1.-(Vgs/(-Vp)))**2.;\n",
"print '%s %.1f %s' %(\"The drain current Id (For Vgs= 2.5V) =\",Id*10**3,\"mA\");\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The drain current Id (For Vgs= -3V) = 46.08 mA\n",
"\n",
"The drain current Id (For Vgs= 2.5V) = 4.5 mA\n"
]
}
],
"prompt_number": 6
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example E6 - Pg 172"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Find the value of drain current\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",
"print '%s %.2f %s' %(\"The drain current is =\",Id*1000,\"mA\");\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The drain current is = 0.25 mA\n"
]
}
],
"prompt_number": 8
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example E7 - Pg 172"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Find the value of Id\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",
"print '%s %.2f %s' %(\"The drain current is =\",Id*1000,\"mA\");\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The drain current is = 10.89 mA\n"
]
}
],
"prompt_number": 7
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example E8 - Pg 174"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Calculate the dynamic drain resistance\n",
"#given\n",
"gm=200.*10.**-6.;#S transconductance\n",
"u=80.;#amplification factor\n",
"rd=u/gm;\n",
"print '%s %.f %s' %(\"The dynamic drain resistance is =\",rd/1000,\"k ohm\");\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The dynamic drain resistance is = 400 k ohm\n"
]
}
],
"prompt_number": 9
}
],
"metadata": {}
}
]
}
|
