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|
{
"metadata": {
"name": ""
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 3: Inverting and Non Inverting Amplifiers"
]
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 3.1 Page No 47"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Given\n",
"Rf=100*(10**3) #Feedback Resistance in ohms\n",
"Ri=10*(10**3) #Input Resistance in ohms\n",
"Ei=1.0 #Input volts\n",
"\n",
"#Calculation \n",
"I=Ei/Ri #Equation for current through Rf\n",
"Vout=-(Rf/Ri)*Ei #Equation for Output Voltage\n",
"Acl=-(Rf/Ri) #Closed loop gain of the amplifier \n",
"\n",
"#Result\n",
"print\"(a) Current through Rf is\",I*1000,\"mA\"\n",
"print\"(b) Value of output voltage is \",Vout,\"V\"\n",
"print\"(c) Value of closed loop gain is\",Acl"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"(a) Current through Rf is 0.1 mA\n",
"(b) Value of output voltage is -10.0 V\n",
"(c) Value of closed loop gain is -10\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 3.2 Page No 47"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Given\n",
"Vout=10.0 #output voltage\n",
"I=0.1*(10**-3) #current through Rf in amperes\n",
"Rl=25*(10**3) #Load resistance in ohms\n",
"\n",
"#Calculation\n",
"Il=Vout/Rl\n",
"Iout=I+Il\n",
"\n",
"#Result\n",
"print\"(a) Value of load current is\",Il*1000,\"m\"\n",
"print\"(b) Total current into the output pin of the opamp is \",Iout*1000,\"mA\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"(a) Value of load current is 0.4 m\n",
"(b) Total current into the output pin of the opamp is 0.5 mA\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 3.3 Page No 49"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Given\n",
"Rf=250*(10**3)#Feedback Resistance in Ohms\n",
"Ri=10*(10**3)#Input Resistance in Ohms\n",
"Ei=0.5#Input voltage\n",
"\n",
"#Calculation\n",
"I=Ei/(Ri)\n",
"VRf=I*Rf\n",
"Ei1=-0.5\n",
"Vout=-(Rf/Ri)*Ei1\n",
"\n",
"#Result\n",
"print\"(a) Value of current through Rf is\",I,\"A\"\n",
"print\"(b) Voltage through Rf is \",VRf,\"V\"\n",
"print\"(c) Output Voltage is\",Vout,\"V\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"(a) Value of current through Rf is 5e-05 A\n",
"(b) Voltage through Rf is 12.5 V\n",
"(c) Output Voltage is 12.5 V\n"
]
}
],
"prompt_number": 5
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 3.4 Page No 49"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Given\n",
"Vout=12.5 #in volts\n",
"Il=2*10**-3 #in amperes\n",
"\n",
"#Calculation\n",
"Rl=Vout/Il #Load resistance in ohm\n",
"I=0.05*10**-3\n",
"Iout=I+Il\n",
"\n",
"#Result\n",
"print\"(a) Value of Load Resistance is\",Rl/1000,\"kohm\"\n",
"print\"(b) Value of output current is \",Iout*1000,\"mA\"\n",
"print\"(c)The circuit input resistance is Rl=10Kohm\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"(a) Value of Load Resistance is 6.25 kohm\n",
"(b) Value of output current is 2.05 mA\n",
"(c)The circuit input resistance is Rl=10Kohm\n"
]
}
],
"prompt_number": 3
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 3.5 Page No 50"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Given\n",
"Rf=20*10**3\n",
"Ri=10*10**3\n",
"\n",
"#calculation\n",
"Acl=-(Rf/Ri) #Voltage Gain\n",
"\n",
"#Result\n",
"print\" Value of Voltage Gain is\",Acl"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" Value of Voltage Gain is -2\n"
]
}
],
"prompt_number": 12
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 3.6 Page No 51"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Given\n",
"Ei=-5 #input voltage\n",
"Acl=-2 #Voltage Gain\n",
"\n",
"#Calculation\n",
"Vout=Ei*Acl #output voltage\n",
"\n",
"#Result\n",
"print\" Value of Output Voltage is\",Vout,\"V\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" Value of Output Voltage is 10 V\n"
]
}
],
"prompt_number": 14
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 3.7 Page No :51"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Given\n",
"Ri=10*10**3 #input resistance in ohm\n",
"Gain= 25\n",
"\n",
"#Calculation\n",
"Rf=Gain*Ri #feedback resistance in ohm\n",
"\n",
"#result\n",
"print\" Value of Rf is\",Rf/1000,\"kohm\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" Value of Rf is 250 kohm\n"
]
}
],
"prompt_number": 5
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 3.8 Page No 52"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Given\n",
"E1=2\n",
"E2=3\n",
"E3=1 #input voltage\n",
"R=10*10**3 #in ohm\n",
"\n",
"#calculation\n",
"Vout=-(E1+E2+E3)\n",
"\n",
"#result\n",
"print\" Value of o/p voltage is \",Vout,\"V\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" Value of o/p voltage is -6 V\n"
]
}
],
"prompt_number": 17
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 3.9 Page No 53"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Given\n",
"E1=2.0\n",
"E2=3.0\n",
"E3=-1.0 #input voltage\n",
"R=10*10**3 #in ohm\n",
"\n",
"#Calculation\n",
"Vout=-(E1+E2+E3)\n",
"\n",
"#result\n",
"print\" Value of o/p voltage is\",Vout"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" Value of o/p voltage is -4.0\n"
]
}
],
"prompt_number": 19
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 3.11 Page No 56"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Given\n",
"#Channel 1\n",
"Ri= 10*10**3 #Choosing Input resistance\n",
"Acl = -10.0 \n",
"\n",
"#Calculation\n",
"Rf1 = - (Acl * Ri)\n",
"#Channel 2\n",
"Acl1 = -5\n",
"Rf2 = - (Acl1 * Ri)\n",
"#channel 3\n",
"Acl2 = -2\n",
"Rf3 = - (Acl2 * Ri)\n",
"\n",
"#result\n",
"print\" Value of Rf1 = \",Rf1/1000,\"kohm\"\n",
"print\" Value of Rf2 = \",Rf2/1000,\"kohm\"\n",
"print\" Value of Rf3 = \",Rf3/1000,\"kohm\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" Value of Rf1 = 100.0 kohm\n",
" Value of Rf2 = 50 kohm\n",
" Value of Rf3 = 20 kohm\n"
]
}
],
"prompt_number": 8
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 3.12 Page No 57"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Given\n",
"R1=100000.0\n",
"R2=100000.0\n",
"R3=100000.0\n",
"R=100000.0 #in ohm\n",
"Rf=33*10**3 #in ohm\n",
"E1=5\n",
"E2=5\n",
"E3=-1 #in volts\n",
"n=3.0 #number of inputs\n",
"\n",
"#Calculation\n",
"Vout=-(E1+E2+E3)/n #output voltage\n",
"\n",
"#Result\n",
"print\" Value of output voltage is\",Vout,\"V\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" Value of output voltage is -3.0 V\n"
]
}
],
"prompt_number": 21
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 3.13 Page No 59"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Given\n",
"Rf = 40*10**3\n",
"R1 = 10*10**3\n",
"\n",
"#calculation\n",
"Acl = (Rf + R1)/R1\n",
"print\"(a) Value of Voltage Gain = \",Acl\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"(a) Value of Voltage Gain = 5\n"
]
}
],
"prompt_number": 15
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 3.14 Page No 59"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Given \n",
"Acl = 10.0 # Gain is positive, so choose noninvering amplifier\n",
"R1 = 10*10**3\n",
"\n",
"#Calculation\n",
"Rf = (Acl*R1)-R1\n",
"\n",
"#Result\n",
"print\" Value of Rf is\",Rf/1000,\"kohm\"\n"
],
"language": "python",
"metadata": {},
"outputs": []
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 3.15 Page No 62"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Given\n",
"Ei=4.0 #in volts\n",
"Rl=10*10**3 #in ohm\n",
"I=0 #in ampere\n",
"\n",
"#Calculation\n",
"Vout=Ei #output voltage\n",
"Il=Vout/Rl #load current\n",
"I0=I+Il #output current \n",
"\n",
"#Result\n",
"print\" Value of o/p voltage is\",Vout,\"V\"\n",
"print\" Value of load current is \",Il*1000,\"mA\"\n",
"print\" Value of output current is \",I0*1000,\"mA\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" Value of o/p voltage is 4.0 V\n",
" Value of load current is 0.4 mA\n",
" Value of output current is 0.4 mA\n"
]
}
],
"prompt_number": 11
}
],
"metadata": {}
}
]
}
|
