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{
"cells": [
{
"cell_type": "markdown",
"metadata": {},
"source": [
"# Chapter 8 Analog Multiplier"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 8.1 Pg 267"
]
},
{
"cell_type": "code",
"execution_count": 1,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"the output voltage of inverting amplifier (V2) is = 3.00 V\n"
]
}
],
"source": [
"from math import sqrt, pi\n",
"from __future__ import division\n",
"# to determine the output voltage of inverting amplifier (V2)\n",
"Vin = 18 # # V\n",
"V1 = -6 # # V\n",
"\n",
"# in the op-amp due to the infinite i/p resiostance the input current is = 0\n",
"# i1+i2 = 0\n",
"# it gives relation\n",
"Vo = -Vin #\n",
"\n",
"# the output of multiplier is defined as\n",
"#Vo = K*V1*V2\n",
"\n",
"K = 1 # # we assume\n",
"\n",
"V2 = (Vo/(K*V1))#\n",
"print 'the output voltage of inverting amplifier (V2) is = %0.2f'%(V2),'V'"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 8.2 Pg 267"
]
},
{
"cell_type": "code",
"execution_count": 2,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"the output voltage of multiplier is = 225.00 V\n"
]
}
],
"source": [
"# to determine the output voltage of multiplier\n",
"Vin = 15 # # V\n",
"\n",
"# the output of multiplier is defined as\n",
"#Vo = K*V1*V2\n",
"# because of i/p terminal the circuit performs mathematical operation squaring\n",
"# i.e V1 = V2 = Vin\n",
"K = 1 # # we assume\n",
"Vo = K*(Vin)**2#\n",
"print 'the output voltage of multiplier is = %0.2f'%(Vo),'V' "
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 8.3 Pg 268"
]
},
{
"cell_type": "code",
"execution_count": 3,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"the output voltage of inverting amplifier is = 4.00 V \n",
"the output voltage of multiplier is = 16.00 V \n"
]
}
],
"source": [
"from math import sqrt, pi\n",
"# to determine the output voltage of multiplier and inverting amplifier\n",
"Vin = 16 #\n",
"# the output of the inverting amplifier\n",
"K =1 # # we assume\n",
"Vos = sqrt(abs(Vin)/K) #\n",
"print 'the output voltage of inverting amplifier is = %0.2f'%(Vos),' V '\n",
"\n",
"# the output of the multiplier\n",
"Vo = K*Vos**2 #\n",
"print 'the output voltage of multiplier is = %0.2f'%(Vo),' V '"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 8.5 Pg 269"
]
},
{
"cell_type": "code",
"execution_count": 6,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"output voltage of RMS detector is = 10.00 V \n"
]
}
],
"source": [
"# output voltage of of RMS detector\n",
"Vin = 10 # \n",
"T = 1 # # we assume that the charging and discharging period of capacitor\n",
"\n",
"# the output voltage of RMS detector\n",
"# Vo =sqrt(1/T*)integrate(Vin**2*(t),t,0,1 [,atol [,rtol]]) #\n",
"Vo = 10 #\n",
"print 'output voltage of RMS detector is = %0.2f'%(Vo),'V '"
]
}
],
"metadata": {
"kernelspec": {
"display_name": "Python 2",
"language": "python",
"name": "python2"
},
"language_info": {
"codemirror_mode": {
"name": "ipython",
"version": 2
},
"file_extension": ".py",
"mimetype": "text/x-python",
"name": "python",
"nbconvert_exporter": "python",
"pygments_lexer": "ipython2",
"version": "2.7.9"
}
},
"nbformat": 4,
"nbformat_minor": 0
}
|
