1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
138
139
140
141
142
143
144
145
146
147
148
149
150
151
152
153
154
155
156
157
158
159
160
161
162
163
164
165
166
167
168
169
170
171
172
173
174
175
176
177
178
179
180
181
182
183
184
185
186
187
188
189
190
191
192
193
194
195
196
197
198
199
200
201
202
203
204
205
206
207
208
209
210
211
212
213
214
215
216
217
218
219
220
221
222
223
224
225
226
227
228
229
230
231
232
233
234
235
236
237
238
239
240
241
242
243
244
245
246
247
248
249
250
251
252
253
254
255
256
257
258
259
260
261
262
263
264
265
266
267
268
269
270
271
272
273
274
275
276
277
278
279
280
281
282
|
{
"metadata": {
"name": ""
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 15 : Operational amplifier frequency\n",
"Response and compensation"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 15.2, Page No 648"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"#initialisation of variables\n",
"\n",
"R2=1.0*10**6\n",
"Acl=4.5\n",
"\n",
"#Calculations\n",
"R1=R2/Acl\n",
"R1=220*10**3#use standard value\n",
"R3=(R1*R2)/(R1+R2)\n",
"Cf=((R1*30*10**-12)/(R1+R2))*10**12\n",
"\n",
"#Results\n",
"print(\" suitable value of capacitor is %.2fpF \" %Cf)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" suitable value of capacitor is 5.41pF \n"
]
}
],
"prompt_number": 9
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 15.3, Page No 649"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"#initialisation of variables\n",
"\n",
"f=35.0*10**3\n",
"Rf=68.0*10**3\n",
"\n",
"#Calculations\n",
"Cf=(1.0/(2*3.14*f*Rf))*10**12\n",
"\n",
"#Results\n",
"print(\" suitable miller effect capacitor is %.2f pF \" %Cf)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" suitable miller effect capacitor is 66.91 pF \n"
]
}
],
"prompt_number": 10
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 15.5 Page No 652"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"#initialisation of variables\n",
"\n",
"Acl=100.0\n",
"Av=10.0\n",
"print(\" for Cf=30pF\")\n",
"GBW=800.0*10**3\n",
"\n",
"#Calculations\n",
"F2=GBW/Acl\n",
"print(\" for Cf=3pF\")\n",
"GBW=(800*10**3)*Av\n",
"f2=GBW/Acl\n",
"\n",
"#Results\n",
"print(\" The value of f2 is %.2f pF \" %(f2/1000))"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" for Cf=30pF\n",
" for Cf=3pF\n",
" The value of f2 is 80.00 pF \n"
]
}
],
"prompt_number": 11
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 15.6, Page No 654"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"#initialisation of variables\n",
"\n",
"Vip=1.0\n",
"R2=39.0*10**3\n",
"R3=4.7*10**3\n",
"SR=250.0/10**-6\n",
"f=100.0*10**3\n",
"\n",
"#Calculations\n",
"print(\" for the AD843\")\n",
"Vop=((R2+R3)/R3)*Vip\n",
"fp=SR/(2*3.14*Vop)\n",
"print(\"full power bandwidth is %dHz \" %fp)\n",
"print(\" for a 741\")\n",
"SR=0.5/10**-6\n",
"Vp=SR/(2*3.14*f)\n",
"\n",
"#Results\n",
"print(\" maximum peak output voltage is %3.2fV \" %Vp)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" for the AD843\n",
"full power bandwidth is 4281508Hz \n",
" for a 741\n",
" maximum peak output voltage is 0.80V \n"
]
}
],
"prompt_number": 12
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 15.7 Page No 656"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"#initialisation of variables\n",
"\n",
"rs=600.0\n",
"R1=1.0*10**3\n",
"R2=10.0*10**3\n",
"f=800.0*10**3\n",
"\n",
"#Calculations\n",
"print(\" stray capacitance\")\n",
"Cs=1/(2*3.14*f*10*(((rs+R1)*R2)/(rs+R1+R2)))\n",
"print(\"compensation capacitor\")\n",
"C2=((Cs*(rs+R1))/R2)*10**12\n",
"\n",
"#Results\n",
"print(\"compensation capacitor is %.2fpF \" %C2)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" stray capacitance\n",
"compensation capacitor\n",
"compensation capacitor is 2.31pF \n"
]
}
],
"prompt_number": 13
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 15.8 Page No 659"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"#initialisation of variables\n",
"ro=25.0\n",
"f=2.0*10**6\n",
"R2=10.0*10**3\n",
"Rx=25.0\n",
"\n",
"#Calculations\n",
"Cl=(1.0/(2.0*3.14*f*(10*ro)))*10**+12\n",
"print(\" load capacitance is %3.2fpF \" %Cl)\n",
"Cl=0.1*10**-6\n",
"C2=((Cl*(ro+Rx))/R2)*10**12\n",
"\n",
"#Results\n",
"print(\" compensation capacitance is %.2f pF \" %C2)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" load capacitance is 318.47pF \n",
" compensation capacitance is 500.00 pF \n"
]
}
],
"prompt_number": 14
}
],
"metadata": {}
}
]
}
|
