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Op amp circuit analysis method
Typology: Exercises
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โ
9
๐
12
9
๐
Problem (3):
Show that
0
๐
=
2
1
(1 +
2
๐บ
)
Nodal @ ๐
๐
๐
1
1
๐
3
2
1
2
๐
1
2
๐
3
1
๐
2
๐
1
1
2
3
1
๐
1
2
3
1
2
1
Nodal @ ๐
๐
๐
2
1
๐
4
2
1
2
๐
2
2
๐
4
1
๐
2
๐
1
2
2
4
1
๐
2
2
4
1
2
1
๐
๐
๐
๐
1
2
3
1
2
1
2
2
4
1
2
1
1
2
3
1
2
2
4
1
1
2
2
3
4
1
1
2
๐๐
๐๐
โ 2 ๐
1
๐
2
3
4
Nodal @ ๐
3
3
๐
2
3
0
2
3
4
๐บ
๐บ
2
๐บ
3
๐
๐บ
3
0
2
3
4
3
๐บ
2
๐บ
๐
๐บ
0
2
4
Nodal @ ๐
4
4
๐
2
4
2
4
3
๐บ
๐บ
2
๐บ
4
๐
๐บ
4
2
4
3
4
๐บ
2
๐บ
๐
2
3
๐
๐
1
2
3
4
๐
๐
Eq.2 โ Eq.
3
๐บ
2
๐บ
๐
๐บ
0
2
4
4
๐บ
2
๐บ
๐
2
3
3
4
๐บ
2
3
4
2
๐บ
0
3
4
๐บ
2
๐บ
0
3
4
๐บ
0
๐บ
2
Substitute in eq. 1
๐๐
1
2
๐บ
0
๐บ
2
๐๐
0
1
2
๐บ
๐บ
2
0
๐๐
2
1
2
๐บ
Problem (5):
Find ๐
Nodal @ ๐ 1
1
โ 3
1
1
๐
1
๐
โ 3
1
๐
โ 7
Nodal @ ๐
1
2
โ 3
2
2
๐
2
๐
โ 3
2
๐
โ 7
Nodal @ ๐ ๐
๐
1
๐
0
๐
1
0
Nodal @ ๐
๐
๐
2
๐
2
2
๐
2
Subs 4 in 2
2
2
โ 7
2
๐
2
๐
๐
๐
Subs. ๐
๐
in 1
1
โ 3
โ 7
1
โ 3
โ 3
0
0
1
2
๐
๐