analysis of 741-op-amp-ic by muhammad irfan yousuf [peon of holy prophet (p.b.u.h)] volume 2
TRANSCRIPT
Muhammad Irfan Yousuf [Peon of Holy Prophet (P.B.U.H)] 2000-E-41 510 Cell: 0300-8454295; Tel: 042-5421893
1 1
0
R1 R2
-1
0
R1
0a36a11a12
a51a52
1 1
-1
R1 R2 R2
0 0
0a41a11a13
a51a53
1 1
-1
R1 R2 R1
0 0
0a42a11a14
a51a54
1 1
0
R1 R2
A
0
Rout
1 1 A
a43 R1 R2 Routa11a15
a51a55
1 1
0
R1 R2
1
0
Rout
1 1 1
a44 R1 R2 Routa11a16
a51a56
1 1
0
R1 R2
-1
0
R3
1 1 -1
a45 R1 R2 R3a11a17
a51a57
1 1
0
R1 R2
AV2 0
Rout
1 1 AV2
a46 R1 R2 Routa11a12
a61a62
1 1
-1
R1 R2 R2
0 0
0
a51a11a13
a61a63
1 1
-1
R1 R2 R1
0 0
0
a52a11a14
a61a64
1 1
0
R1 R2
0 0
0
a53a11a15
a61a65
1 1
0
R1 R2
-1
0
R3
1 1 -1
a54 R1 R2 R3a11a16
a61a66
1 1
0
R1 R2
11
0
R3 R4
1 1 1 1
a55 R1 R2 R3 R4a11a17
a61a67
1 1
0
R1 R2
0 0
0
a56a11a12
a71a72
1 1
-1
R1 R2 R2
0 0
0
a61a11a13
a71a73
1 1
-1
R1 R2 R1
A
0
Rout
1 1 A
a62 R1 R2 Routa11a14
a71a74
1 1
0
R1 R2
0 0
0
a63a11a15
a71a75
1 1
0
R1 R2
0 0
0
a64a11a16
a71a76
1 1
0
R1 R2
-A
0
Rout
1 1 -A
a65 R1 R2 Routa11a17
a71a77
1 1
0
R1 R2
0 0
0
a66
a11 a12 a13 a14 a15 a16
a21 a22 a23 a24 a25 a26
a31 a32 a33 a34 a35 a36 (A( a41 a42 a43 a44 a45 a46
a51 a52 a53 a54 a55 a56
a61 a62 a63 a64 a65 a66
a11 a12 a11 a13 a11 a14 a11 a15 a11 a16
a21 a22 a21 a23 a21 a24 a21 a25 a21 a26
a11 a12 a11 a13 a11 a14 a11 a15 a11 a16
a31 a32 a31 a33 a31 a34 a31 a35 a31 a36
a11 a12 a11 a13 a11 a14 a11 a15 a11 a16
n - 2
1
a41 a42 a41 a43 a41 a44 a41 a45 a41 a46
a11
a11 a12 a11 a13 a11 a14 a11 a15 a11 a16
a51 a52 a51 a53 a51 a54 a51 a55 a51 a56
a11 a12 a11 a13 a11 a14 a11 a15 a11 a16
a61 a62 a61 a63 a61 a64 a61 a65 a61 a66
a11a12
a21a22= a11a22 a21a12
1 1
1 A 1 1 11 1 1 1-1
R1 R2 Rout Rout R2 R1 R2 R3 R4 R1 R2 R3
11 -1A1
R1 R2 R3 Rout R1
a11
a11a13
a21a23= a11a23 a21a13
= 0
a12a11a14
a21a24= a11a24 a21a14
= 0
a13a11a15
a21a25= a11a25 a21a15
= 0
a14a11a16
a21a26= a11a26 a21a16
11 -1 1
1 AV1
a15 R1 R2 R3 R1 R2 Routa11a12
a31a32= a11a32 a31a12
1 1
1 A 1 -1 -1
R1 R2 Rout Rout R2 R12 R1R2
a21 11 -1A1
R1 R2 R3 Rout R1a11a13
a31a33= a11a33 a31a13
2
1 1
1 A 1 1 1
a22 R1 R2 Rout Rout R2 R1 R2
a11a14
a31a34= a11a34 a31a14
1 1
1 A 1 1 1 -1
a23 R1 R2 Rout Rout R2 R1 R2 R2
a11a15
a31a35= a11a35 a31a15
0
a24a11a16
a31a36= a11a36 a31a16
1 11 AV1
a25R1R2 R1 R2 Routa11a12
a41a42= a11a42 a41a12
0
a31a11a13
a41a43= a11a43 a41a13
1 1
1 A 1 1 1 A
a32 R1 R2 Rout Rout R2 R1 R2 Rout
a11a14
a41a44= a11a44 a41a14
1 1
1 A 1 1 1 1
a33 R1 R2 Rout Rout R2 R1 R2 Rout
a11a15
a41a45= a11a45 a41a15
1 1
1 A 1 1 1 -1
a34 R1 R2 Rout Rout R2 R1 R2 R3
a11a16
a41a46= a11a46 a41a16
1 1
1 A 1 1 1 AV2
a35 R1 R2 Rout Rout R2 R1 R2 Rout
a11a12
a51a52= a11a52 a51a12
0
a41a11a13
a51a53= a11a53 a51a13
0
a42a11a14
a51a54= a11a54 a51a14
1 1
1 A 1 1 1 -1
a43 R1 R2 Rout Rout R2 R1 R2 R3
a11a15
a51a55= a11a55 a51a15
1 1
1 A 1 1 11 1
a44 R1 R2 Rout Rout R2 R1 R2 R3 R4 a11a16
a51a56= a11a56 a51a16
0
a45
1 1
1 A 1 1 1 A
a51 R1 R2 Rout Rout R2 R1 R2 Rout
a11a13
a61a63= a11a63 a61a13
0
a52a11a14
a61a64= a11a64 a61a14
0
a53a11a15
a61a65= a11a65 a61a15
1 1
1 A 1 1 1 -A
a54 R1 R2 Rout Rout R2 R1 R2 Rout
a11a16
a61a66= a11a66 a61a16 0
a55
a11 a12 a13 a14 a15
a21 a22 a23 a24 a25 (A( a31 a32 a33 a34 a35 a41 a42 a43 a44 a45
a51 a52 a53 a54 a55
a11 a12 a11 a13 a11 a14 a11 a15
a21 a22 a21 a23 a21 a24 a21 a25
a11 a12 a11 a13 a11 a14 a11 a15
n - 2
1
a31 a32 a31 a33 a31 a34 a31 a35
a11
a11 a12 a11 a13 a11 a14 a11 a15
a41 a42 a41 a43 a41 a44 a41 a45
a11 a12 a11 a13 a11 a14 a11 a15
a51 a52 a51 a53 a51 a54 a51 a55
a11a12
a21a22= a11a22 a21a12
1 1
1 A 1 1 11 1 1 1-1
R1 R2 Rout Rout R2 R1 R2 R3 R4 R1 R2 R3
2
11 -1A1 1 11A1 1 1
R1 R2 R3 Rout R1 R1 R2 Rout Rout R2 R1 R2a11a11a13
a21a23= a11a23 a21a13
1 1
1 A 1 1 11 1 1 1-1
R1 R2 Rout Rout R2 R1 R2 R3 R4 R1 R2 R3
11 -1A1 1 11A1 1 1
R1 R2 R3 Rout R1 R1 R2 Rout Rout R2 R1 R2
-1
R2a12 a11a14
a21a24= a11a24 a21a14
0
a13
a11a15
a21a25= a11a25 a21a15
1 1
1 A 1 1 11 1 1 1-1
R1 R2 Rout Rout R2 R1 R2 R3 R4 R1 R2 R3
11 -1A1111 AV1
R1 R2 R3 Rout R1R1R2 R1 R2Rout
1 1
1 A 1 -1 -1
R1 R2 Rout Rout R2 R12 R1R2
11 -1A11 1-11
1AV1
R1 R2 R3 Rout R1 R1R2 R3 R1 R2 Routa14
a11a12
a31a32= a11a32 a31a12
1 1
1 A 1 1 11 1 1 1-1
R1 R2 Rout Rout R2 R1 R2 R3 R4 R1 R2 R3
11 -1A1 1 11A1 1 1
R1 R2 R3 Rout R1 R1 R2 Rout Rout R2 R1 R2
A
Routa21
a11a13
a31a33= a11a33 a31a13
1 1
1 A 1 1 11 1 1 1-1
R1 R2 Rout Rout R2 R1 R2 R3 R4 R1 R2 R3
11 -1A1 1 11A1 1 1
R1 R2 R3 Rout R1 R1 R2 Rout Rout R2 R1 R2
1
Routa22
a11a14
a31a34= a11a34 a31a14
1 1
1 A 1 1 11 1 1 1-1
R1 R2 Rout Rout R2 R1 R2 R3 R4 R1 R2 R3
11 -1A1 1 11A1 1 1
R1 R2 R3 Rout R1 R1 R2 Rout Rout R2 R1 R2
-1
R3a23
a11a15
a31a35= a11a35 a31a15
1 1
1 A 1 1 11 1 1 1-1
R1 R2 Rout Rout R2 R1 R2 R3 R4 R1 R2 R3
11 -1A1
R1 R2 R3 Rout R1
1 1
1 A 1 1 1 AV2
a24 R1 R2 Rout Rout R2 R1 R2 Rout
a11a12
a41a42= a11a42 a41a12
0
a31
a11a13
a41a43= a11a43 a41a13
1 1
1 A 1 1 11 1 1 1-1
R1 R2 Rout Rout R2 R1 R2 R3 R4 R1 R2 R3
11 -1A1 1 11A1 1 1
R1 R2 R3 Rout R1 R1 R2 Rout Rout R2 R1 R2
-1
R3a32
a11a14
a41a44= a11a44 a41a14
1 1
1 A 1 1 11 1 1 1-1
R1 R2 Rout Rout R2 R1 R2 R3 R4 R1 R2 R3
11 -1A1 1 11A1 1 1
R1 R2 R3 Rout R1 R1 R2 Rout Rout R2 R1 R2
11
R3 R4a33
a11a15
a41a45= a11a45 a41a15
0
a34
a11a12
a51a52= a11a52 a51a12
0
a41
a11a13
a51a53= a11a53 a51a13
0
a42
a11a14
a51a54= a11a54 a51a14
1 1
1 A 1 1 11 1 1 1-1
R1 R2 Rout Rout R2 R1 R2 R3 R4 R1 R2 R3
11 -1A1 1 11A1 1 1
R1 R2 R3 Rout R1 R1 R2 Rout Rout R2 R1 R2
-A
Routa43a11a15
a51a55
= a11a55 a51a15
1 1
1 A 1 1 1 A
R1 R2 Rout Rout R2 R1 R2 Rout
11 -1 1
1 AV1
a44 R1 R2 R3 R1 R2 Rout
a11 a12 a13 a14
a21 a22 a23 a24 (A(
a31 a32 a33 a34 a41 a42 a43 a44
a11 a12 a11 a13 a11 a14
a21 a22 a21 a23 a21 a24
n - 2
1
a11 a12 a11 a13 a11 a14
a11
a31 a32 a31 a33 a31 a34
a11 a12 a11 a13 a11 a14
a41 a42 a41 a43 a41 a44
a11a12
a21a22= a11a22 a21a12
1 1
1 A 1 1 11 1 1 1-1
R1 R2 Rout Rout R2 R1 R2 R3 R4 R1 R2 R3
2
11 -1A1 1 11A1 1 1
R1 R2 R3 Rout R1 R1 R2 Rout Rout R2 R1 R2
1 1
1 A 1 1 11 1 1 1-1
R1 R2 Rout Rout R2 R1 R2 R3 R4 R1 R2 R3
11 -1A1 1 11A1 1 1
R1 R2 R3 Rout R1 R1 R2 Rout Rout R2 R1 R2
1
Rout
1 1
1 A 1 1 11 1 1 1-1
R1 R2 Rout Rout R2 R1 R2 R3 R4 R1 R2 R3
11 -1A1 1 11A1 1 1
R1 R2 R3 Rout R1 R1 R2 Rout Rout R2 R1 R2
A
Rout
1 1
1 A 1 1 11 1 1 1-1
R1 R2 Rout Rout R2 R1 R2 R3 R4 R1 R2 R3
11 -1A1 1 11A1 1 1
R1 R2 R3 Rout R1 R1 R2 Rout Rout R2 R1 R2
-1
R2a11
a11a13
a21a23= a11a23 a21a13
1 1
1 A 1 1 11 1 1 1-1
R1 R2 Rout Rout R2 R1 R2 R3 R4 R1 R2 R3
2
11 -1A1 1 11A1 1 1
R1 R2 R3 Rout R1 R1 R2 Rout Rout R2 R1 R2
1 1
1 A 1 1 11 1 1 1-1
R1 R2 Rout Rout R2 R1 R2 R3 R4 R1 R2 R3
11 -1A1 1 11A1 1 1
R1 R2 R3 Rout R1 R1 R2 Rout Rout R2 R1 R2
-1
R3a12
a11a14
a21a24= a11a24 a21a14
1 1
1 A 1 1 11 1 1 1-1
R1 R2 Rout Rout R2 R1 R2 R3 R4 R1 R2 R3
2
11 -1A1 1 11A1 1 1
R1 R2 R3 Rout R1 R1 R2 Rout Rout R2 R1 R2
1 1
1 A 1 1 11 1 1 1-1
R1 R2 Rout Rout R2 R1 R2 R3 R4 R1 R2 R3
11 -1A1
R1 R2 R3 Rout R1
1 1
1 A 1 1 1 AV2
R1 R2 Rout Rout R2 R1 R2 Rout
1 1
1 A 1 1 11 1 1 1-1
R1 R2 Rout Rout R2 R1 R2 R3 R4 R1 R2 R3
11 -1A1 1 11A1 1 1
R1 R2 R3 Rout R1 R1 R2 Rout Rout R2 R1 R2
A
Rout
1 1
1 A 1 1 11 1 1 1-1
R1 R2 Rout Rout R2 R1 R2 R3 R4 R1 R2 R3
11 -1A1111 AV1
R1 R2 R3 Rout R1R1R2 R1 R2Rout
1 1
1 A 1 -1 -1
R1 R2 Rout Rout R2 R12 R1R2
11 -1A11 1-11
1 AV1
a13 R1 R2 R3 Rout R1 R1R2 R3 R1 R2 Routa11a12
a31a32= a11a32 a31a12
1 1
1 A 1 1 11 1 1 1-1
R1 R2 Rout Rout R2 R1 R2 R3 R4 R1 R2 R3
2
11 -1A1 1 11A1 1 1
R1 R2 R3 Rout R1 R1 R2 Rout Rout R2 R1 R2
1 1
1 A 1 1 11 1 1 1-1
R1 R2 Rout Rout R2 R1 R2 R3 R4 R1 R2 R3
11 -1A1 1 11A1 1 1
R1 R2 R3 Rout R1 R1 R2 Rout Rout R2 R1 R2
-1
R3a21
a11a13
a31a33= a11a33 a31a13
1 1
1 A 1 1 11 1 1 1-1
R1 R2 Rout Rout R2 R1 R2 R3 R4 R1 R2 R3
2
11 -1A1 1 11A1 1 1
R1 R2 R3 Rout R1 R1 R2 Rout Rout R2 R1 R2
1 1
1 A 1 1 11 1 1 1-1
R1 R2 Rout Rout R2 R1 R2 R3 R4 R1 R2 R3
11 -1A1 1 11A1 1 1
R1 R2 R3 Rout R1 R1 R2 Rout Rout R2 R1 R2
11
R3 R4a22
a11a14
a31a34= a11a34 a31a14
0
a23
a11a12
a41a42= a11a42 a41a12
0a31a11a13
a41a43= a11a43 a41a13
1 1
1 A 1 1 11 1 1 1-1
R1 R2 Rout Rout R2 R1 R2 R3 R4 R1 R2 R3
2
11 -1A1 1 11A1 1 1
R1 R2 R3 Rout R1 R1 R2 Rout Rout R2 R1 R2
1 1
1 A 1 1 11 1 1 1-1
R1 R2 Rout Rout R2 R1 R2 R3 R4 R1 R2 R3
11 -1A1 1 11A1 1 1
R1 R2 R3 Rout R1 R1 R2 Rout Rout R2 R1 R2
-A
Routa32a11a14
a41a44= a11a44 a41a14
1 1
1 A 1 1 11 1 1 1-1
R1 R2 Rout Rout R2 R1 R2 R3 R4 R1 R2 R3
2
11 -1A1 1 11A1 1 1
R1 R2 R3 Rout R1 R1 R2 Rout Rout R2 R1 R2
1 1
1 A 1 1 1 A
R1 R2 Rout Rout R2 R1 R2 Rout
11 -1 1
1 AV1
a33 R1 R2 R3 R1 R2 Rout
a11 a12 a13 (A( a21 a22 a23
a31 a32 a33
a11 a12 a11 a13
n - 2
1
a21 a22 a21 a23
a11
a11 a12 a11 a13
a31 a32 a31 a33
a11a12
a21a22= a11a22 a21a12
1 1
1 A 1 1 11 1 1 1-1
R1 R2 Rout Rout R2 R1 R2 R3 R4 R1 R2 R3
2
11 -1A1 1 11A1 1 1
R1 R2 R3 Rout R1 R1 R2 Rout Rout R2 R1 R2
1 1
1 A 1 1 11 1 1 1-1
R1 R2 Rout Rout R2 R1 R2 R3 R4 R1 R2 R3
11 -1A1 1 11A1 1 1
R1 R2 R3 Rout R1 R1 R2 Rout Rout R2 R1 R2
1
Rout
1 1
1 A 1 1 11 1 1 1-1
R1 R2 Rout Rout R2 R1 R2 R3 R4 R1 R2 R3
11 -1A1 1 11A1 1 1
R1 R2 R3 Rout R1 R1 R2 Rout Rout R2 R1 R2
A
Rout
1 1
1 A 1 1 11 1 1 1-1
R1 R2 Rout Rout R2 R1 R2 R3 R4 R1 R2 R3
11 -1A1 1 11A1 1 1
R1 R2 R3 Rout R1 R1 R2 Rout Rout R2 R1 R2
-1
R2
1 1
1 A 1 1 11 1 1 1-1
R1 R2 Rout Rout R2 R1 R2 R3 R4 R1 R2 R3
2
11 -1A1 1 11A1 1 1
R1 R2 R3 Rout R1 R1 R2 Rout Rout R2 R1 R2
1 1
1 A 1 1 11 1 1 1-1
R1 R2 Rout Rout R2 R1 R2 R3 R4 R1 R2 R3
11 -1A1 1 11A1 1 1
R1 R2 R3 Rout R1 R1 R2 Rout Rout R2 R1 R2
11
R3 R4
1 1
1 A 1 1 11 1 1 1-1
R1 R2 Rout Rout R2 R1 R2 R3 R4 R1 R2 R3
2
11 -1A1 1 11A1 1 1
R1 R2 R3 Rout R1 R1 R2 Rout Rout R2 R1 R2
1 1
1 A 1 1 11 1 1 1-1
R1 R2 Rout Rout R2 R1 R2 R3 R4 R1 R2 R3
11 -1A1 1 11A1 1 1
R1 R2 R3 Rout R1 R1 R2 Rout Rout R2 R1 R2
-1
R3
1 1
1 A 1 1 11 1 1 1-1
R1 R2 Rout Rout R2 R1 R2 R3 R4 R1 R2 R3
2
11 -1A1 1 11A1 1 1
R1 R2 R3 Rout R1 R1 R2 Rout Rout R2 R1 R2
1 1
1 A 1 1 11 1 1 1-1
R1 R2 Rout Rout R2 R1 R2 R3 R4 R1 R2 R3
11 -1A1 1 11A1 1 1
R1 R2 R3 Rout R1 R1 R2 Rout Rout R2 R1 R2
-1
R3a11
a11a13
a21a23= a11a23 a21a13
1 1
1 A 1 1 11 1 1 1-1
R1 R2 Rout Rout R2 R1 R2 R3 R4 R1 R2 R3
2
11 -1A1 1 11A1 1 1
R1 R2 R3 Rout R1 R1 R2 Rout Rout R2 R1 R2
1 1
1 A 1 1 11 1 1 1-1
R1 R2 Rout Rout R2 R1 R2 R3 R4 R1 R2 R3
11 -1A1
R1 R2 R3 Rout R1
1 1
1 A 1 1 1 AV2
R1 R2 Rout Rout R2 R1 R2 Rout
1 1
1 A 1 1 11 1 1 1-1
R1 R2 Rout Rout R2 R1 R2 R3 R4 R1 R2 R3
11 -1A1 1 11A1 1 1
R1 R2 R3 Rout R1 R1 R2 Rout Rout R2 R1 R2
A
Rout
1 1
1 A 1 1 11 1 1 1-1
R1 R2 Rout Rout R2 R1 R2 R3 R4 R1 R2 R3
11 -1A1111 AV1
R1 R2 R3 Rout R1R1R2 R1 R2Rout
1 1
1 A 1 -1 -1
R1 R2 Rout Rout R2 R12 R1R2
11 -1A11 1-11
1 AV1
a12 R1 R2 R3 Rout R1 R1R2 R3 R1 R2 Routa11a12
a31a32= a11a32 a31a12
1 1
1 A 1 1 11 1 1 1-1
R1 R2 Rout Rout R2 R1 R2 R3 R4 R1 R2 R3
2
11 -1A1 1 11A1 1 1
R1 R2 R3 Rout R1 R1 R2 Rout Rout R2 R1 R2
1 1
1 A 1 1 11 1 1 1-1
R1 R2 Rout Rout R2 R1 R2 R3 R4 R1 R2 R3
11 -1A1 1 11A1 1 1
R1 R2 R3 Rout R1 R1 R2 Rout Rout R2 R1 R2
1
Rout
1 1
1 A 1 1 11 1 1 1-1
R1 R2 Rout Rout R2 R1 R2 R3 R4 R1 R2 R3
11 -1A1 1 11A1 1 1
R1 R2 R3 Rout R1 R1 R2 Rout Rout R2 R1 R2
A
Rout
1 1
1 A 1 1 11 1 1 1-1
R1 R2 Rout Rout R2 R1 R2 R3 R4 R1 R2 R3
11 -1A1 1 11A1 1 1
R1 R2 R3 Rout R1 R1 R2 Rout Rout R2 R1 R2
-1
R2
1 1
1 A 1 1 11 1 1 1-1
R1 R2 Rout Rout R2 R1 R2 R3 R4 R1 R2 R3
2
11 -1A1 1 11A1 1 1
R1 R2 R3 Rout R1 R1 R2 Rout Rout R2 R1 R2
1 1
1 A 1 1 11 1 1 1-1
R1 R2 Rout Rout R2 R1 R2 R3 R4 R1 R2 R3
11 -1A1 1 11A1 1 1
R1 R2 R3 Rout R1 R1 R2 Rout Rout R2 R1 R2
-A
Rout
a21a11a13
a31a33= a11a33 a31a13
1 1
1 A 1 1 11 1 1 1-1
R1 R2 Rout Rout R2 R1 R2 R3 R4 R1 R2 R3
2
11 -1A1 1 11A1 1 1
R1 R2 R3 Rout R1 R1 R2 Rout Rout R2 R1 R2
1 1
1 A 1 1 11 1 1 1-1
R1 R2 Rout Rout R2 R1 R2 R3 R4 R1 R2 R3
11 -1A1 1 11A1 1 1
R1 R2 R3 Rout R1 R1 R2 Rout Rout R2 R1 R2
1
Rout
1 1
1 A 1 1 11 1 1 1-1
R1 R2 Rout Rout R2 R1 R2 R3 R4 R1 R2 R3
11 -1A1 1 11A1 1 1
R1 R2 R3 Rout R1 R1 R2 Rout Rout R2 R1 R2
A
Rout
1 1
1 A 1 1 11 1 1 1-1
R1 R2 Rout Rout R2 R1 R2 R3 R4 R1 R2 R3
11 -1A1 1 11A1 1 1
R1 R2 R3 Rout R1 R1 R2 Rout Rout R2 R1 R2
-1
R2
1 1
1 A 1 1 11 1 1 1-1
R1 R2 Rout Rout R2 R1 R2 R3 R4 R1 R2 R3
2
11 -1A1 1 11A1 1 1
R1 R2 R3 Rout R1 R1 R2 Rout Rout R2 R1 R2
1 1
1 A 1 1 11 1 1 1-1
R1 R2 Rout Rout R2 R1 R2 R3 R4 R1 R2 R3
11 -1A1 1 11A1 1 1
R1 R2 R3 Rout R1 R1 R2 Rout Rout R2 R1 R2
1 111A 111A
Rout R1 R2 Rout Rout R2 R1 R2 Rout
1 1
1 A 1 1 1 -1
a22
R1 R2 Rout Rout R2 R1 R2 R4
n - 2 n 2 n - 2
n - 2 n 2 n - 2
1
11111 a11
a12(A( a11 a11 a11 a11 a11 a11 a21
a22
7 - 2 6 2 5 - 2
4 - 2 3 2 2 - 2
1
11111 a11
a12(A( a11 a11 a11 a11 a11 a11 a21
a22Q#4.40: The electronic ammeter in Example 4.9 has been modified and is shown in Fig. P4.40. The selector switch allows the user to change the range of the meter. Using values for R1 and R2 from Example 4.9, find the values of RA and RB that will yield a 10-V output when the current being measured is 100 mA and 10 mA, respectively.
Solution:
Unknown current
I+
RA RB RC= 1 k( R2 = 9 k(
V0
R1 = 1 k(
-
fig. (a)
Unknown current
I+
RA = 1 k(
R2 = 9 k(
V0
R1 = 1 k(
-
fig. (a)
OP-AMP MODEL
I
V0 Rin Rout
AVe R2
RA
R1
Fig. (b)
I V1 V2 V0
V+ Rin V- R2
Rout
RAR1
AVe
Using Nodal AnalysisFig. (c)
Applying KCL at node labeled V1
According to KCL
The algebraic sum of all the currents leaving that junction = the algebraic sum of all the currents entering into the junction
V1 V2 V1 I
Rin RAV1 V2 V1 0.1
Rin Rin RAHere
Rin = 108 (Rout = 10 (A = 105
We to bring about a result that
1 1 1 1 11 1
>> & > &( 0
R0 RF RinHere R0 = Rout; Ri = Rin
V1 0.1
RA
V1 = 0.1 RA (i)
Applying KCL at node labeled V2
According to KCL
The algebraic sum of all the currents leaving that junction = the algebraic sum of all the currents entering into the junction
V2 V1 V2 V2 V0 0
Rin R1R2V2 V1 V2 V2 V0 0
Rin Rin R1 R2 R2V2 V2 V0 0
R1 R2 R2
11-1
V2 V0 0 (ii)R1 R2 R2Applying KCL at node labeled V0
According to KCL
The algebraic sum of all the currents leaving that junction = the algebraic sum of all the currents entering into the junction
V0 V2 V0 - AVe 0
R2RoutHere
Ve = V+ - V-
Ve = V1 V2
V0 V2 V0 A(V1 V2)
0
R2RoutV0 V2 V0 AV1 AV2
0
R2 R2 Rout Rout Rout
-A A11 1
V1 V2 V0 0
Rout Rout R2 Rout R2
-A A
1
V1 V2 V0 0 (iii)
Rout Rout Rout Writing equations (i), (ii) & (iii) in matrix form
1 0
0
V1 0.1 RA 1 1 -1
0
R1 R2 R2 V2 0
-AA
1
V0 0
RoutRout RoutA
X B
1 0
0
11
-1
(A( 0 R1 R2 R2
-AA1
Rout RoutRout
111A1
R1 R2 RoutRout R2
1 0
0.1 RA
11
0 R1 R2 0
-AA
0
Rout Rout
V0
(A(
A11
0.1 RA Rout R1R2
111A1
R1 R2 RoutRout R2Substituting the corresponding values
105 1 1
0.1 RA 10 1000 9000
1
1
1 105 1
1000 9000 10 10 9000
0.1 RA (10000) (0.001 + 1.111 ( 10-4)
10
(0.001 + 1.111 ( 10-4) (0.1) + (10000) (1.111 ( 10-4)
1.111 RA
10
1.111
1.111 RA = 11.11
RA = 10 (
Unknown current
I+
RB = 1 k(
R2 = 9 k(
V0
R1 = 1 k(
-
fig. (a)
OP-AMP MODEL
I
V0 Rin Rout
AVe R2
RB
R1
Fig. (b)
I V1 V2 V0
V+ Rin V- R2
Rout
RBR1
AVe
Using Nodal AnalysisFig. (c)
Applying KCL at node labeled V1
According to KCL
The algebraic sum of all the currents leaving that junction = the algebraic sum of all the currents entering into the junction
V1 V2 V1 I
Rin RBV1 V2 V1 0.01
Rin Rin RBHere
Rin = 108 (Rout = 10 (A = 105
We to bring about a result that
1 1 1 1 11 1
>> & > &( 0
R0 RF RinHere R0 = Rout; Ri = Rin
V1 0.01
RB
V1 = 0.01 RB (i)
Applying KCL at node labeled V2
According to KCL
The algebraic sum of all the currents leaving that junction = the algebraic sum of all the currents entering into the junction
V2 V1 V2 V2 V0 0
Rin R1R2V2 V1 V2 V2 V0 0
Rin Rin R1 R2 R2V2 V2 V0 0
R1 R2 R2
11-1
V2 V0 0 (ii)R1 R2 R2Applying KCL at node labeled V0
According to KCL
The algebraic sum of all the currents leaving that junction = the algebraic sum of all the currents entering into the junction
V0 V2 V0 - AVe 0
R2RoutHere
Ve = V+ - V-
Ve = V1 V2
V0 V2 V0 A(V1 V2)
0
R2RoutV0 V2 V0 AV1 AV2
0
R2 R2 Rout Rout Rout
-A A11 1
V1 V2 V0 0
Rout Rout R2 Rout R2
-A A
1
V1 V2 V0 0 (iii)
Rout Rout Rout Writing equations (i), (ii) & (iii) in matrix form
1 0
0
V1 0.01 RB 1 1 -1
0
R1 R2 R2 V2 0
-AA
1
V0 0
RoutRout RoutA
X B
1 0
0
11
-1
(A( 0 R1 R2 R2
-AA1
Rout RoutRout
111A1
R1 R2 RoutRout R2
1 0
0.01 RB
11
0 R1 R2 0
-AA
0
Rout Rout
V0
(A( A 11
0.01 RB Rout R1 R2
111A1
R1 R2 RoutRout R2Substituting the corresponding values
105 1 1
0.01 RB 10 1000 9000
1
1
1 105 1
1000 9000 10 10 9000
0.01 RB (10000) (0.001 + 1.111 ( 10-4)
10
(0.001 + 1.111 ( 10-4) (0.1) + (10000) (1.111 ( 10-4)
0.111 RB
10
1.111
0.111 RB = 11.11
RB = 100.09 (Q#4.41: Given a box of 10-k( resistors and an op-amp, design a circuit that will have an output voltage of V0 = -2V1 4V2.
Solution:
Circuit diagram:
R1R
R2
+
V1 V2
V0
-
fig. (a)
R1
R
OP-AMP MODEL
R2
+
Rin Rout
V1 V2 AVe
V0 -
fig. (b)
V3RV0
V- R1 R2 RinRout
V+
V1 V2AVe
fig. (c)
Applying KCL at node labeled V3
According to KCL
The algebraic sum of all the currents leaving that junction = the algebraic sum of all the currents entering into the junction
V3 V1 V3 V2 V3 V3 V0
0
R1R2 RinR
V3 V1 V3 V2 V3 V3 V0
0
R1 R1 R2 R2 Rin R R
Here
Rin = 108 (Rout = 10 (A = 105
We to bring about a result that
1 1 1 1 11 1
>> & > &( 0
R0 RF RinHere R0 = Rout; Ri = Rin
111 -1
V1 V2 V3 V0 (i) R1 R2 R R R1 R2Applying KCL at node labeled V0
According to KCL
The algebraic sum of all the currents leaving that junction = the algebraic sum of all the currents entering into the junction
V0 V3 V0 - AVe 0
RRoutVe = V+ - V-
Ve = 0 V3
Ve = -V3
V0 V3 V0 A(-V3)
0
RRoutV0 V3 V0 + AV3 0
RRoutV0 V3 V0 AV3
0
R R Rout Rout
A111
V3 V0 0
Rout R Rout R
A 1
V3 V0 0
(ii) Rout Rout Writing equations (i) & (ii) in matrix form
111 -1 V1 V2
V3 R1 R2 RR R1 R2
A 1
V0 0
Rout Rout
A X B
11 1 -1
R1R2R R
(A(
A 1
Rout Rout
1 111A1
R1 R2 R RoutRout R
11 1 V1 V2
R1R2R R1 R2
A
0
Rout
V0
(A(
-A V1 V2
Rout R1 R2
1 111A1
R1 R2 R RoutRout R
-A R2V1 + R1V2
Rout R1R2
RR2 + RR1 + R1R2 A
RR1R2RoutRRout
-A R2V1 + R1V2
Rout R1R2
RR2 + RR1 + R1R2 + AR1R2
RR1R2Rout
-AR (R2V1 + R1V2)
RR2 + RR1 + R1R2 + AR1R2
-AR (R2V1 + R1V2) Hint: A = 105; R, R1 & R2 are in kilo ohms
AR1R2
-RR
V0 V1 V2 (iii) R1 R2V0 = -2V1 4V2 (iv)Comparing equations (iii) & (iv)
R
2
R120 k( 2
10 k(
R
4
R220 k( 4
5 k(
R1 = 10 kR = 20 k
R2 = 5 k10 k 10 k
10 k+
V0 10 k
V1 V2-
fig. (d)
Since signs on gains associated with V1 & V2 are both negative, a simple summer will suffice.
Q#4.42: Design an op-amp circuit that has a gain of 50 using resistors no smaller than 1 k(.
Solution:
Circuit diagram:
Since sign is negative, use inverting configuration.
R2
VS
R1
V0
Fig. (a)
R2OP-AMP MODEL
V0 Rin Rout
R1AVe
VS
Fig. (b)
V1 V0
R1
R2
V-
Rout
VS
Rin
V+
AVe
Fig. (c)
Using nodal analysis:
Applying KCL at node Labeled V1
Assume that all the currents are going away from the junction
According to KCL
The algebraic sum of all the currents leaving that junction = the algebraic sum of all the entering into the junction
V1 - VS V1V1 V0
+ + 0R1 Rin
R2V1 VS V1 V1 V0
+ +0R1 R1 Rin R2 R2
1 11-1 VS + +V1 + V0
R1 Rin R2 R2 R1
1 1 -1 VS V1 V0
(i)R1 R2 R2 R1Applying KCL at node Labeled V0
Assume that all the currents are going away from the junction
According to KCL
The algebraic sum of all the currents leaving that junction = the algebraic sum of all the entering into the junction
V0 V1 V0 - AVe
+ 0
R2Rout
V0 V1 V0 AVe
+ 0
R2 R2 Rout Rout Here
Ve = V+ - V-
Ve = 0 V V1
Ve = -V1
V0 V1 V0 A(-V1)
+ 0
R2 R2 Rout Rout V0 V1 V0 AV1
+ + 0
R2 R2 Rout Rout A11
1
V1 V0 0
Rout R2 Rout R2
A
1
V1 V0 0 (ii)Rout Rout Writing equations (i) & (ii) in matrix form
11 -1VS V1 R1 R2 R2 R1
A1
V0 0
Rout Rout A X B
11-1
R1 R2 R2
(A(
A1
RoutRout
1 11 A
1
R1 R2 Rout Rout R2 11VS
R1 R2 R1
A
0
Rout
V0 1 1 1 A 1
R1 R2 Rout Rout R2
-A VS
Rout R1
R2 + R1A
R1R2RoutR2Rout
-A VS
Rout R1
R2 + R1 + AR1
R1R2Rout
-AR2VS
R2 + R1 + AR1
-AR2VS Hint: A = 105 and R1, R2 are in kilo ohm
AR1
-R2VS
R1
V0 -R2 -100 k
-50
VS R1 2 kR1 = 2 k(; R2 = 100 k(
R2 = 100 k
VS
R1 = 2 k
V0
Fig. (d)
Q#4.43: Design a two-stage op-amp network that has a gain of 50,000 while drawing no current into its input terminal. Use no resistors smaller than 1 k(.
Solution:
Circuit diagram:
VS
VX
R2
R1
fig. (a)OP-AMP MODEL
VX Rin Rout
AVe
R2
VS
R1
fig. (b)
V1 VX
V+ Rin V- R2
Rout
VS
R1
AVe
Using Nodal Analysisfig. (c)Applying KCL at node V1
According to KCL
The algebraic Sum of all the currents leaving that junction = the algebraic sum of all the currents entering into the junction
V1 - VS V1 V1 - VX
0
Rin R1R2V1 VS V1 V1 VX 0
Rin Rin R1 R2 R2
1 1 -1
V1 VX 0
(i)R1 R2 R2 Applying KCL at node Labeled V0
Assume that all the currents are going away from the junction
According to KCL
The algebraic sum of all the currents leaving that junction = the algebraic sum of all the entering into the junction
VX V1 VX - AVe
0
R2
Rout
Here
Ve = V+ - V-
Ve = VS V1
VX V1 VX A (VS V1)
0
R2
Rout
VX V1 VX AVS AV1
0
R2 R2 Rout RoutRout
A111 AVS V1 VX Rout R2Rout R2 Rout
A
1
AVS V1 VX (ii)Rout Rout RoutWriting equations (i) & (ii) in matrix form
11 -1
V1 0
R1 R2 R2
A1 AVS VX
Rout Rout Rout A X B 11-1
R1 R2 R2
(A(
A1
RoutRout
1 11 A
1
R1 R2 Rout Rout R2 11
0
R1 R2
A AVS
Rout Rout V0 1 1 1 A 1
R1 R2 Rout Rout R2
1 1 AVS
R1 R2 Rout
R2 + R1A
R1R2RoutR2Rout
R2 + R1 AVS
R1R2Rout
R2 + R1 + AR1
R1R2Rout
AVS (R2 + R1)
R2 + R1 + AR1
AVS (R2 + R1)
Hint: A = 105 and R1, R2 are in kilo ohm
AR1
VS (R2 + R1)
R1
VX R2 1 +
VS R1
Here R2 = 498 k( and R2 = 2 k(
VX 250 = A1
VS
R2
VX
R1
V0
Fig. (a)
R2OP-AMP MODEL
V0 Rin Rout
R1AVe
VX
Fig. (b)
V1 V0
R1
R2
V-
Rout
VX
Rin
V+
AVe
Fig. (c)
Using nodal analysis:
Applying KCL at node Labeled V1
Assume that all the currents are going away from the junction
According to KCL
The algebraic sum of all the currents leaving that junction = the algebraic sum of all the entering into the junction
V1 VX V1 V1 V0
+ + 0R1 Rin
R2V1 VX V1 V1 V0
+ +0R1 R1 Rin R2 R2
1 11-1 VX + +V1 + V0
R1 Rin R2 R2 R1
1 1 -1 VX V1 V0
(i)R1 R2 R2 R1Applying KCL at node Labeled V0
Assume that all the currents are going away from the junction
According to KCL
The algebraic sum of all the currents leaving that junction = the algebraic sum of all the entering into the junction
V0 V1 V0 - AVe
+ 0
R2Rout
V0 V1 V0 AVe
+ 0
R2 R2 Rout Rout Here
Ve = V+ - V-
Ve = 0 V V1
Ve = -V1
V0 V1 V0 A(-V1)
+ 0
R2 R2 Rout Rout V0 V1 V0 AV1
+ + 0
R2 R2 Rout Rout A11
1
V1 V0 0
Rout R2 Rout R2
A
1
V1 V0 0 (ii)Rout Rout Writing equations (i) & (ii) in matrix form
11 -1VX V1 R1 R2 R2 R1
A1
V0 0
Rout Rout A X B
11-1
R1 R2 R2
(A(
A1
RoutRout
1 11 A
1
R1 R2 Rout Rout R2 11VX
R1 R2 R1
A
0
Rout
V0 1 1 1 A 1
R1 R2 Rout Rout R2
-A VX
Rout R1
R2 + R1A
R1R2RoutR2Rout
-A VX
Rout R1
R2 + R1 + AR1
R1R2Rout
-AR2VX
R2 + R1 + AR1
-AR2VX Hint: A = 105 and R1, R2 are in kilo ohm
AR1
-R2VX
R1
V0 -R2 -400 k
-200 A2VX R1 2 kR1 = 2 k(; R2 = 400 k(
R2 = 400 k
VX
R1 = 2 k
V0
Fig. (d)
A1A2
A = A1A2
A = (250)(-200)
A = -50,000
Q#4.44: Design an op-amp circuit that has the following input/output relationship:
V0 = -5 V1 + 0.5 V2.
Solution:
Circuit diagram:
RF V1
V0 R1
V2
R2
RI
Fig. (a)
RFOP-AMP MODEL
V0 Rin Rout
R1AVe
R2
RI
V1
V2
Fig. (b)
V3 V0
R1
RF
V-
Rout
V1
Rin
V+
AVe
V4
RI R2
V2
Fig. (c)
Using Nodal Analysis
Applying KCL at node labeled V3
According to KCL
The algebraic sum of all the currents leaving that junction = the algebraic sum of all the currents entering into the junction
V3 V1 V3 V4 V3 V0
0
R1RinRFV3 V1 V3 V4V3 V0 0
R1 R1 Rin Rin RF RF
111-1 -1 V1 V3 V4 V0 (i) R1 Rin RF Rin RF R1Applying KCL at node labeled V4
According to KCL
The algebraic sum of all the currents leaving that junction = the algebraic sum of all the currents entering into the junction
V4 V2 V4 V4 V3
0
R2 RI
RinV4 V2 V4 V4 V3 0
R2 R2 RI Rin Rin
-1111 V2 V3 V4 (ii) Rin Rin RI R2 R2Applying KCL at node labeled V0
According to KCL
The algebraic sum of all the currents leaving that junction = the algebraic sum of all the currents entering into the junction
V0 V3 V0 - AVe 0
RFRout
Here Ve = V+ - V-Ve = V4 - V3
V0 V3 V0 A(V4 V3)
0
RFRout
V0 V3 V0 AV4 + AV3
0
RFRout
V0 V3 V0 AV4 AV3
0
RF RF Rout Rout Rout
A1 -A1 1
V3V4 V0 0 (iii)
Rout RF Rout Rout RFWriting equations (i), (ii) & (iii) in matrix form
11 1-1
-1 V1
V3 R1 Rin RF Rin RF R1
-1111V2 0 V4 RinRin RI R2R2
A 1
-A 11
V0 0
Rout RFRout Rout RF A X B
1 1 1 -1 -1
R1 Rin RF Rin RF
-1 1 1 1
(A( 0
Rin Rin RI R2
A1 -A
11
Rout RF RoutRout RF
1 1 1
0
111 Rin RI R2(A(
R1 Rin RF-A1 1
Rout Rout RF
-1 0
1Rin
RinA 11 1
Rout RFRout RF
-111 1
-1
Rin Rin RI R2 RF
A1-A
Rout RFRout
11 1 11 11 11
(A(
R1Rin RF Rin RI R2 Rout RFRin
-111-1 A
A1 1 1 1
Rin Rout RF RF RinRout Rout RF Rin RI R2 1 1 1 -1 V1
R1 Rin RF Rin R1
-1 1 1 1 V2
Rin Rin RI R2 R2
A 1 -A
0
Rout RF Rout
V0(A( 11 1 AV2 1 -V2 A 1 V1
R1 Rin RF R2Rout Rin R2 Rout RF R1
A A 11 1 1
RinRoutRout RF Rin RI R2
11 1 11 11 11
R1Rin RF Rin RI R2 Rout RFRin
-111-1 A
A1 1 1 1
Rin Rout RF RF RinRout Rout RF Rin RI R2Here
Rin = 108 (Rout = 10 (A = 105
We to bring about a result that
1 1 1 1 11 1
>> & > &( 0
R0 RF RinHere R0 = Rout; Ri = Rin
11AV2 V1 -A 1 1
R1 RF R2RoutR1 Rout RI R2
1 11 1 1-1-A1 1
R1 RF RI R2Rout RF Rout RI R2AV2 (RF + R1) AV1 (R2 + RI)
R1RFR2RoutRIR2R1Rout
(RF + R1)(R2 + RI) A (R2 + RI)
R1RFRIR2Rout RIR2RFRoutARIV2 (RF + R1) - ARFV1 (R2 + RI)
RIRFR1R2Rout
(RF + R1)(R2 + RI) + AR1 (R2 + RI)
R2RFRoutRIR1ARIV2 (RF + R1) - ARFV1 (R2 + RI)
(RF + R1)(R2 + RI) + AR1 (R2 + RI)
ARIV2 (RF + R1) - ARFV1 (R2 + RI)
R2RF + R1R2 + RIRF + R1RI + AR1R2 + AR1RIARIV2 (RF + R1) - ARFV1 (R2 + RI)
AR1R2 + AR1RI Hint: A = 105 and R1, R2, RI & RF are in kilo ohm
ARIV2 (RF + R1) - ARFV1 (R2 + RI)
AR1 (R2 + RI)
V2 (RF + R1) RF V1
(R2 + RI) R1By comparison, we get
RF = 5 k(R1 = 1 k(R2 = 11 k(RI = 1 k(
RF = 5 k
V1
V0 R1 = 1 k
V2
R2 = 11 k
RI = 1 k
Fig. (d)
Q#4.45: A voltage waveform with a maximum value of 200 mV must be amplified to a maximum of 10 V and inverted. However, the circuit that produces the waveform can provide no more than 100 (A. Design the required amplifier.
Solution:
Circuit diagram:
R2
VX
R1
V0
Fig. (a)
R2OP-AMP MODEL
V0 Rin Rout
R1AVe
VX
Fig. (b) V1 V0
R1
R2
V-
Rout
VX
Rin
V+
AVe
Fig. (c)
Using nodal analysis:
Applying KCL at node Labeled V1
Assume that all the currents are going away from the junction
According to KCL
The algebraic sum of all the currents leaving that junction = the algebraic sum of all the entering into the junction
V1 VX V1 V1 V0
+ + 0R1 Rin
R2V1 VX V1 V1 V0
+ +0R1 R1 Rin R2 R2
1 11-1 VX + +V1 + V0
R1 Rin R2 R2 R1
1 1 -1 VX V1 V0
(i)R1 R2 R2 R1Applying KCL at node Labeled V0
Assume that all the currents are going away from the junction
According to KCL
The algebraic sum of all the currents leaving that junction = the algebraic sum of all the entering into the junction
V0 V1 V0 - AVe
+ 0
R2Rout
V0 V1 V0 AVe
+ 0
R2 R2 Rout Rout Here
Ve = V+ - V-
Ve = 0 V V1
Ve = -V1
V0 V1 V0 A(-V1)
+ 0
R2 R2 Rout Rout V0 V1 V0 AV1
+ + 0
R2 R2 Rout Rout A11
1
V1 V0 0
Rout R2 Rout R2
A
1
V1 V0 0 (ii)Rout Rout Writing equations (i) & (ii) in matrix form
11 -1VX V1 R1 R2 R2 R1
A1
V0 0
Rout Rout A X B
11-1
R1 R2 R2
(A(
A1
RoutRout
1 11 A
1
R1 R2 Rout Rout R2 11VX
R1 R2 R1
A
0
Rout
V0 1 1 1 A 1
R1 R2 Rout Rout R2
-A VX
Rout R1
R2 + R1A
R1R2RoutR2Rout
-A VX
Rout R1
R2 + R1 + AR1
R1R2Rout
-AR2VX
R2 + R1 + AR1
-AR2VX Hint: A = 105 and R1, R2 are in kilo ohm
AR1
-R2VX
R1
V0 -R2
VX R1
10 -R2 -50
0.2 R1
R1 = 1 k(; R2 = 50 k(
R2 = 50 k
VX
R1 = 1 k
V0
Fig. (d)
Q#4.46: An amplifier with a gain of ( ( 1( is needed. Using resistor values from table 2.1, design the amplifier. Use as few resistors as possible.
Solution:
Circuit diagram:
VS
VX
R2
R1
fig. (a)OP-AMP MODEL
VX Rin Rout
AVe
R2
VS
R1
fig. (b)
V1 VX
V+ Rin V- R2
Rout
VS
R1
AVe
Using Nodal Analysisfig. (c)Applying KCL at node V1
According to KCL
The algebraic Sum of all the currents leaving that junction = the algebraic sum of all the currents entering into the junction
V1 - VS V1 V1 - VX
0
Rin R1R2V1 VS V1 V1 VX 0
Rin Rin R1 R2 R2
1 1 -1
V1 VX 0
(i)R1 R2 R2 Applying KCL at node Labeled V0
Assume that all the currents are going away from the junction
According to KCL
The algebraic sum of all the currents leaving that junction = the algebraic sum of all the entering into the junction
VX V1 VX - AVe
0
R2
Rout
Here
Ve = V+ - V-
Ve = VS V1
VX V1 VX A (VS V1)
0
R2
Rout
VX V1 VX AVS AV1
0
R2 R2 Rout RoutRout
A111 AVS V1 VX Rout R2Rout R2 Rout
A
1
AVS V1 VX (ii)Rout Rout RoutWriting equations (i) & (ii) in matrix form
11 -1
V1 0
R1 R2 R2
A1 AVS VX
Rout Rout Rout A X B 11-1
R1 R2 R2
(A(
A1
RoutRout
1 11 A
1
R1 R2 Rout Rout R2 11
0
R1 R2
A AVS
Rout Rout V0 1 1 1 A 1
R1 R2 Rout Rout R2
1 1 AVS
R1 R2 Rout
R2 + R1A
R1R2RoutR2Rout
R2 + R1 AVS
R1R2Rout
R2 + R1 + AR1
R1R2Rout
AVS (R2 + R1)
R2 + R1 + AR1
AVS (R2 + R1)
Hint: A = 105 and R1, R2 are in kilo ohm
AR1
VS (R2 + R1)
R1
VX R2 1 +
VS R1
( = 3.143
( + 0.01 = 3.153
( - 0.01 = 3.133
R23.153 1 +
R1
R22.153
R1 R23.133 1 +
R1
R22.133
R1 R22.133 ( ( 2.153
R1
Here R1 = 20 k( and R2 = 43 k(Q#4.47: Design an op-amp based circuit to produce the function V0 = 5 V1 4 V2Solution:
Circuit diagram:
RF V2
V0 R1
V1
R2
RI
Fig. (a)
RFOP-AMP MODEL
V0 Rin Rout
R1AVe
R2
RI
V2
V1
Fig. (b)
V3 V0
R1
RF
V-
Rout
V2
Rin
V+
AVe
V4
RI R2
V1
Fig. (c)
Using Nodal Analysis
Applying KCL at node labeled V3
According to KCL
The algebraic sum of all the currents leaving that junction = the algebraic sum of all the currents entering into the junction
V3 V2 V3 V4 V3 V0
0
R1RinRFV3 V2 V3 V4V3 V0 0
R1 R1 Rin Rin RF RF
111-1 -1 V2 V3 V4 V0 (i) R1 Rin RF Rin RF R1Applying KCL at node labeled V4
According to KCL
The algebraic sum of all the currents leaving that junction = the algebraic sum of all the currents entering into the junction
V4 V1 V4 V4 V3
0
R2 RI
RinV4 V1 V4 V4 V3 0
R2 R2 RI Rin Rin
-1111 V1 V3 V4 (ii) Rin Rin RI R2 R2Applying KCL at node labeled V0
According to KCL
The algebraic sum of all the currents leaving that junction = the algebraic sum of all the currents entering into the junction
V0 V3 V0 - AVe 0
RFRout
Here Ve = V+ - V-Ve = V4 - V3
V0 V3 V0 A(V4 V3)
0
RFRout
V0 V3 V0 AV4 + AV3
0
RFRout
V0 V3 V0 AV4 AV3
0
RF RF Rout Rout Rout
A1 -A1 1
V3V4 V0 0 (iii)
Rout RF Rout Rout RFWriting equations (i), (ii) & (iii) in matrix form
11 1-1
-1 V2
V3 R1 Rin RF Rin RF R1
-1111V1 0 V4 RinRin RI R2R2
A 1
-A 11
V0 0
Rout RFRout Rout RF A X B
1 1 1 -1 -1
R1 Rin RF Rin RF
-1 1 1 1
(A( 0
Rin Rin RI R2
A1 -A
11
Rout RF RoutRout RF
1 1 1
0
111 Rin RI R2(A(
R1 Rin RF-A1 1
Rout Rout RF
-1 0
1Rin
RinA 11 1
Rout RFRout RF
-111 1
-1
Rin Rin RI R2 RF
A1-A
Rout RFRout
11 1 11 11 11
(A(
R1Rin RF Rin RI