electric circuits 2 sheet 6, Exercises of Electrical Circuit Analysis

Op amp circuit analysis method

Typology: Exercises

2018/2019

Uploaded on 05/04/2019

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Problem (1):
a)Find ๐‘ฝ๐’ as function of ๐‘ฝ๐’”
๐‘๐‘œ๐‘‘๐‘Ž๐‘™ @๐‘‰๐‘›
๐‘‰๐‘›
7๐‘˜+๐‘‰๐‘›โˆ’๐‘‰๐‘œ
56๐‘˜= 0
๐‘‰๐‘›(1
7๐‘˜+1
56๐‘˜) = ๐‘‰๐‘œ
56๐‘˜
๐‘‰๐‘›=1
9๐‘‰๐‘œ
๐‘๐‘œ๐‘‘๐‘Ž๐‘™ @๐‘‰๐‘
๐‘‰๐‘
8๐‘˜+๐‘‰๐‘โˆ’๐‘‰๐‘ 
32๐‘˜= 0
๐‘‰๐‘(1
8๐‘˜+1
32๐‘˜) = ๐‘‰๐‘ 
32๐‘˜
๐‘‰๐‘=1
9๐‘‰๐‘ 
๐‘‰๐‘›= ๐‘‰๐‘
1
9๐‘‰๐‘œ=1
9๐‘‰๐‘ 
๐‘‰๐‘œ=9
5๐‘‰๐‘ 
b) Find the range of values for ๐‘ฝ๐’” such that not saturate and the op amp remains in its linear
region of operation. โˆ’15 โ‰ค ๐‘‰๐‘œโ‰ค12
โˆ’15 โ‰ค9
5๐‘‰๐‘ โ‰ค12
โˆ’25
3โ‰ค ๐‘‰๐‘ โ‰ค20
3
Faculty of IET
Dr. Tarek Saad
Dr. Rania Sweif
Spring 2018
Electric circuits II
Sheet Nr. (6)โ€“Solution
Operational Amplifier (Opamp)
๐‘‰๐‘›
๐‘‰๐‘
pf3
pf4
pf5

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Problem (1):

a)Find ๐‘ฝ

๐’

as function of ๐‘ฝ

๐’”

๐‘›

๐‘›

๐‘›

๐‘œ

๐‘›

๐‘œ

๐‘›

๐‘œ

๐‘

๐‘

๐‘

๐‘ 

๐‘

๐‘ 

๐‘

๐‘ 

๐‘›

๐‘

๐‘œ

๐‘ 

๐‘œ

๐‘ 

b) Find the range of values for ๐‘ฝ

๐’”

such that not saturate and the op amp remains in its linear

region of operation.

๐‘œ

๐‘ 

๐‘ 

Faculty of IET

Dr. Tarek Saad

Dr. Rania Sweif

Spring 2018

Electric circuits II

Sheet Nr. ( 6 )โ€“Solution

Operational Amplifier (Opamp)

๐‘›

๐‘

C) Find the range of values of ๐’Š

๐’‚

๐‘Ž

๐‘›

๐‘œ

๐‘œ

๐‘›

๐‘œ

๐‘›

๐‘›

๐‘œ

โˆ’

9

๐‘Ž

12

9

๐‘Ž

Problem (3):

Show that

0

๐‘–

=

2

1

(1 +

2

๐บ

)

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

๐‘–๐‘›

๐‘–๐‘›

โˆ’ 2 ๐‘…

1

๐‘…

2

3

4

Nodal @ ๐‘‰

3

3

๐‘

2

3

0

2

3

4

๐บ

๐บ

2

๐บ

3

๐‘

๐บ

3

0

2

3

4

3

๐บ

2

๐บ

๐‘

๐บ

0

2

4

Nodal @ ๐‘‰

4

4

๐‘

2

4

2

4

3

๐บ

๐บ

2

๐บ

4

๐‘

๐บ

4

2

4

3

4

๐บ

2

๐บ

๐‘

2

3

๐‘

๐‘

1

2

3

4

๐‘

๐‘

Eq.2 โ€“ Eq.

3

๐บ

2

๐บ

๐‘

๐บ

0

2

4

4

๐บ

2

๐บ

๐‘

2

3

3

4

๐บ

2

3

4

2

๐บ

0

3

4

๐บ

2

๐บ

0

3

4

๐บ

0

๐บ

2

Substitute in eq. 1

๐‘–๐‘›

1

2

๐บ

0

๐บ

2

๐‘–๐‘›

0

1

2

๐บ

๐บ

2

0

๐‘–๐‘›

2

1

2

๐บ

Problem (5):

Find ๐‘‰

Nodal @ ๐‘‰ 1

1

โˆ’ 3

1

1

๐‘

1

๐‘

โˆ’ 3

1

๐‘

โˆ’ 7

Nodal @ ๐‘‰

1

2

โˆ’ 3

2

2

๐‘

2

๐‘

โˆ’ 3

2

๐‘

โˆ’ 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

โˆ’ 7

2

๐‘

2

๐‘

๐‘

๐‘

Subs. ๐‘‰

๐‘

in 1

1

โˆ’ 3

โˆ’ 7

1

โˆ’ 3

โˆ’ 3

0

0

1

2

๐‘

๐‘