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Program Studi Teknik Elektro Konsentrasi Mekatronika Universitas Katolik Parahyangan DASAR RANGKAIAN LISTRIK Ali Sadiyoko, S.T., M.T. Faisal Wahab, S.Pd., M.T. FAKULTAS TEKNOLOGI INDUSTRI PROGRAM STUDI TEKNIK ELEKTRO KONSENTRASI MEKATRONIKA 2016

05. operational amplifier

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Page 1: 05. operational amplifier

Program Studi Teknik Elektro Konsentrasi Mekatronika Universitas Katolik Parahyangan

DASAR RANGKAIAN LISTRIK

Ali Sadiyoko, S.T., M.T. Faisal Wahab, S.Pd., M.T.

FAKULTAS TEKNOLOGI INDUSTRI

PROGRAM STUDI TEKNIK ELEKTRO KONSENTRASI MEKATRONIKA

2016

Page 2: 05. operational amplifier

Program Studi Teknik Elektro Konsentrasi Mekatronika Universitas Katolik Parahyangan

Topik 5 Operational Amplifier

β€’ Gelombang Sinusoidal

β€’ Pengantar

β€’ Ideal Op-Amp

β€’ Inverter Op-Amp

β€’ Non-Inverter Op-Amp

β€’ Summing Op-Amp

β€’ Diferensial Op-Amp

β€’ Cascade Op-Amp

Page 3: 05. operational amplifier

Program Studi Teknik Elektro Konsentrasi Mekatronika Universitas Katolik Parahyangan

Gelombang Sinusoidal

Dimana : A = Amplituda πœ” = 2πœ‹π‘“ = π‘“π‘Ÿπ‘’π‘˜π‘’π‘’π‘›π‘ π‘– 𝑠𝑒𝑑𝑒𝑑

𝑓 =1

𝑇= π‘“π‘Ÿπ‘’π‘˜π‘’π‘’π‘›π‘ π‘–

𝑑 = π‘ π‘Žπ‘‘π‘’π‘Žπ‘› π‘‘π‘’π‘‘π‘–π‘˜

T

A

πœ‹ πœ”π‘‘ 2πœ‹ πœ‹

2

3πœ‹

2

)2

(sin tT

Ay

)(sin tAy

)2(sin tfAy

A

Page 4: 05. operational amplifier

Program Studi Teknik Elektro Konsentrasi Mekatronika Universitas Katolik Parahyangan

Gambarkan Sinyal sinusoidal y= 2 sin (6280t)

Jawab : Amplitudo A = 2 perioda

πœ” =2πœ‹

𝑇

𝑇 =2πœ‹

πœ”=

2π‘₯3,14

6280= 0.001 𝑠

frekuensi πœ” = 2πœ‹π‘“

𝑓 =πœ”

2πœ‹=

6280

2π‘₯3.14= 1000 𝐻𝑧/1π‘˜π»π‘§

Page 5: 05. operational amplifier

Program Studi Teknik Elektro Konsentrasi Mekatronika Universitas Katolik Parahyangan

Gain(A)

𝑉𝑠

𝑅𝑠

𝑉𝑙

+

βˆ’

Op-Amp

𝑽𝒍 = 𝑨𝑽𝒔

Tegangan output 𝑉𝑙 adalah penguatan dari Tegangan sumber 𝑉𝑠

Amplifier Load Source

Page 6: 05. operational amplifier

Program Studi Teknik Elektro Konsentrasi Mekatronika Universitas Katolik Parahyangan

+

-𝑣𝑠

𝑅𝑠

𝑅𝑖𝑛 𝐴𝑣𝑖𝑛

π‘…π‘œπ‘’π‘‘

+

βˆ’

𝑣𝑖𝑛

𝑣𝑖𝑛 =𝑅𝑖𝑛

𝑅𝑠 + 𝑅𝑖𝑛𝑣𝑠 𝑣𝑙 =

𝑅𝑙

π‘…π‘œπ‘’π‘‘ + 𝑅𝑙𝐴𝑣𝑖𝑛

+

βˆ’

𝑣𝑙

π’—π’Šπ’ = 𝒗𝒔 Idealnya 𝒗𝒍 = π‘¨π’—π’Šπ’

𝒗𝒍 = 𝑨 𝒗𝒔

𝑅𝑙

𝑅𝑖𝑛= ∞ (open loop)

π‘…π‘œπ‘’π‘‘ = 0 (close loop)

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Program Studi Teknik Elektro Konsentrasi Mekatronika Universitas Katolik Parahyangan

+

-𝑅𝑖𝑛

𝐴𝑣𝑖𝑛

π‘…π‘œπ‘’π‘‘

+

βˆ’

𝑣𝑖𝑛

𝑣𝑝

𝑣𝑛

Karakteristik Ideal Op-Amp 1. 𝑅𝑖𝑛= ∞, (open circuit) 2. π‘…π‘œ = 0, (short circuit) 3. 𝐴 = ∞,

𝑣𝑝 = 𝑣𝑛

𝑖𝑝 = 𝑖𝑛 = 0

𝑖𝑝

𝑖𝑛

Circuit Model

𝑣𝑖𝑛 = 𝑣𝑝 βˆ’ 𝑣𝑛

Dengan gain 𝐴 = ∞, sedangkan

𝑣𝑐𝑐 dibatasi 24 V, Maka, 𝑣𝑐𝑐

𝐴= 0

Pada Op-Amp ideal 𝑅𝑖𝑛= ∞ sehingga arus yang masuk 𝑣𝑝 dan 𝑣𝑛 dianggap nol

π‘£π‘œ

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Program Studi Teknik Elektro Konsentrasi Mekatronika Universitas Katolik Parahyangan

+

-

𝑣𝑠

𝑅𝑠

𝑅𝑓

+

βˆ’

π‘£π‘œ 𝑣𝑝

𝑣𝑛

Inverting Amplifier

Circuit Model

𝑣𝑠

𝑅𝑠

𝑅𝑓

+

βˆ’

π‘£π‘œ

𝑣𝑝

𝑣𝑛

Tentukan close loop voltage

gain 𝐴 =π‘‰π‘œ

𝑉𝑠 !

𝐴𝑣𝑖𝑛

+

βˆ’

𝑣𝑖𝑛 𝑅𝑖𝑛

π‘…π‘œπ‘’π‘‘

Page 9: 05. operational amplifier

Program Studi Teknik Elektro Konsentrasi Mekatronika Universitas Katolik Parahyangan

+

-Terminal non-inversi dihubungkan ke ground, sehingga 𝑣𝑝 = 0 maka

𝑣𝑝 = 𝑣𝑛 = 0 dan

𝑣𝑠

𝑅𝑠

𝑅𝑓

+

βˆ’

π‘£π‘œ

𝑣𝑝

𝑣𝑛

𝑣𝑠 βˆ’ 𝑣𝑛

π‘…π‘ βˆ’

𝑣𝑛 βˆ’ π‘£π‘œ

π‘…π‘“βˆ’ 𝑖𝑛 = 0

𝑣𝑠 βˆ’ 𝑣𝑛

𝑅𝑠=

𝑣𝑛 βˆ’ π‘£π‘œ

𝑅𝑓

𝑣𝑠𝑅𝑓 βˆ’ 𝑣𝑛𝑅𝑓 = 𝑣𝑛𝑅𝑠 βˆ’ π‘£π‘œπ‘…π‘ 

𝑣𝑠𝑅𝑓 = 𝑣𝑛(𝑅𝑠 + 𝑅𝑓) βˆ’ π‘£π‘œπ‘…π‘ 

karena 𝑣𝑛 = 0

𝑣𝑠𝑅𝑓 = βˆ’ π‘£π‘œπ‘…π‘ 

π‘£π‘œ

𝑣𝑠= βˆ’

𝑅𝑓

𝑅𝑠

𝐴 =π‘£π‘œ

𝑣𝑠= βˆ’

𝑅𝑓

𝑅𝑠

Tentukan close loop

voltage gain 𝐴 =π‘£π‘œ

𝑣𝑠 !

Inverting Amplifier

KCL di node 𝑣𝑛: 𝑖1 βˆ’ 𝑖2 βˆ’ 𝑖𝑛 = 0

𝑖𝑝 = 𝑖𝑛 = 0

𝑖𝑝

𝑖𝑛

𝐴𝑣𝑖𝑛

+ βˆ’ 𝑖1

+ βˆ’ 𝑖2

Page 10: 05. operational amplifier

Program Studi Teknik Elektro Konsentrasi Mekatronika Universitas Katolik Parahyangan

𝑣𝑠 = 0.5

10Ξ©

+

βˆ’

π‘£π‘œ 𝑣𝑝

𝑣𝑛

Contoh Inverting Amplifier

25Ξ©

1. Tentukan π‘£π‘œ 2. Tentukan arus i

π‘£π‘œ = βˆ’π‘…π‘“

𝑅𝑠𝑣𝑠

π‘£π‘œ = βˆ’25

100.5

π‘£π‘œ = βˆ’1.25

+ βˆ’ 𝑖

𝑖 =𝑣𝑖 βˆ’ 𝑣𝑛

𝑅𝑠

𝑖 =0.5 βˆ’ 0

10

𝑖 = 0.05 𝐴

Page 11: 05. operational amplifier

Program Studi Teknik Elektro Konsentrasi Mekatronika Universitas Katolik Parahyangan

𝑣𝑠

𝑅𝑠

𝑅𝑓

+

βˆ’

π‘£π‘œ

𝑣𝑝

𝑣𝑛

Tentukan close loop voltage

gain 𝐴 =π‘£π‘œ

𝑣𝑠 !

0 βˆ’ 𝑣𝑛

π‘…π‘ βˆ’

𝑣𝑛 βˆ’ π‘£π‘œ

π‘…π‘“βˆ’ 𝑖𝑛 = 0

βˆ’π‘£π‘›

𝑅𝑠=

𝑣𝑛 βˆ’ π‘£π‘œ

𝑅𝑓

βˆ’π‘£π‘ π‘…π‘“ = 𝑣𝑠𝑅𝑠 βˆ’ π‘£π‘œπ‘…π‘ 

π‘£π‘œ

𝑣𝑠=

𝑅𝑓 + 𝑅𝑠

𝑅𝑠

𝑣𝑠(𝑅𝑓 + 𝑅𝑠) = π‘£π‘œπ‘…π‘ 

Non-Inverting Amplifier

Terminal non-inversi dihubungkan ke sumber, sehingga 𝑣𝑝 = 𝑣𝑠 maka 𝑣𝑝 = 𝑣𝑛 = 𝑣𝑠

Op-amp ideal 𝑖𝑝 = 𝑖𝑛 = 0

βˆ’π‘£π‘ 

𝑅𝑠=

𝑣𝑠 βˆ’ π‘£π‘œ

𝑅𝑓 dan 𝑣𝑝 = 𝑣𝑛 = 𝑣𝑠

𝑖𝑛

𝑖1

𝑖2

Page 12: 05. operational amplifier

Program Studi Teknik Elektro Konsentrasi Mekatronika Universitas Katolik Parahyangan

Tentukan close loop voltage

gain 𝐴 =π‘£π‘œ

𝑣𝑠 !

0 βˆ’ 𝑣𝑛

π‘…π‘ βˆ’

𝑣𝑛 βˆ’ π‘£π‘œ

π‘…π‘“βˆ’ 𝑖𝑛 = 0

βˆ’π‘£π‘›

𝑅𝑠=

𝑣𝑛 βˆ’ π‘£π‘œ

𝑅𝑓

βˆ’π‘£π‘ π‘…π‘“ = 𝑣𝑠𝑅𝑠 βˆ’ π‘£π‘œπ‘…π‘ 

π‘£π‘œ

𝑣𝑠=

𝑅𝑓 + 𝑅𝑠

𝑅𝑠

𝑣𝑠(𝑅𝑓 + 𝑅𝑠) = π‘£π‘œπ‘…π‘ 

Non-Inverting Amplifier

Terminal non-inversi dihubungkan ke sumber, sehingga 𝑣𝑝 = 𝑣𝑠 maka 𝑣𝑝 = 𝑣𝑛 = 𝑣𝑠

Op-amp ideal 𝑖𝑝 = 𝑖𝑛 = 0

βˆ’π‘£π‘ 

𝑅𝑠=

𝑣𝑠 βˆ’ π‘£π‘œ

𝑅𝑓 dan 𝑣𝑝 = 𝑣𝑛 = 𝑣𝑠

𝑅𝑠

𝑅𝑓

𝑣𝑝

𝑣𝑛

+

βˆ’ π‘£π‘œ

Page 13: 05. operational amplifier

Program Studi Teknik Elektro Konsentrasi Mekatronika Universitas Katolik Parahyangan

Contoh non-Inverting Amplifier

6𝑉

4Ξ©

10Ξ©

+

βˆ’

π‘£π‘œ 𝑣𝑝

𝑣𝑛

4𝑉

6 βˆ’ 𝑣𝑛

π‘…π‘ βˆ’

𝑣𝑛 βˆ’ π‘£π‘œ

π‘…π‘“βˆ’ 𝑖𝑛 = 0

Karena 𝑣𝑝 = 𝑣𝑛 = 𝑣𝑠 dan 𝑖𝑝 = 𝑖𝑛 = 0

6 βˆ’ 4

4βˆ’

4 βˆ’ π‘£π‘œ

10βˆ’ 0 = 0

6 βˆ’ 4

4=

4 βˆ’ π‘£π‘œ

10

2

4=

4 βˆ’ π‘£π‘œ

10

5 = 4 βˆ’ π‘£π‘œ

π‘£π‘œ = βˆ’1

1. Tentukan π‘£π‘œ

KCL di node 𝑣𝑛

Page 14: 05. operational amplifier

Program Studi Teknik Elektro Konsentrasi Mekatronika Universitas Katolik Parahyangan

𝑣𝑠

+

βˆ’

π‘£π‘œ

𝑣𝑝

𝑣𝑛

Buffer Amplifier

Terminal non-inversi (𝑣𝑝) dihubungkan ke sumber,

sehingga 𝒗𝒑= 𝒗𝒏 maka 𝒗𝒑 = 𝒗𝒏 = 𝒗𝒔

Dan terminal 𝑣𝑛 terhubung langsung dengan 𝑣𝑠 𝑣𝑛 = 𝑣𝑠 Sehingga 𝒗𝒑 = 𝒗𝒏 = 𝒗𝒔 = 𝒗𝒐

Jadi gain loop tertutupnya adalah 1.

Page 15: 05. operational amplifier

Program Studi Teknik Elektro Konsentrasi Mekatronika Universitas Katolik Parahyangan

Summing Amplifier 𝑣1

𝑅1 𝑅𝑓

+

βˆ’

π‘£π‘œ 𝑣𝑝

𝑣𝑛 𝑣2

𝑣3

𝑅2

𝑅3

𝑖1 + 𝑖2 + 𝑖3 βˆ’ 𝑖 = 0

𝑖1

𝑖2

𝑖3

𝑖

𝑣1 βˆ’ 𝑣𝑛

𝑅1+

𝑣2 βˆ’ 𝑣𝑛

𝑅2+

𝑣3 βˆ’ 𝑣𝑛

𝑅3=

𝑣𝑛 βˆ’ π‘£π‘œ

𝑅𝑓

𝑣1

𝑅1+

𝑣2

𝑅2+

𝑣3

𝑅3=

βˆ’π‘£π‘œ

𝑅𝑓

βˆ’π‘…π‘“

𝑅1𝑣1 +

𝑅𝑓

𝑅2𝑣2 +

𝑅𝑓

𝑅3𝑣3 = π‘£π‘œ

Terminal non-inversi dihubungkan ke ground, sehingga 𝑣𝑝 = 0 maka 𝑣𝑝 = 𝑣𝑛 = 0

KCL di node 𝑣𝑛

Page 16: 05. operational amplifier

Program Studi Teknik Elektro Konsentrasi Mekatronika Universitas Katolik Parahyangan

2 𝑉

5Ξ© 10 Ξ©

+

βˆ’

π‘£π‘œ 𝑣𝑝

𝑣𝑛

1 𝑉

2.5Ξ©

𝑖1

𝑖2

Contoh Summing Amplifier

2 Ξ©

βˆ’10

52 +

10

2,51 = π‘£π‘œ

βˆ’ 4 + 4 = π‘£π‘œ

1. Tentukan π‘£π‘œ 2. Tentukan π‘–π‘œ

βˆ’8 = π‘£π‘œ

π‘–π‘œ =π‘£π‘œ βˆ’ 0

10+

π‘£π‘œ βˆ’ 0

2

π‘–π‘œ = βˆ’8

10βˆ’

8

2

π‘–π‘œ = βˆ’0.8 βˆ’ 4

π‘–π‘œ = βˆ’4.8 𝐴

1. Mentukan π‘£π‘œ

1. Mentukan π‘–π‘œ

𝑖0

Page 17: 05. operational amplifier

Program Studi Teknik Elektro Konsentrasi Mekatronika Universitas Katolik Parahyangan

Difference Amplifier

𝑉1

𝑅1

𝑅4

+

βˆ’

π‘‰π‘œ 𝑉𝑝

𝑉𝑛

𝑉2

𝑅2

𝑅3

𝑖1

𝑖2

Page 18: 05. operational amplifier

Program Studi Teknik Elektro Konsentrasi Mekatronika Universitas Katolik Parahyangan

π‘‰π‘œ =𝑅2 + 𝑅1

𝑅1𝑉𝑛 βˆ’

𝑅2

𝑅1𝑉1

karena 𝑉𝑝 = 𝑉𝑛

π‘‰π‘œ =𝑅2 + 𝑅1

𝑅1

𝑅4

𝑅3 + 𝑅4𝑉2 βˆ’

𝑅2

𝑅1𝑉1

𝑉𝑝 =𝑅4

𝑅3 + 𝑅4𝑉2

𝑉2 βˆ’ 𝑉𝑝

𝑅3=

𝑉𝑝

𝑅4

𝑉2𝑅4 βˆ’ 𝑉𝑝𝑅4 = 𝑉𝑝𝑅3

𝑉2𝑅4 = 𝑉𝑝𝑅3 + 𝑉𝑝𝑅4

𝑉2𝑅4 = 𝑉𝑝(𝑅3 + 𝑅4)

𝑉𝑝 =𝑅4

𝑅3 + 𝑅4𝑉2

KCL di node 𝑣𝑝

𝑉1 βˆ’ 𝑉𝑛

𝑅1=

𝑉𝑛 βˆ’ π‘‰π‘œ

𝑅2

𝑉1𝑅2 βˆ’ 𝑉𝑛𝑅2 = 𝑉𝑛𝑅1 βˆ’ π‘‰π‘œπ‘…1

π‘‰π‘œπ‘…1 = 𝑉𝑛 𝑅2 + 𝑅1 βˆ’ 𝑉1𝑅2

π‘‰π‘œ =𝑅2 + 𝑅1

𝑅1𝑉𝑛 βˆ’

𝑅2

𝑅1𝑉1

KCL di node 𝑣𝑛

Page 19: 05. operational amplifier

Program Studi Teknik Elektro Konsentrasi Mekatronika Universitas Katolik Parahyangan

Cascaded Op Amp Circuits

𝑉𝑠

𝑅𝑠 𝑅𝑓

+

βˆ’

π‘‰π‘œ

𝑉𝑝

𝑉𝑛

𝑉𝑝

𝑉𝑛

𝑅𝑠 𝑅𝑓

𝑉𝑠

𝑅𝑠

𝑅𝑓

+

βˆ’

π‘‰π‘œ 𝑉𝑝

𝑉𝑛

𝑉𝑝

𝑉𝑛 𝑅𝑠

𝑅𝑓

Inverting Amplifier

Non-Inverting Amplifier

Page 20: 05. operational amplifier

Program Studi Teknik Elektro Konsentrasi Mekatronika Universitas Katolik Parahyangan

20mV

3 12

+

βˆ’

π‘‰π‘œ

𝑉𝑝

𝑉𝑛

𝑉𝑝

𝑉𝑛

4 10

Op-amp pertama Op-amp kedua

π‘£π‘œ =12 + 3

320

π‘£π‘œ = 100π‘šπ‘‰

π‘£π‘œ =10 + 4

4100

π‘£π‘œ = 350π‘šπ‘‰