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


Topik 5 Operational Amplifier

• Gelombang Sinusoidal

• Pengantar

• Ideal Op-Amp

• Inverter Op-Amp

• Non-Inverter Op-Amp

• Summing Op-Amp

• Diferensial Op-Amp

• Cascade Op-Amp


Gelombang Sinusoidal

Dimana : A = Amplituda 𝜔 = 2𝜋𝑓 = 𝑓𝑟𝑒𝑘𝑢𝑒𝑛𝑠𝑖 𝑠𝑢𝑑𝑢𝑡

𝑓 =1

𝑇= 𝑓𝑟𝑒𝑘𝑢𝑒𝑛𝑠𝑖

𝑡 = 𝑠𝑎𝑡𝑢𝑎𝑛 𝑑𝑒𝑡𝑖𝑘

T

A

𝜋 𝜔𝑡 2𝜋 𝜋

2

3𝜋

2

)2

(sin tT

Ay

)(sin tAy

)2(sin tfAy

A


Gambarkan Sinyal sinusoidal y= 2 sin (6280t)

Jawab : Amplitudo A = 2 perioda

𝜔 =2𝜋

𝑇

𝑇 =2𝜋

𝜔=

2𝑥3,14

6280= 0.001 𝑠

frekuensi 𝜔 = 2𝜋𝑓

𝑓 =𝜔

2𝜋=

6280

2𝑥3.14= 1000 𝐻𝑧/1𝑘𝐻𝑧


Gain(A)

𝑉𝑠

𝑅𝑠

𝑉𝑙

+

−

Op-Amp

𝑽𝒍 = 𝑨𝑽𝒔

Tegangan output 𝑉𝑙 adalah penguatan dari Tegangan sumber 𝑉𝑠

Amplifier Load Source


+

-𝑣𝑠

𝑅𝑠

𝑅𝑖𝑛 𝐴𝑣𝑖𝑛

𝑅𝑜𝑢𝑡

+

−

𝑣𝑖𝑛

𝑣𝑖𝑛 =𝑅𝑖𝑛

𝑅𝑠 + 𝑅𝑖𝑛𝑣𝑠 𝑣𝑙 =

𝑅𝑙

𝑅𝑜𝑢𝑡 + 𝑅𝑙𝐴𝑣𝑖𝑛

+

−

𝑣𝑙

𝒗𝒊𝒏 = 𝒗𝒔 Idealnya 𝒗𝒍 = 𝑨𝒗𝒊𝒏

𝒗𝒍 = 𝑨 𝒗𝒔

𝑅𝑙

𝑅𝑖𝑛= ∞ (open loop)

𝑅𝑜𝑢𝑡 = 0 (close loop)


+

-𝑅𝑖𝑛

𝐴𝑣𝑖𝑛

𝑅𝑜𝑢𝑡

+

−

𝑣𝑖𝑛

𝑣𝑝

𝑣𝑛

Karakteristik Ideal Op-Amp 1. 𝑅𝑖𝑛= ∞, (open circuit) 2. 𝑅𝑜 = 0, (short circuit) 3. 𝐴 = ∞,

𝑣𝑝 = 𝑣𝑛

𝑖𝑝 = 𝑖𝑛 = 0

𝑖𝑝

𝑖𝑛

Circuit Model

𝑣𝑖𝑛 = 𝑣𝑝 − 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

𝑣𝑜


+

-

𝑣𝑠

𝑅𝑠

𝑅𝑓

+

−

𝑣𝑜 𝑣𝑝

𝑣𝑛

Inverting Amplifier

Circuit Model

𝑣𝑠

𝑅𝑠

𝑅𝑓

+

−

𝑣𝑜

𝑣𝑝

𝑣𝑛

Tentukan close loop voltage

gain 𝐴 =𝑉𝑜

𝑉𝑠 !

𝐴𝑣𝑖𝑛

+

−

𝑣𝑖𝑛 𝑅𝑖𝑛

𝑅𝑜𝑢𝑡


+

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

𝑣𝑝 = 𝑣𝑛 = 0 dan

𝑣𝑠

𝑅𝑠

𝑅𝑓

+

−

𝑣𝑜

𝑣𝑝

𝑣𝑛

𝑣𝑠 − 𝑣𝑛

𝑅𝑠−

𝑣𝑛 − 𝑣𝑜

𝑅𝑓− 𝑖𝑛 = 0

𝑣𝑠 − 𝑣𝑛

𝑅𝑠=


𝑅𝑓

𝑣𝑠𝑅𝑓 − 𝑣𝑛𝑅𝑓 = 𝑣𝑛𝑅𝑠 − 𝑣𝑜𝑅𝑠

𝑣𝑠𝑅𝑓 = 𝑣𝑛(𝑅𝑠 + 𝑅𝑓) − 𝑣𝑜𝑅𝑠

karena 𝑣𝑛 = 0

𝑣𝑠𝑅𝑓 = − 𝑣𝑜𝑅𝑠

𝑣𝑜

𝑣𝑠= −

𝑅𝑓

𝑅𝑠

𝐴 =𝑣𝑜

𝑣𝑠= −

𝑅𝑓

𝑅𝑠

Tentukan close loop

voltage gain 𝐴 =𝑣𝑜

𝑣𝑠 !

Inverting Amplifier

KCL di node 𝑣𝑛: 𝑖1 − 𝑖2 − 𝑖𝑛 = 0

𝑖𝑝 = 𝑖𝑛 = 0

𝑖𝑝

𝑖𝑛

𝐴𝑣𝑖𝑛

+ − 𝑖1

+ − 𝑖2


𝑣𝑠 = 0.5

10Ω

+

−

𝑣𝑜 𝑣𝑝

𝑣𝑛

Contoh Inverting Amplifier

25Ω

1. Tentukan 𝑣𝑜 2. Tentukan arus i

𝑣𝑜 = −𝑅𝑓

𝑅𝑠𝑣𝑠

𝑣𝑜 = −25

100.5

𝑣𝑜 = −1.25

+ − 𝑖

𝑖 =𝑣𝑖 − 𝑣𝑛

𝑅𝑠

𝑖 =0.5 − 0

10

𝑖 = 0.05 𝐴


𝑣𝑠

𝑅𝑠

𝑅𝑓

+

−

𝑣𝑜

𝑣𝑝

𝑣𝑛


gain 𝐴 =𝑣𝑜

𝑣𝑠 !

0 − 𝑣𝑛

𝑅𝑠−



−𝑣𝑛

𝑅𝑠=


𝑅𝑓

−𝑣𝑠𝑅𝑓 = 𝑣𝑠𝑅𝑠 − 𝑣𝑜𝑅𝑠

𝑣𝑜

𝑣𝑠=

𝑅𝑓 + 𝑅𝑠

𝑅𝑠

𝑣𝑠(𝑅𝑓 + 𝑅𝑠) = 𝑣𝑜𝑅𝑠

Non-Inverting Amplifier

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

Op-amp ideal 𝑖𝑝 = 𝑖𝑛 = 0

−𝑣𝑠

𝑅𝑠=

𝑣𝑠 − 𝑣𝑜

𝑅𝑓 dan 𝑣𝑝 = 𝑣𝑛 = 𝑣𝑠

𝑖𝑛

𝑖1

𝑖2



gain 𝐴 =𝑣𝑜

𝑣𝑠 !

0 − 𝑣𝑛

𝑅𝑠−



−𝑣𝑛

𝑅𝑠=


𝑅𝑓

−𝑣𝑠𝑅𝑓 = 𝑣𝑠𝑅𝑠 − 𝑣𝑜𝑅𝑠

𝑣𝑜

𝑣𝑠=

𝑅𝑓 + 𝑅𝑠

𝑅𝑠

𝑣𝑠(𝑅𝑓 + 𝑅𝑠) = 𝑣𝑜𝑅𝑠


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

Op-amp ideal 𝑖𝑝 = 𝑖𝑛 = 0

−𝑣𝑠

𝑅𝑠=

𝑣𝑠 − 𝑣𝑜

𝑅𝑓 dan 𝑣𝑝 = 𝑣𝑛 = 𝑣𝑠

𝑅𝑠

𝑅𝑓

𝑣𝑝

𝑣𝑛

+

− 𝑣𝑜


Contoh non-Inverting Amplifier

6𝑉

4Ω

10Ω

+

−

𝑣𝑜 𝑣𝑝

𝑣𝑛

4𝑉

6 − 𝑣𝑛

𝑅𝑠−



Karena 𝑣𝑝 = 𝑣𝑛 = 𝑣𝑠 dan 𝑖𝑝 = 𝑖𝑛 = 0

6 − 4

4−

4 − 𝑣𝑜

10− 0 = 0

6 − 4

4=

4 − 𝑣𝑜

10

2

4=

4 − 𝑣𝑜

10

5 = 4 − 𝑣𝑜

𝑣𝑜 = −1

1. Tentukan 𝑣𝑜

KCL di node 𝑣𝑛


𝑣𝑠

+

−

𝑣𝑜

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.


Summing Amplifier 𝑣1

𝑅1 𝑅𝑓

+

−

𝑣𝑜 𝑣𝑝

𝑣𝑛 𝑣2

𝑣3

𝑅2

𝑅3

𝑖1 + 𝑖2 + 𝑖3 − 𝑖 = 0

𝑖1

𝑖2

𝑖3

𝑖

𝑣1 − 𝑣𝑛

𝑅1+

𝑣2 − 𝑣𝑛

𝑅2+

𝑣3 − 𝑣𝑛

𝑅3=


𝑅𝑓

𝑣1

𝑅1+

𝑣2

𝑅2+

𝑣3

𝑅3=

−𝑣𝑜

𝑅𝑓

−𝑅𝑓

𝑅1𝑣1 +

𝑅𝑓

𝑅2𝑣2 +

𝑅𝑓

𝑅3𝑣3 = 𝑣𝑜

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



2 𝑉

5Ω 10 Ω

+

−

𝑣𝑜 𝑣𝑝

𝑣𝑛

1 𝑉

2.5Ω

𝑖1

𝑖2

Contoh Summing Amplifier

2 Ω

−10

52 +

10

2,51 = 𝑣𝑜

− 4 + 4 = 𝑣𝑜

1. Tentukan 𝑣𝑜 2. Tentukan 𝑖𝑜

−8 = 𝑣𝑜

𝑖𝑜 =𝑣𝑜 − 0

10+

𝑣𝑜 − 0

2

𝑖𝑜 = −8

10−

8

2

𝑖𝑜 = −0.8 − 4

𝑖𝑜 = −4.8 𝐴

1. Mentukan 𝑣𝑜

1. Mentukan 𝑖𝑜

𝑖0


Difference Amplifier

𝑉1

𝑅1

𝑅4

+

−

𝑉𝑜 𝑉𝑝

𝑉𝑛

𝑉2

𝑅2

𝑅3

𝑖1

𝑖2


𝑉𝑜 =𝑅2 + 𝑅1

𝑅1𝑉𝑛 −

𝑅2

𝑅1𝑉1

karena 𝑉𝑝 = 𝑉𝑛

𝑉𝑜 =𝑅2 + 𝑅1

𝑅1

𝑅4

𝑅3 + 𝑅4𝑉2 −

𝑅2

𝑅1𝑉1

𝑉𝑝 =𝑅4

𝑅3 + 𝑅4𝑉2

𝑉2 − 𝑉𝑝

𝑅3=

𝑉𝑝

𝑅4

𝑉2𝑅4 − 𝑉𝑝𝑅4 = 𝑉𝑝𝑅3

𝑉2𝑅4 = 𝑉𝑝𝑅3 + 𝑉𝑝𝑅4

𝑉2𝑅4 = 𝑉𝑝(𝑅3 + 𝑅4)

𝑉𝑝 =𝑅4

𝑅3 + 𝑅4𝑉2

KCL di node 𝑣𝑝

𝑉1 − 𝑉𝑛

𝑅1=

𝑉𝑛 − 𝑉𝑜

𝑅2

𝑉1𝑅2 − 𝑉𝑛𝑅2 = 𝑉𝑛𝑅1 − 𝑉𝑜𝑅1

𝑉𝑜𝑅1 = 𝑉𝑛 𝑅2 + 𝑅1 − 𝑉1𝑅2

𝑉𝑜 =𝑅2 + 𝑅1

𝑅1𝑉𝑛 −

𝑅2

𝑅1𝑉1



Cascaded Op Amp Circuits

𝑉𝑠

𝑅𝑠 𝑅𝑓

+

−

𝑉𝑜

𝑉𝑝

𝑉𝑛

𝑉𝑝

𝑉𝑛

𝑅𝑠 𝑅𝑓

𝑉𝑠

𝑅𝑠

𝑅𝑓

+

−

𝑉𝑜 𝑉𝑝

𝑉𝑛

𝑉𝑝

𝑉𝑛 𝑅𝑠

𝑅𝑓

Inverting Amplifier



20mV

3 12

+

−

𝑉𝑜

𝑉𝑝

𝑉𝑛

𝑉𝑝

𝑉𝑛

4 10

Op-amp pertama Op-amp kedua

𝑣𝑜 =12 + 3

320

𝑣𝑜 = 100𝑚𝑉

𝑣𝑜 =10 + 4

4100

𝑣𝑜 = 350𝑚𝑉

05. operational amplifier

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