Dasar Opamp
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Transcript of Dasar Opamp
KULIAH MINGGU ke 9
Elektronika Dasar
1
OPERASIONAL AMPLIFIEROP-AMP
Jurusan Teknik Elektro2012
2
Op-amp : suatu IC analogOp-amp : suatu IC analog
+
_
Input 1
Input 2 output
+ VCC
- VEE
SIMBOL
SIFAT IDEAL
3
Ideally,
1.No current can enter terminals V+ or V-. Called infinite input impedance.
2.Vout=A(V+ - V-) with A → ∞
3. In a circuit
V+ is forced equal to V-
4.An opamp needs two voltages to power it Vcc and -Vee.
A
VOUT = (AV - AV ) = A (V - V )
+
+
--
4
Operational Amplifier (Op Amp)Operational Amplifier (Op Amp)
An An operational amplifieroperational amplifier (Op Amp) (Op Amp)
is an integrated circuit of a is an integrated circuit of a
complete amplifier circuit.complete amplifier circuit.
Op ampsOp amps have an extremely high have an extremely high
gain (gain (A=10A=1055 typically typically).).
Op ampsOp amps also have a high input also have a high input
impedance (impedance (R=4 MR=4 MΩΩ , typically , typically) )
and a low output impedanceand a low output impedance
((in order of 100 in order of 100 ΩΩ , typically , typically) .) .
-
+
Vi1 VoutA
B
Vi2
AVVV iiout 12
5
Characters of Operational Characters of Operational Amplifiers Amplifiers
high open loop gainhigh open loop gain
high input impedancehigh input impedance
low output impedancelow output impedance
low input offset voltagelow input offset voltage
low temperature coefficient low temperature coefficient of input offset voltageof input offset voltage
low input bias currentlow input bias current
wide bandwidthwide bandwidth
large common mode large common mode rejection ratio (CMRR)rejection ratio (CMRR)
1 2 3 4
8 7 6 5
Offset null
Offset nullNot used
6
Q1 Q2D1
D2
Q3
R2
R1
R3
R4
Q5
C1
R5
Q4
7
8
Voltage Output from an AmplifierVoltage Output from an AmplifierThe linear range of an
amplifier is finite, and limited
by the supply voltage and the
characteristics of the
amplifier.A
Linearregion
Non-linearregion
Vout
Vin
Daerah Linier ini sangat Kecil
If an amplifier is driven
beyond the linear range
(overdriven), serious errors
can result if the gain is
treated as a constant.
Kalau A = 106 dan VCC = 12 Volt maka daerah linier = 24 μV
Vin= V2-V1
OPAMP: COMPARATOR(bekerja di daerah jenuh)
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Vout=A(Vin – Vref)
If Vin>Vref, Vout = +∞ but practically hits +ve power supply = Vcc
If Vin<Vref, Vout = -∞ but practically hits –ve power supply = -Vee
Vcc
-Vee VIN
VREF
Application: detection of a complex signal in ECG
A (gain) very high
Vout
A
10
OPAMP: ANALYSISOPAMP: ANALYSIS
The key to op amp analysis is simple
1.No current can enter op amp input terminals.
=> Because of infinite input impedance
2.The +ve and –ve (non-inverting and inverting) inputs are forced to be at the same potential.
=> Because of infinite open loop gain
3.Use the ideal op amp property in all your analysis
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Inverting AmplifierInverting Amplifier((bekerja di daerah linierbekerja di daerah linier))
Point Point BB is grounded, point is grounded, point A A
is called Virtual is called Virtual
Grounded.Grounded.
Voltage across Voltage across RR11 is is VVinin, and , and
across across RRFF is is VVout.out.
The output node voltage The output node voltage
determined by determined by Kirchhoff'sKirchhoff's
Current Law (KCL).Current Law (KCL).
Circuit voltage gainCircuit voltage gain
determined by the determined by the ratio of ratio of
RR11 and and RRF.F.
1R
R
V
VG F
in
out
-
+
Vin
Vout
R1
RF
A
B
F
F
RRRR
R
1
13
12
PENGUAT INVERTINGPENGUAT INVERTING
((bekerja di daerah linierbekerja di daerah linier))
Kondisi fisikKondisi fisik
1R2
R1
2
3
4
8
7
6
5
input
outputR3
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OPAMP: INVERTING OPAMP: INVERTING AMPLIFIERAMPLIFIER
1. V- = V+
2. As V+ = 0, V- = 0 (VG)
3. As no current can enter V- and from Kirchoff’s Ist law, I1=I2.
4. I1 = (VIN - V-)/R1 = VIN/R1
5. I2 = (0 - VOUT)/R2 = -VOUT/R2 => VOUT = -I2R2
6. From 3 and 6, VOUT = -I2R2 = -I1R2 = -VINR2/R1 (NEG)
7. Therefore VOUT = (-R2/R1)VIN
14
Analysis of Inverting AmplifierAnalysis of Inverting Amplifier
Ideal transfer characteristics:
-
+
Vin
Vout
R1
RF
A
BR
ii++
iiFF
ii11
ii--
--
0 ii
VV
FF iiii 1
F
outF
IN
R
VViand
R
VVi
11
000 VVi
F
outIN
R
V
R
V
1
oror1R
R
V
V F
in
out
VIN
KCL at A:
15
OPAMP: NON – INVERTING OPAMP: NON – INVERTING AMPLIFIERAMPLIFIER
((bekerja di daerah linierbekerja di daerah linier))
1. V- = V+
2. As V+ = VIN, V- = VIN
3. As no current can enter V- and from Kirchoff’s Ist law, I1=I2.
4. I1 = Vx/R1=VIN/R1
5. I2 = (VOUT - VIN)/R2 VOUT = VIN + I2R2
6. VOUT = I1R1 + I2R2 = (R1+R2)I1 = (R1+R2)VIN/R1
7. Therefore VOUT = (1 + R2/R1)VIN (tak berlawanan)
Vx
16
Op-amp circuit is a voltage Op-amp circuit is a voltage
divider.divider.
Noninverting AmplifierNoninverting Amplifier
-
+Vin
Vout
R1
RF
A
BF
outA RR
RVV
1
1
1
1R
R
V
VG F
in
out
Circuit voltage gain determined by the ratio of R1 and RF.
Point VA equals to Vin .
OPAMP : VOLTAGE FOLLOWER (BUFER) (bekerja di daerah linier)
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V+ = VIN.
V- = V+
Thus Vout = V- = V+ = VIN !!!!
So what’s the point ?
The point is, due to the infinite input impedance of an op amp, no current at all can be drawn from the circuit before VIN. Thus this part is effectively isolated.
Very useful for interfacing to high impedance sensors such as microelectrode, microphone…
i = 0
18
Differential AmplifierDifferential Amplifier
Point Point BB is grounded, so does is grounded, so does
point point A A (very small).(very small).
Voltage across Voltage across RR11 is is VV11, and , and
across across RR22 is is VV2.2.
Normally: Normally: RR11 = = RR22, , andand R RFF
= = RR3.3.
Commonly used as a single Commonly used as a single
op-amp instrumentation op-amp instrumentation
amplifier.amplifier.)( 12
1
VVR
RV F
out
RF
-
+
V1
Vout
R1
A
BR3
V2
R2
19
Analysis of an Instrumentation Analysis of an Instrumentation AmplifierAmplifier
Design a single op-amp Design a single op-amp instrumentation amplifier. instrumentation amplifier.
RR11 = R = R22, R, RFF = R = R33
Determine the instrumentation gain.Determine the instrumentation gain.-
+
V1
Vout
R1
RF
A
B
R3
V2
R2A
F
OUTAA iR
VV
R
VV
1
1
32
2
R
Vi
R
VV BB
B
0 BA ii
2
2
31
1
R
VV
R
V
R
VV
R
VV BB
F
OUTAA
1
12
R
VVVV
R
VVV BA
F
BAOUT
)( 121
VVR
RV F
out
BA VV
SUMMING AMPLIFIER
3
32
21
1
VR
RV
R
RV
R
RV FFF
out
20
VOUT = -Rf (V1/R1 + V2/R2 + … + Vn/Rn)
If
Recall inverting amplifier and If = I1 + I2 + … + In
Summing amplifier is a good example of analog circuits serving as analog computing amplifiers (analog computers)!
Note: analog circuits can add, subtract, multiply/divide (using logarithmic components, differentiat and integrate – in real time and continuously.
21
For the following circuit, calculate the input resistance.
R1Rf
R2
Vin
Vout
22
INSTRUMENTATIINSTRUMENTATION AMPLIFIERON AMPLIFIER
23
INSTRUMENTATION INSTRUMENTATION AMPLIFIERAMPLIFIER Inverting
amplifier
Non-inverting amplifier
very high input impedance
- So, you can connect to sensors
Differential amplifier -> it rejects common-mode interference -> so you can reject noise
Gain in the multiple stages: i.e. High Gain – so, you can amplify small signals
INSTRUMENTATION AMPLIFIER: STAGE 1
24
I1
Recall virtual ground of opamps
I1 = (V1 – V2)/R1
Recall no current can enter opamps and Kirchoff’s current law
I2 = I3 = I1
Recall Kirchoff’s voltage law
VOUT = (R1 + 2R2)(V1 – V2)/R1
= (V1 – V2)(1+2R2/R1)
I2
I3I1
INSTRUMENTATION AMPLIFIER: STAGE 2
25
I1
Recall virtual ground of opamps and voltage divider
V- = V+ = VBR4/(R3 + R4)
Recall no current can enter opamps
(VA – V-)/R3 = (V- – VOUT)/R4
Solving,
VOUT = – (VA – VB)R4/R3
I2
I3
VA
VB
26
INSTRUMENTATION INSTRUMENTATION AMPLIFIER: COMPLETEAMPLIFIER: COMPLETE
VOUT = – (V1 – V2)(1 + 2R2/R1)(R4/R3)
27
As force is applied on the sensor, the value of the variable resistor changes which results in a specific voltage output.
Gain = Vout/Vin = 1 Resistor Values for the Inverting OP-Amp can be changed to modify gain of converter or to amplify the signal of interest.
Analog Signal Conditioner Analog Signal Conditioner (Current to Voltage Converter, (Current to Voltage Converter, LM-324)LM-324)
+5 VDC
Sensor – Variable Resistor
4.7k
10k
10k
5k
Vout
0-5 VDC
I1I2
I3
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KOMPARATORKOMPARATOR
Rangkaian komparator digunakan untuk membandingkan tegangan masukan dan tegangan referensi.
Tegangan keluaran hanya ada dua kondisi yaitu tegangan tinggi atau rendah (negatif). Kondisi ini ditentukan oleh besarnya tegangan masukan apakah lebih tinggi terhadap tegangan referensi atau lebih rendah.
Persoalan dalam komparator sederhana adalah stabilitas. Bila tegangan masukan bervariasi sekitar tegangan referensi maka tegangan keluaran akan berubah-ubah tidak stabil.
Hal tersebut dapat dihilangkan dengan rangkaian schmitt.
29
KOMPARATOR KOMPARATOR SEDERHANASEDERHANA
30
KURVA HUBUNGAN TEGANGANKURVA HUBUNGAN TEGANGAN
Vi < Vr
Vo = +Vsat
Vi > Vr
Vo = -Vsat
Vr>0
31
KURVA HUBUNGAN TEGANGANKURVA HUBUNGAN TEGANGAN
Vi < Vr
Vo = +Vsat
Vi > Vr
Vo = -Vsat
Vi
Vo
Vr=0
32
KURVA HUBUNGAN TEGANGANKURVA HUBUNGAN TEGANGAN
Vi < Vr
Vo = +Vsat
Vi > Vr
Vo = -Vsat
Vi
Vo
Vr<0
33
STABILITAS KOMPARATOR STABILITAS KOMPARATOR SEDERHANASEDERHANA
Vcc/2
Vin
R
Vcc
R
Tegangan
masukan
Tegangan keluaran
VCC/2
34
RANGKAIAN SCHMITT Positive Feedback Positive Feedback
Rangkaian ini disebut komparator Schmitt trigger.
Rangkaian resistor membuat positive feedback.
R
Vout
R
Vout /2
Vin
35
CARA KERJA Schmitt triggerCARA KERJA Schmitt trigger
Anggap tegangan masukan kecil, tegangan keluaran menjadi tinggi.
Bila Vout is 4 V, maka masukan non-inverting V+ adalah 2 Volt.
Kondisi keluaran tetap selama Vin kurang dari 2 Volt.
Bila Vin diperbesar sehingga lebih besar dari 2 V, maka Vout akan nol, dan V+ akan nol juga. Kondisi output ini akan tetap, selama Vin lebih besar 2 V.
R
Vout
R
Vout /2
Vin
36
TAK STABILTAK STABIL
37
STABILSTABILhisterisis
38
STABILSTABIL
39
STABILSTABIL
40
RANGKAIAN dan RANGKAIAN dan OUTPUTOUTPUT
41
KETERANGAN SCHMITTKETERANGAN SCHMITT
Schmitt trigger adalah sebuah aplikasi Schmitt trigger adalah sebuah aplikasi comparator yang mengubah tagangan yang mengubah tagangan keluaran menjadi negatif bila mtegangan keluaran menjadi negatif bila mtegangan masukan lebih besar tegangan referensi. masukan lebih besar tegangan referensi.
Kemudian menggunakan Kemudian menggunakan negative feedback untuk mencegah agar untuk mencegah agar tegangan keluaran tdk kembali ke kondisi tegangan keluaran tdk kembali ke kondisi semula saat tegangan kembali kurang semula saat tegangan kembali kurang dari tegangan referensi, sampai nanti dari tegangan referensi, sampai nanti masukan lebih kecil dari yang ditentukan. masukan lebih kecil dari yang ditentukan.
42
AAPPLLIIKKAASSII
43
PERHITUNGAN PERHITUNGAN Kerja Kerja Schmitt trigger merupakan proses merupakan proses
komparasi dengan threshold ganda. komparasi dengan threshold ganda. Persamaan arus di titik A:Persamaan arus di titik A:
Karena hanya 2 pers, maka harus ada satu R Karena hanya 2 pers, maka harus ada satu R yang ditentukan dulu. yang ditentukan dulu.
Ingat : Vout = VCC saat Vin diatas batas atas (V2)Vout = -VEE saat Vin dibawah batas bawah (V2’).
44
uu
LM741
45
NOMOR KAKI
46
47
48
49
50
51
52
KESIMPULANOp-amp dapat digunakan sebagai :
1. Penguat INVERTING2. Penguat NON INVERTING3. BUFER4. Penguat PENJUMLAH5. Penguat INSTRUMENTASI6. Pengubah ARUS KE TEGANGAN atau
sebaliknya7. KOMPARATOR
53
PR
• Buktikan rumus untuk menghitung R1, R2 dan R3 pada komparator Schmitt (slide 41), bila batas atas dan batas bawah diketahui.
• Rencanakanlah sebuah komparator Schmitt dengan menggunakan sebuah op-amp, yang menggunakan single supply 5 Volt. Batas tegangan yang dideteksi adalah diatas 3 Volt memberikan tegangan output tinggi, dan dibawah 1 Volt menghasilkan tegangan output rendah. Tentukan nilai R yang diperlukan.
54