Download kul-eldas2-07 - IFI TALKS SOMETHING

KULIAH MINGGU ke 2
Elektronika Dasar
OPERASIONAL AMPLIFIER
OP-AMP
Jurusan Teknik Elektro
2007
1
Op-amp : suatu IC analog
+ VCC
Input 1
+
output
Input 2
_
- VEE
SIMBOL
2
SIFAT IDEAL
Ideally,
1. No current can enter terminals
V+ or V-. Called infinite input
impedance.
A
2. Vout=A(V+ - V-) with A → ∞
3. In a circuit
VOUT = (AV+ - AV )
= A (V+ - V )
V+ is forced equal to V4. An opamp needs two voltages
to power it Vcc and -Vee.
3
-
Operational Amplifier (Op Amp)
Vi1
A
Vi2
B
-
Vout
+
An operational amplifier (Op Amp)
is an integrated circuit of a
complete amplifier circuit.
Op amps have an extremely high
gain (A=105 typically).
Vout  Vi 2  Vi1 A
Op amps also have a high input
impedance (R=4 MΩ , typically)
and a low output impedance
(in order of 100 Ω , typically) .
4
Characters of Operational
Amplifiers
Not used
8
7
6
5
Offset null

high open loop gain

high input impedance

low output impedance

low input offset voltage

Offset null
1
2
3
4
low temperature coefficient of
input offset voltage

low input bias current

wide bandwidth

large common mode rejection
ratio (CMRR)
5
Voltage Output from an Amplifier
Vout
A
Non-linear
region
Linear
region
Vin= V2-V1
Daerah Linier ini sangat Kecil
Vin
The linear range of an
amplifier is finite, and limited
by the supply voltage and the
characteristics of the
amplifier.
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 6
OPAMP: COMPARATOR
(bekerja di daerah jenuh)
Vout=A(Vin – Vref)
A
If Vin>Vref, Vout = +∞ but practically
hits +ve power supply = Vcc
A (gain)
very high
If Vin<Vref, Vout = -∞ but practically
hits –ve power supply = -Vee
Application: detection of a complex signal in ECG
VREF
VIN
Vcc
Vout
-Vee
7
OPAMP: 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
8
Inverting Amplifier
(bekerja di daerah linier)
Point B is grounded, point A is
called Virtual Grounded.
RF
R1
Vin
A
B
Vout
+
R3 
G
Voltage across R1 is Vin, and
across RF is Vout.
-
R1 RF
R1  RF
Vout
R
 F
Vin
R1
The output node voltage
determined by Kirchhoff's
Current Law (KCL).
Circuit voltage gain determined
by the ratio of R1 and RF.
9
PENGUAT INVERTING
(bekerja di daerah linier)
Kondisi fisik
R1
R3
1
8
2
7
3
6
R2
output
4
5
input
10
OPAMP: INVERTING
AMPLIFIER
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
11
Analysis of Inverting Amplifier
RF
KCL at A:
iF
R1
VIN Vin
-
i-A
i1
-
i+
B
+
Vout
R
i1  i  iF  iF
VIN  V
i1 
R1
and
V  Vout
iF 
RF
i  0  V  0  V  0
Ideal transfer characteristics:
i  i  0
V  V
Vout
VIN

or
R1
RF
Vout
RF

Vin
R1
12
OPAMP: NON – INVERTING AMPLIFIER
(bekerja di daerah linier)
1. V- = V+
2. As V+ = VIN, V- = VIN
Vx
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)
13
Noninverting Amplifier
Point VA equals to Vin .
RF
Op-amp circuit is a voltage divider.
R1
A
Vin
B
+
Vout
R1
VA  Vout 
R1  RF
Circuit voltage gain determined by the ratio of R1 and RF.
Vout
RF
G
 1
Vin
R1
14
OPAMP : VOLTAGE FOLLOWER (BUFER)
(bekerja di daerah linier)
V+ = VIN.
i=0
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…
15
Differential Amplifier
RF
R1
V1
V2
A
-
R2
B
Point B is grounded, so does point A
(very small).
+
Vout
Voltage across R1 is V1, and across R2
is V2.
Normally: R1 = R2, and RF = R3.
R3
Commonly used as a single op-amp
instrumentation amplifier.
Vout 
RF
(V2  V1 )
R1
16
Analysis of an Instrumentation
Amplifier
Design a single op-amp instrumentation
amplifier.
R1 = R2, RF = R3
Determine the instrumentation gain.
RF
R1
-
V1
A
R2
V2
B
+
R3
Vout
V  VOUT
V1  V A
 A
 iA
R1
RF
V2  VB
V
 iB  B
R2
R3
i A  iB  0
VA  VB
Vout 
RF
(V2  V1 )
R1
V  VOUT VB V2  VB
V1  V A
 A


R1
RF
R3
R2
VOUT  V A  VB  V2  V1   V A  VB 

RF
R1
17
SUMMING AMPLIFIER
Recall inverting
amplifier and
If = I1 + I2 + … + In
If
V
= -R (V1/R1 + V2/R2 + … + Vn/Rn)
R

RF
R
Vout   F V1  OUT
V2  F V3 f
R2
R3 
 R1
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.
18
For the following circuit, calculate the
input resistance.
Vin
R1
Rf
Vout
R2
19
INSTRUMENTATION
AMPLIFIER
20
INSTRUMENTATION AMPLIFIER
Gain in the multiple stages: i.e.
High Gain – so, you can
amplify small signals
Inverting
amplifier
very high
input
impedance
- So, you can
connect to
sensors
Differential amplifier ->
it rejects common-mode
interference -> so you
can reject noise
Non-inverting
amplifier
21
INSTRUMENTATION AMPLIFIER:
STAGE 1
Recall virtual ground of opamps
I1 = (V1 – V2)/R1
I1I1
I2
I3
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)
22
INSTRUMENTATION AMPLIFIER:
STAGE 2
Recall virtual ground of opamps
and voltage divider
V- = V+ = VBR4/(R3 + R4)
VA
VB
I1
I2
I3
Recall no current can enter
opamps
(VA – V-)/R3 = (V- – VOUT)/R4
Solving,
VOUT = – (VA – VB)R4/R3
23
INSTRUMENTATION AMPLIFIER:
COMPLETE
VOUT = – (V1 – V2)(1 + 2R2/R1)(R4/R3)
24
Analog Signal Conditioner
(Current to Voltage Converter, LM-324)
+5 VDC
Sensor –
Variable Resistor
4.7k
I1
10k
I2
10k
I3
5k
Vout
0-5
VDC
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.
25
Single-Ended Input
+
~ Vi
V
o

+
V
o

~
Ref:080114HKN
• + terminal : Source
• – terminal : Ground
• 0o phase change
• + terminal : Ground
• – terminal : Source
• 180o phase change
V
i
Operational Amplifier
26
Double-Ended Input
• Differential input
V
+
d
~
V
•
o
• 0o phase shift change

+
~ V1
between Vo and Vd
V
o
Qu: What Vo should be if,
V
2

~
Vd  V  V
V
2
V
1
Ans: (A or B) ?
(A)
Ref:080114HKN
Operational Amplifier
(B)
27
Distortion
+V =+5V
cc
+
V
V
+5V
o
d

0
5V
V =5V
cc
The output voltage never excess the DC voltage supply
of the Op-Amp
Ref:080114HKN
Operational Amplifier
28
Common-Mode Operation
+
• Same voltage source is applied
at both terminals
o
• Ideally, two input are equally
amplified
• Output voltage is ideally zero
due to differential voltage is
zero
• Practically, a small output
signal can still be measured
V

V
i
~
Note for differential circuits:
Opposite inputs : highly amplified
Common inputs : slightly amplified
 Common-Mode Rejection
Ref:080114HKN
Operational Amplifier
29
Common-Mode Rejection Ratio (CMRR)
Differential voltage input :
Vd  V  V
Noninverting
+
Input
Output
Common voltage input :
1
Vc  (V  V )
2
Output voltage :
Vo  Gd Vd  GcVc
Gd : Differential gain
Gc : Common mode gain
Ref:080114HKN
Inverting
Input

Common-mode rejection ratio:
CMRR 
Gd
G
 20 log 10 d (dB)
Gc
Gc
Note:
When Gd >> Gc or CMRR 
Vo = GdVd
Operational Amplifier
30
CMRR Example
What is the CMRR?
100V
100V
+
60700V
80600V
20V

+
40V

Solution :
Vd 1  100  20  80V
(1)
Vd 2  100  40  60V
100  20
100  40
 60V
Vc 2 
 70V
2
2
From (1)
Vo  80Gd  60Gc  80600V
Vc1 
From (2)
Gd  1000
(2)
Vo  60Gd  70Gc  60700V
and
Gc  10
 CMRR  20 log( 1000 / 10)  40dB
NB: This method is Not work! Why?
Ref:080114HKN
Operational Amplifier
31
Op-Amp Properties
V1
(1) Infinite Open Loop gain
-
The gain without feedback
Equal to differential gain
Zero common-mode gain
Pratically, Gd = 20,000 to 200,000
(2) Infinite Input impedance
-
Input current ii ~0A
T- in high-grade op-amp
m-A input current in low-grade op-amp
(3) Zero Output Impedance
-
act as perfect internal voltage source
No internal resistance
Output impedance in series with load
Reducing output voltage to the load
Practically, Rout ~ 20-100 
Ref:080114HKN
Operational Amplifier
+
V2
Vo

i1~0
+
i2~0

Vo
Rout
Vo' +
Vload
Rload
Rload
 Vo
Rload  Rout
32
Frequency-Gain Relation
•
•
•
•
•
Ideally, signals are amplified from
DC to the highest AC frequency
(Voltage Gain)
Practically, bandwidth is limited
Gd
741 family op-amp have an limit 0.707Gd
bandwidth of few KHz.
Unity Gain frequency f1: the gain at
unity
Cutoff frequency fc: the gain drop by
3dB from dc gain Gd
20log(0.707)=3dB
1
0
fc
f1
(frequency)
GB Product : f1 = Gd fc
Ref:080114HKN
Operational Amplifier
33
GB Product
Example: Determine the cutoff frequency of an op-amp having a unit gain
frequency f1 = 10 MHz and voltage differential gain Gd = 20V/mV
(Voltage Gain)
Sol:
Gd
0.707Gd
Since f1 = 10 MHz
? Hz
By using GB production equation
f1 = Gd fc
fc = f1 / Gd = 10 MHz / 20 V/mV
= 10  106 / 20  103
10MHz
1
= 500 Hz
0
fc
f1
(frequency)
Ref:080114HKN
Operational Amplifier
34
Ideal Vs Practical Op-Amp
Ideal
Practical
Open Loop gain A

105
Bandwidth BW

10-100Hz
Input Impedance Zin

>1M
0
10-100 
Depends only
on Vd = (V+V)
Differential
mode signal
Depends slightly
on average input
Vc = (V++V)/2
Common-Mode
signal

10-100dB
Output Impedance Zout
Output Voltage Vout
CMRR
Ref:080114HKN
Operational Amplifier
Ideal op-amp
+ AVin
Vin
Vout
~

Zout=0
Practical op-amp
+
Vin
Zout
Zin
~
Vout
 AVin
35
Ideal Op-Amp Applications
Analysis Method :
Two ideal Op-Amp Properties:
(1) The voltage between V+ and V is zero V+ = V
(2) The current into both V+ and V termainals is zero
For ideal Op-Amp circuit:
(1) Write the kirchhoff node equation at the noninverting
terminal V+
(2) Write the kirchhoff node eqaution at the inverting
terminal V
(3) Set V+ = V and solve for the desired closed-loop gain
Ref:080114HKN
Operational Amplifier
36
Noninverting Amplifier
(1)
Kirchhoff node equation at V+
yields, V  V

V
+
in
V
o
i

(2)
Kirchhoff node equation at V
yields, V  0 V  Vo
Ra
(3)

Rf
0
Ra
Rf
Setting V+ = V– yields
Vi Vi  Vo

 0 or
Ra
Rf
Ref:080114HKN
Rf
Vo
 1
Vi
Ra
Operational Amplifier
37
v+
vi
v-
+
vo
R
1
Rf
v+
v-
Rf
Ra
2
v-

Voltage follower
o
R
f
Noninverting input with voltage divider
Rf
R2
vo  (1  )(
)vi
Ra R1  R2
)vi
+
v

a
Rf
v+
v
i
R
vo
1
R
2
v-
+
v
o

R
Less than unity gain
f
vo  vi
Ref:080114HKN
R
R
Noninverting amplifier
vo  (1 
+
i

Ra
vi
v+
v
R2
vo 
vi
R1  R2
Operational Amplifier
38
Inverting Amplifier
(1)
Rf
Kirchhoff node equation at V+
yields, V  0
Ra

(2)
Kirchhoff node equation at V
yields, Vin  V_ V  V
 o  0
Ra
Rf
(3)
Setting V+ = V– yields
Vo  R f

Vin
Ra
Ref:080114HKN
V ~
in

V
o
+
Notice: The closed-loop gain Vo/Vin is
dependent upon the ratio of two resistors,
and is independent of the open-loop gain.
This is caused by the use of feedback output
voltage to subtract from the input voltage.
Operational Amplifier
39
Multiple Inputs
(1)
Kirchhoff node equation at V+
yields, V  0

(2)
Kirchhoff node equation at V
yields,
V_  Vo
Rf
(3)

Rf
Va
Vb
Vc
Ra
Rb
Rc

V
o
+
V  Va V  Vb V  Vc


0
Ra
Rb
Rc
Setting V+ = V– yields
c V
 Va Vb Vc 
j
Vo   R f       R f 
j a R j
 Ra Rb Rc 
Ref:080114HKN
Operational Amplifier
40
Inverting Integrator
Zf
Now replace resistors Ra and Rf by complex
 ZZ
f , respectively, therefore
componentsV Za and
f in
V
o
Za
in
Supposing
(i) The feedback1 component is a capacitor C, i.e.,
Zf 
jC
(ii) The input component is a resistor R, Za = R
Therefore, the closed-loop gain (Vo/Vin) become:
Za
where
What happens if Za = 1/jC whereas, Zf
Inverting differentiator
Ref:080114HKN
+
C
R
vi (t )  Vi e jt
~
V
= R?
Operational Amplifier
in
V
o
V ~
1
vo (t ) 
vi (t )dt

RC


V
o
+
41
Example:
(a)
Op-Amp Integrator
C
Determine the rate of change
of the output voltage.
R
+5V
0
100s
(b) Draw the output waveform.
V
i
10 k
0.01F

V
o
+
Vo(max)=10 V
Solution:
(a) Rate of change of the output voltage
Vo
V
5V
 i 
t
RC (10 k)(0.01 F)
 50 mV/ s
+5V
0
V
i
0
(b) In 100 s, the voltage decrease
Vo  (50 mV/ s)(100μs)  5V
Ref:080114HKN
-5V
-10V
Operational Amplifier
V
o
42
Op-Amp Differentiator
R
C
0
to
t1
t2
V

i
V
o
0
+
to
t1
t2
 dV 
vo   i  RC
 dt 
Ref:080114HKN
Operational Amplifier
43
Non-ideal case (Inverting Amplifier)
Rf
Ra
V
in
+

V
o
Vin ~

Vin
Zout
Zin
~
Vout
 AVin
+
Equivalent Circuit
3 categories are considering
Rf
Ra

R
 R
V
+ +

Ref:080114HKN
Practical op-amp
V
o
 Close-Loop Voltage Gain
 Input impedance
 Output impedance
-AV
Operational Amplifier
44
Close-Loop Gain
Applied KCL at V– terminal,
Rf
Vin  V  V Vo  V


0
Ra
R
Rf
V
Ra

in
By using the open loop gain,
V
R R
+ +

Vo   AV


Vin Vo
V
V
V

 o  o  o 0
Vin
Ra ARa AR R f AR f
R R f  Ra R f  Ra R  ARa R
Vin
 Vo
Ra
ARa R R f
Ra
V
o
-AV
Rf
V
o
V R
The Close-Loop Gain, Av
 AR R f
Vo
Av 

Vin R R f  Ra R f  Ra R  ARa R
Ref:080114HKN
Operational Amplifier
45
Close-Loop Gain
When the open loop gain is very large, the above equation become,
Av ~
 Rf
Ra
Note : The close-loop gain now reduce to the same form
as an ideal case
Ref:080114HKN
Operational Amplifier
46
Input Impedance
Rf
Input Impedance can be regarded as,
V
Rin  Ra  R // R

in
R
where R is the equivalent impedance
of the red box circuit, that is
R 
Ra
V R
+ +

V
if
V
o
-AV
R'
However, with the below circuit,
V  ( AV )  i f ( R f  Ro )
V R f  Ro
 R 

if
1 A
Ref:080114HKN
if
Rf
R
V
+

Operational Amplifier
-AV
47
Input Impedance
Finally, we find the input impedance as,
1
1 A 
Rin  Ra   

R
R

R

f
o
 
Since,
1

Rin  Ra 
R ( R f  Ro )
R f  Ro  (1  A) R
R f  Ro  (1  A) R , Rin become,
Rin ~ Ra 
Again with
( R f  Ro )
(1  A)
R f  Ro  (1  A)
Rin ~ Ra
Note: The op-amp can provide an impedance isolated from
input to output
Ref:080114HKN
Operational Amplifier
48
Output Impedance
Only source-free output impedance would be considered,
i.e. Vi is assumed to be 0
Firstly, with figure (a),
Ra
Ra // R
Ra R
Vo  V 
Vo
R f  Ra // R
Ra R f  Ra R  R f R
By using KCL, io = i1+ i2
V 
io 
Rf
R
V
Vo
V  ( AV )
 o
R f  Ra // R f
Ro
io
R
V
+
 -AV
By substitute the equation from Fig. (a),
The output impedance, Rout is
Ro ( Ra R f  Ra R  R f R )
Vo

io (1  Ro )( Ra R f  Ra R  R f R )  (1  A) Ra R
R and A comparably large,
Ro ( Ra  R f )
Rout ~
ARa
Ref:080114HKN
Rf
V
R
V
Ra
R
(a)
Operational Amplifier
i2
i1
V
+

-AV
(b)
49
o
KOMPARATOR




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.
50
KOMPARATOR SEDERHANA
51
KURVA HUBUNGAN TEGANGAN
Vr>0
Vi < Vr 
Vi > Vr 
Vo = +Vsat
Vo = -Vsat
52
KURVA HUBUNGAN TEGANGAN
Vo
Vr=0
Vi
Vi < Vr 
Vi > Vr 
Vo = +Vsat
Vo = -Vsat
53
KURVA HUBUNGAN TEGANGAN
Vo
Vr<0
Vi
Vi < Vr 
Vi > Vr 
Vo = +Vsat
Vo = -Vsat
54
STABILITAS KOMPARATOR
SEDERHANA
Vcc
Vin
VCC/2
Teganga
n
masukan
R
Vcc/2
R
Tegangan
keluaran
55
RANGKAIAN SCHMITT
Positive Feedback
Vin
Vout
Vout /2
R
R
Rangkaian ini disebut komparator
Schmitt trigger.
Rangkaian resistor membuat positive
feedback.
56
CARA KERJA Schmitt trigger
Vin
Vout
Vout /2
R
R
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.
57
TAK STABIL
58
STABIL
histerisis
59
STABIL
60
STABIL
61
RANGKAIAN dan OUTPUT
62
KETERANGAN SCHMITT


Schmitt trigger adalah sebuah aplikasi comparator yang
mengubah tagangan keluaran menjadi negatif bila
mtegangan masukan lebih besar tegangan referensi.
Kemudian menggunakan negative feedback untuk
mencegah agar tegangan keluaran tdk kembali ke
kondisi semula saat tegangan kembali kurang dari
tegangan referensi, sampai nanti masukan lebih kecil
dari yang ditentukan.
63
A
P
L
I
K
A
S
I
64
PERHITUNGAN

Kerja Schmitt trigger merupakan proses komparasi dengan
threshold ganda. Persamaan arus di titik A:

Karena hanya 2 pers, maka harus ada satu R yang ditentukan
dulu.
Ingat : Vout = VCC saat Vin diatas batas atas (V2)
Vout = -VEE saat Vin dibawah batas bawah (V2’).
65
u
66
LM741
67
NOMOR KAKI
68
69
70
71
72
73
74
KESIMPULAN
Op-amp dapat digunakan sebagai :
1.
2.
3.
4.
5.
6.
Penguat INVERTING
Penguat NON INVERTING
BUFER
Penguat PENJUMLAH
Penguat INSTRUMENTASI
Pengubah ARUS KE TEGANGAN atau
sebaliknya
7. KOMPARATOR
75
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.
76