operational amplifier
DESCRIPTION
Tugas presentasi mata kuliah Elektronika Dasar..TRANSCRIPT
Rizki Anugrah Wibowo 1004405117
I Made Agung Pranata 1204405030
Made Danu Widana 1204405033
I Kadek Arya Wiratama 1204405038
Ketut Alit Sukertha Winaya 1204405043
Bhrama Sakti Karenda Putra 1204405082
Electrical Engineering Department
Udayana University
Summarized by:
Introduction of Operation Amplifier (Op-Amp)
Analysis of ideal Op-Amp applications
Comparison of ideal and non-ideal Op-Amp
Outlines
Non-ideal Op-Amp consideration
Vd
+
Vo
R in~inf Rout~0
Input 1
Input 2
Output
+Vcc
-Vcc
Operational Amplifier (Op-Amp)
Very high differential gain
High input impedance
Low output impedance
Provide voltage changes
(amplitude and polarity)
Used in oscillator, filter
and instrumentation
Accumulate a very high
gain by multiple stages
ddo VGV
510say large,y ver
normally gain aldifferenti : dG
IC Product
+
1
2
3
4
8
7
6
5
OFFSET
NULL
-IN
+IN
V
N.C.
V+
OUTPUT
OFFSET
NULL
+
1
2
3
4
8
7
6
5
OUTPUT A
-IN A
+IN A
V
V+
OUTPUT B
-IN B
+IN B+
DIP-741 Dual op-amp 1458 device
Single-Ended Input
+
Vo
~ V i
+
Vo
~ V i
• + terminal : Source
• – terminal : Ground
• 0o phase change
• + terminal : Ground
• – terminal : Source
• 180o phase change
Double-Ended Input
~ V1
+
Vo
~ V2
+
Vo
~Vd
• Differential input
•
• 0o phase shift change
between Vo and Vd
VVVd
Qu: What Vo should be if,
V1
V2
(A) (B)
Ans: (A or B) ?
Distortion
Vd
+
Vo
+Vcc=+5V
Vcc=5V
0
+5V
5V
The output voltage never excess the DC voltage
supply of the Op-Amp
Common-Mode Operation
+
Vo
V i ~
• Same voltage source is applied
at both terminals
• 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 Note for differential circuits:
Opposite inputs : highly
amplified
Common inputs : slightly
amplified
Common-Mode Rejection
Common-Mode Rejection Ratio (CMRR)
Differential voltage input :
VVVd
Common voltage input :
)(2
1 VVVc
Output voltage :
ccddo VGVGV
Gd : Differential gain
Gc : Common mode gain
)dB(log20CMRR 10
c
d
c
d
G
G
G
G
Common-mode rejection ratio:
Note:
When Gd >> Gc or CMRR
Vo = GdVd
+
Noninverting
Input
Inverting
Input
Output
CMRR Example
What is the CMRR?
Solution :
dBCMRR and
V (2) From
V (1) From
V V
V V
40)10/1000log(20101000
607007060
806006080
702
4010060
2
20100
60401008020100
21
21
cd
cdo
cdo
cc
dd
GG
GGV
GGV
VV
VV
+
100V
20V
80600V
+
100V
40V
60700V
(1) (2)
Op-Amp Properties
(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
+
V1
V2
Vo
+
Vo
i1~0
i2~0
+
Rout
Vo'Rload
outload
load
oloadRR
RVV
Frequency-Gain Relation
• Ideally, signals are amplified
from DC to the highest AC
frequency
• Practically, bandwidth is limited
• 741 family op-amp have an limit
bandwidth of few KHz.
• Unity Gain frequency f1: the
gain at unity
• Cutoff frequency fc: the gain
drop by 3dB from dc gain Gd
(Voltage Gain)
(frequency)
f1
Gd
0.707Gd
fc0
1
GB Product : f1 = Gd fc
20log(0.707)=3dB
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
Sol:
Since f1 = 10 MHz
By using GB production equation
f1 = Gd fc
fc = f1 / Gd = 10 MHz / 20 V/mV
= 10 106 / 20 103
= 500 Hz
(Voltage Gain)
(frequency)
f1
Gd
0.707Gd
fc0
1
10MHz
? Hz
Ideal Vs Practical Op-Amp
Ideal Practical
Open Loop gain A 105
Bandwidth BW 10-100Hz
Input Impedance Zin >1M
Output Impedance Zout 0 10-100
Output Voltage VoutDepends only
on Vd = (V+V)
Differential
mode signal
Depends slightly
on average input
Vc = (V++V)/2
Common-Mode
signal
CMRR 10-100dB
+
~
AVin
Vin Vout
Zout=0
Ideal op-amp
+
AVin
Vin Vout
Zout
~
Zin
Practical op-amp
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
Noninverting Amplifier
(1) Kirchhoff node equation at V+
yields,
(2) Kirchhoff node equation at V
yields,
(3) Setting V+ = V– yields
or
+V in
Vo
RaRf
iVV
00
f
o
a R
VV
R
V
0
f
oi
a
i
R
VV
R
V
a
f
i
o
R
R
V
V 1
+
vi
Ra
vov-
v+
Rf
+
vi
Ra
vov-
v+
Rf
R2R1
+
vi
vov-
v+
Rf
+
vi
vov-
v+
R f
R2R1
Noninverting amplifier Noninverting input with voltage divider
i
a
f
o vRR
R
R
Rv ))(1(
21
2
Voltage follower
io vv
i
a
f
o vR
Rv )1(
Less than unity gain
io vRR
Rv
21
2
Inverting Amplifier
(1) Kirchhoff node equation at V+
yields,
(2) Kirchhoff node equation at V
yields,
(3) Setting V+ = V– yields
0V
0_
f
o
a
in
R
VV
R
VV
a
f
in
o
R
R
V
V
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.
+
~
Rf
Ra
Vin
Vo
Multiple Inputs
(1) Kirchhoff node equation at V+
yields,
(2) Kirchhoff node equation at V
yields,
(3) Setting V+ = V– yields
0V
0_
c
c
b
b
a
a
f
o
R
VV
R
VV
R
VV
R
VV
c
aj j
j
f
c
c
b
b
a
a
foR
VR
R
V
R
V
R
VRV
+
Rf
Va
Vo
Rb
Ra
Rc
Vb
Vc
Inverting Integrator
Now replace resistors Ra and Rf by complex
components Za and Zf, respectively, therefore
Supposing
(i) The feedback component is a capacitor C,
i.e.,
(ii) The input component is a resistor R, Za = R
Therefore, the closed-loop gain (Vo/Vin) become:
where
What happens if Za = 1/jC whereas, Zf = R?
Inverting differentiator
in
a
f
oV
Z
ZV
CjZ f
1
dttvRC
tv io )(1
)(
tj
iieVtv
)(
+
~
Zf
Za
V in
Vo
+
~
R
V in
Vo
C
Op-Amp Integrator
Example:
(a) Determine the rate of change
of the output voltage.
(b) Draw the output waveform.
Solution:
+
R
VoV i
0
+5V
10 k
0.01FC
Vo(max)=10 V
100s
(a) Rate of change of the output voltage
smV/
F k
V
50
)01.0)(10(
5
RC
V
t
V io
(b) In 100 s, the voltage decrease
VμssmV/ 5)100)(50( oV
0
+5V
0
-5V
-10V
V i
Vo
Op-Amp Differentiator
RCdt
dVv i
o
+
R
C
VoVi
0to t1 t2
0
to t1 t2
Non-ideal case (Inverting Amplifier)
+
~
Rf
Ra
Vin
Vo
3 categories are considering
Close-Loop Voltage Gain
Input impedance
Output impedance
Equivalent Circuit
Rf
Ra
V in
Vo
+
+
R R
V
-AV
+
AVin
Vin Vout
Zout
~
Zin
Practical op-amp
Close-Loop Gain
Rf
Ra
V in
Vo
+
+
R R
-AV
Ra Rf
R
V
V
V in Vo
Applied KCL at V– terminal,
0
f
o
a
in
R
VV
R
V
R
VV
By using the open loop gain,
AVVo
0f
o
f
oo
a
o
a
in
AR
V
R
V
AR
V
AR
V
R
V
fa
aafaf
o
a
in
RRAR
RARRRRRRRV
R
V
The Close-Loop Gain, Av
RARRRRRRR
RAR
V
VA
aafaf
f
in
o
v
Close-Loop Gain
When the open loop gain is very large, the above equation become,
a
f
vR
RA
~
Note : The close-loop gain now reduce to the same form
as an ideal case
Input Impedance
Rf
Ra
Vin
Vo
+
+
R
-AV
R'
R
V
Rf
+
R
-AV
if
V
Input Impedance can be regarded as,
RRRR ain //
where R is the equivalent impedance
of the red box circuit, that is
fi
VR
However, with the below circuit,
A
RR
i
VR
RRiAVV
of
f
off
1
)()(
Input Impedance
Finally, we find the input impedance as,
1
11
of
ainRR
A
RRR
RARR
RRRRR
of
of
ain)1(
)(
Since,RARR of )1( , Rin become,
)1(
)(~
A
RRRR
of
ain
Again with )1( ARR of
ain RR ~
Note: The op-amp can provide an impedance isolated from
input to output
Output Impedance
Only source-free output impedance would be considered,
i.e. Vi is assumed to be 0Rf
Ra
Vo
+
R
-AV
R
V
io
V
V
Rf
Ra R
+
R
-AV
V
i2 i1
(a) (b)
Firstly, with figure (a),
o
fafa
a
o
af
a VRRRRRR
RRVV
RRR
RRV
//
//
By using KCL, io = i1+ i2
o
o
faf
o
oR
AVV
RRR
Vi
)(
//
By substitute the equation from Fig. (a),
RRARRRRRRR
RRRRRRR
i
V
R
afafao
fafao
o
o
out
)1())(1(
)(
is impedance,output The
R and A comparably large,
a
fao
outAR
RRRR
)(~
