Download Operational Amplifier

Lecture 1 Op-Amp
• Introduction of Operation Amplifier (OpAmp)
• Analysis of ideal Op-Amp applications
• Comparison of ideal and non-ideal Op-Amp
• Non-ideal Op-Amp consideration
Operational Amplifier (Op-Amp)
+Vcc
Very high differential gain
Input 1
+
High input impedance
Vo
Vd
Low output impedance
Output

Provide voltage changes
Input 2
(amplitude and polarity)
Rin~inf -Vcc Rout~0
• Used in oscillator, filter
Vo  GdVd
and instrumentation
• Accumulate a very high
Gd : differenti al gain normally
gain by multiple stages
5
•
•
•
•
ver y large, say 10
IC Product
OFFSET
NULL
-IN
1
8
N.C.
2
7
V+
6
OUTPUT
5
OFFSET
NULL
+IN
3
V
4

+
DIP-741
OUTPUT A
1
-IN A
2
+IN A
3
V
4
8 V+

+
7 OUTPUT B

6
+
5 +IN B
-IN B
Dual op-amp 1458 device
Single-Ended Input
+
~ Vi
V
o

+
V
o

~
V
i
• + terminal : Source
• – terminal : Ground
• 0o phase change
• + terminal : Ground
• – terminal : Source
• 180o phase change
Double-Ended Input
V
+
d
~
• Differential input
V
o
•

Vd  V  V
• 0o phase shift change
between Vo and Vd
+
~ V1
V
o
Qu: What Vo should be if,
V
2

~
V
2
V
1
Ans: (A or B) ?
(A)
(B)
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
Common-Mode Operation
• 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
V
o
V
i
~
Note for differential circuits:
Opposite inputs : highly amplified
Common inputs : slightly amplified
 Common-Mode Rejection
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
Inverting
Input

Common-mode rejection ratio:
CMRR 
Gd
G
 20 log 10 d (dB)
Gc
Gc
Note:
When Gd >> Gc or CMRR 
Vo = GdVd
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?
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
V1
+
V2

i1~0
+
i2~0

(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 
Vo
Vo
Rout
Vo' +
Vload
Rload
Rload
 Vo
Rload  Rout
Frequency-Gain Relation
•
•
•
•
•
Ideally, signals are amplified
from DC to the highest AC
(Voltage Gain)
frequency
Gd
Practically, bandwidth is limited 0.707Gd
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
20log(0.707)=3dB
1
0
fc
f1
(frequency)
GB Product : f1 = Gd fc
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:
Since f1 = 10 MHz
Gd
0.707Gd
? Hz
By using GB production equation
f1 = Gd fc
10MHz
fc = f1 / Gd = 10 MHz / 20 V/mV
= 10  106 / 20  103
= 500 Hz
1
0
fc
f1
(frequency)
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
Ideal op-amp
+ AVin
Vin
Vout
~

Zout=0
Practical op-amp
+
Vin
Zout
Zin
~
 AVin
Vout
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, V  V

V
+
in
V
o
i

(2)
Kirchhoff node equation at V
yields, V  0 V  Vo
Ra
(3)

Rf
0
Setting V+ = V– yields
Vi Vi  Vo

 0 or
Ra
Rf
Rf
Vo
 1
Vi
Ra
Ra
Rf
v+
vi
v-
+
v+
vo
R
1
Rf
v+
v-
Rf
Ra
2
v-
)vi

Rf
Voltage follower
o
R
f
Noninverting input with voltage divider
Rf
R2
vo  (1  )(
)vi
Ra R1  R2
v
v+
i
vo
v

a
+
vo  vi
R
R
Noninverting amplifier
vo  (1 
+
i

Ra
vi
v
R
1
R
2
v-
+
v
o

R
f
Less than unity gain
R2
vo 
vi
R1  R2
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
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.
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
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 
Ra
Rb
Rc

V
o
+
Inverting Integrator
Now replace resistors Ra and Rf by complex
components Za and Zf, respectively, therefore
Zf
Vo 
Vin
Supposing
Za
(i) The feedback component
is a capacitor C, in
i.e.,
1
Zf 
(ii) The input component
is a resistor R, Za = R
jC
Therefore, the closed-loop gain (Vo/Vin) become:
Zf
Za
where
vi (t )  Vi e jt
What happens if Za = 1/jC whereas, Zf = R?
Inverting differentiator
+
C
R
V ~
in
V
o
V ~
1
vo (t ) 
vi (t )dt

RC


V
o
+
Op-Amp Integrator
Example:
(a) Determine the rate of change
of the output voltage.
C
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
-5V
-10V
V
o
Op-Amp Differentiator
R
C
0
to
t1
t2
V

i
V
o
0
+
to
 dV 
vo   i  RC
 dt 
t1
t2
Non-ideal case (Inverting Amplifier)
Rf
Ra
Vin ~

V
in
Practical op-amp
+

V
o
+
Equivalent Circuit
Rf
Ra

R
 R
V
+ +

-AV
Vin
Zout
Zin
~
Vout
 AVin
3 categories are considering
V
o
 Close-Loop Voltage Gain
 Input impedance
 Output impedance
Close-Loop Gain
Applied KCL at V– terminal,
Vin  V  V Vo  V


0
Ra
R
Rf
By using the open loop gain,
Rf
V
Ra

in
V
R R
+ +

Vo   AV
Vin Vo
V
V
V

 o  o  o 0
Vin
Ra ARa AR R f AR f
R R  Ra R f  Ra R  ARa R
 Vin  Vo  f
Ra
ARa R R f

The Close-Loop Gain, Av
 AR R f
Vo
Av 

Vin R R f  Ra R f  Ra R  ARa R
Ra
V
o
-AV
Rf
V
o
V R
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
Input Impedance
Rf
Input Impedance can be regarded as,
Rin  Ra  R // R
where R is the equivalent impedance
of the red box circuit, that is
V
R  
if
However, with the below circuit,
V  ( AV )  i f ( R f  Ro )
V R f  Ro
 R 

if
1 A
V
Ra

in
R
V R
+ +

-AV
R'
if
Rf
R
V
+

-AV
V
o
Input Impedance
Finally, we find the input impedance as,
1
1 A 
Rin  Ra   

R
R

R

f
o
 
1

Rin  Ra 
R ( R f  Ro )
R f  Ro  (1  A) R
Since, R f  Ro  (1  A) R , Rin become,
Rin ~ Ra 
( R f  Ro )
(1  A)
Again with R f  Ro  (1  A)
Rin ~ Ra
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 0
Firstly, with figure (a),
V 
Rf
Ra
Ra // R
Ra R
Vo  V 
Vo
R f  Ra // R
Ra R f  Ra R  R f R
R
V
By using KCL, io = i1+ i2
Vo
V  ( AV )
io 
 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
Rf
V
i2
R
V
Ra
R
(a)
i1
V
+

-AV
(b)
o