### Operational Amplifiers

```The Ideal Op-amp
(Operational amplifier)
v+
+
v–
–
+
–
+15V
15
+
VOUT
–
–15V
VOUT=A(v+–v–)
A~105
VOUT [V]
VIN
VOUT
saturation
10
5
0
-400
-200
0
-5
-10
-15
200
400
VIN [μV]
Op-amp Feedback
VIN
v+
v–
R2
v+
+
v–
–
VOUT
VOUT
+

VO U T  A v
–
R1
Non-Inverting
Amplifier Circuit
VO U T

 v



R2
 A  V I N  V O U T
R1  R2


R2
V O U T  1  A
R1  R2

VO U T
VI N

A
1
AR2
R1 R2


  AV I N


R1  R2
R2




Op-amp Feedback
v+
+
v–
–
Assumptions:
Gain is very large (A)
Inputs draw no current (ZIN=)
Output attempts to make input voltage
difference zero (v+=v–)
VIN
v+
v–
R2
VOUT
+
–
R1
Non-Inverting
Amplifier Circuit
Inverting Amplifier Circuit
R2
i
VIN
R1
v–
i
V1
V2
V3
R1
R1

i 
–
v+
VOUT
+
v–
v+
 v
R2
i 
V1
R1
VO U T  
–
+
VOUT
V2
R1
R2
R1

 0
 
R1
VO U T


VI N
VI N
i
R1
v
 
V3
R1
V1
VO U T
R2
R2
R1
 
VO U T
R2
 V2  V3 
Summing amplifier
RF
i
V1
V2
V3
R1
v–
R2
v+
–
+
VOUT
R3
VO U T
 V1
V3 
V2

  R F 



R
R
R
2
3 
 1
Difference amplifier
R1
V1
v–
v+
R1
V2
R2
–
VOUT
+
VO U T  v
R2
v

v

 V2
 v

 v


 iR 2
 V1  v 
 
 R1

R 2



R 
R

 v  1  2   V1 2
R1 
R1

R2
R1  R2
V O U T  V2
R2
R1  R2
 V2  V1 
R2
R1
 R1  R2


R1


R
  V1 2

R1

Integrator
 dt
VIN
VOUT
Capacitor as integrator
R
Vi
C
Vin t  VC 
Vint
VC 

1
RC
1

C

t
0
1
C

t
0
Id t '
t
Vi  V C
0
R
Vi d t '
dt '
If RC>>t
VC<<Vi
Op-amp Integrator
C
i
VIN
R
v–
i
v+
–
+
VOUT
v

 v

 0
VO U T   VC
VO U T  
VO U T  
1
C
1
C

t
0
idt
t
VI N
0
R

dt
Differentiation
Vi
V diff  VR
C
R
Vdiff
dV R 
 dV i
 RI  RC 


dt 
 dt
Small RC 
dV i
dt

V diff  RC
dV R
dV i
dt
dt
Op-amp Differentiator
d
VIN
VIN
R
i
C
v–
i
v+
dt
–
VOUT
+
v

 v

 0
V O U T   iR
V O U T   RC
dV I N
dt
VOUT
Complex analysis
C
V0
R2
i
R1
v–
i
ejωt
V O U T   iR 2 

High pass filter
v+
 V0 e
R1 
 j  CR 2 e
1  j  CR
–
VOUT
+
j t
j
R2
C
j t
V0
1
  0
  
VO U T  0
VO U T  
R2
R1
V0 e
j t
Exploiting op-amp saturation
v+
+
v–
–
VOUT =
A
VOUT
A(v+–v–)
Saturation voltage
VOUT = +VSat
v+>v–
VOUT = –VSat
v+<v–
Bridge circuits
Z1
V0
VA
Z3
VB
+
–
Z2
Z4
Bridge balanced when VAVB=0
Z 2Z 3  Z 4Z 1
VOUT
V0
+
R
–
+
R
VIN
–
+
R
–
+
R
–
Oscillator
R
VOUT

+
C
R1
R1
Op-amp applications
Building block of analogue electronics
Signal amplifiers
Audio amplifiers
Integrators / differentiators
Voltage / current sources
Active filters
Oscillators
Digital-analogue and analogue-digital convertors
```