Circuits and Analog Electronics
Ch7 Operational Amplifiers and Op Amp Circuits
7.1 Operational Amplifiers
7.2 Op Amp Circuits
7.3 Active Filter
7.4 Op Amp Positive Feedback
References: Floyd-Ch6; Gao-Ch7, 9;
Ch7 Operational Amplifiers and Op Amp Circuits
7.1 Operational Amplifiers
Key Words:
Op Amp Model
Ideal Op Amp
Op Amp transfer characteristic
Feedback
Virtual short
Ch7 Operational Amplifiers and Op Amp Circuits
7.1 Operational Amplifiers (Op Amp )
Symbol
Non-inverting input
Inverting input
Positive voltage supply
+
Output
-
Negative voltage supply
• At a minimum, op amps have 3 terminals: 2 input and 1 output.
• An op amp also requires dc power to operate. Often, the op
amp requires both positive and negative voltage supplies (V+
and V-).
Ch7 Operational Amplifiers and Op Amp Circuits
7.1 Operational Amplifiers
Symbol
• One of the input terminals (1) is called an inverting input
terminal denoted by ‘-’
• The other input terminal (2) is called a non-inverting input
terminal denoted by ‘+’
Ch7 Operational Amplifiers and Op Amp Circuits
7.1 Operational Amplifiers
The Op Amp Model
v+
Non-inverting input
Inverting input
v-
+
Rin
-
Ro
+
A(v+ -v- )
• The op amp is designed to sense the difference between the voltage
signals applied to the two input terminals and then multiply it by a
gain factor A such that the voltage at the output terminal is A(v+-v-).
• The voltage gain A is very large (practically infinite). The gain A is
often referred to as the differential gain or open-loop gain.
• The input resistance Rin is very large (practically infinite). The output
resistance Ro is very small (practically zero).
vo
Ch7 Operational Amplifiers and Op Amp Circuits
7.1 Operational Amplifiers
Ideal Op Amp
Circuit model (ideal)
• We can model an ideal amplifier as a voltage-controlled voltage
source (VCVS)
• The input resistance is infinite. Ri    i  0, i  0
• The output resistance is zero.
RO  0  vO  A(v  v )
• The gain A is infinite.
Ch7 Operational Amplifiers and Op Amp Circuits
7.1 Operational Amplifiers
Non-inverting input
Inverting input
v+
+
Rin
v-
-
Ro
+
-
For A741, A = 100dB=105， if vo=10V，
10
 0.1mV
5
10
v  v 0.110 3
i 

 0.05nA
6
Ri
2 10
Then v   v  
A(v+ -v- )
vo
Ch7 Operational Amplifiers and Op Amp Circuits
7.1 Operational Amplifiers
Op Amp transfer characteristic curve
saturation
active region
vo  A vi  A ( v  v )
Ch7 Operational Amplifiers and Op Amp Circuits
7.1 Operational Amplifiers
Op Amp transfer characteristic curve
So far, we have been looking at
the amplification that can be
achieved for relatively small
(amplitude) signals.
For a fixed gain, as we increase
the input signal amplitude, there is
a limit to how large the output
signal can be. The output saturates
as it approaches the positive and
negative power supply voltages.
In other words, there is limited
range across which the gain is
linear.
v  v
v  v
v o  V
v o  V
Ch7 Operational Amplifiers and Op Amp Circuits
7.1 Operational Amplifiers
Review
Ideal op amp characteristics:
• Does not draw input current so that the input impedance
is infinite (i.e., i1=0 and i2=0)
• The output terminal can supply an arbitrary amount of
current (ideal VCVS) and the output impedance is zero
• The op amp only responds to the voltage difference
between the signals at the two input terminals and ignores
any voltages common to both inputs. In other words, an
ideal op amp has infinite common-mode rejection.
• A is or can be treated as being infinite.
Ch7 Operational Amplifiers and Op Amp Circuits
7.1 Operational Amplifiers
Ch7 Operational Amplifiers and Op Amp Circuits
7.1 Operational Amplifiers
R2
vo  A(v  v )  A(vIN  vo
)
R1  R2


AvIN
vo 
AR2
1
R1  R2
What happens when “A” is
very large?
Ch7 Operational Amplifiers and Op Amp Circuits
7.1 Operational Amplifiers
Closed-loop gain
Af=vo/vin
AvIN
vo 
AR2
1
R1  R2
 vIN
R1  R2
 Af vIN
R2
Gain
Suppose
A=106,
R1=9R, R2=R,
vo  vIN 10
R1
Af  (1  )
R2
Closed-loop gain: determined by resistor ratio
insensitive to A, temperature
Ch7 Operational Amplifiers and Op Amp Circuits
7.1 Operational Amplifiers
Negative feedback
Why did this happen?
Ch7 Operational Amplifiers and Op Amp Circuits
7.1 Operational Amplifiers
Negative feedback
Observe, under negative feedback,
 R1  R2 

vIN
vo  R1 
v  v  
0
A
A
v  v
i  0, i  0
analysis method under negative feedback!
– Hence, we say there is a virtual short between the two terminals
(“+” and “-”) .
Ch7 Operational Amplifiers and Op Amp Circuits
7.1 Operational Amplifiers
Negative feedback
v  v
i  0, i  0
Ch7 Operational Amplifiers and Op Amp Circuits
7.1 Operational Amplifiers
Negative feedback
v  v
i  0, i  0
R1  R2  A v
vo  v IN
f IN
R2
R1
Af  (1  )
R2
When R1=0, R2=,
vo  vIN
Buffer: voltage gain = 1
Voltage Follower
Ch7 Operational Amplifiers and Op Amp Circuits
7.1 Operational Amplifiers
Negative feedback
v  v
i  0, i  0
R1
Af  (1  )
R2
vi  v vi
i1 

R1
R1
vo  v vo
i2 

R2
R2
i  i1  i2  0
vo
R2
Af 

vi
R1
Ch7 Operational Amplifiers and Op Amp Circuits
7.1 Operational Amplifiers
Negative feedback
vo
R1
Af   (1  )
vi
R2
Af 
vo
R
 2
vi
R1
• We can adjust the closed-loop gain by changing the ratio of R2
and R1.
• The closed-loop gain is (ideally) independent of op amp openloop gain A (if A is large enough) and we can make it arbitrarily
large or small and with the desired accuracy depending on the
accuracy of the resistors.
Ch7 Operational Amplifiers and Op Amp Circuits
7.1 Operational Amplifiers
Negative feedback
The terminal 1 is a virtual ground
since terminal 2 is grounded.
Inverting configuration,
This is a classic example of what negative feedback does. It
takes an amplifier with very large gain and through negative
feedback, obtain a gain that is smaller, stable, and predictable. In
effect, we have traded gain for accuracy. This kind of trade off is
common in electronic circuit design… as we will see more later.
Ch7 Operational Amplifiers and Op Amp Circuits
7.1 Operational Amplifiers
Negative feedback
vi
vi
Rin  
 R1
i1 vi / R1
Inverting configuration,
Input Resistance:
Assuming an ideal op amp (open-loop gain A = infinity), in the
closed-loop inverting configuration, the input resistance is R1.
Ch7 Operational Amplifiers and Op Amp Circuits
7.1 Operational Amplifiers
Negative feedback
Inverting configuration,
Output Resistance:
Roa is usually small and
so Rout is negligible when
A is large
vt
R2  R1
R1
vt
R1  R2
AR1
vt (1 
)
vt  ( Av1 )
R1  R2
i2 

Roa
Roa
v
Roa
R t 
i2 1  A R1
R1  R2
i1 
Rout  R ( R1  R2 )
v1 
Rout 
Roa
1 A
R1
R1  R2
Ch7 Operational Amplifiers and Op Amp Circuits
7.1 Operational Amplifiers
Negative feedback
vo
R2
Af 

vi
R1
Rin 
vi
 R1
i1
Ro  0
We can model the closed-loop
inverting amplifier (with A =
infinite) with the following
equivalent circuit using a
voltage-controlled voltage
source…
Inverting configuration,
Ch7 Operational Amplifiers and Op Amp Circuits
7.1 Operational Amplifiers
Homework
Vo
 0.5
1) Design a circuit to Af 
Vi
2) Find the vo=?
Ch7 Operational Amplifiers and Op Amp Circuits
7.1 Operational Amplifiers
Review: Two fundamental Op Amp
Structure
Af
Input voltage
( )terminal
Feed back
( )terminal
Inverting
Amp
_
_
Non
inverting
Amp
+
_
Ch7 Operational Amplifiers and Op Amp Circuits
7.2 Op Amp Circuits
Key Words:
Subtracting Amplifiers
Summing Amplifiers
Intergrator
Differentiator
Ch7 Operational Amplifiers and Op Amp Circuits
7.2 Op Amp Circuits
Consider this circuit:
v  v
i1  i2  0
vO  v  iR2
vIN  v1
i
R2
 v
R2  R1
v2  v 
R1
Subtraction!
 v 
v2  v 
R2
R1

R2 
R
 v 1    v2 2
R1 
R1

R
 2 v1  v2 
R1
Let R1  R2 ,
 vO  (v1  v2 )
Ch7 Operational Amplifiers and Op Amp Circuits
7.2 Op Amp Circuits
Subtracting Amplifiers
v  v
i1  i2  0
Another way of solving —use superposition
Ch7 Operational Amplifiers and Op Amp Circuits
7.2 Op Amp Circuits
Subtracting Amplifiers
Another way of solving —use superposition
R2
vo2   v 2
R1
Ch7 Operational Amplifiers and Op Amp Circuits
7.2 Op Amp Circuits
Subtracting Amplifiers
Another way of solving —use superposition
R1 + R2
vo1 = v+
R1
R2 R1 + R2
= v1
R1 + R2 R1
R2
= v1
R1
Ch7 Operational Amplifiers and Op Amp Circuits
7.2 Op Amp Circuits
Subtracting Amplifiers
Another way of solving —use superposition
vo1 = v+
R
vo2   2 v 2
R1
vo  vo1  vo 2 
R2
(v1  v2 )
R1
R1 + R2
R1
= v1
R2 R1 + R2
R1 + R2 R1
= v1
R2
R1
Still subtracts!
Ch7 Operational Amplifiers and Op Amp Circuits
7.2 Op Amp Circuits
Subtracting Amplifiers
vo1
v O  (
Rf 2
R3
 Rf 2 (
Rf1
vO1

v S1
R1
vO1  
Rf1
R1
vO1 
Rf 1
R1 R3
Let R f 1  R1 , vO  R f 2 (
v S1
Rf 2
R5
vS 1 
vS 2 )
1
vS 2 )
R5
v S1 1

vS 2 )
R3 R5
Let R f 2  R3  R5 , vO  (vS1  vS 2 )
Ch7 Operational Amplifiers and Op Amp Circuits
7.2 Op Amp Circuits
Summing Amplifiers
vo  k1vi1  k 2 vi 2  k n vin
For node N，
vS 3 vS 2 vS 1
v


 o
R3 R2 R1
Rf
v o  (
Let
Rf
R1
v S1 
Rf
R2
vS 2 
R1  R2  R3
vo  
Rf
R1
(v S 1  v S 2  v S 3 )
R f  R1
 vo  (vS1  vS 2  vS 3 )
Rf
R3
vS 3 )
Ch7 Operational Amplifiers and Op Amp Circuits
7.2 Op Amp Circuits
Weighted Summer
We can also build a summer:
i1 
v
v1
v
, i2  2 ,  , in  n
R1
R2
Rn
i  i1  i2    in
vo  0  iR f
vo  (
Rf
R1
v1 
Rf
R2
v2   
Rf
Rn
vn )
Ch7 Operational Amplifiers and Op Amp Circuits
7.2 Op Amp Circuits
Example 1 Design a summer which has an output voltage given
by vO=1.5vs1-5vs2+0.1vs3。
Solution 1:
R4
R3
R2
we have，
Rf1
R1
vO1  (
 1.5 ,
Rf1
R1
Rf1
R3
v S1 
Rf1
R3
 0.1
Let R1  2K , R f 1  3K ，R3  30K
R5  R1 // R3 // R f 1  2 // 30 // 3  1.15
vS 3 )
Rf 2 
 Rf 2
vO  
vO1 
vS 2 
R2
 R4

Rf 2
R2
5
Rf 2
R4
1
Let R2  2K ， R f 2  10K ，R4  10K
R6  R2 // R4 // R f 2  2 // 10 // 10  1.43
Ch7 Operational Amplifiers and Op Amp Circuits
7.2 Op Amp Circuits
Example 1 Design an summer which has an output voltage given
by vO=1.5vs1-5vs2+0.1vs3。
Solution 2:
Because
R2 // R f  R4 // R1 // R3
vO 
Rf
R1
Let R2  2K
Rf
R1
v S1 
 1.5 ,
Rf
R2
Rf
R2
R f  10K ，R3  100K ，R1 
vS 2 
5
R3
R3
vS 3
 0.1
10  2
 6.7 K
3
2 // 10  100 // 6.7 // R4
R4 
Rf
Rf
20  670
 2.27 K
12  670  20  106.7
Ch7 Operational Amplifiers and Op Amp Circuits
7.2 Op Amp Circuits
Let’s build an integrator…
vO  k  vI dt
Let’s start with the following insight:
vI
vO  vC 
1
i dt

C
But we need to somehow convert voltage vI to current.
Ch7 Operational Amplifiers and Op Amp Circuits
7.2 Op Amp Circuits
First try… use resistor
When is v small compared to vR?
O
i C
dvo
dt
vI  vR  vO  iR  vO  RC
When vR >>vO ,
dvO
 vO
dt
larger the RC,
smaller the v
O
for good integrator
ωRC >> 1
Ch7 Operational Amplifiers and Op Amp Circuits
7.2 Op Amp Circuits
There’s a better way…
Ch7 Operational Amplifiers and Op Amp Circuits
7.2 Op Amp Circuits
There’s a better way…
v
i I
R
B
u
t
,
v
O
m
u
Ch7 Operational Amplifiers and Op Amp Circuits
7.2 Op Amp Circuits
Integrator
vR  vI
1 t
vC  vC (0)   iC ( )d
C 0
vI  0
R
1
vO (t )  
RC
iC 
How about in the frequency domain?
Vo ( j )
1 jC
1



Vi ( j )
R
jCR
Vo ( j )
1


Vi ( j ) CR
t
 v ( )d  v
0
I
C
(0)
Vo ( j )
1 jC
1


Vi ( j )
R
jCR
Ch7 Operational Amplifiers and Op Amp Circuits
7.2 Op Amp Circuits
Integrator
vR  vI
1 t
vC  vC (0)   iC ( )d
C 0
vI  0
iC 
R
1 t
vO (t )  
 vI ( )d  vC (0)
RC
0
Vo ( j )
1 jC
1


Vi ( j )
R
jCR
Ch7 Operational Amplifiers and Op Amp Circuits
7.2 Op Amp Circuits
Integrator
v S  Vm sin  t
vO  
Vm
Vm
1
V
sin

tdt

cos

t

sin( 90   t )
m

RC
RC
RC
Ch7 Operational Amplifiers and Op Amp Circuits
7.2 Op Amp Circuits
Now, let’s build a differentiator…
Let’s start with the following insights:
But we need to somehow convert current to voltage.
Ch7 Operational Amplifiers and Op Amp Circuits
7.2 Op Amp Circuits
Differentiator
vo=-iR
Ch7 Operational Amplifiers and Op Amp Circuits
7.3 Active Filter
Key Words:
Basic Filter Responses
Low-Pass Filter
High-Pass Filter
Band-Pass Filter
Band-Stop Filter
Ch7 Operational Amplifiers and Op Amp Circuits
7.3 Active Filter
Basic Filter Responses
.
.V (t)
Filter
Vi(t)
o
S  j
Basic Filter Responses
voltage gain
Transition region
stopband region
cutoff frequency
vO ( s )
vi ( s )
Vo ( j )
A( j ) 
 A( j )  ( j )

Vi ( j )

Low-Pass Filter
bandwidth
A( s) 
Ch7 Operational Amplifiers and Op Amp Circuits
7.3 Active Filter
Ch7 Operational Amplifiers and Op Amp Circuits
7.3 Active Filter
Low-Pass Filter

A
1 j

VO

Vi


0
1 / j c
1


2
1




R
1 j

1

j c
0
  
 0
1
(O 
)
RC
  tg 1
  
 45
0
0
Ch7 Operational Amplifiers and Op Amp Circuits
7.3 Active Filter
High-Pass Filter

A

VO

Vi
R

R
1
j c
1

1
1
jRc

1
1 j
o

Ch7 Operational Amplifiers and Op Amp Circuits
7.3 Active Filter
Advantages of Filter

| A |max  1
1
 1

RL
1



j

c
V
(
R
//
)
i
L
R


1  jcRL
j c
L


A


1

1
RL


(R 
// RL ) Vi R   1  jc 
R
j c
1  jcRL
R

 L

RL

R Rj  c
( R  RL )(1  L
)
RL  R



RL
RL /( R  RL )
Av

1  j cR'L 1  j 
O '

where AV  RL ( R  RL )
O ' 
1
RL ' C
Ch7 Operational Amplifiers and Op Amp Circuits
7.3 Active Filter
Low-Pass Filter
-20dB/decade
A( j ) 
Vi ( j )
1
VO ( j )
1
1
j c




1

Vi ( j )
(R 
)Vi ( j ) 1  j cR 1  j
j c
O
Ch7 Operational Amplifiers and Op Amp Circuits
7.3 Active Filter
Low-Pass Filter
AV
1
Rf
R1

-20dB/decade
0

f0 
Vi 
1
R  
R 

j c
V0  V   1  f    1  f 
R1  
R1  R  1

j c
1
2RC
1
A( j ) 
Rf
VO ( j )
AVF
R1


Vi ( j ) 1  j cR 1  j 
First-order (one-pole) Filter
O
Ch7 Operational Amplifiers and Op Amp Circuits
7.3 Active Filter
Low-Pass Filter
AV
1
Rf
R1

-20dB/decade

-40dB/decade
0
R 

V0  V  1  f 
R1 

 R
 1  f
R1


f0 
Vi 
1

jc

R  1 / /R 1 

j c  
jc 

A( j ) 
1
2RC
VO ( j )
1
 AVF 
Vi ( j )
1  3 jRC  ( jRC ) 2
Second-order (two-pole) Filter
Ch7 Operational Amplifiers and Op Amp Circuits
7.3 Active Filter
Low-Pass Filter
Voltage-controlled voltage source
(VCVS) filter
V  V 
R3
R3  R f
A
For simplicity, R1  R2  R C1  C2  C


A
Vo

Vi


Af
1  (3  A f ) jRC  ( jRC ) 2
Af
f 2
1 f
1 ( )  j
f0
Q f0
Q
1
3  Af
Af  1 
Rf
R3
Ch7 Operational Amplifiers and Op Amp Circuits
7.3 Active Filter
Low-Pass Filter


 R f R3
1    V  Af
V o VV  V
 RR33  R f
Voltage-controlled voltage source
(VCVS) filter
A
For simplicity, R1  R2  R C1  C2  C
 Rf
V o  V   1 
R3


  V  Af

1
Using super position: R //  R  1 


jc 
jc

V  1  V 0 


1
1  R 1

 R //  R 
jc
jc
j

c


1


jc
V 2   V 1 

1
1 

R
//  R 
jc 
jc 


V o  V1  V2   Af

Ch7 Operational Amplifiers and Op Amp Circuits
7.3 Active Filter
High-Pass Filter
• Transfer functions:
• Circuit: R↔C
• Frequency domain
AL 
1
1  j RC
SRC
1
SRC
1
 
1
1
SRC
 AH
Ch7 Operational Amplifiers and Op Amp Circuits
7.3 Active Filter
Band-Pass Filter
Vi
High-Pass
Low-Pass
Vo
A
A
A
AAff
A
A
AAf f
1
A
Aff
A
1
ωH
ω
ωL
ω
ωL
Lower-frequency
ωH
ω
Upper-frequency
Ch7 Operational Amplifiers and Op Amp Circuits
7.3 Active Filter
Band-Stop Filter
Low-Pass
Vi
Vo
High-Pass
A
A
A
A
AA
f f 1
A
A
Aff
ωh
A
AA
f f
ωL
ω
ωh
ωL
ω
Ch7 Operational Amplifiers and Op Amp Circuits
7.3 Active Filter
Example 2 For the circuit shown, show that what it is filter?


1

A
vO
v
 i
R f // Z C R1
Vo


 R f // Z C
R1
Vi
(a)

Rf

Rf
R1

j c
1
Rf 
j c
1
1
 AVF 

R1 1  jR f c
1 j

The Inverting First-order Low-Pass Filter.
o
Ch7 Operational Amplifiers and Op Amp Circuits
7.3 Active Filter
Example 2 For the circuit shown, show that what it is filter?
vo
vi


Rf
R1  Z C
Rf
 Vi 

Vo
1
R1  Z C
A  
 Rf 
1
Vi
Vi
R1 
j c
R
1
1
 f 
 AVF 

R1 1  1 / jR1C
1 j o

(b)
The Inverting First-order High-Pass Filter.

Ch7 Operational Amplifiers and Op Amp Circuits
7.3 Active Filter
Example 2 For the circuit shown, show that what it is filter?
(c)
The Non-Inverting Band-Stop Filter(Second-order).
Ch7 Operational Amplifiers and Op Amp Circuits
7.3 Active Filter
Example 2 For the circuit shown, show that what it is filter?
The Inverting Band-Pass Filter.
(Second-order)
The Inverting High-Pass Filter.
(Second-order)
Ch7 Operational Amplifiers and Op Amp Circuits
7.4 Op Amp Positive Feedback
Key Words:
Positive Feedback
The Comparator
Oscillator
Ch7 Operational Amplifiers and Op Amp Circuits
7.4 Op Amp Positive Feedback
Positive Feedback
What’s the difference?
Positive feedback drives op amp
into saturation: VoutVsaturation
Ch7 Operational Amplifiers and Op Amp Circuits
7.4 Op Amp Positive Feedback
Positive Feedback
vout  A( v  v )  Av
vout  vIN
 A(
 R1  v IN )
R1  R2
R1 

1



R1  R2 

vout  AvIN

AR1 

 R R 
1
2 

R2
R2
R1  R2
 AvIN 
  vIN
AR1
R1

R1  R2
Ch7 Operational Amplifiers and Op Amp Circuits
7.4 Op Amp Positive Feedback
The Comparator
The op amp is often used as a comparator.
The output voltage exhibits two stable states. The output
state depends on the relative value of one input voltage
compared to the other input voltage.
Threshold voltages
vi  VR ,
VO  VO L
vi  VR ,
VO  VO H
vi ( VR )  VTH
Ch7 Operational Amplifiers and Op Amp Circuits
7.4 Op Amp Positive Feedback
The Comparator
Ch7 Operational Amplifiers and Op Amp Circuits
7.4 Op Amp Positive Feedback
The Comparator
Ch7 Operational Amplifiers and Op Amp Circuits
7.4 Op Amp Positive Feedback
The Comparator
R1
-
vi
vO
t
+
vO
R1
VOH
+
t
vo’
vO
VOL
C
vo ’
R
t
R1
+
v’O
vO
vL
C
R
v
vLL
RL
t
Ch7 Operational Amplifiers and Op Amp Circuits
7.4 Op Amp Positive Feedback
The Comparator
vi  0 ,
vO  (VZ  VD )
vi  0 ,
vO  VZ  VD
( VO  VZ )
Ch7 Operational Amplifiers and Op Amp Circuits
7.4 Op Amp Positive Feedback
The Comparator
Transmission characteristics
Ch7 Operational Amplifiers and Op Amp Circuits
7.4 Op Amp Positive Feedback
The Comparator with Positive Feedback
Positive feedback is often used with comparator circuits. The
feedback is applied from the output to the non-inverting input
of the op amp.
Ch7 Operational Amplifiers and Op Amp Circuits
7.4 Op Amp Positive Feedback
The Comparator with Positive Feedback
Ch7 Operational Amplifiers and Op Amp Circuits
7.4 Op Amp Positive Feedback
The Comparator (Schmidt trigger)
The input has to change sufficiently to
trigger a change. e.g.( -7.5V 7.5V)
Only at vi  7.5V , v0 is switched from
15V to -15V.
hysteresis
Ch7 Operational Amplifiers and Op Amp Circuits
7.4 Op Amp Positive Feedback
The Comparator (Schmidt trigger)
When vi（<0）<V+，vO=VO+ >0，
V1  VTH 1 

RF
(VR  VO )  VO
R2  RF
1
( RFVR  R2Vo )
R2  R F
When vi>VTH1，vOVO- <0，
VTH2
VTH1
V 2  VTH 2 

RF
(VR  VO )  VO
R2  RF
1

( RFVR  R2VO )
R2  R F
VH  VTH 1  VTH 2 
R2
(VO  V O )
R2  RF
Ch7 Operational Amplifiers and Op Amp Circuits
7.4 Op Amp Positive Feedback
The Comparator (Schmidt trigger)
Why is hysteresis useful?
e.g., analog to digital
Ch7 Operational Amplifiers and Op Amp Circuits
7.4 Op Amp Positive Feedback
The Comparator (Schmidt trigger)
Without hysteresis
Ch7 Operational Amplifiers and Op Amp Circuits
7.4 Op Amp Positive Feedback
The Comparator (Schmidt trigger)
Oscillator — can create a clock
Ch7 Operational Amplifiers and Op Amp Circuits
7.4 Op Amp Positive Feedback
The Comparator (Schmidt trigger)
There’s a better way…----triangular-wave generator
Ch7 Operational Amplifiers and Op Amp Circuits
7.4 Op Amp Positive Feedback
The Wien-Bridge Oscillator (RC Oscillator )
Op Amp Circuits
Positive Feedback
Lead-lag network
Ch7 Operational Amplifiers and Op Amp Circuits
7.4 Op Amp Positive Feedback
The Wien-Bridge Oscillator
1
 R
c
Ch7 Operational Amplifiers and Op Amp Circuits
7.4 Op Amp Positive Feedback
The Wien-Bridge Oscillator
1
 R
c
-
Ch7 Operational Amplifiers and Op Amp Circuits
7.4 Op Amp Positive Feedback
The Wien-Bridge Oscillator
Resonant frequency?
-
Ch7 Operational Amplifiers and Op Amp Circuits
7.4 Op Amp Positive Feedback
The Wien-Bridge Oscillator



max
Vf
R //
1
j c
1
1

1

1
1
1
1

VO R
 R //
R
3  j ( cR 
)
3
 cR
j c
j c
j c
1
1
R //
j c
1
The phase shift through the network is 0 for  cR 
 cR
FV 
Vf
VO



2 f o Rc 
fo 

f  0
F
V
f

VO

1
2 Rc



1



f
f )

3  j(  o
fo
f



1
2 f o Rc

V
f

Vo

1
f
f
32  (  o ) 2
fo
f
1 f
f
 o)
3 fo
f
 f  arctg (
Ch7 Operational Amplifiers and Op Amp Circuits
7.4 Op Amp Positive Feedback
The Wien-Bridge Oscillator
 a   f  2n

AVF 


VO


V R1
VO
R1  R2
R2


1

R1 
R1
R1
VO
R1  R2
Loop gain of
1 causes a sustained constant

output | AVF F |  1

1
1
When f o 
FV 
2RC
3

 AVF
1
R2
3
R1
1
 3
F
R2  2R1
Ch7 Operational Amplifiers and Op Amp Circuits
7.4 Op Amp Positive Feedback
The Wien-Bridge Oscillator
All practical methods to achieve
stability for feedback oscillators
require the gain to be selfadjusting. This requirement is a
form of automatic gain control.


| AVF F |  1

AVF 

VO

V R1
R
 1 2
R1
1
R2
3
R1

FV 
1
3
Negative temperature
coefficient
Ch7 Operational Amplifiers and Op Amp Circuits
7.4 Op Amp Positive Feedback
The LC Oscillator
Admittance y 
1
R
L
 2

j
[

c

]
2
2
2
Z R  ( L)
R  ( L)
1
f0 
ZO 
2 LC
1 R  (0 L)


y0
R
2
Q
0 L
R
2
1
 j L
j

c
j

c
Impedance Z 

(1 j c )  ( R  j L) R  j ( L  1 )
c
L
ZO
RC


1
L  
f
f
1 j
 (  o ) 1  jQ(  o )
fo f
LC R o 
1
R2  (
L
 R2 )
L
L
C

Q
R
RC
C
--Quality Factor
| Ic || IL |  | I |
( R  jL)
ZO

Z


f
f

1  Q 2 (  o )2
fo f


f
f
  arctgQ(  o )

fo f
Ch7 Operational Amplifiers and Op Amp Circuits
7.4 Op Amp Positive Feedback
The LC Oscillator
Frequency response curve
ZO

Z


f
f

1  Q 2 (  o )2
fo f


f
f
  arctgQ(  o )

fo f
larger
smaller
Resistors Circuit
larger
Inductance Circuit
smaller
1
f
 o2 )
fo f
 f f  fo  
1
f
1  Q 2 (  o2 )2
fo f
Q(
f  f0
 Q
2
fo
Capacitance Circuit
Ch7 Operational Amplifiers and Op Amp Circuits
7.4 Op Amp Positive Feedback
The RC(Phase-Shift) Oscillator
Ch7 Operational Amplifiers and Op Amp Circuits
7.4 Op Amp Positive Feedback
The LC Oscillator
Ch7 Operational Amplifiers and Op Amp Circuits
7.4 Op Amp Positive Feedback
The LC Oscillator
 +Vcc
 +Vcc

Vo

 Vo

Vi

i
V
 Vo

i
V