Transcript Why Op Amps Have Low Bandwidth

Why Op Amps Have Low
Bandwidth
Define gain of non-inverting amplifier
Open loop gain of op amp
(a simplification)
A O L (s) 
A0

s 
1



ω0 

3
Let A0=105, ω0=106
Closed Loop Gain
vo
vi

A O L (s)
1  β A O L (s)
;
β 
R1
R1  R 2
β =1 for follower.
Characteristic equation
1  β A O L (s)  1  β
A0

s 
1



ω0 

3
Note that for this circuit
β≤1.
 0
Since denominator is
1+Aβ, the condition for
osicllation is Aβ=-1.
Gain of Op-Amp vs Frequency
Bode plot of A(s), A 0=1E5, three poles at w 0=1E6
100
Magnitude (dB)
80
60
40
20
0
-20
-40
0
Phase (deg)
-45
-90
-135
-180
Phase goes to 180⁰ here
-225
-270
0
10
1
10
2
10
3
10
4
10
Frequency (rad/sec)
5
10
6
10
7
10
8
10
Find closed loop gain (low freq)
Closed Loop Gain
vo
A C L (s) 

vi
A O L (s) 
A O L (s)
1  β A O L (s)
A0

s 
1 

ω0 

3
(A0=105, ω0=106)
For large AOL(0)=A0; A0β>>1
A C L (0) 
vo
vi

A0
βA 0

F o r th is c irc u it β 
1
β
R1
R1  R 2
, s o A C L ( 0) 
1
β

R1  R 2
R1
 1
R2
R1
For resistive circuits, the extreme value is β→1 (R2 →0, Ri→∞); a follower.
Find where roots are on axis
Characteristic equation
1  β A O L (s)  1  β
1  β
A0

s 
1



ω0 

3
 0
A0

s 
1



ω0 

3
A0
1  β
1 3
s
ω0
2
 s 
 s 
 3




ω
ω
 0
 0
2
3
Find where roots are on jω axis
3
 s 
 s 
s
1 3
 3
  
  βA 0
ω0
ω
ω
 0
 0
2
3
 jω 
 jω 
jω
1 3
 3
  
  βA 0
ω0
ω
ω
 0
 0
Find conditions for oscillation
1 3
jω
ω0
1  j3
ω
ω0
2
3
 jω 
 jω 
 3



  βA 0
 ω0 
 ω0 
2
 ω 
ω
3
 

ω0
ω
 0
ω 
1 3
2
3
ω
ω0
R e al
3ω 0
 
3
2
 βA 0
1  9  βA 0
8  βA 0
3
 ω 
 
  0
 ω0 
 0
3
 ω 
 ω 
 3

j


  βA 0
ω
ω
 0
 0
 ω 
1  3
  βA 0
 ω0 
32
Im a g
Restrictions on closed loop gain
For oscillation
ω 
This is obviously only useful if you
require large gain. It is certainly
not useable for a follower.
3ω 0
8  βA 0
(A0=105, ω0=106)
8  β  10
5
To make it useable we need to
decrease the loop gain. Since we
want to use a full range of β, our only
choice is to decrease the open loop
gain of the open loop gain.
For stability
β 
8
10
5
 82d B
or
A C L (0) 
A0
1  A 0β

10
8
5
 11, 000
Graphical explanation
Bode plot of A(s), A 0=1E5, three poles at w 0=1E6
100
Magnitude (dB)
80
60
40
• When phase is -180⁰, mag is 82dB.
•|Aβ|=1 at this frequency for stability
• So we need to make β=-82dB (this
is only marginally stable)
20
0
-20
-40
0
Phase (deg)
-45
-90
-135
-180
Phase goes to -180⁰ here
ω = √3·ω0 = 1.7E6
-225
-270
0
10
1
10
2
10
3
10
4
10
Frequency (rad/sec)
5
10
6
10
7
10
8
10
Increasing Stability (1)
To decrease the gain, redesign op amp so one of it’s open loop poles
is at a much lower frequency, to decrease gain at high frequencies
(where oscillations occur)
A O L (s) 


A0

s 
s 
1

1




ωd  
ω0 

2
;
ωd  ω 0
Characteristic equation
0  1  β A O L (s)
A0

 s  
s 
s
 
1 
 1  2
 

ω
ω
ω
d 
0

 0  

2
A0

 1
2 
2
1 
1 
2 
3


s


s
1  s 


2 
2 
ω
ω
ω
ω
ω
ω
ω
d
0
d
0
0
d
0






A O L (s) 
A0

1
2
1 
2
3
1

s

s

s

2 
ωd
ω dω 0
ω dω 0 

0 1 β
A0

1
2
1 
2
3
1

s

s

s

2 
ωd
ω dω 0
ω dω 0 

1  β
A0

1
2
1 
2
3
1

s

s

s

2 
ω
ω
ω
ω
ω
d
d
0
d
0



1
2
1 
2
3
βA 0  1  s
 s
 s
2 
ω
ω
ω
ω
ω
d
d
0
d
0


Increasing Stability (2)
Let’s find the conditions for instability
βA 0  1  s
1
ωd
 s
2
2
ω dω 0
 s
1
3
ω dω
Let s=jω
0  ω
1
ωd
 ω
ωd
 ω
1
32
2
ωd ω0
ω  ω0
 β A 0  1  jω
βA 0  1  ω
1
0  ω
2
0
2
1
ωd
 ω
2
ω dω 0
 jω
3
βA 0  1  ω 0
1
2
2
2
ω dω 0
ωd ω0
ω d  1  β A 0   2ω 0
2
ω dω 0
3
2
1
2
ω dω 0
R e al
Im a g
ωd 
2ω 0
1  β A 0 
Let’s find the value of ωd that
creates marginal stability with β=1.
ωd 
2  10
6
1  1  1 0 
5
 20
ra d
se c
This is
quite low…
Graphical explanation
Bode plot of A(s), A 0=1E5, two poles at w 0=1E6 and one at w d=20
100
Magnitude (dB)
50
0
• When phase is -180⁰, mag is 0 dB (i.e., 1).
• |Aβ|=1 at this frequency for stability
• So circuit is stable for β≤1
(marginally stable for β=1)
-50
-100
-150
0
Phase (deg)
-45
-90
-135
-180
Phase goes to -180⁰ here
ω = ω0 = 1E6
-225
-270
0
10
1
10
2
10
3
10
4
10
Frequency (rad/sec)
5
10
6
10
7
10
8
10