Transcript Document

Quiz: Find an expression for
G s 
Vo  s 
Vi  s 
in terms of the component symbols.
C2
Vi(s)
C3
C1
U1A
2
3
1
+
R3
R2
-
R1
1

R1
Vo(s)
1
1 
1 
1
||  R2 
R


3
Z F  s  sC2 
sC1 
sC3
- G s 


1
ZI  s

1 
sC

R1 ||  R3 
2

1
sC
3 

R2 
sC1
Lets see if we can express this in a more revealing form …
Manipulate transfer function to rational fraction form . . .
sC 3
sR3C3  1  sR1C3
1

R sR3C3  1
R1 sR3C3  1
sR2 C1  1
1 sR3  R1 C3  1
 Gs   1


sC1
sC 2 sR2 C1  1  sC1 sR1C 2 sR3C3  1 sR2 C1  1  C1 C 2
sC 2 
sR2 C1  1
sR2 C1  1
 Gs  
1 sR3  R1 C3  1
sR1C 2 sR3C3  1
sR3  R1 C3  1 sR2 C1  1
sR2 C1  1
1

RCC
sR3C3  1
 C C 2  sR1 C1 C 2 
s 2 1 2 1

sR2 C1   1
C1 C 2
 C2 
k s  z1  1s  z 2  1
 Gs   C
s s  p1  1s  p 2  1
This represents a second order
lead/lag network with integration. The
following symbolic associations apply:
kC 
It is often desirable to
make the zero frequencies
the same, and the pole
frequencies the same:
Then . .
 z1   z 2   z
 p1   p 2   p
1
R1 C1 C 2 
1
 z1 
R3  R1 C3
z2
1

R2 C1
2
k  s  z  1 
 Gs   C 
 kC
s  s  p  1 
Linear gain
 p1 
1
R3C3
 p2 
C1 C 2
R2 C1C 2
1  s  z  1  s  z  1 
s  s  p  1  s  p  1 
Integration
Two identical first
order lead/lag terms
 s z  1 

Dissecting the effects of a first order lead/lag term : G LL    
 s  1
p


The phase at any frequency (s = j is:
Maximum phase shift occurs
when the derivative is zero:
 
 

    tan1    tan1 
 

 z
 p
d
1
1
1
1


2
2
d
   z
   p



1  
1 


 z 
p 

z
 z2   2

ˆ   z  p
2
2
 z2  ˆ 2  p  ˆ

z
p
ˆ 2   z  p
fˆ 2  f z f p
p
z
ˆ
ˆ

;

z
z
p
p
p
The maximum phase shift, given z and p (or the ratio
p / z) can be expressed:
 p2   2
ˆ   ˆ   tan1  ˆ


 z
Max Phase Lead (degrees)
100
80
 ˆ

  tan1 


 p
 p 



  tan1   z 
  tan1 

 p 
 z 





 p  
  p 
1 




 tan

 tan

 z 
 z 

 2


1
60
40
20
0
1
2
4
8
p
z
16
32

 p   
1 

 
ˆ
  2 tan

 z  4 




 s z  1 

G LL    
 s  1
p


Dissecting the effects of a first order lead/lag term (cont) :

 p   
1 

 
ˆ
  2 tan

 z  4 




If the desired maximum phase shift is specified,
we can determine the required ratio p / z :
 p z 
Gain Considerations:
fp
j  z  1
G LL   

j p 1
The magnitude of the transfer function
(gain) of the standard form first order
lead/lag network at any frequency  is:
The gain magnitude at the
frequency of maximum phase shift
is determined when   ˆ :
ˆ 
   2

f z  tan

4


GLL ˆ  
ˆ z 
2
1
ˆ    1
1    
2

p


p
z
z
  z 2  1
  
2
p
1
 p z  1
z  p  1
p
z  p  1
GLL ˆ    p  z 
fp
ˆ 
   2
f z  tan 

4


Summary:
A lead/lag network is used to compensate feedback loops that exhibit too much phase lag
(approaching or exceeding 180o) at frequencies where the magnitude of the loop gain is greater
than 0 dB.
The strategy is to identify a critical frequency where the phase lag is to be adjusted, and then to adjust
the phase by a specified amount in order to meet some desired phase margin specification.
Since phase margin is defined at the crossover (0 dB) frequency of the open loop transfer function, it is
common practice to choose a desired crossover frequency , A, (usually, but not necessarily, near the
low-pass break frequency of the uncompensated loop gain) as the critical frequency for establishing how
much phase adjustment the lead/lag compensation should provide.
Since the gain magnitude of the compensating network (kc) can usually be set independently, the overall
loop gain and crossover frequency can also be determined by our choice of kc .
The relations developed above allow the designer to choose the break frequencies ( z and p) of
the lead/lag network in order to provide the required phase adjustment, A at a specific
frequency, A:
   2 A 

4


 p  z  tan
   2 A 

4


 p   A tan
 A   p z
   2 A 

4


 z   A cot
Application: Buck Converter Control
VREF
 t 
+
Loop
Compensation
GC(s) (
)
GPWM
-
We examine stability of the closed loop system by determining
the gain and phase margins of the Open Loop Transfer Function:
From example 4.1:
dD 100%

 0.556 v 1
dv
1.8v
(Linearized
Power Stage
Model)
Load
GL  s   kFB GC  s  GPWM GPS  s 
GPS  s  
v +(t)
GPWM 
GPS(s)
From example 4.2, for the Buck Converter (eq 4.15):
1.8 v
VB
d s
vo  s 
kFB
DT
VIN
vo  s 
d s

VIN
1  srC
LC s 2  s 1 RC  r L   1 LC
Digression: Relating Closed Loop behavior to
Open Loop transfer function (“Loop Gain”).
~
VD s
 t 
+
GOL(s)
(Controller)
-
vo  s 
GCL  s  
GOL  s 
GOL  s   1
GOL(s) usually exhibits a low pass characteristic with considerable gain at low frequencies. As a
result, at low frequencies when | GOL(s) | >> 1, GCL(s) ~ 1, and the output follows the input
variations very closely.
At higher frequencies when | GOL(s) | << 1, GCL(s) ~ GOL(s), and the high frequency variations in
the input appear at the output highly attenuated. To say it another way, changes to the input (or
other disturbances) that occur at high frequencies do not materially affect the output.
In between these two frequency ranges, when | GOL(s) | ~ 1, the response of the system
transitions from following changes in the input, to not following changes in the input. This is the
critical range where gain and phase margin are determined. The frequency c, where |GOL (jc)|
= 1 (0 dB) is called the crossover frequency, and the phase lag must be less than  radians below
c or the system will exhibit instability.
Referring to the Bode plot for the Power Stage, we can conclude that a simple controller
exhibiting only linear gain will be stable, but will also exhibit a steady state output error.
Example 4.3 invokes the requirement that there be zero steady state output error, which implies
an integrator in the controller. An integrator always introduces /2 phase lag at all frequencies,
and the additional phase lag will cause a total phase lag  >  when |GOL (s)| > 1. BAD JUJU.
Example 4.3
•Zero Steady State Error
•Phase Margin of /3
•kFB = 0.2
•Crossover frequency greater than low pass corner of LC filter.
Design the Compensator:
GC  s  
kC
s

? ?

The unknown part of the controller transfer function must introduce additional leading phase in
the vicinity of the resonance frequency of the LC filter where its phase lag is greatest.
Low pass corner (resonance)
of LC filter, ~ 600 hz
Choose loop crossover
frequency to be 1000 hz.
= 17.1
This just happens to be where the Power Stage
phase lag is maximum ~ 138o .
With the additional 90o lag of the integrator,
this will give a total phase lag of 228o.
In order to have a phase margin of 60o, our
compensator must introduce 108o of phase lead
at the crossover frequency.
Since a first order lead/lag compensator cannot
provide even 90o of lead, we must use a second
order lead/lag . . . .
k
Gs   C
s
 s z  1 


 s   1
p


2
. . . Which just happens to be consistent with the transfer
function of the previously analyzed circuit. Who’d have guessed?
The second order network must provide 108o of
phase lead at fA = 1000 hz, which is twice the
phase shift of a first order network. Therefore,
the maximum phase shift, A , of each of the
first order networks must be 54o.
GL  j A   k F GC  j A  k d GPS  j 2000   1
At 1000 hz, ( = A = 2000), the
compensated loop gain should be
equal to 1 (0 dB).
0.526  GC  j A 
k
 C
s
kC f p
0.526 
2f A f z
 180o  108o 
f p  f A tan 
  3078 hz
4


 180o  108o 
f z  f A cot 
  325 hz
4


GC  j A  
 s z  1 


 s  1
p


2
1
 0.526
0.20.55617.1
Recalling that the magnitude of the first
order lead/ lag term when  = A is
s  j A
G LL  A  
f p fz

f 
325
k C  0.526 2f A z   0.5262000 
 348.9


fp 
3078

We can now use the cookbook formulae from the circuit analysis to solve for component values. We
have five equations in six unknowns.
kc  348.9 
1
R1  C1  C2 
z
1
 324.9 
2
2  R3  R1  C3
z
1
 324.9 
2
2 R2 C1
p
1
 3078 
2
2 R3 C3
p
C C2
 3078  1
2
2 R2 C1C 2
If we choose R1 to be 10,000 ohms, we get the following solution:
R1 = 10,000 W
R2 = 1910 W
R3 = 1180 W
C1 = 0.256 mF
C2 = 0.0302 mF
C3 = 0.0438 mF
Multiplying all resistor values and dividing all capacitor values by any scaling constant
will yield valid solutions.