Type 3 Compensator Design
Type 3 Compensator Design
Type 3 Compensator Design
The Type III compensator has three poles (one at the origin) and two zeros. In
practice, it is usually arranged to have two coincident zeros and two coincident poles,
and the loop crossover frequency is placed somewhere between the zeros and poles.
For this kind of design, the transfer function in Equation (1) can be rewritten as:
(3)
2
Z
2
P
s
K 1
C(s)
s
s 1
and the constant gain K is given by:
A detailed analysis of the
type III compensator derives
the appropriate equations
and guarantees the targeted
bandwidth and phase margin,
as well as an unconditionally
stable control loop.
LIYU CAO
Ametek Programmable Power
Type III Compensator Design
for Power Converters
C3
R3 R2
C2
C1
OUT
R1
V
O
V
I
+
Fig. 1. A type III error amplifier configuration employs six passive circuit compo-
nents and has three poles (one at the origin) and two zeros.
www.powerelectronics.com January 2011 | Power Electronics Technology 21
(5)
1 1 2
1
K
R C C
The phase of the transfer function in Equation (3) at a given
frequency t can be calculated as:
(7)
2
2
z p
z p
K
C( j ) 1 j 1 j
j
2 1 j 2 1 j
2
As can be seen, the phase of C(jt) has two parts: a con-
stant part of -/2 due to the pole at the origin, and a variable
part as a function of frequency t given by:
(8)
v
z p
1 1
z p
( ) 2 1 j 2 1 j
2 tan tan
Or equivalently, we have:
(24)
k 1
b
2 k
Define:
(26)
x k
Then, from Equation (24) we get the following quadratic
equation in terms of x:
(27)
2
x 2bx 1 0
The solutions to Equation (27) are given by:
(28)
2
x b b 1 (
From Equation (26) you can see that x is positive, therefore
the solution we need is given by:
(29)
2
k x b b 1
Given tm and k, we can get the zero t
z
and pole t
p
based
on Equation (15):
(30)
m
z p m
, k
k
From Equation (6), we can get the compensators gain at
the crossover frequency, t
c
= t
m
:
POWER CONVERTERcompensation
60
40
20
0
0
20
40
M
a
g
n
i
t
u
d
e
(
d
B
)
180
135
90
45
P
h
a
s
e
(
d
e
g
)
Bode Diagram
Frequency (Hz)
10
2
10
3
10
4
10
5
10
6
Fig. 4. Control plants Bode plot for a synchronous buck converter, including
the PWM modulator.
www.powerelectronics.com January 2011 | Power Electronics Technology 23
(31)
m
z
m 2
m m m
m
p
1
K K 1 k Kk
C j
1
1
1
k
(46)
3 1
3 z
1
R R
C
7. Calculate C
1
:
(52)
z
1
p 1
C
R K
8. Calculate C
2
:
(53)
2 1
1
1
C C
R K
9. Calculate R
2
:
(54)
1 2
2
1 2 p
C C
R
C C
10. Calculate R
3
:
(55)
3 1
3 z
1
R R
C
Or, equivalently:
(59)
c p z m p
2
c p z
tan
2 4
60
40
20
0
0
20
40
M
a
g
n
i
t
u
d
e
(
d
B
)
180
135
90
45
45
P
h
a
s
e
(
d
e
g
)
Frequency (Hz)
10
2
10
3
10
4
10
5
10
6
Fig. 6. Compensator (red) and loop (blue) Bode plots from Procedure 2 is
unconditionally stable.
www.powerelectronics.com January 2011 | Power Electronics Technology 25
POWER CONVERTERcompensation
(61)
p z d
where t
d
is defined by:
(62)
m p
d c mp
tan
2 4
Note that t
d
is known with the given parameters q
m
, q
p
,
and t
p
, and the selected frequency t
c
.
We can solve Equation (60) and (61) and get the com-
pensators zero and pole frequencies:
(63)
2 2
z d m d
0.5 4
(64)
2 2
p d m d
0.5 4
The separation factor can be calculated as:
(65)
2 2
d m d
2 2
d m d
4
k
4
With t
z
, t
p
and k determined based on the above equa-
tions, the compensators components can be determined in
the same way as in Procedure I. The design procedure that
accounts for unconditional stability is summarized below.
Given desired crossover frequency t
c
and phase margin
q
m
, and the control plants gain and phase at t
c
as G
p
and
t
p
. Also given the frequency t
mp
at which the plant has the
maximum phase lag.
Based on Equation (56) determine the compensators
maximum phase frequency by choosing a value for |.
Usually you can start with | =1, and adjust it based on the
loops Bode plot resulted from the design procedure.
1. Calculate the difference between the zeros frequency
and poles frequency using Equation (62).
2. Calculate the zeros frequency t
z
and poles frequency
t
p
using Equation (63) and Equation (64).
3. Calculate the separation factor k using Equation (65).
4. From Equation (6) we have
(66)
2
c
z
c 2
c
c
p
1
K
C( j )
1