APW1172
Application Description (Cont.)
Frequency Compensation (Cont.)
sented as Figure-5:
The problem is the phase nearly –180 degrees at the
natural frequency especially in the high Q situation. If
the Q factor is high, the phase decrease vary sharp at
the location of the double poles. This problem leads
the regulator oscillating when use ceramic one as the
output capacitor without compensation. The purpose
of the compensation is saving the phase. The manner
is added additional zeros to achieve the goal. A zero
have theability that contribute the maximum phase of
90 degrees. According this characteristic, needs two
zeros to compensate the phase loss. The PID com-
slope=-20db/
decade
0db
270d
180d
phase
90d
45d
0d
f
pensator is good for this.It shows as Figure-4.
C3
Zero3 Pole3 Pole4
Figure-5
Zero2
The assumption is10(zero2)<zero3,10(zero3)<pole3,
10(pole3)<pole4.In order to compensate the phase,
place the two zeros closely and located before the
natural frequency. In general
C2
R3
R2
C1
FB
R1
zero2 @ zero3 = k × pole1,2
(11)
EA
COMP
Where k is a constant, the value of k is almost 0.7 to
0.8.
Vref
Figure-4
The useful rules are:
The transfer function H(s) is
(1) Determine the value of C2,the value must smaller
than 5nF to get fast response time.
(2) Find R3 by the equation
SC2R3 +1
SC1
R + R2
+1
1
H(s) =
(
)
(
)
]
S SC1R2 +1
[
SC2C3R3 + C2 + C3
R3 = (2p ×C2 ×k × pole )-1
1
2p ×C2 R3
1
1,2
zero2 =
(3) Determine thevalue of C1from 470pF to 1uF. This
range of C1 is for reference.
zero3 =
pole3 =
pole4 =
2p ×C1(R + R2 )
(4) The range of pole3 is from 150KHz to 300KHz.
Use this range to find the value of R2.
(5) Find R1 by the equation
1
1
2p ×C1R2
C2 + C3
R = (2p×C1 ×k × pole )- 1 - R2
1
1,2
2p ×C2C3R3
(6) The location of pole4 is 5 times pole3. Use this
result to find the value of R3.
The frequency response of the PID compensator pre-
Copyright ã ANPEC Electronics Corp.
16
www.anpec.com.tw
Rev. A.4 - Aug., 2005
ANPEC [ ANPEC ELECTRONICS COROPRATION ]
