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參數(shù)資料
| 型號(hào)： | RT9206GS |
| 廠商： | Richtek Technology Corporation |
| 英文描述： | High Efficiency, Synchronous Buck with Dual Linear Controllers |
| 中文描述： | 高效率，雙直線同步降壓控制器 |
| 文件頁數(shù)： | 19/21頁 |
| 文件大?。?/td> | 351K |
| 代理商： | RT9206GS |
RT9206
DS9206-11 March 2007
www.richtek.com
19
Step2. Determine the zero crossover frequency and
compensated type.
Select desired zero-crossover frequency :
Select f
C
= 20kHz
Step3. Determine desired location of poles and zeros for
type2 compensator.
Select:
Assume
Step4. Calculate the real parameters-resistor and
capacitors for type2 compensator.
From equation (21), the R
C1
is calculated as following :
The design procedure as following :
(1). Selecting the zero crossover frequency f
C
is 1/10 ~
1/20 switching frequency. Then according equation (24)
set the resistor R
C1
to determine the zero crossover
frequency.
×
IN
V
gm
(
) (24)
(2). Place the zero of compensator is 70% fp that is
resonance frequency of power stage. The compensator
capacitor Cc1 can be selected to set the zero. The
equation is shown in following :
×
C
REF
r
V
O
C1
C1
L C
0.7 R
C
=
p
p
×
×
×
×
C1
C1
C1
1
R
fs
R
fs-
C
Design example
Design example of type 2 compensator: the schematic is
shown in Figure 4, where the parameters as following:
V
IN
=12V,
V
OUT
=5V,
frequency=200kHz, L=15
μ
H, C
O
=940
μ
F, r
C
=22m
, the
parameters of RT9206 as following: gm=1.6ms, ramp
amplitude=1.9V,and reference voltage Vref=0.8V.
I
OUT
=5A,
switching
Step1. Determine the power stage poles and zeros. The
pole caused by the output inductor and output capacitor is
calculated as :
p
p
m
m
p
p
m
×
×
×
×
×
P
O
Z
C
O
1
1
f =
=
=1.34kHz
2
L C
2
15
940
1
r
1
f =
=
= 7.7kHz
2
C
2
22m
Ω
940 F
(3). Set a second pole to suppress the switching noise.
Assume the pole is one half of switching frequency
f
s
, which results in capacitor Cc2 as shows in following:
(F) (26)
≤
C
S
S
f
f /10 ~ f /20
×
CZ
P
f
= 0.7
f = 0.7 1.34kHz = 938Hz
S
CP
f
f
2
=
=100kHz
m
×
×
V
×
×
×
×
×
12V
×
×
×
C
OUT
C1
C
IN
REF
f
L
Vr
gm
V
V
R
=
r
20kHz
22m
15 H
1.9
5V
0.8V
=
= 8.4k
1.6ms
m
m
×
O
C1
C1
L C
0.7 R
15
0.7 8.2k
940
C
=
=
= 20.7nF
Select C
C1
= 22nF
Second capacitor C
C2
can be calculated using equation
(26)
Select C
C2
= 220pF
Select R
C1
= 8.2k
Calculate C
C1
from equation (25)
(F) (25)
194pF
200kHz
8.2k
1
f
R
1
C
S
C1
C2
=
×
×
=
×
×
=
p
p
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