參數(shù)資料
型號: L5989D
廠商: STMICROELECTRONICS
元件分類: 穩(wěn)壓器
英文描述: 5.06 A SWITCHING REGULATOR, 1000 kHz SWITCHING FREQ-MAX, PDSO16
封裝: ROHS COMPLIANT, HTSSOP-16
文件頁數(shù): 21/51頁
文件大?。?/td> 1947K
代理商: L5989D
Application information
L5989D
28/51
Doc ID 15778 Rev 3
The PWM modulator gain is a function of the switching frequency:
The transfer function on the LC filter is given by:
where:
Two different kind of networks can compensate the loop depending on the output capacitor.
Type II network is used to compensate the loop with high ESR output capacitors, type III
with low ESR output capacitors (MLCC). In the two following paragraph the guidelines to
select the Type II and Type III compensation network are illustrated.
6.4.1
Type III compensation network
The methodology to stabilize the loop consists of placing two zeros to compensate the effect
of the LC double pole, so increasing phase margin; then to place one pole in the origin to
minimize the dc error on regulated output voltage; finally to place other poles far away the
zero dB frequency.
In Figure 12 the type III compensation network is shown. This network introduces two zeros
(fZ1, fZ2) and three poles (fP0, fP1, fP2). They expression are:
G
PW0 fSW
()
G
PW0 400 kHz
()
f
SW kHz
()
400
--------------------------
9
f
SW kHz
()
400
--------------------------
==
G
LC s
()
1
s
2
π f
zESR
--------------------------
+
1
s
2
π Qf
LC
----------------------------
s
2
π f
LC
-------------------
2
++
-------------------------------------------------------------------------
=
f
LC
1
2
π
LC
OUT
1
ESR
R
OUT
---------------
+
------------------------------------------------------------------------
=
f
zESR
1
2
π ESR C
OUT
--------------------------------------------
=
,
Q
R
OUT
LC
OUT
R
OUT
ESR
+
()
LC
OUT
R
OUT
ESR
+
------------------------------------------------------------------------------------------
R
OUT
V
OUT
I
OUT
--------------
=
,
=
f
Z1
1
2
π C
7
R
8
R
7
+
()
------------------------------------------------
=
f
Z2
1
2
π R
5
C
5
------------------------------
=
,
f
P0
0
=
f
P1
1
2
π R
7
C
7
------------------------------
=
f
P2
1
2
π R
5
C
5
C
6
C
5
C
6
+
--------------------
--------------------------------------------
=
,,
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