參數(shù)資料
型號(hào): L6911DTR
廠商: 意法半導(dǎo)體
英文描述: 5 BIT PROGRAMMABLE STEP DOWN CONTROLLER WITH SYNCHRONOUS RECTIFICATION
中文描述: 5位可編程降壓控制器采用同步整流
文件頁數(shù): 13/17頁
文件大小: 233K
代理商: L6911DTR
13/17
L6911D
Where
Where Z
C
(s) and Z
L
(s) are the output capacitor and inductor impedance respectively.
The expression of Z
I
(s) may be simplified as follow:
1
--
C25
1
s
--
Where:
τ
1
= R4×C20,
τ
2
= (R4+R3)×C20 and
τ
d
= Rd×C25.
The regulator transfer function became now:
Figure 8 shows a method to select the regulator components (please note that the frequencies f
EC
and f
CC
cor-
responds to the singularities introduced by additional ceramic capacitors in parallel to the output main electro-
lytic capacitor).
I
To obtain a flat frequency response of the output impedance, the droop time constant
τ
d
has to be equal
to the inductor time constant (see the note at the end of the section):
I
To obtain a constant -20dB/dec Gloop(s) shape the singularity f
1
and f
2
are placed in proximity of f
CE
and f
LC
respectively. This implies that:
I
To obtain a Gloop bandwidth of f
C
, results:
Note.
To understand the reason of the previous assumption, the scheme in figure 9 must be considered.
In this scheme, the inductor current has been substituted by the load current, because in the frequencies range
of interest for the Droop function these current are substantially the same and it was supposed that the droop
network don't represent a charge for the inductor.
Gloop s
Av s
R s
Av s
( )
Zi s
-------------
=
=
Av s
V
osc
---------------
Z
s
( )
Z
L
s
+
Z
C
s
( )
------------------------------------
=
Z
I
s
( )
Rd
---------------------------------
Rd
+
C25
R4
-----------------------------------------------------
1
s
--
C20
1
s
--
C20
+
R3
R4
+
R3
+
+
Rd 1
--------------------------------------------------------------------------------------------------
=
s
τ
1
τ
d
+
(
)
s
2
R3
-------
τ
1
τ
d
+
+
1
s
τ
2
+
(
)
1
s
τ
d
+
(
)
=
=
Rd
1
sR3
R
s
τ
2
+
τ
d
+
--------------------------------------------------------------------
1
s
τ
1
+
(
)
1
(
)
1
s
τ
d
+
(
)
=
R s
1
s
τ
+
(
)
sR3
R
d
)
s C18 R
d
1
τ
d
+
1
s
τ
1
+
(
)
--------------------------------------------------------------+
τ
d
R
d
C25
R
L
------
τ
L
C25
R
L
R
d
(
)
-----------------------
=
=
=
=
f
2
f
1
---
f
f
CE
--------
R4
R3
f
f
CE
--------
1
=
=
f1
f
CE
C20
1
2
--
π
R4 f
CE
=
=
G
0
f
LC
1 f
C
G
0
A
0
R
0
-----------------
-----------------------------
f
C
f
LC
-------
C18
-----------------
+
C25
----------------------------
f
C
-------
=
=
=
=
=
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