E-L6911C STMicroelectronics, E-L6911C Datasheet - Page 13
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E-L6911C
Manufacturer Part Number
E-L6911C
Description
IC CTRLR 5BIT PROGR STPDN 20SOIC
Manufacturer
STMicroelectronics
Type
Step-Down (Buck)r
Datasheet
1.E-L6911C.pdf
(20 pages)
Specifications of E-L6911C
Internal Switch(s)
No
Synchronous Rectifier
Yes
Number Of Outputs
1
Voltage - Output
1.3 ~ 3.5 V
Current - Output
1.3A
Frequency - Switching
200kHz
Voltage - Input
5 ~ 12 V
Operating Temperature
-40°C ~ 150°C
Mounting Type
Surface Mount
Package / Case
20-SOIC (7.5mm Width)
Lead Free Status / RoHS Status
Lead free / RoHS Compliant
Power - Output
-
Lead Free Status / Rohs Status
Lead free / RoHS Compliant
Available stocks
Company
Part Number
Manufacturer
Quantity
Price
Where Z
The expression of Z
Where:
The regulator transfer function became now:
Figure 8 shows a method to select the regulator components (please note that the frequencies f
responds to the singularities introduced by additional ceramic capacitors in parallel to the output main electro-
lytic capacitor).
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.
To obtain a flat frequency response of the output impedance, the droop time constant
to the inductor time constant (see the note at the end of the section):
To obtain a constant -20dB/dec Gloop(s) shape the singularity f
and f
To obtain a Gloop bandwidth of f
G
0
LC
C
Z
1
f
LC
(s) and Z
I
= R4×C20,
G loo p s
respectively. This implies that:
s
=
1 f
=
--------------------------------- -
Rd
Rd
C
L
I
(s) are the output capacitor and inductor impedance respectively.
(s) may be simplified as follow:
+
=
1
-- - C 25
2
s
1
-- -
s
= (R4+R3)×C20 and
Av s
C25
G
d
0
R s
=
=
+
R s
R
f1
----------------------------------------------------- -
A
f
--- -
f
R 4
d
R4
0
2
1
=
C
------------------------------------------------------------------------------------------------------- -
s C 18 R
=
=
C 25
R
, results:
f
+
=
+
CE
0
-------- -
f
R d
f
1
-- - C20
s
CE
1
-- - C20
LC
s
=
Av s
=
----------------- -
-------------------------------------------------------------------- -
V IN
Vosc
1
------ -
R
1
d
+
L
1
L
= Rd×C25.
+
s
d
+
Zf s
------------- -
Zi s
+
R3
------- -
R
=
s
C20
s
C20 // C25
---------------------------- -
R 3
R3
d
1
R4
L
2
+
C18
2
d
Where
=
s
=
R3
------- -
R
=
Rd 1
-------------------------------------------------------------------------------------------------- - =
1
d
R3
1
1
-- -
2
1
+
+
=
+
s
s
d
C25
s
------- -
f
Av s
+
f
LC
-------- - 1
f
f
C
CE
LC
R 4 f
s
d
1
d
1
1
1
+
=
–
and f
+
s
-----------------------
1
=
CE
s
R
+
C 18
L
--------------- -
2
2
L
Vin
V
1
d
are placed in proximity of f
R
o sc
=
+
d
1
----------------- -
s
+
2
VIN
Vosc
------------------------------------ -
Z
s
C
R3
------- -
R
s
Z
d
d
C
---------------------------- -
C20
C20 C25
+
d
s
1
Z
has to be equal
L
+
EC
C25
s
d
and f
L6911C
f
------- -
f
LC
CC
C
13/20
CE
cor-