BL8532B
White LED PFM Step-Up DC-DC Converter
Outline:
The BL8532B is a PFM Step-up DC/DC
driver IC with invariant current, design for LED
applications. Thought the external resistance,
output current reach 0mA~500mA.
A low ripple, high efficiency step-up DC/DC
converter can be constructed of BL8532Bxx
with only three external components. Also
available is a CE (chip enable) function that
reduce power consumption
Features:
•
•
•
•
•
•
•
•
Low start voltage：0.8V ( at Iout=1mA)；
Output Current range：0~ 500mA；
Output Current accuracy：±10%；
High Efficiency：82%(Type)；
PACKAGE： SOT-89-5.
Applications
：
Power source for white LED
Supply constant current
Power source for a single or dual-cell
battery-powered equipments
Selection guide：
：
BL8532B
-
XX X XX
Pin configuration：
：
BL8532B-
Package:
SN：SOT-89-5
Temperature range:
C：standard
P：standard
（Pb
free）
Feedback voltage:
10……100mV
20……200mV
25……250mV
30……300mV
35……350mV
40……400mV
□□
PSN
Pin No.
1
2
3
4
5
Symbol
IFB
Vout
CE
Lx
GND
Description
Current feedback pin
Output pin, power supply
for internal circuits
Chip enable pin (active
high)
Switching pin(Nch open
drain)
Ground pin
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BL8532B
Typical applications:
Constant output current application
(
A) Drive only one white LED
(B) Drive two white-LED in series and several in parallel connection
Suggestions:
1. you can increase the output capacitor properly for improve to the characteristic of
the output invariant current ( for example：150uF or 200uF).
2. to avoid the feedback voltage excursion, please wear the ring with static electricity
and the electric iron connect to ground when the soldering
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BL8532B
Detailed description:
The BL8532B is a DC/DC step-up converter with voltage type , PFM control mode ,
output invariablenes current. It has only three Peripheral components that is a inductor
、a
output capacitor、a schottky diode and a resistor with set-up output current, that can afford to
a invariant output current between 0 and 500mA .
Rc set-up method：
If the output current is Iout，that Rc=
V
IFB
Iout
。
For example, want to the current value of 100mA , select the chip of VIFB=200mV,
that Rc=200mV/100mA=2
。
Selection of the external components:
Thus it can be seen, the inductor and shottky diode affect the conversion efficiency greatly.
The inductor and the capacitor also have great influence on the output voltage ripple of the
converter. So it is necessary to choose a suitable inductor, a capacitor and a right shottky diode,
to obtain high efficiency, low ripple and low noise.
D
≡
Before discussion，we define:
（
1）Inductor Selection
）
Vout
−
Vin
Vout
.
Make sure DC-DC can natural work firstly in the model of the minimum continuous current that
D
(1
−
D
)
2
R
L
.
is Lmin，Lmin≧
2
f
This formula deduce that ignoring the autoecious resistor and a diode with the forward voltage
drop, but the actual value is still big. If the inductance less than Lmin, inductor will reach
magnetic saturation, efficiency will greatly drop, and hardly output steady voltage.
Secondly , considering the current ripple of the inductor, ignoring the autoecious parameter in
the mode of continuous current.
∆I=
Vin
D
•
Vin
DVin
+
，
Imax=
(1
−
D
)
2
R
L
2
Lf
Lf
.
When “L” is too small, will lead to high current ripple of the inductor , and the maximum
current of the inductor , schottky diode , power tube of the chip are excessive .
Thirdly, generally speaking, not considering efficiency , small inductor can drive load more
then large inductor .But in the same load conditions , large inductor with the current ripple and
the maximum current value are small. So the large inductor should be able to start up circuit in
the low input voltage.
Use inductor with an inductance of 10μH or more ,it’s ensure to normal work . If output port
has load with output large current (for example: output current is more than 50mA), for
improving to efficiency, suggest to use large inductor. At the same time, in the large load, the
resistor is in series with the inductor that will affect the switching efficiency. Supposed , the
resistor is rL，Rload，that the power
consumption
as follow：
∆η ≈
r
L
2
R
load
1
−
D
）
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