SD42509

COMPONENTS SELECTION

The input capacitor provides the pulse current when the power MOSFET is on, and charge the capacitor when

the power MOSFET is off, thus to keep the stability of the input voltage. The input capacitor is recommended to

be more than 10

μF, which can reduce the peak current drawn from input source and the switch noise. The input

capacitor should be near to the input pin when in real routing.

2． Output Capacitor Selection

Parallel connecting a capacitor between the two ends of LED can reduce the output voltage ripple accordingly

reduce the ripple current of LED, while this capacitor will not effect the operating frequency and efficiency, but the

start time will be longer by reducing the rising speed of the voltage on LED. The larger the output capacitor is, the

smaller the current ripple on LED is. It is recommend use the capacitor of 4.7

μF or larger.

3． Inductance Selection

The inductance is used to keep the output current constant, the bigger the inductance is, the smaller the output

current ripple is; while the bigger the physical size is, the bigger the series-wound resistor is. The selected

effective current (RMS current rating)of inductance current should be bigger than the maximum output current,

and the saturation current should be 30% bigger than maximum output current. In order to improve the efficiency,

the series-wound resistor (DCR) of inductance should be smaller than 0.2

Ω.

The relation between inductance and ripple:

ΔI

*

f

*

V

V

V

*

V

L

IN

OUT

IN

OUT

−

=

Where: L: Inductance value

F: Oscillator frequency

ΔI is ripple current

When select inductance, you should consider the combination of various factors to select suitable inductance.

When output current is 1A, 47

μH inductance is recommended.

4． Diode Selection

SD42509 is a non-synchronous step-down adjuster, so the diode should provide continuous current when the

power MOSFET is off. Because the forward voltage of Schottky diode is small, and the reverse continuous

current time is short, so it is usually used for continuous current. During the power MOSFET is conducting, the

diode will withstand high voltage, so the reverse voltage of selected diode should be bigger than the input voltage.

LED

D

I

*

D)

(1

I

−

=

When the input voltage is high with a small duty, ID increases, so the selected maximum continuous current

diode should be bigger than the output current.

HANGZHOU SILAN MICROELECTRONICS CO.,LTD

REV:1.4

2009.02.26

Http://www.silan.com.cn

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