Datasheet

19
MSL2021 [DATASHEET]
42062A–LED–02/2013
This topology does not require an output capacitor, C
o
in Figure 11-1 on page 17. When used, C
o
steers the inductor’s
ripple current away from the LEDs but reduces the accuracy of PWM dimming because the voltage across it cannot
change quickly. When using C
o
, a ceramic capacitor of between 1.0µF and 10µF is adequate, with a voltage rating higher
than V
BUCK
.
The output capacitor of the AC/DC converter that produces the main string voltage, C
i
in Figure 11-1 on page 17, doubles
as the buck’s input capacitor. The capacitor’s function is to provide a smooth voltage to the buck converter. It should be
able to handle the R.M.S. ripple current of the buck converter, which is approximately equal to
A
This ripple current peaks at a duty ratio of D = 0.5.
Select an N-channel MOSFET for Q with a maximum drain-source voltage at least 25% above V
LED
. The R.M.S. current
in the MOSFET is approximately equal to
A
The MOSFET conduction power loss due to this current is
W
where R
DS
is the hot on-resistance of the MOSFET, which can be found in the MOSFET datasheet, and is typically 1.5 to
1.8 times greater than the cold resistance. The MOSFET will also incur switching losses, which can be difficult to
calculate exactly. A good rule-of-thumb is to choose a MOSFET in a package that dissipates at least four times P
CON
.
The average current in the output rectifier D
1
is
A
and the power dissipated in the rectifier due to conduction is
W
where V
on
is the voltage drop across the rectifier at the forward current of I
D1
. Pick a rectifier with an average current
rating at least 50% higher than I
D1
. Use a Schottky rectifier if the LED voltage is less than 50V. The Schottky rectifier’s
voltage rating should be at least 25% higher than V
LED
. Schottky rectifiers have very low on-state voltage and very fast
switching speed, but at high voltage and high temperatures their leakage current becomes significant. The power
dissipated in the Schottky rectifier due to the leakage current at any temperature and duty ratio is
W
where I
r
is the reverse leakage current, found in the diode’s datasheet. This power must be added to the conduction
power loss.
W
Make sure that the rectifier’s total power dissipation is within the rectifier’s specifications.
I
C
i
I
AVE
D 1 D–=
I
Q
I
AVE
D=
DRIRIP
DSAVEDSQCON
22

I
D
i
I
AVE
1 D–=
P
CON
D
1
I
D
1
V
on
=
P
lkg
V
LED
I
r
D=
P
D
1
P
CON
D
P
lkg
+=