LM3444
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SNVS682C –NOVEMBER 2010–REVISED MAY 2013
OUTPUT CAPACITOR
A capacitor placed in parallel with the LED or array of LEDs can be used to reduce the LED current ripple while
keeping the same average current through both the inductor and the LED array. With a buck topology the output
inductance (L2) can now be lowered, making the magnetics smaller and less expensive. With a well designed
converter, you can assume that all of the ripple will be seen by the capacitor, and not the LEDs. One must
ensure that the capacitor you choose can handle the RMS current of the inductor. Refer to manufacture’s
datasheets to ensure compliance. Usually an X5R or X7R capacitor between 1 µF and 10 µF of the proper
voltage rating will be sufficient.
SWITCHING MOSFET
The main switching MOSFET should be chosen with efficiency and robustness in mind. The maximum voltage
across the switching MOSFET will equal:
VDS(MAX) = VAC-RMS(MAX)
2
(30)
The average current rating should be greater than:
IDS-MAX = ILED(-AVE)(DMAX
)
(31)
RE-CIRCULATING DIODE
The LM3444 Buck converter requires a re-circulating diode D10 (see the Typical Application circuit Figure 11) to
carry the inductor current during the MOSFET Q2 off-time. The most efficient choice for D10 is a diode with a low
forward drop and near-zero reverse recovery time that can withstand a reverse voltage of the maximum voltage
seen at VBUCK. For a common 110VAC ± 20% line, the reverse voltage could be as high as 190V.
VD
í VAC-RMS(MAX)
2
(32)
The current rating must be at least:
ID = 1 - (DMIN) x ILED(AVE)
(33)
Or:
VLED(MIN)
1 -
x ILED(AVE)
ID =
VBUCK(MAX)
(34)
Design Example
The following design example illustrates the process of calculating external component values.
Known:
1. Input voltage range (90VAC – 135VAC
)
2. Number of LEDs in series = 7
3. Forward voltage drop of a single LED = 3.6V
4. LED stack voltage = (7 x 3.6V) = 25.2V
Choose:
1. Nominal switching frequency, fSW-TARGET = 250 kHz
2. ILED(AVE) = 400 mA
3. Δi (usually 15% - 30% of ILED(AVE)) = (0.30 x 400 mA) = 120 mA
4. Valley fill stages (1,2, or 3) = 2
5. Assumed minimum efficiency = 80%
Calculate:
1. Calculate minimum voltage VBUCK equals:
x SIN(135o)
2
2
90
= 45V
VBUCK(MIN)
=
(35)
19
2. Calculate maximum voltage VBUCK equals:
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