参数资料
型号: LM3402_07
厂商: National Semiconductor Corporation
英文描述: 0.5A Constant Current Buck Regulator for Driving High Power LEDs
中文描述: 0.5A的恒流降压稳压器用于驱动高功率LED
文件页数: 9/24页
文件大小: 431K
代理商: LM3402_07
For this example the peak current rating of the inductor should
be greater than 479 mA. In the case of a short circuit across
the LED array, the LM3402 will continue to deliver rated cur-
rent through the short but will reduce the output voltage to
equal the CS pin voltage of 200 mV. Worst-case peak current
in this condition is equal to:
Δi
L(LED-SHORT) = [(26.4 – 0.2) x 300 x 10
-9
] / 26.4 x 10-6
= 298 mA
P-P
I
L(PEAK) = 0.35 + 0.149 = 499 mA
In the case of a short at the switch node, the output, or from
the CS pin to ground the short circuit current limit will engage
at a typical peak current of 735 mA. In order to prevent in-
ductor saturation during these short circuits the inductor’s
peak current rating must be above 735 mA. The device se-
lected is an off-the-shelf inductor rated 33 H ±20% with a
DCR of 96 m
and a peak current rating of 0.82A. The phys-
ical dimensions of this inductor are 7.0 x 7.0 x 4.5 mm.
R
SNS
The current sensing resistor value can be determined by re-
arranging the expression for average LED current from the
LED Current Accuracy section:
R
SNS = 0.74, tSNS = 220 ns
Sub-1
resistors are available in both 1% and 5% tolerance.
A 1%, 0.75
resistor will give the best accuracy of the aver-
age LED current. To determine the resistor size the power
dissipation can be calculated as:
P
SNS = (IF)
2
x R
SNS
P
SNS = 0.35
2
x 0.75 = 92 mW
Standard 0805 size resistors are rated to 125 mW and will be
suitable for this application.
To select the proper output capacitor the equation from Buck
Regulators with Output Capacitors is re-arranged to yield the
following:
The target tolerance for LED ripple current is ±5% or 10%
P-
P = 35 mAP-P, and the LED datasheet gives a typical value for
r
D of 1.0 at 350 mA. The required capacitor impedance to
reduce the worst-case inductor ripple current of 258 mA
P-P is
therefore:
Z
C = [0.035 / (0.258 - 0.035] x 1.0 = 0.157
A ceramic capacitor will be used and the required capacitance
is selected based on the impedance at 468 kHz:
C
O = 1/(2 x π x 0.157 x 4.68 x 10
5
) = 2.18 F
This calculation assumes that impedance due to the equiva-
lent series resistance (ESR) and equivalent series inductance
(ESL) of C
O is negligible. The closest 10% tolerance capacitor
value is 2.2 F. The capacitor used should be rated to 10V or
more and have an X7R dielectric. Several manufacturers pro-
duce ceramic capacitors with these specifications in the 0805
case size. A typical value for ESR of 1 m
can be read from
the curve of impedance vs. frequency in the product
datasheet.
INPUT CAPACITOR
Following the calculations from the Input Capacitor section,
Δv
IN(MAX) will be 1%P-P = 240 mV. The minimum required ca-
pacitance is:
C
IN(MIN) = (0.35 x 300 x 10
-9
) / 0.24 = 438 nF
In expectation that more capacitance will be needed to pre-
vent power supply interaction a 1.0 F ceramic capacitor
rated to 50V with X7R dielectric in a 1206 case size will be
used. From the Design Considerations section, input rms cur-
rent is:
I
IN-RMS = 0.35 x Sqrt(0.154 x 0.846) = 126 mA
Ripple current ratings for 1206 size ceramic capacitors are
typically higher than 1A, more than enough for this design.
RECIRCULATING DIODE
The first parameter for D1 which must be determined is the
reverse voltage rating. Schottky diodes are available at re-
verse ratings of 30V and 40V, often in the same package, with
the same forward current rating. To account for ringing a 40V
Schottky will be used.
The next parameters to be determined are the forward current
rating and case size. In this example the low duty cycle (D =
3.7 / 24 = 15%) requires the recirculating diode D1 to carry
the load current much longer than the internal power MOS-
FET of the LM3402. The estimated average diode current is:
I
D = 0.35 x 0.85 = 298 mA
Schottky diodes are available at forward current ratings of
0.5A, however the current rating often assumes a 25°C am-
bient temperature and does not take into account the appli-
cation restrictions on temperature rise. A diode rated for
higher current may be needed to keep the temperature rise
below 40°C.To determine the proper case size, the dissipa-
tion and temperature rise in D1 can be calculated as shown
in the Design Considerations section. V
D for a small case size
such as SOD-123 in a 40V, 0.5A Schottky diode at 350 mA is
approximately 0.4V and the
θ
JA is 206°C/W. Power dissipa-
tion and temperature rise can be calculated as:
P
D = 0.298 x 0.4 = 119 mW
T
RISE = 0.119 x 206 = 24.5°C
According to these calculations the SOD-123 diode will meet
the requirements. Heating and dissipation are among the fac-
17
www.national.com
LM3402/LM3402HV
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