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: 10/24页
文件大小: 431K
代理商: LM3402_07
tors most difficult to predict in converter design. If possible, a
footprint should be used that is capable of accepting both
SOD-123 and a larger case size, such as SMA. A larger diode
with a higher forward current rating will generally have a lower
forward voltage, reducing dissipation, as well as having a
lower
θ
JA, reducing temperature rise.
C
B and CF
The bootstrap capacitor C
B should always be a 10 nF ceramic
capacitor with X7R dielectric. A 25V rating is appropriate for
all application circuits. The linear regulator filter capacitor C
F
should always be a 100 nF ceramic capacitor, also with X7R
dielectric and a 25V rating.
EFFICIENCY
To estimate the electrical efficiency of this example the power
dissipation in each current carrying element can be calculated
and summed. This term should not be confused with the op-
tical efficacy of the circuit, which depends upon the LEDs
themselves.
Total output power, P
O, is calculated as:
P
O = IF x VO = 0.35 x 3.7 = 1.295W
Conduction loss, P
C, in the internal MOSFET:
P
C = (IF
2
x R
DSON) x D = (0.35
2
x 1.5) x 0.154 = 28 mW
Gate charging and VCC loss, P
G, in the gate drive and linear
regulator:
P
G = (IIN-OP + fSW x QG) x VIN
P
G = (600 x 10
-6
+ 468000 x 3 x 10-9) x 24 = 48 mW
Switching loss, P
S, in the internal MOSFET:
P
S = 0.5 x VIN x IF x (tR + tF) x fSW
P
S = 0.5 x 24 x 0.35 x (40 x 10
-9
) x 468000 = 78 mW
AC rms current loss, P
CIN, in the input capacitor:
P
CIN = IIN(rms)
2
x ESR = (0.126)2 x 0.006 = 0.1 mW (negligible)
DCR loss, P
L, in the inductor
P
L = IF
2
x DCR = 0.352 x 0.096 = 11.8 mW
Recirculating diode loss, P
D = 119 mW
Current Sense Resistor Loss, P
SNS = 92 mW
Electrical efficiency,
η = P
O / (PO + Sum of all loss terms) =
1.295 / (1.295 + 0.377) = 77%
DIE TEMPERATURE
T
LM3402 = (PC + PG + PS) x θJA
T
LM3402 = (0.028 + 0.05 + 0.078) x 200 = 31°C
Design Example 2: LM3402HV
The second example application is an RGB backlight for a flat
screen monitor. A separate boost regulator provides a 60V
±5% DC input rail that feeds three LM3402HV current regu-
lators to drive one series array each of red, green, and blue
1W LEDs. The target for average LED current is 350 mA ±5%
in each string. The monitor will adjust the color temperature
dynamically, requiring fast PWM dimming of each string with
external, parallel MOSFETs. 1W green and blue InGaN LEDs
have a typical forward voltage of 3.5V, however red LEDs use
AlInGaP technology with a typical forward voltage of 2.9V. In
order to match color properly the design requires 14 green
LEDs, twice as many as needed for the red and blue LEDs.
This example will follow the design for the green LED array,
providing the necessary information to repeat the exercise for
the blue and red LED arrays. The circuit schematic for Design
Example 2 is the same as the Typical Application on the front
page. The bill of materials (green array only) can be found in
Table 2 at the end of this datasheet.
OUTPUT VOLTAGE
Green Array: V
O(G) = 14 x 3.5 + 0.2 = 49.2V
Blue Array: V
O(B) = 7 x 3.5 + 0.2 = 24.7V
Red Array: V
O(R) = 7 x 2.9 + 0.2 = 20.5V
R
ON and tON
A compromise in switching frequency is needed in this appli-
cation to balance the requirements of magnetics size and
efficiency. The high duty cycle translates into large conduc-
tion losses and high temperature rise in the IC. For best
response to a PWM dimming signal this circuit will not use an
output capacitor; hence a moderate switching frequency of
300 kHz will keep the inductance from becoming so large that
a custom-wound inductor is needed. This design will use only
surface mount components, and the selection of off-the-shelf
SMT inductors for switching regulators is poor at 1000 H and
above. R
ON is selected from the equation for switching fre-
quency as follows:
R
ON = 49.2 / (1.34 x 10
-10
x 3 x 105) = 1224 k
The closest 1% tolerance resistor is 1.21 M
. The switching
frequency and on-time of the circuit can then be found using
the equations relating R
ON and tON to fSW:
f
SW = 49.2 / (1210000 x 1.34 x 10
-10
) = 303 kHz
t
ON = (1.34 x 10
-10
x 1210000) / 60 = 2.7 s
USING AN OUTPUT CAPACITOR
This application is dominated by the need for fast PWM dim-
ming, requiring a circuit without any output capacitance.
OUTPUT INDUCTOR
In this example the ripple current through the LED array and
the inductor are equal. Inductance is selected to give the
smallest ripple current possible while still providing enough
Δv
SNS signal for the CS comparator to operate correctly. De-
www.national.com
18
LM3402/LM3402HV
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LM3402EVAL 制造商:Texas Instruments 功能描述:EVAL BOARD
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LM3402HV 制造商:Texas Instruments 功能描述:Buck Regulator ,0.5A ,LED Drivers,MSOP8
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