參數(shù)資料
型號: LM3402
廠商: National Semiconductor Corporation
英文描述: 0.5A Constant Current Buck Regulator for Driving High Power LEDs
中文描述: 0.5A的恒流降壓穩(wěn)壓器用于驅(qū)動高功率LED
文件頁數(shù): 17/22頁
文件大?。?/td> 840K
代理商: LM3402
Design Example 1: LM3402
(Continued)
size such as SOD-123 in a 40V, 0.5A Schottky diode at 350
mA is approximately 0.4V and the
θ
JA
is 206C/W. Power
dissipation and temperature rise can be calculated as:
P
D
= 0.298 x 0.4 = 119 mW
T
RISE
= 0.119 x 206 = 24.5C
According to these calculations the SOD-123 diode will meet
the requirements. Heating and dissipation are among the
factors most difficult to predict in converter design. If pos-
sible, a footprint should be used that is capable of accepting
both SOD-123 and a larger case size, such as SMA.Alarger
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 C
F
The bootstrap capacitor C
should always be a 10 nF ce-
ramic capacitor with X7R dielectric. A 25V rating is appropri-
ate for all application circuits. The linear regulator filter ca-
pacitor 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 optical efficacy of the circuit, which depends upon
the LEDs themselves.
Total output power, P
O
, is calculated as:
P
O
= I
F
x V
O
= 0.35 x 3.7 = 1.295W
Conduction loss, P
C
, in the internal MOSFET:
P
C
= (I
F2
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
= (I
+ f
x Q
) x V
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 V
IN
x I
F
x (t
R
+ t
F
) x f
SW
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
= I
IN(rms)2
x ESR = (0.126)
2
x 0.006 = 0.1 mW (negli-
gible)
DCR loss, P
L
, in the inductor
P
L
= I
F2
x DCR = 0.35
2
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
/ (P
O
+ Sum of all loss terms) =
1.295 / (1.295 + 0.377) = 77%
DIE TEMPERATURE
T
= (P
+ P
+ P
) x
θ
T
LM3402
= (0.028 + 0.05 + 0.078) x 200 = 31C
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
regulators 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 t
ON
A compromise in switching frequency is needed in this ap-
plication 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
μH and above. R
is selected from the equation for switch-
ing frequency as follows:
R
ON
= 49.2 / (1.34 x 10
-10
x 3 x 10
5
) = 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 t
ON
to f
SW
:
f
SW
= 49.2 / (1210000 x 1.34 x 10
-10
) = 303 kHz
L
www.national.com
17
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