LM3402HVMR/NOPB National Semiconductor, LM3402HVMR/NOPB Datasheet - Page 16

IC LED DRVR HP CONST CURR 8-PSOP

LM3402HVMR/NOPB

Manufacturer Part Number
LM3402HVMR/NOPB
Description
IC LED DRVR HP CONST CURR 8-PSOP
Manufacturer
National Semiconductor
Series
PowerWise®r
Type
High Power, Constant Currentr
Datasheets

Specifications of LM3402HVMR/NOPB

Constant Current
Yes
Topology
PWM, Step-Down (Buck)
Number Of Outputs
1
Internal Driver
Yes
Type - Primary
Automotive
Type - Secondary
High Brightness LED (HBLED), White LED
Frequency
1MHz
Voltage - Supply
6 V ~ 75 V
Mounting Type
Surface Mount
Package / Case
8-PSOP
Operating Temperature
-40°C ~ 125°C
Current - Output / Channel
500mA
Internal Switch(s)
Yes
Efficiency
96%
Primary Input Voltage
75V
No. Of Outputs
1
Output Voltage
73V
Output Current
500mA
Voltage Regulator Case Style
PSOP
No. Of Pins
8
Operating Temperature Range
-40°C To +125°C
Svhc
No SVHC (15-Dec-2010)
Rohs Compliant
Yes
For Use With
551600000-001A/NOPB - BOARD WEBENCH SO8/SOP LM3404/2551600003-001A - BOARD WEBENCH MSOP LM3402LM3402HVEVAL - BOARD EVALUATION FOR LM3402HV
Lead Free Status / RoHS Status
Lead free / RoHS Compliant
Voltage - Output
-
Other names
LM3402HVMR

Available stocks

Company
Part Number
Manufacturer
Quantity
Price
Part Number:
LM3402HVMR/NOPB
Manufacturer:
NS/国半
Quantity:
20 000
Company:
Part Number:
LM3402HVMR/NOPB
Quantity:
275
www.national.com
R
To select R
the Controlled On-time Overview section can be re-written as:
Minimum on-time occurs at the maximum V
110% = 26.4V. R
The closest 1% tolerance resistor is 59.0 kΩ. The switching
frequency of the circuit can then be found using the equation
relating R
USING AN OUTPUT CAPACITOR
The inductor will be the largest component used in this design.
Because the application does not require any PWM dimming,
an output capacitor can be used to greatly reduce the induc-
tance needed without worry of slowing the potential PWM
dimming frequency. The total solution size will be reduced by
using an output capacitor and small inductor as opposed to
one large inductor.
OUTPUT INDUCTOR
Knowing that an output capacitor will be used, the inductor
can be selected for a larger current ripple. The desired max-
imum value for Δi
Minimum inductance is selected at the maximum input volt-
age. Re-arranging the equation for current ripple selection
yields the following:
ON
L
MIN
and t
R
ON
= [(26.4 – 3.7) x 300 x 10
f
SW
ON
= (300 x 10
ON
ON
= 3.7 / (59000 x 1.34 x 10
to f
the expression relating t
SW
ON
L
:
is ±30%, or 0.6 x 350 mA = 210 mA
is therefore calculated as:
-9
x 26.4) / 1.34 x 10
-9
] / (0.6 x 0.35) = 32.4 µH
ON
-10
to input voltage from
) = 468 kHz
FIGURE 5. Schematic for Design Example 1
-10
IN
= 59105 Ω
, which is 24V x
P-P
.
16
The closest standard inductor value is 33 µH. Off-the-shelf
inductors rated at 33 µH are available from many magnetics
manufacturers.
Inductor datasheets should contain three specifications which
are used to select the inductor. The first of these is the aver-
age current rating, which for a buck regulator is equal to the
average load current, or I
by a specified temperature rise in the inductor, normally 40°
C. For this example, the average current rating should be
greater than 350 mA to ensure that heat from the inductor
does not reduce the lifetime of the LED or cause the LM3402
to enter thermal shutdown.
The second specification is the tolerance of the inductance
itself, typically ±10% to ±30% of the rated inductance. In this
example an inductor with a tolerance of ±20% will be used.
With this tolerance the typical, minimum, and maximum in-
ductor current ripples can be calculated:
The third specification for an inductor is the peak current rat-
ing, normally given as the point at which the inductance drops
off by a given percentage due to saturation of the core. The
worst-case peak current occurs at maximum input voltage
and at minimum inductance, and can be determined with the
equation from the Design Considerations section:
Δi
Δi
Δi
L(MAX)
L(MIN)
L(TYP)
I
L(PEAK)
= [(26.4 – 3.7) x 300 x 10
= [(26.4 – 3.7) x 300 x 10
= [(26.4 – 3.7) x 300 x 10
= 0.35 + 0.258 / 2 = 479 mA
= 206 mA
= 172 mA
= 258 mA
F
. The average current rating is given
P-P
P-P
P-P
-9
-9
-9
] / 39.6 x 10
] / 26.4 x 10
] / 33 x 10
20192119
-6
-6
-6

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