Design Example 2: LM3402HV
(Continued)
t
ON
= (1.34 x 10
-10
x 1210000) / 60 = 2.7 μs
USING AN OUTPUT CAPACITOR
This application is dominated by the need for fast PWM
dimming, 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
v
signal for the CS comparator to operate correctly.
Designing to a desired
v
of 25 mV and assuming that
the average inductor current will equal the desired average
LED current of 350 mA yields the target current ripple in the
inductor and LEDs:
i
F
=
i
L
=
v
SNS
/ R
SNS
, R
SNS
= V
SNS
/ I
F
i
F
= 0.025 / 0.57 = 43.8 mA
With the target ripple current determined the inductance can
be chosen:
L
MIN
= [(60 – 49.2) x 2.7 x 10
-6
] / (0.044) = 663 μH
The closest standard inductor value is 680 μH. As with the
previous example, the average current rating should be
greater than 350 mA. Separation between the LM3402HV
drivers and the LED arrays mean that heat from the inductor
will not threaten the lifetime of the LEDs, but an overheated
inductor could still cause the LM3402HV to enter thermal
shutdown.
The inductance itself of the standard part chosen is
±
20%.
With this tolerance the typical, minimum, and maximum in-
ductor current ripples can be calculated:
i
F(TYP)
= [(60 - 49.2) x 2.7 x 10
-6
] / 680 x 10
-6
= 43 mA
P-P
i
F(MIN)
= [(60 - 49.2) x 2.7 x 10
-6
] / 816 x 10
-6
= 36 mA
P-P
i
F(MAX)
= [(60 - 49.2) x 2.7 x 10
-6
] / 544 x 10
-6
= 54 mA
P-P
The peak LED/inductor current is then estimated:
I
L(PEAK)
= I
L
+ [
i
L(MAX)
] / 2
I
L(PEAK)
= 0.35 + 0.027 = 377 mA
In the case of a short circuit across the LED array, the
LM3402HV will continue to deliver rated current 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
would be equal to:
i
F(LED-SHORT)
= [(63 – 0.2) x 2.7 x 10
-6
] / 544 x 10
-6
= 314 mA
I
F(PEAK)
= 0.35 + 0.156 = 506 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
inductor saturation during these fault conditions the induc-
tor’s peak current rating must be above 735 mA. A 680 μH
off-the shelf inductor rated to 1.2A (peak) and 0.72A (aver-
age) with a DCR of 1.1
will be used for the green LED
array.
R
SNS
Apreliminary value for R
SNS
was determined in selecting
i
L
.
This value should be re-evaluated based on the calculations
for
i
F
:
Sub-1
resistors are available in both 1% and 5% tolerance.
A1%, 0.56
device is the closest value, and a 0.125W, 0805
size device will handle the power dissipation of 69 mW. With
the resistance selected, the average value of LED current is
re-calculated to ensure that current is within the
±
5% toler-
ance requirement. From the expression for LED current
accuracy:
I
F
= 0.19 / 0.56 + 0.043 / 2 = 361 mA, 3% above 350 mA
INPUT CAPACITOR
Following the calculations from the Input Capacitor section,
v
will be 1%
P-P
= 600 mV. The minimum required
capacitance is:
C
IN(MIN)
= (0.35 x 2.7 x 10
-6
) / 0.6 = 1.6 μF
In expectation that more capacitance will be needed to pre-
vent power supply interaction a 2.2 μF ceramic capacitor
rated to 100V with X7R dielectric in an 1812 case size will be
used. From the Design Considerations section, input rms
current is:
I
IN-RMS
= 0.35 x Sqrt(0.82 x 0.18) = 134 mA
Ripple current ratings for 1812 size ceramic capacitors are
typically higher than 2A, more than enough for this design.
RECIRCULATING DIODE
The input voltage of 60V
±
5% requires Schottky diodes with
a reverse voltage rating greater than 60V. Some manufac-
turers provide Schottky diodes with ratings of 70, 80 or 90V;
however the next highest standard voltage rating is 100V.
Selecting a 100V rated diode provides a large safety margin
for the ringing of the switch node and also makes cross-
referencing of diodes from different vendors easier.
L
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