LM3406MH National Semiconductor, LM3406MH Datasheet - Page 19

• Manufacturer Part Number: LM3406MH
• Description: LM3406/06HV 1.5A Constant Current Buck Regulator for Driving High Power LEDs; Package: TSSOP EXP PAD; No of Pins: 14; Qty per Container: 94; Container: Rail
• Manufacturer: National Semiconductor

Extending the tangent line to the ends of the plot yields values
for ΔV
resistance is then:
The most filtering (and therefore the highest output capaci-
tance) is needed when r
only one LED. Inductor ripple current with one LED is 478
mA
A ceramic capacitor will be used and the required capacitance
is selected based on the impedance at 362 kHz:
This calculation assumes that C
and therefore impedance due to the equivalent series resis-
tance (ESR) and equivalent series inductance (ESL) of of the
device is negligible. The closest 10% tolerance capacitor val-
ue is 4.7 µF. The capacitor used should be rated to 25V or
more and have an X7R dielectric. Several manufacturers pro-
duce ceramic capacitors with these specifications in the 1206
case size. A typical value for ESR of 3 mΩ can be read from
the curve of impedance vs. frequency in the product
datasheet.
R
Using the expression for R
Sub-1Ω resistors are available in both 1% and 5% tolerance.
A 1%, 0.13Ω device is the closest value, and a 0.33W, 1210
size device will handle the power dissipation of 290 mW. With
the resistance selected, the average value of LED current is
SNS
FIGURE 14. Calculating r
P-P
. The required impedance of C
F
and ΔI
C
Z
C
O
= [0.15 / (0.478 - 0.15] x 0.35 = 0.114Ω
= 1/(2 x
r
D
F
= ΔV
of 0.7V and 2000 mA, respectively. Dynamic
R
SNS
π
F
/ ΔI
= 0.2 / 1.5 = 0.133Ω
x 0.16 x 3.62 x 10
R
SNS
D
F
SNS
is lowest, which is when there is
= 0.5V / 2A = 0.25Ω
= 0.2 / I
D
:
O
from the V
will be a ceramic capacitor,
F
O
is calculated:
5
) = 3.9 µF
F
vs. I
F
30020324
Curve
19
re-calculated to ensure that current is within the ±5% toler-
ance requirement. From the expression for average LED
current:
INPUT CAPACITOR
Following the calculations from the Input Capacitor section,
Δv
quired capacitance is calculated for the largest t
sponding to five LEDs:
As with the output capacitor, this required value is low enough
to use a ceramic capacitor, and again the effective capaci-
tance will be lower than the rated value with 24V across C
Reviewing plots of %C vs. DC Bias for several capacitors re-
veals that a 4.7 µF, 1812-size capacitor in X7R rated to 50V
loses about 40% of its rated capacitance at 24V, hence two
such caps are needed.
Input rms current is high in buck regulators, and the worst-
case is when the duty cycle is 50%. Duty cycle in a buck
regulator can be estimated as D = V
converter drives three LEDs the duty cycle will be nearly 50%.
Ripple current ratings for 1812 size ceramic capacitors are
typically higher than 2A, so two of them in parallel can tolerate
more than enough for this design.
RECIRCULATING DIODE
The input voltage of 24V ±5% requires Schottky diodes with
a reverse voltage rating greater than 30V. The next highest
standard voltage rating is 40V. Selecting a 40V rated diode
provides a large safety margin for the ringing of the switch
node and also makes cross-referencing of diodes from differ-
ent vendors easier.
The next parameters to be determined are the forward current
rating and case size. The lower the duty cycle the more ther-
mal stress is placed on the recirculating diode. When driving
one LED the duty cycle can be estimated as:
The estimated average diode current is then:
A 2A-rated diode will be used. To determine the proper case
size, the dissipation and temperature rise in D1 can be cal-
culated as shown in the Design Considerations section. V
for a case size such as SMB in a 40V, 2A Schottky diode at
1.5A is approximately 0.4V and the θ
sipation and temperature rise can be calculated as:
IN(MAX)
I
F
= 0.2 / 0.13 = 1.54A, 3% above the target current
C
will be 24V x 2%
I
IN(MIN)
IN-RMS
I
D
P
= (1.5 x 1.5 x 10
= 1.5 x Sqrt(0.5 x 0.5) = 750 mA
T
= (1 - 0.17) x 1.54 = 1.28A
D
RISE
= 1.28 x 0.4 = 512 mW
D = 4.1 / 24 = 0.17
= 0.51 x 75 = 38°C
P-P
= 480 mV. The minimum re-
-6
) / 0.48 = 4.7 µF
JA
O
is 75°C/W. Power dis-
/ V
IN
, and when this
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ON
, corre-
IN
D
.