LM3406_09 Datasheet, PDF (15/26 Pages) National Semiconductor (TI) – 1.5A Constant Current Buck Regulator for Driving High Power LEDs

The peak LED/inductor current is then estimated. This calcu-
lation uses the worst-case ripple current which occurs with
three LEDs.
I_L(PEAK) = I_L + 0.5 x Δi_L(MAX)
I_L(PEAK) = 1.5 + 0.5 x 0.56 = 1.78A
In order to prevent inductor saturation the inductor’s peak
current rating must be above 1.8A. A 22 µH off-the shelf in-
ductor rated to 2.1A (peak) and 1.9A (average) with a DCR of
59 mΩ will be used.
USING AN OUTPUT CAPACITOR
This application does not require high frequency PWM dim-
ming, allowing the use of an output capacitor to reduce the
size and cost of the output inductor while still meeting the 10%
P-P ^{target for LED ripple current. To select the proper output}
capacitor the equation from Buck Regulators with Output Ca-
pacitors is re-arranged to yield the following:
The dynamic resistance, r_D,of one LED can be calculated by
taking the tangent line to the V_F vs. I_F curve in the LED
datasheet. Figure 8 shows an example r_D calculation.
30020324
FIGURE 8. Calculating r_D from the V_F vs. I_F Curve
Extending the tangent line to the ends of the plot yields values
for ΔV_F and ΔI_F of 0.7V and 2000 mA, respectively. Dynamic
resistance is then:
r_D = ΔV_F / ΔI_F = 0.5V / 2A = 0.25Ω
The most filtering (and therefore the highest output capaci-
tance) is needed when r_D is lowest, which is when there is
only one LED. Inductor ripple current with one LED is 478
mA_P-P. The required impedance of C_O is calculated:
Z_C = [0.15 / (0.478 - 0.15] x 0.35 = 0.114Ω
A ceramic capacitor will be used and the required capacitance
is selected based on the impedance at 362 kHz:
C_O = 1/(2 x π x 0.16 x 3.62 x 10⁵) = 3.9 µF
This calculation assumes that C_O will be a ceramic capacitor,
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_SNS
Using the expression for R_SNS:
R_SNS = 0.2 / I_F
R_SNS = 0.2 / 1.5 = 0.133Ω
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
re-calculated to ensure that current is within the ±5% toler-
ance requirement. From the expression for average LED
current:
I_F = 0.2 / 0.13 = 1.54A, 3% above the target current
INPUT CAPACITOR
Following the calculations from the Input Capacitor section,
Δv_IN(MAX) will be 24V x 2%_P-P = 480 mV. The minimum re-
quired capacitance is calculated for the largest t_ON, corre-
sponding to five LEDs:
C_IN(MIN) = (1.5 x 1.5 x 10^-6) / 0.48 = 4.7 µF
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_IN.
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_O / V_IN, and when this
converter drives three LEDs the duty cycle will be nearly 50%.
I_IN-RMS = 1.5 x Sqrt(0.5 x 0.5) = 750 mA
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.
15
www.national.com