TLC5970
SBVS140 –MARCH 2010
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2. Inductor value and current selection:
VFB Maximum Voltage
PH On-Duty Ratio2 (%) =
´ 100
VCC Minimum Input Voltage
(5)
Example: VCC = 23 V (minimum), 24 V (typical), and 25 V (maximum). LED forward voltage (VF) = 3.2 V
(minimum), 3.5 V (typical), and 3.8 V (maximum). Two LEDs are connected in series.
Thus, VFB = 2 × 3.8 + 1 = 8.6 V. The PH on-duty ratio 2 (%) = 8.6/23 = 37.4% in this case.
Calculate inductor peak current (mA):
IOUT + IFBn
´ 2
Selected PH On-Duty
PH On-Duty Ratio2
Inductor Peak Current (mA) =
h
100
Where:
IOUT (mA) = Total current of LEDs connected to OUT0/1/2.
IFBn (mA) = Maximum input current of IFB pin.
h (%) = Efficiency of TLC5970 buck converter (recommended to use 90).
(6)
Example: In case all LED currents are set to 60 mA by the 2.50-kΩ external resistor and total current is 180
mA. IFB3 in this data sheet is used when the differential interface output drives the next TLC5970 without a
resistor between SDTA/SDTB and SCKA/SCKB. Therefore:
180 + 115
29
37.4
´ 2
ILPK (mA) =
= 845.5 mA
90
100
(7)
A 25% margin for inductor variation is required. Thus, ILPK = 845.5 × 1.25 = 1057 mA. The maximum inductor
current should be larger than 1057 mA. However, the TLC5970 PH peak current must be less than 2 A in
any case.
3. Calculate inductor value (µH) for minimum inductor value:
1
Inductor Value (mH) = VCC Voltage (V, Minimum) ´
Maximum PH Switching Frequency (MHz, Maximum)
(8)
Selected PH On-Duty (%)
ILPH (mA) ´ 1000
(9)
Example: VCC = 23 V (minimum), 24 V (typical), and 25 V (maximum). Maximum PH switching frequency is
1.5 MHz. The selected PH on-duty ratio as calculated by Equation 4 is 29%. ILPK (mA) is 1057 mA as
calculated by Equation 6.
1
1.5 1057 ´ 1000
0.29
Therefore, the inductor value (mH)
= 23 ´
´
0.29
= 23 ´ 0.67 ´
= 4.2 mH
1057 ´ 1000
(10)
26
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