LM3424 Datasheet, PDF (41/69 Pages) National Semiconductor (TI) – Constant Current N-Channel Controller with Thermal Foldback for Driving LEDs

www.ti.com
LM3424, LM3424-Q1
SNVS603C – AUGUST 2009 – REVISED AUGUST 2015
Typical Applications (continued)
R_SNS
=
V_SNS
I_LED
=
100 mV
1A
=
0.1:
Assume R_CSH = 12.4 kΩ and solve for R_HSP:
R_HSP
=
I_LED
x
R_CSH x
1.24V
R_SNS
=
1A
x
12.4 k:
1.24V
x
0.1:
= 1.0 k:
The closest standard resistor for R_SNS is actually 0.1Ω and for R_HSP is actually 1 kΩ therefore I_LED is:
I_LED
=
1.24V x R_HSP
R_SNS x R_CSH
=
1.24V x 1.0
0.1: x12.4
k:
k:
= 1.0A
The chosen components from step 3 are:
R_SNS = 0.1:
R_CSH =12.4 k:
R_HSP = R_HSN = 1k:
(110)
(111)
(112)
(113)
8.2.2.2.4 Thermal Foldback
Find the resistances corresponding to T_BK and T_END (R_NTC-BK = 24.3 kΩ and R_NTC-END = 7.15 kΩ) from the
manufacturer's datasheet. Assuming R_REF1 = R_REF2 = 49.9 kΩ, then R_BIAS = R_NTC-BK= 24.3 kΩ.
Solve for R_GAIN:
R_{GAIN =}
¨¨©§_RREF^R₁ ^R₊^E_R^F1REF2
-
R_{NTC -}
R_{NTC - END}
END
+ R_BIAS
¸¸¹·
x
2.45V
I_CSH
R_GAIN =
¨¨©§
1
2
-
7.15 k:
7.15 k: + 24.3k:
¸¸¹·
x
100 PA
2.45V
= 6.68 k:
(114)
The chosen components from step 4 are:
R_GAIN = 6.81kΩ
R_BIAS = 24.3 k Ω
R_REF1 = R _REF2 = 49.9 kΩ
(115)
8.2.2.2.5 Inductor Ripple Current
Solve for L1:
L1=
V_IN x D
i_{L- PP} x f_SW
=
24V x 0.467
700 mA x 504 kHz
=
32
2H
The closest standard inductor is 33 µH therefore Δi_L-PP is:
i_{L- PP}
=
V_IN x D
L1x f_SW
=
24V x 0.467
33 2H x 504 kHz
=
674
mA
Determine minimum allowable RMS current rating:
I_L-RMS =
I_LED
Dc
x
1+
1
12
x
¨¨©§'^iL
-PP ^x
I_LED
Dc¸¸¹·²
I_{L- RMS}
=
1A
0. 533
x
1+
1
12
x
¨¨©§⁶⁷⁴
mA x
1A
0.533_{¸¸¹·}²
=
1.89A
The chosen component from step 5 is:
(116)
(117)
(118)
Copyright © 2009–2015, Texas Instruments Incorporated
Product Folder Links: LM3424 LM3424-Q1
Submit Documentation Feedback
41