LM3424 Datasheet, PDF (44/69 Pages) National Semiconductor (TI) – Constant Current N-Channel Controller with Thermal Foldback for Driving LEDs

LM3424, LM3424-Q1
SNVS603C – AUGUST 2009 – REVISED AUGUST 2015
www.ti.com
Typical Applications (continued)
I_{T- MAX}
=
0.677
1- 0.677
x 1A
=
2.1A
(141)
A 100V NFET is chosen with a current rating of 32A due to the low R_DS-ON = 50 mΩ. Determine I_T-RMS and P_T:
I_{T- RMS}
=
I_LED
D?
x
D
=
1A
0.533
x
0.467 = 1.28A
(142)
P_T
=
I_T-
2
RMS
x
R_DSON
= 1.28A²
x
50
m:
=
82
mW
(143)
The chosen component from step 11 is:
Q1 o 32A, 100V, DPAK
(144)
8.2.2.2.12 Diode
Determine minimum D1 voltage rating and current rating:
V_RD-MAX = V_IN-MAX + V_O = 70V + 21V = 91V
I_D-MAX = I_LED = 1A
A 100V diode is chosen with a current rating of 12A and V_DF = 600 mV. Determine P_D:
P_D = I_D x V_FD = 1A x 600 mV = 600 mW
The chosen component from step 12 is:
D1 o 12A, 100V, DPAK
(145)
(146)
(147)
(148)
8.2.2.2.13 Input UVLO
Solve for R_UV2:
R_UV2
=
V_HYS
20 PA
=
3V
20 PA
=150 k:
The closest standard resistor is 150 kΩ therefore V_HYS is:
V_HYS = R_UV2 x 20 PA = 150 k: x 20 PA = 3V
Solve for R_UV1:
R_UV1 =
1.24Vx R_UV2
V_{TURN- ON} - 1.24V
=
1.24V x150 k:
10V -1.24V
= 21.2 k:
The closest standard resistor is 21 kΩ making V_TURN-ON:
( ) _{VTURN- ON}
=
1.24V
x
R _UV1
R _UV1
+
R _UV2
1.24Vx (21 k: +150 k:)
V_{TURN- ON} =
21 k:
= 10.1V
The chosen components from step 13 are:
R_UV1 = 21k:
R_UV2 = 150 k:
(149)
(150)
(151)
(152)
(153)
8.2.2.2.14 Output OVLO
Solve for R_OV2:
R_OV2
=
V_HYSO
20 PA
=
10V
20 PA
=
500
k:
The closest standard resistor is 499 kΩ therefore V_HYSO is:
(154)
44
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