LM3429_14 Datasheet, PDF (34/50 Pages) Texas Instruments – LM3429Q1 N-Channel Controller for Constant Current LED Drivers

LM3429
SNVS616F – APRIL 2009 – REVISED JANUARY 2010
www.ti.com
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
Dc
x
D
=
1A
0.533
x
0.467 = 1.28A
(125)
P_T
=
I_T-
2
RMS
x
R_DSON
= 1.28A²
x
50
m:
=
82
mW
(126)
The chosen component from step 9 is:
Q1 o 32A, 100V, DPAK
(127)
10. 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_D = 600 mV. Determine P_D:
P_D = I_D x V_FD = 1A x 600 mV = 600 mW
The chosen component from step 10 is:
D1 o 12A, 100V, DPAK
(128)
(129)
(130)
(131)
11. 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 11 are:
R_UV1 = 21k:
R_UV2 = 150 k:
(132)
(133)
(134)
(135)
(136)
12. 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:
V_HYSO = R_OV2 x 20PA = 499 k: x 20 PA = 9.98V
(137)
(138)
34
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