LM3429_14 Datasheet, PDF (31/50 Pages) Texas Instruments – LM3429Q1 N-Channel Controller for Constant Current LED Drivers

LM3429
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V_O = N x V_LED = 6 x 3.5V = 21V
r_D = N x r_LED = 6 x 325 m: = 1.95:
Solve for D, D', D_MAX, and D_MIN:
D
=
V_O
V_O + V_IN
=
21V
21V + 24V
=
0.467
D' = 1- D = 1- 0.467= 0.533
V_O
21V
D_MIN = _{VO + VIN-MAX} = _{21V + 70V} = 0.231
D_MAX =
V_O
V_O + V_IN-MIN
=
21V
21V +10V
= 0.677
SNVS616F – APRIL 2009 – REVISED JANUARY 2010
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2. SWITCHING FREQUENCY
Assume C_T = 1 nF and solve for R_T:
25
25
R_T = _{fSW x CT} = _{700 kHz x 1 nF} = 35.7 k:
The closest standard resistor is actually 35.7 kΩ therefore the f_SW is:
f_SW
=
25
R_T x C_T
=
35.7
25
k: x
1
nF
=
700
kHz
The chosen components from step 2 are:
C_T = 1 nF
R_T = 35.7 k:
3. AVERAGE LED CURRENT
Solve for R_SNS:
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
x12.4 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:
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4. INDUCTOR RIPPLE CURRENT
Solve for L1:
L1=
V_IN x D
'i_{L- PP} x f_SW
=
24V x 0.467
500 mA x 700 kHz
=
32
PH
The closest standard inductor is 33 µH therefore the actual Δi_L-PP is:
'i_{L- PP}
=
V_IN x D
L1x f_SW
=
24V x 0.467
33 PH x 700 kHz
=
485 mA
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