LM3429_14 Datasheet, PDF (33/50 Pages) Texas Instruments – LM3429Q1 N-Channel Controller for Constant Current LED Drivers

LM3429
www.ti.com
Z_Z1
=
r_D x
Dx
Dc²
L1
=
1.95:
0.467
x
x
0.533²
33 PH
=
37k
rad
sec
T_U0 is approximated:
Dc x 620V
0.533 x 620V
T_U0 = _{(1+D) x ILED x RLIM} = _{1.467x1A x 0.04:} = 5630
To ensure stability, calculate ω_P2:
Z_P2
=
min(Z_P1,Z_Z1)
5 x T_U0
=
5
Z_Z1
x 5630
=
37k
rad
sec
5 x 5630
=
1.173
rad
sec
Solve for C_CMP:
1
1
C_CMP
=
Z_P2
x
5e⁶:
=
1.173
rad
sec
x
5e⁶:
=
0.17 PF
To attenuate switching noise, calculate ω_P3:
Z_P3 = max Z_P1, Z_Z1 x 10 = Z_P1 x 10
Z_P3
=
110k
rad
sec
x10
=
1.1M
rad
sec
Assume R_FS = 10Ω and solve for C_FS:
1
1
C_FS = _{10:x ZP3} = _{10: x1.1M rad} = 0.091PF
sec
The chosen components from step 7 are:
C_COMP = 0.22 PF
R_FS = 10:
C_FS = 0.1 PF
SNVS616F – APRIL 2009 – REVISED JANUARY 2010
(113)
(114)
(115)
(116)
(117)
(118)
(119)
8. INPUT CAPACITANCE
Solve for the minimum C_IN:
C_IN =
I_LED x D
'v_IN-PP x f_SW
1A x 0.467
= _{100 mV x 700 kHz}
=
6.66
PF
(120)
To minimize power supply interaction a 200% larger capacitance of approximately 14 µF is used, therefore the
actual Δv_IN-PP is much lower. Since high voltage ceramic capacitor selection is limited, three 4.7 µF X7R
capacitors are chosen.
Determine minimum allowable RMS current rating:
I_IN-RMS = I_LED x
D_{MAX = 1A x}
1- D_MAX
0.677
1- 0.677
=
1.45A
(121)
The chosen components from step 8 are:
C_IN = 3 x 4.7 PF
(122)
9. NFET
Determine minimum Q1 voltage rating and current rating:
V_T-MAX = V_{IN- MAX} + V_O = 70V + 21V = 91V
I_{T- MAX}
=
0.677
1- 0.677
x 1A
=
2.1A
(123)
(124)
Copyright © 2009–2010, Texas Instruments Incorporated
Product Folder Links: LM3429
Submit Documentation Feedback
33