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LM3421, LM3421-Q1
LM3423, LM3423-Q1
www.ti.com
SNVS574E – JULY 2008 – REVISED MAY 2013
T U 0 =
D c x 620 V
( 1+ D ) x I LED x R LIM
=
0 . 533 x 620 V
1 . 467 x 1 A x 0 . 04 :
= 5630
(121)
To ensure stability, calculate ω P2 :
Z P1
Z P2 =
min( Z P1 , Z Z1 )
5 x T U0
=
rad
19 k
sec
=
5 x 5630 5 x 5630
= 0 . 675
rad
sec
(122)
Solve for C CMP :
Z P 2 x 5 e :
C CMP =
1
6
=
0 . 675
1
rad
sec
x 5 e 6 :
= 0 . 30 P F
(123)
To attenuate switching noise, calculate ω P3 :
Z P3 = (max Z P1 , Z Z1 ) x 10 = Z Z1 x 10
rad           rad
Z P3 = 36k sec x 10 = 360k sec
Assume R FS = 10 ? and solve for C FS :
(124)
C FS =
1
10 : x Z P 3
=
1
10 : x 360 k
rad
sec
= 0 . 28 P F
(125)
The chosen components from step 7 are:
C CMP = 0.33 P F
R FS = 10 :
C FS = 0 . 27 P F
8. INPUT CAPACITANCE
Solve for the minimum C IN :
(126)
C IN =
I LED x D
' v IN- PP x f SW
=
1 A x 0 . 467
100 mV x 504 kHz
= 9 . 27 P F
(127)
To minimize power supply interaction a 200% larger capacitance of approximately 20 μF is used, therefore the
actual Δ v IN-PP is much lower. Since high voltage ceramic capacitor selection is limited, four 4.7 μF X7R capacitors
are chosen.
Determine minimum allowable RMS current rating:
I IN- RMS = I LED x
D MAX
1- D MAX
= 1 A x
0 . 677
1- 0.677
= 1.45A
(128)
The chosen components from step 8 are:
C IN = 4 x 4.7 P F
9. NFET
Determine minimum Q1 voltage rating and current rating:
V T - MAX = V IN - MAX + V O = 70 V + 21 V = 91 V
(129)
(130)
Copyright ? 2008–2013, Texas Instruments Incorporated
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