LM3409MY-NOPB Datasheet, PDF (24/43 Pages) Texas Instruments – PFET Buck Controller for High Power LED Drivers

LM3409, LM3409HV, LM3409-Q1
SNVS602J – MARCH 2009 – REVISED MAY 2013
www.ti.com
R_OFF
=
(C_OFF
+
-
¨¨©§^1-
K
V_O
x V_IN
¸¸¹·
20 pF) x f_SW x ln
¨¨©§^{1- 1.2V4}_O^V¸¸¹·
- ¨§1- ^35V ¸·
R_OFF
=
490
pF
x
© ^{0.95 x 48V}¹
525 kHz x ln ¨§1- ^1.24V¸·
=
25.1 k:
© ^35V ¹
(45)
The closest 1% tolerance resistor is 24.9 kΩ therefore the actual t_OFF and target f_SW are:
t_OFF = -(C_OFF + 20 pF) x R_OFF x ln _{¨¨©§}1- ^1.2V4_O^V_{¸¸¹·}
t_OFF = - 490 pF x 24.9 k: x ln _{¨¨©§}1- ^13.254VV_{¸¸¹·} = 440 ns
(46)
f_SW
=
1^-
¨¨©§K
V_O
x V_IN
t _OFF
¸¸¹·
=
1^-
¨§ ^35V ¸·
©^{0.95 x48V}¹
440 ns
=
528
kHz
(47)
The chosen components from step 1 are:
C_OFF = 470 pF
R_OFF = 24.9 k:
(48)
2. INDUCTOR RIPPLE CURRENT
Solve for L1:
L1
=
V_O x t _OFF
'i_{L - PP}
=
35V x 440 ns
1A
=
15.4 PH
(49)
The closest standard inductor value is 15 µH therefore the actual Δi_L-PP is:
'i_L-PP =
V_O x t_OFF
L1
=
35V x 440 ns
15 PH
= 1.027A
(50)
The chosen component from step 2 is:
L1 = 15 PH
(51)
3. AVERAGE LED CURRENT
Determine I_L-MAX:
I_L-MAX = I_LED +
'i_{L - PP}
2
=
2A +
1.027A
2
=
2.51A
(52)
Assume V_ADJ = 1.24V and solve for R_SNS:
R_{SNS =}
5
V_ADJ
x I_{L- MAX}
=
5
1.24V
x 2.51A
=
0.099:
(53)
The closest 1% tolerance resistor is 0.1 Ω therefore the I_LED is:
24
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