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PP-
==
L
DV
IN
x
f1L
SW
x
kHz045H33 xP
467.0V42 x
mA674
=
'i
==
DV
IN
x
f
SW
x
467.0V42 x
PH
32
=
1L
PP-
'i
L
kHz045700 mA x
R15 = 10 k:
R19 = R21 = R22 = 49.9 k:
R
GAIN
=
R19
x 2.45V
100 PA
R19 + R21
-
¸
¸
¹
·
¸
¸
¹
·
R
NTC-END
R
NTC-END
+ R22
I
CSH
= 9.49 k:
R
GAIN
=
1
x 2.45V
2
-
¸
¸
¹
·
¸
¸
¹
·
6.34 k: + 49.9 k:
6.34 k:
R8 = R7 = 1 k:
R1 = 12.4 k:
R9 = 0.1:
I
LED
=
= = 1.0A
1.24V x R8
R9 x R1
1.24V x 1.0 k:
0.1: x 1.24 k:
R8 = =
= 1.0 k:
I
LED
x R1 x R9
1.24V
1A x 1.24 k: x 0.1:
1.24V
R9 = =
= 0.1:
V
SNS
100 mV
1A
I
LED
R10 = 14.3 k:
f
SW
=
= 504 kHz
1.40e
-10
x 14.3 k: - 1.95e
-8
1
f
SW
=
1.40e
-10
x R10 - 1.95e
-8
1
www.ti.com Design Procedure
The closest standard resistor is 14.3 kΩ therefore f
SW
is:
(8)
The chosen component from step 2 is:
(9)
8.4 Average LED Current
Solve for R
SNS
:
(10)
Assume R
CSH
= 12.4 kΩ and solve for R
HSP
:
(11)
The closest standard resistor for R
SNS
is actually 0.1Ω and for R
HSP
is actually 1 kΩ therefore I
LED
is:
(12)
The chosen components from step 3 are:
(13)
8.5 Thermal Foldback
Using a standard 100k NTC thermistor (connected to pins 4 and 11 of J7), find the resistances
corresponding to T
BK
and T
END
(R
NTC-BK
= 243 kΩ and R
NTC-END
= 71.5 kΩ) from the manufacturer's
datasheet. Assuming R
REF1
= R
REF2
= 49.9 kΩ, then R
BIAS
= R
NTC-BK
= 243 kΩ.
Solve for R
GAIN
:
(14)
The chosen components from step 4 are:
(15)
8.6 Inductor Ripple Current
Solve for L1:
(16)
The closest standard inductor is 33 µH therefore Δi
L-PP
is:
(17)
9
SNVA397A–August 2009–Revised May 2013 AN-1967 LM3424 Buck-Boost Evaluation Board
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