Datasheet

:
k243R
BIAS
=
:
k49.9RR
REF2
==
REF1
6.81R
GAIN
=
:
k
:
=
x
-
=
x-
=
k68.6
2
1
R
R
GAIN
GAIN
¸
¸
¹
·
¸
¸
¹
·
¨
¨
©
§
¨
¨
©
§
V45.2
V45.2
PA100
I
CSH
+
RR
2REF1REF
+
-
RR
BIASENDNTC
-
R
ENDNTC
R
1REF
:+: k24.3k7.15
:k7.15
:
k4.21R
CSH
=
:
k1RR
HSN
==
HSP
0.1R
NSS
=
:
I
LED
= =
k0.11.24V :x
A0.1=
k4.121.0 :x:
RR
CSHSNS
x
R1.24V
HSP
x
=
1.24V1.24V
=
R
HSP
:
=
k0.1
:
x
:
x
0.1k12.4A1
xx
RRI
SNSCSHLED
:
===
1.0
1A
R
SNS
I
LED
mV100
V
SNS
R
T
= 14.3 k:
kHz504
1
1
f
SW
==
=
e95.1k3.14e40.1
810
-
:x
-
-
e95.1Re40.1
8
T
10
-
x
--
f
SW
:
==
k4.14
=
R
T
x+
-
f
e
95.11
SW
8
x+
-
kHz500e95.11
8
x
-
fe40.1
SW
10
x
-
kHz500e40.1
10
LM3424
SNVS603B –AUGUST 2009–REVISED OCTOBER 2009 www.ti.com
2. SWITCHING FREQUENCY
Solve for R
T
:
(107)
The closest standard resistor is 14.3 kΩ therefore f
SW
is:
(108)
The chosen component from step 2 is:
(109)
3. AVERAGE LED CURRENT
Solve for R
SNS
:
(110)
Assume R
CSH
= 12.4 kΩ and solve for R
HSP
:
(111)
The closest standard resistor for R
SNS
is actually 0.1Ω and for R
HSP
is actually 1 kΩ therefore I
LED
is:
(112)
The chosen components from step 3 are:
(113)
4. THERMAL FOLDBACK
Find the resistances corresponding to T
BK
and T
END
(R
NTC-BK
= 24.3 kΩ and R
NTC-END
= 7.15 kΩ) from the
manufacturer's datasheet. Assuming R
REF1
= R
REF2
= 49.9 kΩ, then R
BIAS
= R
NTC-BK
= 24.3 kΩ.
Solve for R
GAIN
:
(114)
The chosen components from step 4 are:
(115)
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