 Semiconductor Components Industries, LLC, 2003
January, 2003 – Rev. 0
1 Publication Order Number:
AND8109/D
C0065C0078C0068C0056C0049C0048C0057C0047C0068
C0076C0069C0068 C0067C0111C0110C0115C0116C0097C0110C0116 C0067C0117C0114C0114C0101C0110C0116
C0083C0111C0117C0114C0099C0101 C0083C0099C0104C0101C0109C0101
Theory of Operation
Prepared by: Mark Scholten
Senior Applications Engineer
The LM317L wants to see 1.25 V between its V
OUT
pin
and the V
adj
pin, and it will do whatever it can to keep that
voltage differential between them. So if a resistor is put in
series with the output, and the V
adj
pin is connected to the
load side of the sense resistor, a current source is set up
which follows Equation 1:
I
OUT
C0043
1.25
R
sense
(Approximately)
(eq. 1)
LM317L
V
IN V
OUT
I
OUT
V
adj
R
sense
R
load
Figure 1.
In order to set up a constant current source for an LED
string, the same circuit can be used by simply substituting
R
load
with an LED string, as shown below.
LM317L
LED1
LED2
LED3
Figure 2.
R
sense
V
IN V
OUT
I
OUT
V
adj
But what if a series/parallel combination of LEDs is
wanted? The following circuit works fine, assuming that
there are not large variations in the forward voltage drop of
the LEDs. There could be a problem however if one of the
strings opens up for some reason. The LEDs that remain
would have 50% more current flowing through them, which
could cause them to be driven too hard and fail. So how does
one get around this problem?
LED6 LED9
LM317L
LED1 LED4 LED7
LED8LED5LED2
LED3
V
IN
V
OUT
I
OUT
V
adj
R
sense
Figure 3.
The following circuit takes care of this issue by splitting
up the current sense resistor into three resistors, one for each
leg of the LED string. The current through any one of the
LED strings is set by Equation 2:
I
OUT
C0043
1.25 VC0041V
sat
R1
(Approximately)
(eq. 2)
http://onsemi.com