E
ehsjr
- Jan 1, 1970
- 0
Tom said:Um, the 1k resistor I suggested goes between the top of the LED string
and the LM317 adj pin. You yank down directly on the adj pin. The
LM317 output will follow 1.25V above that. So if you manage to fully
ground the adj pin, the output voltage will be 1.25V. 1.25V across a
series string of 3 LEDs will cause a current of perhaps nanoamps to
flow in the LEDs, depending on what kind they are. But it will be at
most a tiny current, even it they are pretty hot.
When the adj pin is allowed to "float", it will be held by the 1k
resistor to very nearly the voltage at the top of the LED string. The
LM317 output pin will be 1.25V above that. So to get 300mA current in
the LEDs, you put a resistor between the output pin and the top of the
LED string: 1.25V/300mA is about 4 ohms.
When you pull the adj pin low, the thing which pulls it low must
handle the current out of the adj pin of the LM317, at most 100uA
(from the data sheet), plus the current in the 1k resistor. Since the
LED current is very low now, with the adj pin held low, there will be
essentially no drop in the 4-ohm resistor, so 1.25V drops across the
1k resistor, causing 1.25mA current. So the net current in the "low-
puller" (whatever it may be) is 1.35mA at most. You do want to make
sure you only pull DOWN on the adj pin to turn things off. Do NOT
connect it directly to the output (pin 3) of a 555, that can pull it
UP. That would try to turn on the LEDs harder than you intend.
Instead, put a diode in (anode to the adj pin) so you can only pull
down if you have any doubts about your circuit. If the LEDs you're
driving have even a moderately high voltage drop, you can safely use
just about any silicon signal diode. You don't need a diode if you're
driving the adj pin low using an open-collector or open-drain
transistor.
You shouldn't need any capacitors on the output end of the LM317 for
all this to work.
Cheers,
Tom
It can't work the way I understand your description - I'm
obviously not getting it. A schematic would help. Here's
a possibility for yanking down - feel free to modify
it to show what you have in mind. The 1K resistor from
the top of the LED string to the adj pin escapes me, so
it's not in the diagram, but it should work as drawn.
Ed
I -> ~300 mA
-----
Vcc + ----Vin|LM317|Vout-----+ <-- 1.25 v higher than
----- | the top of the LEDs.
Adj [4R]
| |
+------------+ <-- total LED vF when the NPN
| |a is off, Vce when the NPN
| [LED] is conducting.
/c |
YDC >---| [LED]
\e |
| [LED]
| |
Gnd ------------+------------+
YDC = "Yank Down Control"