Fred said:
James said:
Here's a "more-efficient" approach, in that all of the current
used (except a few hundred uA) goes through the LED: [see revision below]
VR is a 3-terminal regulator, like LM317, although I used
a LX8383, having a bunch on hand. I used: R1 = 11 ohms,
R2 = 5K pot, R3 = 1K, D = a diode.
-jiw
How do you figure that? If the LED goes to 3.3V, you end up with
3.3-0.7=2.3V across R3 for 2.3mA. And you need to make R1 more like 39
ohms, since the LEDs do not like continuous operation at more than 30mA.
Then because 'D' has an orientation, you might show it.
I think you're right, the loss would be a couple of mA rather than
a few hundred uA. The 100 uA I was thinking of (by mistake) was the
current in the Adjust pin of the VR.
The circuit I built is shown more accurately in a post I made about
the same time you were posting your comment, and still more accurately
below. (My other post showed D1 and D2 but not D3, which was included
to isolate multiple LED's)
.. VR
.. +---+---I O-----------+---+
.. | | A | |
.. | | | R1 |
.. | | | | |
.. +| | | +--+-D1-+ |
.. Bat | | | | | |
.. -| | | | D2 | |
.. | | +---->R2 | | | D1: |<
.. | | | | LED |
.. | | +--+ | | D2, D3: v
.. | C1 R3 | C2 -
.. | | D3 | |
.. +---+-----------+-------+---+
Here, the drop across D1-D3 is about 2V, so there would be
1.3V across R3 (for a 3.3V LED), hence 1.3mA for R3=1K.
To cut down that loss without D3 in the circuit, increase R3
a bunch.
You suggested 39 ohms for R1, vs the 11 ohms I used. With
R2 slid to the D1 end, and 1.25 V at the A pin, and .65V
across D1, there's .4 V on R1, or 36mA through it, so about
34mA through the LED, a bit high. 15 ohms would limit LED
current to 25mA; 39 ohms limits to 8mA. So 11 is too low
and 39 too high, for the circuit above. With D1 not in there,
your 39 limits LED current to about 30mA, so is good for the
earlier circuit.
Using D1 removes about .65V overhead from the LED current path,
allowing the circuit to work on that much less battery voltage.
-jiw