I guess 2ma of loss. I don't think it would mess with mode switching or anything, but I'm just an amateur. I'd wait for a real electronic savvy member weigh it.
Makes sense, but I wouldn’t know which resistor to use.
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That’s the problem, other than a few heavyweights, I don’t know who to listen to on these questions…
^ There are calculators online where you put in the various parameters.
I just realized you would need a limiting resistor in series with the LED like in your diagram. If not, when the driver turns, the LED will be allowed to draw all the current it could take.
EDIT: Here is a calculator.
Where would the positive side of the led actually receive positive? The body would be negative whenever the switch is on and dead whenever the switch is off? It’s still DC and still has to have positive and negative for the circuit right? The circuit, with the switch on, is the driver/led circuit, not in the body. With the switch off, there would not be one. At least I would believe so.
Right. I misunderstood your first statement.
So I would just need to add a resistor on the driver bridging Pos and Neg, and another one on the switch PCB maybe?
I may have an improper understanding of this, but…
The current has to flow in a complete circuit, which includes the tailcap. The current doesn’t just flow to the driver, but through it as well. Normally the current would flow through the switch when it is on, but this would be a bypass for the switch so that the circuit is always complete by going throught the limiting resistor and led.
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I figure if I can keep the amp draw down to 2ma, it would take 40days to completely drain a 2000ma battery, and I can always partially unscrew the tail to lockout/disable the “locator” led
^ Correct. The body of the light is much more connected to positive (via the driver circuits) then it is to negative when the tail cap switch is off.
Pilotdog68 wrote:
. . . So I would just need to add a resistor on the driver bridging Pos and Neg, and another one on the switch PCB maybe?
You may not need one on the driver if it has enough parasitic drain. If it didn't, then yes. You have me intrigued. I may try to make a simulation of this (just not stuffed into a tail cap) to see if it works. I too would like to have illuminated tail caps on 2 of my lights. What FW and driver would you use?
The tailcap light would obviously be completely independent of FW, but for what it’s worth I always use some version of STAR on one of wights drivers.
I think I will try to wire a driver, led, switch with an led tonight to see how it acts. I'm wondering if a limiting resistor would be needed in the tail cap. I would think that once the switch is closed, the tail cap led will essentially be bypassed (not the word I want). It will a have negative charge on both sides. So only positive resistance in the driver would be needed. Not sure.
The resistor would be needed for an smd led to drop the voltage I think. Most of the best options need ~1.7-2.0v
Oh, there would be resistance in series with the LED. It would just be in the driver.
To “make nails with heads” as we are saying here in Germany I did a quick test to see if the circuit from OP works. I’ve connected a LED with an 470 Ohm resistor as protection between body and negative BAT (as would the switch do) and I can say: it depends on the driver!
First one I’ve tried was the one I’ve built-in today in the Convoy M1 Host. It is the “SKU 1127403 - 8* AMC7135 4-Group 2~5 Modes LED Flashlight Driver Circuit (Generic) https://www.fasttech.com/product/1127403” that I’ve used. And the LED’s lid up! But not much. For testing I used ultra low current 5mm LED Red 620-625nm and Green 515-530nm and the green one got about 75uA (MicroAmpere!) and another type 50uA; the red about 150uA.
I had these LED because some time ago I researched the innards of an USB-controlled 230V power strip which had LED’s connected via a 1MegaOhm resistor directly to 230V AC. I couldn’t believe at first that this was possible (I’ve added some 1N4148 anti-parallel later as protection). The thing I’ve learned was - there are LED’s today whith visible shine at very low currents. A great help were the pages of Don Klipstein found here: Lo-mA LEDs
Another Light (Ultrafire SK-98 clone) with the typical generic China driver unfortunately did not work. The LED shortly flashed and the current was only about 8uA.
So, you have to test it to see if it works. Of course, one could add a resistor in parallel to the driver to increase the quiescent current. The 470 Ohm resistor is presumably not necessary but was the first at hand. For the intended function it has no influence at all.
So far my tests suggest that it would be best to test the light in question by simply put a LED between BAT and Body with tailcap taken off and see if it lights up. Worst case would be that it burns up 
If successful, you only have to find a suited LED which mets the criteria above and fits the designated place…
One further aspect would be however to find out if the “standby” LED affects mode-switching which is based on short interruptions of the switch.
I love a proof of concept!
Good point about mode-switching, I hadn’t thought about that
One approach is to use a trit, but they have disadvantages. The circuit in the OP will work if the driver/LED combination will let some current pass at very low voltage. Some house light switches light up and I suppose that is how it is done. I think it has to be either a boost driver, that doesn’t send all current through the LED or else the LED has to have some leakage.
I do not see why the circuit posted in OP would not work. If a green LED with a Vf of around 2V (example ) was used, a 1.1k ohm resistor could be used to limit the current (one battery @ 4.2V). Really would not even need to add to the circuit board, just tap into the positive side with a wire and put a resistor (example ) from the cathode of the LED to a constant ground. I would think that routing a wire around the batteries would be the hard part.
Yes, you can run a small SMD LED with just a series resistor. You see that all the time. If you know the forward voltage of the LED at a given current and the input voltage it is a fairly straightforward calculation. If you're not sure, then just start with a high value resistor and work your way down.
Going body positive solves your problem of getting power to the tailcap to run stuff back there, but then you are left with the same problem as before, just in reverse. How then do you get ground up top to the LED without an extra electrical path? You can't.
Yes, the only way to effectively use the body as an electrical path is to put everything in series together - the Driver and LED for normal use would be in series with the standby LED. That isn’t going to work out very well for you, I think.
But it would be much easier to put the resistor (if necessary) on the driver between plus (spring) and minus (e.g. at the bottom as an SMD) instead of adding a wire through the whole light. The LED can then be in parallel with the Switch as planned. And with the right (ultrabright) LED the value of the resistor should be higher than 1k1 for a current as low as possible. With my red 5mm LED @3,7V from powersource I get with a 3k9 resistor (3,9KOhm) a current of ~0,5 mA - more then enough for such a LED to be visible in the dark. After all you want to avoid to drain the battery more than necessary when the light is not in use!
I think it makes more sense to put a button cell and e-switch in the tail cap. Current flows through the body and tail cap but all the voltage drop has been accounted for, there isn’t any left to power another led.