the 805 SMD red leds that I found are all 2-2.2V, my first thought is that two in series would give a 'your battery is not completely full' warning :-) . But at the sub-mA currents the voltage may be well lower than that...
So if the light goes out, it cuts the circuit? I thought it still allowed current to flow even when the led isn’t lit
The green and blue leds i bought have a 3v Vf
There is a relation between led colour and forward voltage, the voltages for these leds below are quite typical for almost all colour leds:
That is correct, at least in theory there is no current when the voltage is too low for the led to emit light
Well hey, I’ve already got a 3v green Led installed, I’ll test it with a 2.8v battery tonight.
A typical Si diode (1N4148) would add about 0.6 Volt to the LED’s own. For the real cutoff voltage of the different options (1 or 2 LED and 0…2 diodes in series) it would be best to test it (outside the flashlight?) with a variable power supply. So you can find out when the LED goes off by decreasing the power to lower levels - the current should go to zero if the LED don’t shine anymore.
You definitely can’t discharge the cell to a lower voltage level than the sum of the forward voltage of the diodes. The forward voltage in the data sheets is taken at the nominal current, the minimum voltage to lit up is lower! Unfortunately for this application there is no abrupte transision between “on” and “off” - the led will get dimmer more and more with lower voltage until the minimum is reached.
I quickly did some measurements to find this minimum for different LED’s:
With 4k7 resistor (a bit lower value could be better if the diodes claim more of the battery voltage for their own) in series:
white 5mm LED: below ~2.5V to shut off
green 5mm LED: below ~2V to s.o.
red 5mm LED: below ~1.6V to s.o.
2x red 5mm LED in series: below ~3.2…3.3 V to shut off
As I suggested, you should try 2 red LED’s in series to get best results:
- no discharge to critical levels by the tailcap feature
- optimal use of energy by two LED’s instead of one (losses in the resistor decreased)
- signaling of “low voltage” if the LED’s go dim/out
- one LED on each side of the switch for a more uniform illumination
The red XP-E2 starts emitting at about 1.7V, perhaps this is similar for a red 805 smd led (I have a few on order, I can test that ). Adding two diodes would make that 2.9V. If there are no further significant voltage eaters in the circuit (?), that could be quite ideal as a voltage indicator.
A little bit of the battery voltage is claimed by the 560 Ohm resistor on the driver. As I said, it would be best to take measurements with the actual setup.
Hmm so they actually shut off a bit below what the claimed Vf is?
I had actually planned to use 2 for more uniform light, but I was planning to wire them in parallel.
I think having them shut off at around 3.2v is ideal so that you still have some usable charge in the battery left.
I don’t have a variable power supply though, so if it varies much with the different set-ups it could take a lot of trial-and-error.
I’m planning to use 2 blue led’s for my first permanent installation, I assume you don’t have any of those to test?
Blue LED’s are similar to white ones - the little voltmeter modules I just bought are shutting off below 2.7V with the blue display. You could add one 1N4148 (or LL4148 for SOD80 package Redirect Notice) in series with the LED’s.
Are you fairly certain that would be the one I need? If so, i can order them on ebay for super cheap.
Yes, these are multi-purpose for smaller currents up to 200mA. They can be used on different applications, to reduce the voltage by 0.6 Volts, as rectifier or protection diode. But you could try almost any other Si diode first if you got some in your spare parts box or salvage it from an obsolate circuit and order them later when the setup is finished.
I don’t have many spare parts. The only diodes I have are LED’s, SS34 schottkeys, zeners, and the reverse polarity diodes used on all of wight’s drivers.
You could use one the zeners you have, just in the forward direction (rather than the usual reverse for zeners), its Vf will be ~0.7V just like a standard diode. Obviously the reverse voltage will be whatever the zener is rated at, but it won't matter as it won't see any reverse voltage.
Just a thought...
Do you think the standard reverse polarity diodes would work the same way?
Reverse polarity diodes are most likely Schottkys, too. You want a forward voltage as low as possible with these, something Schottkys deliver (~0.1…0.2V) what makes them unsuitable to get a notable drop in voltage. But the zeners should work as Crux said.
Ok. I’ll give the zener a shot when I get the 20mm boards (oshpark marked them ’shipped” today).
I was hoping to use the other ones because I have plenty of them, wheras I only have 2 zeners left.
EDIT: I want to save my zeners, and I’ll probably need more than one, so I just ordered these.
Yep, those will work fine.
Thanks for the tips guys. It’ll make this idea even better. Now there will be no worries about over-draining cells.
I also updated the OP to make it easier for others to try this.