dealgrabber2002
Flashlight Enthusiast
If I have one of those cheap 9 LED light, if I remove 8 LEDs; with just 1 LED left. Does that increase runtime?
A question regarding multiple LED lights
- Thread starter dealgrabber2002
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Thr3Evo
Enlightened
If it's one of those 3xaaa lights then I think so since the last one I took apart was wired in parallel.
Remove one and see if it still works then go from there.
Remove one and see if it still works then go from there.
LEDninja
Flashlight Enthusiast
LEDs like to see 3.3V to 3.6V max. 3 fresh alkalines is 4.5V.
With the current spread over 9 LEDs each LED gets less allowing them to survive. With a single LED all the current goes through it and will cause the LED to have a short life.
With the current spread over 9 LEDs each LED gets less allowing them to survive. With a single LED all the current goes through it and will cause the LED to have a short life.
dealgrabber2002
Flashlight Enthusiast
Darn, thanks guys.
Justin Case
Flashlight Enthusiast
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- Mar 19, 2008
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- 3,797
As Thr3Evo writes, the multiple LEDs are probably in parallel. Also, these inexpensive lights are undoubtedly direct drive. So the parallel LEDs will all see the same voltage from the batteries, which are presumably 3xAAA. Your typical 5mm high brightness white LED draws around 25mA at 3.4V. If there are nine of these LEDs, the total current sent to the nine LEDs in parallel is nominally 9*25mA = 0.225A. Under that sort of load, a AAA alkaline might quickly droop to below 1.4V and hold above 1.2V for tens of minutes.
It looks like 3xAAA could deliver more than 3.4V to the 5mm LEDs for some appreciable amount of time. This type of LED probably can take a max of about 100mA forward current at 4V or so.
The issue isn't that "with a single LED all the current goes through it". It's not a constant current driver setup. What can happen as you remove more and more LEDs is that the 3xAAAs are under a lighter and lighter load. Thus, they suffer less voltage drop and deliver more voltage to the remaining LEDs. The current draw by the LEDs will be dictated by their Vf-If curves. At 3.2V, maybe the forward current is 20mA. At 3.4V, it's 25mA. At 4.0V, it's 100mA. Whatever voltage you send to the LED, it'll try to draw the corresponding amount of current. At some point that Vf/If draw will exceed the max specs and the LED will cook itself.
My guess is that with just one LED, 3xAAA will hold above 4V for a "long" time, and perhaps close to 4.2V-4.5V, and thus you risk frying the LED. The LEDs in those inexpensive lights are not heat sinked as far as I can tell because they normally don't draw much power and don't generate much heat. But if you start driving one of them at 4.2V and perhaps 150mA, you are looking at a lot of waste heat that the LED has to dissipate.
AnAppleSnail
Flashlight Enthusiast
Put a resistor in series with the batteries. Ten ohms should do.
AnAppleSnail
Flashlight Enthusiast
Unfortunately, these don't account for battery voltage droop or sag. Ten ohms will generally give about ten percent output and greatly increased runtime. One hundred makes a good night-marker light. Ten thousand a dim marker.
Illum
Flashaholic
heh, not even, more like 4.2V after loss, cheap 3AAA lights have quite a bit of linear resistance in its circuit
tam17
Enlightened
- Joined
- Jun 9, 2011
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True, but you need a baseline to start from. From my experience with homemade resistored 9V flashlights, ultra-long runtimes are achieved with very high ohm values, but it's also important not to fry the LED on first power-up :)
Cheers
