I have recently bought a new (Chinese) 3 LED (XM-L T6) bike light which came with an 8.4V 3200mAh (allegedly) battery.
I have measured the current and voltage between the battery and the light driver:
Battery Voltage with No Load – 7.57V (Note Battery Not Fully Charged).
Low Power – 0.33A @ 7.29V
Medium Power – 0.74A @ 7.11V
High Power – 1.21A @ 6.83V
My question is, how can I determine how much current and voltage each LED is getting?
I am unsure if the LEDs are mounted in Series or Parallel either.
Are there any calculations for these values?
I am not great with electronics and from what I can see, there is a microchip (unlabelled), an inductor, many resistors and many capacitors.
Im guessing the light uses a Buck driver with the leds in parallel. Giving the leds about 1amps at high. The amps are a little low to be a boost driver with leds in series. Best way to make sure is by opening the light and getting your DMM out. Im not a mathwizard myself but im sure someone else will give you some more help
Without pictures showing if the LEDs are series or parallel we’d be guessing. With losses it could be ~1A to each emitter for buck/parallel. Best would be to unsolder one of the emitter wire and put a DMM in the circuit there.
If the leds are in series the required voltage would be 3.3ish x 3 = 9.9v
The battery pack is pretty clearly 2S
As bigger (non AAA or AA sized) boost drivers are unusual and probably cost more it is unlikely one of them. Especially when in a bike light long runtimes are typical.
If the leds are parallel then 3.3v for the group but amperage is the total of all three
Then it is a buck driver (funny because it seems like a good place to run the battery pack in parallel) I guess they are running cheap cells and need two to keep the voltage up.
So your leds are in parallel as noted above.
“My question is, how can I determine how much current and voltage each LED is getting?”
Well the specs are clear but it depends on what you mean by “is getting”. (I feel a bit like Bill Clinton).
first fully charge your cells and then take you top amp reading again and divide by three (3). that will give you a before losses to that point consumption of amps per emitter. (if you can take it right from the led board that is best, read fewest other losses possible)
Oops! Missed the 3 led part, thought it was two. Without a meter all you can do is guess but figuring a buck driver with losses around 20% then wattage out equals wattage in - 20% and each led gets 1/3 of the wattage out. Wattage in on high is x 6.83 x 1.21 = 8.26W
-.2 x 8.26 = 6.61W
/3 = 2.2 W per led
A look at the cree data sheet Vf vs current for the xml2 puts that a bit under 2.9V and a bit under 800mA. Not really accurate as Vf vs current charts are only gross approximations but that’s what a guess looks like. At 3.8V per cell they are about half full. If well regulated you won’t see much more than this in output
Thank you for that information, it is helpful but also a little confusing.
I now understand that power out is equal to roughly 80% of power in, therefore as you said, 2.2W per emmitter.
How did you get to the 800mA output? Is the 2.9V an educated guess?
Secondly, if I have an 8.4V battery with a 1600mAh capacity, is the following calculation for run time?
Battery = 8.4 x 1.6 = 13.44 Whr
Driver = 6.83 x 1.21 = 8.26 W
Runtime = 13.44 / 8.26 = 1.63 hr
I presume this should all be done with the battery at full charge to make different calculations as the LED voltage may change? So maybe this is more accurate.