Electrical Math quiz for XML dive light design(help needed!)

Candle Power Forums

Help Support Candle Power:

BlackAngel-SCUBA

Newly Enlightened
Joined
Oct 19, 2012
Messages
23
City & State/Province
Lebanon, Pa
OK here is what im looking to do, i want a dive light that runs 100-80% brightness for 1.5 hours @1000 lumens because thats about how long my night dives are lasting right now. i have a tektite expedition star 175 lumen and its just not cutting it. also i want to stay with NiMH and not li-ion because i use C cells for everything and i dont like the huge cost of li-ion

i think this is the correct math but i forget if this is the right way to calculate it

Using: XML U2 LED
Kaidomain 8x7135 V2 driver
3x Tenergy premiums NiMH cells
Reflector? suggestions for tight beam

Power required to run LED: xml u2 LED 3.7V @ 3A......(3.7x3)= 11.1 watts
Power available at batteries: 3x tenergy premiums 5000mAH @1.2V......3x(5x1.2)=18 watts
Power available to LED: power available at batteries-power loss at driver: kd 8x7135 V2 @ 90%......18x0.9=16.2 watts
Approximate run time: = power available to LED/power to run LED......16.2/11.1=1.46 hours
the only question i have is that with combined 3 C cells its only 3.6V, am i not driving the LED to 1000 lumens then?:banghead:
 
Nice topic start!

Few things to consider:
- an XM-L U2 usually runs at a lower voltage than 3.7v (more like 3.3v)
- The 7135V2 driver is a linear driver, the loss is not fixed at 90%. Therefor, the power calculations are not needed and incorrect.

In essence, what the driver does, is limit the amount of current going through.
Current IN = Current OUT

Instead of buck driver which have:
Power IN * efficiency = Power OUT

This means that as long as the power source can supply the needed voltage (3.3v of the led + 0.1v drop for the 7135 chips or so), the current will be maxed. If the power source can't supply enough current at the required voltage of course, the output will drop.

So assuming the batteries can supply 1.2v each loaded (which is more like 1.35v when fresh and 1.2v for a long time and then dropping), the equation will be as follows:
5000mAh = 3000mA * 1.6h
with 3.6v in and 3.3v out with means an efficiency of 91.7%

If you deside to use four cells, the result will be almost the same. The burntime on maximum brightness will only slightly increase.
5000mAh = 3000mA * 1.6h
4.8v in and 3.3v out means an efficieny of ~69%
 
As an Amazon Associate we earn from qualifying purchases. Product prices and availability are accurate as of the date/time indicated and are subject to change.
Epsilon and Justin Case, thank you for putting time into your very thorough threads. they have helped a lot. now with a few more threads else were for answers ill be on my way to completing my build
 
Back
Top