Values For Run Time Calculations

I'm using LED Pro to calculate run times and am not getting results close to published data. I think I am using incorrect values for Battery Input Voltage and Battery Capacity.

What values would you use for TWOJ, DB917 & 2x123?

This is more of a question for you but Led Pro says 2hr 13min, what run time are you getting? I don't have this set-up but am curious.

hello there,

LED_Pro was only ment to give you an idea on runtime. There are too many variables to be really accurate.
Here are a few tips...

Here is the BIN code info.
http://home.comcast.net/~theledguy/bin_codes/index.htm

J Bin is 3.27V - 3.51V, so lets say 3.32V

2x123 you would think are 6V, but not really. Here is the variable that makes your runtime calculation go off the most.
Actually, if you just test a battery, they would read alittle over 3V each.
But, under load, the Voltage drops. And thats the actual voltage that matters in the runtime.
This is hard to compinsate for, sence you really dont know exactly how far it will drop.
Saying 6V for 2x123 isnt TOO far off though, seeing how they are over 6V total, when not under load, so, under load, 6V should be fairly close.

The rating for a 123 cell is about 1300 mAh, though I have heard some ppl say, in real life, its not that high.

Also, LED-Pro has an efficiency of 88%, I found that about 83% is alittle more accurate.

I havent had a chance to test it against real life runtimes yet /ubbthreads/images/graemlins/frown.gif

Hope that helped alittle, let me know if you have any other questions.

-John

Thanks John. I found a run time chart that included most of the relevant measurements. It's kj's chart for the Aleph w/ DB1000:

http://www.candlepowerforums.com/ubbthreads/showflat.php?Cat=&Number=630619&page=0&view=collapsed&sb=5&o=365&fpart=1&vc=1

I used your suggestion of 83% for the DB and the measurements listed by kj for the rest of the stuff. Came up with 800 mAh for the batteries.

If this sounds about right for real world numbers, I can start playing. What do you think?

Roguesoul, I'm not getting anything. I have the parts ready to assemble, but have seen some pretty short runtimes posted - 45 minutes or less. It caused me to investigate further.

kj's chart is looking good. I'm trying to establish a baseline on something so I can get a reasonable runtime estimate using a DB917. Looks like about 1:15 using John and kj's data and using LED Pro to get mAh.

lgl_olophart,

The 800 mAh sounds about right. Thats actually the exact rating someone told me the other day.

So, entering on 1000mA to the LED, 3.32Vf, 6V input from batts, and 800 mAh, your runtime was close to kj's chart?

Hmm, thanks that gives me a good comparison, I may be able to adjust for that, or at the least, make ppl aware of it.

If anyone else can varify LED-Pros runtimes against some of the charts here on CPF, that would be great.

-John

You may try reading datasheets and calculating the runtime by yourself.

Current to LED
===========
It should be 917 mA /ubbthreads/images/graemlins/smile.gif

Converter Efficiency
===============
It should be around 80% at that driving current. Usually, the efficiency will decrease as input/output increases. But it's hard for non-electronics experts (including me /ubbthreads/images/graemlins/wink.gif) to estimate the efficiency from the datasheet so I would just believe this number.

Vf of LED
=======
The Vf bin of your LED is J. According to Luxeon Product Binning and Labeling document (P.13), the Vf of your LED will be between 3.27 and 3.51 volts at 700 mA. (IIRC, the Vf on the datasheet is measured at 700 mA, junction temperature 25 C. Correct me if wrong.)

The Vf will increase as the current increases. Thus, you need to adjust the Vf because the driving current is 917 mA. As I see Luxeon III Emitter datasheet (P.8), the Vf will increase approx 0.1 volts at 917 mA. Because the average Vf of J bin at 700 mA is (3.27 + 3.51) / 2 = 3.39 volts, add 0.1 volts to calculate the average Vf at 917 mA and you will get 3.49 volts.

Required Power From Battery
======================
The LED consumes 3.49 volts x .917 amps = 3.2 watts. Assuming the converter efficiency is 80%, the batteries need to provide 3.2 watts / 80% = 4 watts. You are using two batteries so the power drain from one battery will be 2 watts.

Delivered Energy vs. Power Drain
=========================
The total energy, which a battery can deliver, will decrease as the power drain increases. According to Duracell's 123 battery datasheet, when the power drain is 2 watts, the total energy will be approx 2.6 Wh. (Please be aware that that is a log-log chart.)

Runtime
======
2.6 Wh / 2 W = 1.3 hours = 78 minutes.

As a rough estimation, I think it's not so bad /ubbthreads/images/graemlins/smile.gif

Very nice, kj. Thanks for taking the time to explain. Just what I wanted to know. Easy to understand, too. I'm feeling dangerous.