battery capacity calculations metal nickle halide

Nov 21, 2004 6 Replies

really confused about this rating



i am using a 3,500 mah c cell to powe a 20 ma led



how may days ( about--i understand the fall offf of volatage) can i run the led--i amusing a vanson smart charger to charge the c cells (2) in series.



i would think the calculation would go like this


3500/20 == how many hours the led should light


((3500)/20) /24 how many days the led sould be lite



i get about 5 days running the led right now after fully charging the cells--about 4 hours



what is wrong with my calculation



tia peter



Most cells are rated based on a 10 hour rate which means if your cell is rated for 3500mAH, you should be able to draw 350mA for 10 hours. In my experience, when you draww less than the 10 hour rate you usually get more than rated capacity. Since you say you are only drawing 20ma, your calculations appear to be correct and come out to a little under 7.5 days and I would expect a little more than this based on what I said above concerning more capacity at loads under the 10 hour rate. If you are not getting that then you must either have more current draw than you think you do or the battery is not delivering rated capacity (which would not surprise me).

What is the rated voltage drop across the LED with 20mA going through it?

The OP didn't describe the circuit (single LED? series resistor?).So another reason for the apparent discrepancy could be that the voltage across the battery is falling below the level at which the LED illuminates.

Terry Pinnell Hobbyist, West Sussex, UK

circuit is simply a led across the + and _ pole of the battery i think what is happening is the voltage is dropping such that the led cannot light

thanks for all the info

ilaboo wrote: [...]

formatting link
That is a small device which should keep you going some longer... Good luck with it:)

Indeed!

Terry Pinnell Hobbyist, West Sussex, UK

Join the Discussion

Have something to add? Share your thoughts — no account required.

Didn't find your answer?

Ask the community — no account required