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battery capacity calculations metal nickle halide

I

ilaboo

Jan 1, 1970
0
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
 
B

Bill

Jan 1, 1970
0
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).
 
T

Terry Pinnell

Jan 1, 1970
0
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).

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.
 
I

ilaboo

Jan 1, 1970
0
Terry said:
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.


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
 
J

John Fields

Jan 1, 1970
0
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
 
T

Terry Pinnell

Jan 1, 1970
0
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

Indeed!
 
R

Rich Grise

Jan 1, 1970
0
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

Not only that, but it's surprising that the LED has not already been
destroyed - or is this a 1.2V cell and red LED?

One thing about LEDs, or any diode, for that matter, is that it has a very
low dynamic resistance - very little change in voltage can cause a much
greater change in current.

Please tell us everything you know about your battery, your circuit, your
LED - we aren't mind readers, after all! :)

Thanks,
Rich
 
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