MrCjolsen

Ok, so I'm still learning. I want to make my own light.

If a battery is 12 volts and rated at 5 amp hours, then ...

12x5=60. Therefore, such a batter will power a 60 watt light for 1 hour, a 30 watt light for 2 hours etc.

And that means until the battery is totally dead. I really only want to power down the battery no more than 50%.

So if I'm using a 12 volt, 35watt light, and have a 10 ahr battery, then ...

12x10=120 120/35 = 3.4 3.4/2 = 1.7

So a 35w light powered by a 12v 10ahr would give me about 1.7 hours of light.

Am I right, or should history majors stay away from electricity.

cyccommute 

That's correct

A battery is dead when you reach the minimum cell voltage for the pack. The minimum cell voltage is dependent on the cell chemistry. Ni chemistry batteries have a minimum cell voltage of 1V per cell. For a pack of 12 cells (a 14.4V battery) the minimum pack voltage would be 12V. Below that and you risk damaging the cell.

I'm not sure what the minimum cell voltage for Li chemistry batteries is but they are protected against going below a threshold voltage by an internal circuit. That's because you can form lithium metal which is vigorously reactive with water and tends to burn

But capacity...the Ah rating...of the battery is figured down to that minimum voltage. If your light dims, you are at the minimum voltage and should turn the light off. For lead chemistry, it's very important to stop using the battery when the lights brown out. Further usage will kill the battery. Ni chemistries can go a little lower than minimum but don't do it all the time. Li...never!



The formula you can use is


where H = hours, Ah = amp-hour rating (divide by 1000 in most cases), V = voltage of the system, and W = wattage. This will give you an estimate of the run time for a new battery. You can still be about 10% off depending on a number of factors. Consider this number to be the 'best' you can get and plan accordingly.

Your 1.7 hr estimate is correct, by the way.

MrCjolsen

I'm generally looking at lead acid batteries due to the cheap factor.

Ngchen

You're right, although you didn't do it in the order that I would have done it :-).
You have a 5 Amp-hour battery (ignore the volts for the moment), which means that theoretically if you drew 5 amps from it, it would last 1 hour. 10 amps, 1/2 hour, and so on. (In practice batteries can put out only finite loads, you can't get 1,000,000 amps for 0.000005 hours (0.018 s), but that's quite extreme).

So where does voltage come in? Voltage dictates how "energetic" each amp you get is. Your light will only tolerate electricity at a narrow range of voltages. Put too much, and it burns out (sometimes instantly). Put too little, and you get basically no light. Your light is rated a certain power (how much energy is drawn from the battery/unit time). Power is voltage*current. So a 12 V battery has twice as much stored energy as a 6 V battery, if they have the same amp-hour rating. However, the voltage you can feed your bulb is dictated by the bulb (see above). In practice, batteries are formed by chaining cells together, and for the same kind of cell, the 12 V battery needs twice as many cells (hence twice the volume and weight) as the 6 V one for the same Amp-hour rating (can't get something for nothing).

Hydrated

You calculated well, Gwasshoppuh...

But we engineers would still prefer that history majors stay out of our backyard!

cyccommute 

Lead acid does not tolerate deep cycling. If I were to use it (but I wouldn't), I'd calculate the run time and subtract 20%. I'd then be careful not to go below that 20% time. If you have to, you can go as low as 10% but be very careful. Carry a back up light

MrCjolsen

It just so happened that last night I scored (for $25) a cygolite night rover just like the one I bought a few weeks ago (for a lot more than $25). The batteries and chargers are entirely interchangeable. So now I have a backup battery and I can ride with both bulbs lit for my whole commute. Half way through, I just disconnect one battery and reconnect the other.

Mr. Fly

I would add that the capacity rating of rechargeable cells are measured at a relatively low discharge rate – typically 0.1C (C=capacity, not coulombs). That means your 10Ah battery is really only 10Ah when discharging at a rate of 0.1*10 = 1A, or lower. In your hypothetical case, this translates to observing only the rated battery capacity when running a 1A*12V = 12W lamp. Higher discharge rates will lower effective capacity.

Thus, in reference to your calculation, I would expect shorter than 1.7 hours to the halfway point.