# mah=amps/ formula



## bobhere (Nov 5, 2006)

anyone know of a thread that explains amp mah volts etc i want to know how to figure out the amp/mah specs when i look at a bulb/power solution..
exp..bulb says 6v 3.2amp so its drawing 3.2 amp per hour is that right? then the figure would be 3.2amp/ 100 to figure the mah? then i would have to figure my voltage etc.. trying to learn lots of knowledge in lil time here


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## TMorita (Nov 5, 2006)

mah is milliampere-hour(s).

If you draw one milliamp for one hour, then that's one milliamp-hour.

If you draw two milliamps for two hours, then that's four milliamp-hours.

If you have a bulb that says "3.2 amps" then it's drawing 3.2 amps, and in one hour it will draw 3.2 amp-hours, or in two hours it will draw 6.4 amp-hours, etc assuming it hasn't burned out.

Toshi


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## DonShock (Nov 5, 2006)

In your example, it is drawing 3.2 amps at all times: per second, per minute, per hour is not an issue. I think what is confusing you is when you see batteries marked in mah. That's just the way they are expressing the total power contained in the battery. So, if you have a 1600 mah battery supplying your example's 3.2 amp bulb (3200ma) then it would last about 1/2 hour.


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## Handlobraesing (Nov 5, 2006)

bobhere said:


> anyone know of a thread that explains amp mah volts etc i want to know how to figure out the amp/mah specs when i look at a bulb/power solution..
> exp..bulb says 6v 3.2amp so its drawing 3.2 amp per hour is that right? then the figure would be 3.2amp/ 100 to figure the mah? then i would have to figure my voltage etc.. trying to learn lots of knowledge in lil time here



Ah consumed is simply ampere x (time in hours), hence ampere-hour. 

As far as batteries, current drawn and duration are perfectly inversely proportional on paper, but in real life, you lose usable mAh. 

If you have 2500mAh NiMH batteries, it means it will deliver that much Ah capacity if the energy is extracted over a five hour period, so in this case, 500mA x 5 hrs = 2500mAh. The five hour rating method is just something used by the industry. 

In theory, if you were to increase current draw to 5000mA, the discharge duration would be 0.5 hours. In rea life, the higher the discharge rate, lower the usable capacity.

Standards are different for alkaline batteries. Energizer rates their AA alkaline battery at 2.85Ah at 0.025A or ~114 hour rate (that is, energy used up evenly through a 114 hour period).

It delivers 2.85Ah at a current draw of 0.025Ah before voltage drops to cut-off point. 

At 0.5A current draw, it's usable capacity is somewhere around 1.4Ah before cut-off point. 

2.5Ah rated NiMH on the other hand delivers full 2.5Ah at 0.5A(1Ah at 0.2A, 5Ah at 1A, and so forth) current draw assuming the battery is fresh off the charger.


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## leduk (Nov 5, 2006)

I don't think anyone has explicitly stated yet.

There are 1000 miliamps in an amp. Mili means one thousandth.
1000 mAh in 1 Ah (amp hour).

Cheers


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## monkeyboy (Nov 5, 2006)

Lets clarify what some of these quantities actually mean with regards to an electrical circuit. I find that following words are frequently misused or confused with each other. 

1) *charge*: This is a fundamental electrical unit relating to the total number of charge carriers. SI unit = coulombs.

2) *current*: this is the rate at which these charge carriers pass a certain point in the circuit. SI unit = amps = coulombs per second. Therefore coulomb = amp second

3) *energy*: if 1 coulomb of charge flows through a potential difference of 1 volt then it gains/outputs 1 joule of energy.

4) *power*: the rate of energy output. SI unit = watt = joule per second. (from the above relations you can see that power = current x voltage)



The (m)Ah rating on batteries is just a non-SI unit of *charge* capacity (1Ah = 3600 Coulombs). To calculate the *energy* stored in the battery you must multiply this by voltage to give VAh = Wh ( 1Wh = 3600Joules)


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## bobhere (Nov 6, 2006)

still not gettin it..if a cell puts out 5000 mah it would take 5 amps to kill the cell over ? 5 hrs? dont seem right.. Bi-Pin WA 01111 6.0v 3.35a how many amps does it pull in an hour? and if i have a cell thats rated at 3350 mah how long will it last? im gonna have to get a book huh?


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## bobhere (Nov 6, 2006)

ok heres one 5 C cells rated at 5000 mah they are 1.2v.. thats 6v total right? so these will power that bulb right? now how long will it last? the bulb is pulling 3.35a right? there must be an easy explanation


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## DonShock (Nov 6, 2006)

Yes, 5 cells x 1.2v each = 6v total, which is what the 1111 bulb needs to run at specification. And it will pull 3.35a when running at that voltage. If you are feeding it with five C cells rated at 5000 mah each, your capacity is still only 5000 mah since the current is pulled from all cells equally. So, a capacity of 5000 mah / 3350 ma current being drawn = 1.5 hours run time (approximately).


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## iamerror (Nov 6, 2006)

DonShock said:


> Yes, 5 cells x 1.2v each = 6v total, which is what the 1111 bulb needs to run at specification. And it will pull 3.35a when running at that voltage. If you are feeding it with five C cells rated at 5000 mah each, your capacity is still only 5000 mah since the current is pulled from all cells equally. So, a capacity of 5000 mah / 3350 ma current being drawn = 1.5 hours run time (approximately).



In your example, is the 1.5 hour runtime you stated mean until 50% brightness? Or would it mean you get less than 1.5 hours? (because you can't run the batteries to zero)


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## Handlobraesing (Nov 6, 2006)

bobhere said:


> still not gettin it..if a cell puts out 5000 mah it would take 5 amps to kill the cell over ? 5 hrs? dont seem right.. Bi-Pin WA 01111 6.0v 3.35a how many amps does it pull in an hour? and if i have a cell thats rated at 3350 mah how long will it last? im gonna have to get a book huh?



No. If you have x A * y h, the Ah is simply x*y. If you have 5,000mAh, it would take 1A x 5 hrs to kill it.

Here's another way to look at it. 

Ah = capacity 
A = flow rate

3.35A for one hour is "3.35 ampere-hour" or 3.35Ah. 

The 01111 may draw 3.35A, but that's when voltage across the bulb is exactly 6.00v. NiMH usually start around 1.4v fresh off the charger and its considered "end of useful charge" at 0.8 to 0.9v and current varies during useful charge. Even if the bulb was a constant current load, the real mAh the battery can deliver depends on current relative to its capacity.
Unfortunately, the simplest way to figure the real life runtime is by experiment.


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## bobhere (Nov 6, 2006)

ty thats a lil more help..see im trying to figure recipies for mag mods but i can not have any 20 min run times...that just wont work for me...i dont like shelf queens..i buy em and USE em..


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