What does it mean that the battery I have is rated to discharge at a 20 amp/hour rate ?

I have a 12 volt 5 amp/hour sla battery. It says it is rated at a "20 hour discharge rate".. What does that mean. I think that means it is the shortest amount of time the battery could be discharged before damaging the battery. Is that right ?

Battery capacity in amp*hours is the product of current in amperes and time in hours. Please, please, stop using a division operator, "/", and if you are using the equivalent phrase "amps per hour" when you speak or think about this quantity, please stop doing that too. The underlying operation is multiplication not division.

Also charge/discharge rates are given in units of amperes, or milliamps. For example the "20 hour discharge rate" for your 5 amp*hour battery is: (5 A*hour)/(20 hour) = 0.25 A = 250 mA

You might wonder how such a quantity can qualify as a rate, when it does not explicitly contain a per-unit-time. The answer to this is that electric current really is a rate. One ampere is one coulomb of charge per second. 1 A = 1C / s

In fact, if you wanted to, you could specify the capacity of your battery in coulombs, since: 1 A*hour = (1 A)*(3600 s) = (1 C/s)*(3600 s) = 3600 C

But for whatever reason, battery manufacturers all over the world use the ampere*hour, as the unit to specify battery capacity.

Anyway, the basic model for battery capacity says that you've got some quantity of charge, some amount of current*time, and, within reason, you can pretty much discharge the battery as fast, or as slow, as you want to.

For example: at a 10 mA rate: 10 mA * 500 hours = 5 A*hour at a 50 mA rate: 10 mA * 100 hours = 5 A*hour at a 100 mA rate: 10 mA * 50 hours = 5 A*hour

However this model, just multiplying two numbers together, is naive. The ugly truth is that for higher discharge currents, the battery time is not proportional. That is to you cannot expect to get:

1A * 5 hours = 5 A*hours

That just wont happen. The battery will supply 1 ampere, but it won't be able to sustain that for 5 hours. Instead the battery will be exhausted in maybe 2 hours, or 1 hour.

Another way to say this is that you get less battery capacity, less total charge delivered, if you discharge the battery quickly. To account for this effect, battery manufacturers will quote a capacity for a some total amount of discharge time. E.g.

Capacity at a "40 hour discharge rate" is 6 A*hour. Capacity at a "20 hour discharge rate" is 5 A*hour. Capacity at a "10 hour discharge rate" is 2 A*hour.

I just made those numbers up, but the trend is always the same, a quick (short time) discharge gives you less battery capacity (in A*hours).

Regarding the question of how fast of a discharge rate has to be to actually damage to the battery; i.e. how much current is too much, I don't know the answer to that question. If you see smoke, or smell burning plastic, that's probably bad. ;-)

No, its how the "5Amp hour" bit is derived. So although the battery has a CAPACITY of 5Ah, you'll only GET that if you pull it out at 5/20, or 250mA, faster, and the capacity is less.

it means that you should not use the batery to a circuit or device that has a total resistance lower than 20 ohms with a continuous use. else it will lower the battery life or even damage it

Basically, 5Amph means that it can give a current of 5 amps for an hour, before it 'dies'. But the real battery can not give you such currents, because it will heat, or get spoiled or whatever..., however you can trust the 5Amph capacity on currents smaller than 250mA. Hope it's logical :)

active| newest | oldestamp*hoursis the product of current in amperes and time in hours. Please, please, stop using a division operator, "/", and if you are using the equivalent phrase "amps per hour" when you speak or think about this quantity, please stop doing that too.The underlying operation is multiplication not division.Also charge/discharge rates are given in units of amperes, or milliamps. For example the "20 hour discharge rate" for your 5 amp*hour battery is:

(5 A*hour)/(20 hour) = 0.25 A = 250 mA

You might wonder how such a quantity can qualify as a rate, when it does not explicitly contain a per-unit-time. The answer to this is that electric current really is a rate. One ampere is one coulomb of charge per second.

1 A = 1C / s

In fact, if you wanted to, you could specify the capacity of your battery in coulombs, since:

1 A*hour = (1 A)*(3600 s) = (1 C/s)*(3600 s) = 3600 C

But for whatever reason, battery manufacturers all over the world use the ampere*hour, as

the unitto specify battery capacity.Anyway, the basic model for battery capacity says that you've got some quantity of charge, some amount of current*time, and, within reason, you can pretty much discharge the battery as fast, or as slow, as you want to.

For example:

at a 10 mA rate: 10 mA * 500 hours = 5 A*hour

at a 50 mA rate: 10 mA * 100 hours = 5 A*hour

at a 100 mA rate: 10 mA * 50 hours = 5 A*hour

However this model, just multiplying two numbers together, is naive.

The ugly truth is that for higher discharge currents, the battery time is not proportional. That is to you cannot expect to get:

1A * 5 hours = 5 A*hours

That just wont happen. The battery will supply 1 ampere,

but it won't be able to sustain that for 5 hours. Instead the battery will be exhausted in maybe 2 hours, or 1 hour.Another way to say this is that you get less battery capacity, less total charge delivered, if you discharge the battery quickly. To account for this effect, battery manufacturers will quote a capacity for a some total amount of discharge time. E.g.

Capacity at a "40 hour discharge rate" is 6 A*hour.

Capacity at a "20 hour discharge rate" is 5 A*hour.

Capacity at a "10 hour discharge rate" is 2 A*hour.

I just made those numbers up, but the trend is always the same, a quick (short time) discharge gives you less battery capacity (in A*hours).

Regarding the question of how fast of a discharge rate has to be to actually damage to the battery; i.e. how much current is too much, I don't know the answer to that question. If you see smoke, or smell burning plastic, that's probably bad.

;-)

at a 10 mA rate: 10 mA * 500 hours = 5 A*hour

at a 50 mA rate: 50 mA * 100 hours = 5 A*hour

at a 100 mA rate: 100 mA * 50 hours = 5 A*hour

Steve

Hope it's logical :)