Hello.
I was experimenting the other day and found out that the maximum current of a single 9V battery is a bit less than 3amps. Then, if 2 batteries are in series, producing 18V, my guess is that the current will still remain with a max of about 3amps. But if the batteries are hookedf in parallell, my guess is that the max current will double.
Can anyone confirm or correct my guess?
Thanks

 

Yes, but 9 volt batteries are poor source of power unless they are freshly charged and only used briefly. Mind you the voltage will still only be 9 volts with two in parallel, and since each 9 volt cell is actually a battery in it's own right you can't put too many in parallel effectively. If you need this kind of current frequently even over the longer term you'll want a real power supply.

 

Also, the 9V battery terminal voltage has probably fallen to a very low value (if not zero) at 3A current load - due to the battery's internal resistance voltage drop.

 

The OP's original question hasn't been answered yet. I'll state that in series, the current should be about the same and in parallel, the current should about double. But that's the theory and you might find some interesting tidbits by actually doing the experiment.

You apparently have the facilities to make these measurements and it would have taken only a minute or so to make the actual measurements and confirm or reject your two hypotheses. Of course, since the battery might cost a buck or two, it might not be worth the money, as your tests will likely greatly reduce the subsequent life of the battery. However, if you have a programmable load (or can make something similar), it's not hard to only load the battery for a fraction of a second. Then you'd still have most of the battery life left. Or, even simpler, hook up a test circuit with a digital storage scope and capture the current discharge (e.g., using a shunt) for a fraction of a second.

As a typical alkaline 9 V battery only has a capacity of around 550 mA*hr, most applications draw far less than the 3 A you chose to draw. I would imagine the battery would only supply this current for a matter of seconds before it started dropping off (and it will get pretty warm because of the internal resistance).

 

Ahh, wait a second.

If the two batteries are in parallel, the current output will double, but the voltage will stay the same.

If the two batteries are in series, the current output will stay the same, but the voltage will double.

However, under such high load conditions, all bets are off. You're talking dead-short current, which you should not subject ANY batteries to, unless they are specifically designed for such heavy loads. The closest you might get to in off-the-shelf products available locally is automotive batteries. You can get literally hundreds of Amperes of current from them for short periods of time without damage, as long as you charge them back up right away.

 

Voltboy, is this an 'interest only' test, or do you have some wicked plan in mind?

I suggest that all bets are on - it is double and half - for the situation where you don't deplete the capacity or alter the chemistry of the battery - but keep in mind that no two batteries that you happen to pick up are exactly the same.

You can use a power FET and a 555 one-shot with CRO to do a good 'scientific' test, as per someonesdad post.

Its an appropriate test for sizing fuses and prospective current - but you would also need to do the test with an 'end-of-life' battery as well.

Ciao, Tim

 

Voltboy, is this an 'interest only' test, or do you have some wicked plan in mind?

I suggest that all bets are on - it is double and half - for the situation where you don't deplete the capacity or alter the chemistry of the battery - but keep in mind that no two batteries that you happen to pick up are exactly the same.

You can use a power FET and a 555 one-shot with CRO to do a good 'scientific' test, as per someonesdad post.

Its an appropriate test for sizing fuses and prospective current - but you would also need to do the test with an 'end-of-life' battery as well.

Ciao, Tim

Actually, theres some wicked plan behind, but maybe not of the type you're thinking of.
So lets say I short out the battery, or maybe through a 1-ohm load, is there a possibility a battery will explode if left that way? I dont mind the battery getting useless, I just need about 30 seconds or so of full power.

 

30 seconds is a lot to ask from a 9volt, but it all depends on your needs. What amperage do your need, over the time period?

Lipo batteries are more apt to deliver high currents over the range of thier capacity.

If you need 3 amps over 30 seconds, I'd be looking at paralleling a number of Lipo style cells.

 

Yes there is a possibility the battery can explode, but the 'explosion' character is very dependant on a lot of factors.

You can test the voltage and current on the battery over time, by logging, and using a shunt as load. Measuring external temp will give some info, but you probably need to set up an equivalent experiment to get a feel for internal temp (eg. dissipate same power profile into a 5W resistor and measure temp of resistor skin with a thermocouple).

Imho, you certainly wouldn't want to risk applying this form of loading to a battery inside a normal 'product'. You would need to consider restricting the ejection of 'guts', and containing the chemicals ejected, and determining if the chemicals change toxicity/corrosive risk (a la msds).

Ciao, Tim

 

A woman gave a speech to a crowd using a wireless microphone when she began jumping and crying. The spare 9V alkaline battery in her pocket was shorted by coins or keys and was burning her.

The high current is for only a few seconds because the battery internal resistance increases as it runs down so the voltage also runs down.