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Battery discharge computation

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scripted13

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please good sirs, i need to find out on how to compute on how long will a battery last, specifically a 9v zinc-carbon battery if it will give out 6.3mA? thanks in advance
 
AudioGuru is the forum expert on this...

According to Wikipedia, a Zinc–carbon 9V battery (6F22, 1604D) has a capacity of 400mAh. You will be lucky if the Chinese-made batteries come close to this. Will your device work on 5V? That is what the 400mAh end-of-life is based on...

If the discharge rate is a constant-current of 6.3mA, divide 400mAh by 6.3mA. If the load current decreases as the battery voltage drops, it will last a bit longer.
 
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Use Cah triangle C being capacity (in this case 9v) a being amps (in this csae 6.3ma) and h being hours. So we divide 9v by 6.3ma wich is 1428.6 your battery should last 1428.8 hours.
 
... your battery should last 1428.8 hours.

Is this new math? You divide the battery capacity (400 mAh) by the load current (6.3mA) to find the run time... , about 65 hours
 
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I have not seen a 9V carbon-zinc battery for about 40 years. I have used 9V alkaline batteries for 40 years.

When you need spec's for a battery then why don't you look at the datasheet on a website of a battery manufacturer??
Energizer's website has a Technical Info button that shows detailed datasheets for all their batteries and even for their obsolete batteries.

They do not show 6.3mA but they show a 620 ohm resistor load that is 14.5mA when the battery is new. The battery will last longer than my graph. 14.5mA/6.3mA= 2.46 times longer.
 

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Use Cah triangle C being capacity (in this case 9v) a being amps (in this csae 6.3ma) and h being hours. So we divide 9v by 6.3ma wich is 1428.6 your battery should last 1428.8 hours.

Hi,

Wow, where did you get that battery? I'd like to get one of those for sure! :)

On the more serious side...

The rough estimate is based on the Ah (or mAh) capacity of the cell and the load current as you noted, but when working in mAh and mA load just divide the mAh rating by the load current in mA, and that gives you the run time in hours.
So to make this simple a few examples:
400/4=100 hours (4ma load)
400/40=10 hours (40ma load)
400/400=1 hour (400ma load)

However, that 400mAh rating is probably given at the 20 hour discharge rate, so anything shorter than 20 hours will show less capacity and anything over 20 hours will show more capacity.

The 20 hour discharge current would have to be:
400/20=20ma

so for a 10ma load we estimate:
400/10=40 hours

and it will actually be a little more then that. It's hard to say exactly how much without having at least two discharge curves for the battery.

But another key factor is what the minimum voltage the application load can take without failure or going into undervoltage mode. If the load can take a serious undervoltage then you're good to go, but if the load can only take a small decrease from the starting voltage of around 9 volts, then the run time could be much, much shorter. By much, much shorter i mean as low as 25 percent of the estimation above. That's because the manufacturer uses it's own idea of what the *ending* voltage should be at the end of the 20 hour discharge period used to rate the battery. If they chose a high voltage that's good, but if they chose a low voltage then the run time will be much shorter.

The best bet is to buy a few new batteries and test it. Run it at the mode you intend to run it at and see how long it stays 'on'. If you intend to use a different brand of battery later, you may have to repeat the test to be sure because different manu's use different techniques to rate the battery. Of course this is if the application is critical in some way.
Also, for critical apps, you should always have some more batteries on hand just in case the first one happened to have much lower capacity than the rest for some unknown reason like shelf life or accidental discharge, etc.
 
Why use an obsolete 9V "carbon-zinc" battery (from China?) that has a short shelf life but might have been on the shelf for a long time before being sold?
Instead why not use a modern Energizer or Duracell alkaline battery that are now guaranteed to last 10 years on the shelf?? Also their working lifetime is much higher.
 
AudioGuru said:
... Instead why not use a modern Energizer or Duracell alkaline battery that are now guaranteed to last 10 years on the shelf?? ...

Probably for the same reason I buy cheap ones; that you can buy a packet of 10 cheapies in the supermarket for much less than a dollar a battery, while good brand-name alkalines cost $4 a battery.

If you have a device using only a few mA (like the OP's 4.3mA) a cheap one will last plenty long enough to be cost efficient and an expensive one might just die from shelf life or corrosion before you get your money's worth.

(Sorry to hear about your lifetime ban on AAC too A.G, everyone is missing you)
 
I use Energizer reliable 9V batteries in my smoke detectors and in my Fluke multimeter that use an extremely low current. I would never use a cheapo battery in one.

AAC has too many little school kids as "members". It is boring when every kid in the class asks the same questions over and over in a forum. I told one kid to ask his teacher and I reported that a circuit in their tutorial was wrong then I was banned.
 
ok..here's another question:

We know that Lead acid batts are rated in RC minutes @ a constant 25A discharge to 10.5V.
I'd like to expedite that a bit...so what's the terminal voltage for 1/2 RC @ 25 A?

Digging around in discharge curves etc...seem to indicate about 11.8V might be the magic number.
I'm not looking for great precision just a workable number +/- 10%.
thx
 
AAC has too many little school kids as "members". It is boring when every kid in the class asks the same questions over and over in a forum. I told one kid to ask his teacher and I reported that a circuit in their tutorial was wrong then I was banned.


I know exactly how you feel AG. One gets sick and tired of repeating the same stuff over and over to people that will not learn....

For you it is Internet advise. For me it is the guy I work 0.5M away from. Every working day of my life.

Gets to a person eventually and you feel that everything you have tried to teach is a waste of your time.

Glad you are back here. Stay strong. And keep us all on top of things with your no nonsense Technical approach. We are, after all, a Technical Forum first and foremost. Not a playground.

Had a quick look at AAC today after reading about your stuff. I have been a member there for a while. Never posted anything though. They seem very relaxed with everything. Threads can drift off topic etc...

Different here.

It's like....ETO gives facts. Always. And I like that.

Members lounge here is the place I relax. No Technical repercussions.

ETO Forum though is another thing. I know that whatever I post there must be bang on target and accurate. Otherwise some knowledgeable member will shoot me down. Properly.

I like that too. If I am Technically wrong with anything I post, I need to know. So I can fix and learn.

You see now why ETO is my home. Still. Even after AAC people wanted me to post there. Not interested.

From what I saw today......I will stay here. I thought I had issues with things/people/stupidity etc :p
I am not alone. Some Members that never show their Emotions here....willingly post their Emotional issues there.

Now that is interesting....Forums....every Posters individual choice where they let off steam.

We all have it. It is how Mods deal with it that either makes or breaks a Forum.

Fortunately, Mods here are not Kids. Eric and Nigel and Ian know stuff. They have been around the block a few times.

Before they punish.....they think.

And that my friends makes the World of difference ;)

That is why ETO is my home.

Regards,
tvtech
 
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Mosaic;1114141 [quote said:
We know that Lead acid batts are rated in RC minutes @ a constant 25A discharge to 10.5V.
I'd like to expedite that a bit...so what's the terminal voltage for 1/2 RC @ 25 A?

Digging around in discharge curves etc...seem to indicate about 11.8V might be the magic number.
I'm not looking for great precision just a workable number +/- 10%.
thx

You can't get there with high precision, but you might get close if all the batteries are the same. Take a look towards the end of this thread.

https://www.electro-tech-online.com/threads/automated-load-tester.125069/
 
I thought there was on why you couldn't if the batteries were different. Maybe I forgot.:eek:

Anyway if you always test the same size battery you could come close by using amp hours instead of RC.

The reason for the difficulty is that the RC method is only speced for a "dead battery" and takes into accout the perket number.

I think this post summarizes it towards the end.

https://www.electro-tech-online.com/threads/automated-load-tester.125069/#post1043627
 
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I agree what the other posters have advised; that the best way to obtain data from an electronic device, is to read the datasheet.

And precisely that my friends, is one of the biggest differences between a brand-name component and a cheap Chinese knockoff.

A good vendor will statistically sample many components during its initial runs to both validate the design and to validate the datasheet's published specifications. And then it will test to those specifications, ensuring that the components you purchase will work very similarly today, tomorrow, and five years from today.
 
Use Cah triangle C being capacity (in this case 9v) a being amps (in this csae 6.3ma) and h being hours. So we divide 9v by 6.3ma wich is 1428.6 your battery should last 1428.8 hours.

I think someone should do some after school maths and electronics lessons ;)

*edit - actually the maths was shockingly correct, the electronics theory was somewhat lacking :p
 
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