How to calculate the Ampere Hour Ratings?

Hi, I have another one doubt, I have made some gadget, It draws 60 mA, 5 volts, I'm using 5AH battery so How to calculate the ampere hour for this. And also, give me exactly battery working time.

Thank you.

Your battery has 5 Ah, that is 5000 mAh. Divide 5000 mAh by 60 mA, you get 83 hours

EXACT working time is not possible. The above is a calculated estimate.

This is because the battery may have partly discharged due to age or bad storage.

Weedpharma

Ok, Now I having a doubt suppose my load is 12V, 60 mA, but I have 6V battery but 4AH battery what will happens.

And When I use a sensor system with arduino uno and Ethernet shield with 12 volt battery it gives different answer, and when I use 6 V battery it gives some other answer. Here I did not give separate power supply for Ethernet shield and arduino and sensors.

While give 12 volt also I didn't convert it to 5 volt, but UNO didn't burst why.

Venki: While give 12 volt also I didn't convert it to 5 volt, but UNO didn't burst why.

Assuming you supplied the 12V via the DC jack plug on your UNO board, there is a built in voltage regulator to convert to 5V.

If you have a multimeter or ammeter (amp meter), you could measure the current taken by your whole project by measuring the current being taken from the battery. To get the mAh, multiply volts and amps. Apply a nominal factor of say 75% of useable battery capacity would give a reasonable guess at your battery life.

Regards,

Graham

ghlawrence2000: If you have a multimeter or ammeter (amp meter), you could measure the current taken by your whole project by measuring the current being taken from the battery. To get the mAh, multiply volts and amps. Apply a nominal factor of say 75% of useable battery capacity would give a reasonable guess at your battery life.

Graham

but, while giving 4Ah 6 volt battery and 12 volt battery why analog readings change can you suggest me about this

The analog input readings are based on Analog Reference. The Default is 5V . When you use a 6V battery, the onboard regulator has at least a 1.5V headroom and your 5V pin on the UNO is probably reading slightly less that 5V , which would account for the difference in analog readings. You need to post a schematic, or tell us what your not telling us. Draw a schematic on a piece of paper and take a photo of it and post that. Don't give us any more information without giving us the voltage on the UNO 5V pin because without that the rest of it is meaningless. The Ah calculation is straight forward. The reason you don't know how to convert it is that you don't know Ohms Law. Think about it;

60mA @ 5V => 0.060A @ 5V => P = I x V = 0.06 * 5 = 0.30 W or 300 mW.

If you "load" (why are you NOT telling us WHAT your LOAD is ? !) is 0.30W, At 6V => P = I x V => I = P/V = 0.30W/6= 0.05 A (50 mA)

4 Ah = 4000 mA / 50 mA => 80 hrs.

Is that so hard to understand ?

It is 30 years ago since I did the dc circuit theory and equivalent circuits and stuff at college, and I have been to bed since then. So somebody else might like to step in with actual facts and figures?

But a general explanation is because the voltage regulator burns off the extra voltage as heat..... you want 5V, you are giving it 12V, your circuit is drawing 60mA, you are wasting 7V*60mA = 420W, but when you are supplying your circuit with 6V, you are only wasting 1V*60mA=60W.

That will have a measurable effect over several hours on the life expectancy of your fully charged battery.

Hopefully grumpy mike is watching. ;) :D

Regards,

Graham

Edit: mAh edited to W. Thanks to alnath! ;)

This could actually become unduly complicated if you wanted it to...... You could factor in the type of battery you are using. For example a 12V lead acid battery may easily start off freshly charged at 13.8V rapidly falling to 12.4-12.6 for quite a while, before starting to taper off towards fully flat at approx 10.8V is considered flat for a 12V lead acid battery. Therefor you cannot calculate in an easy linear way the life expectancy of your fully charged battery based on a given fixed load, because as the voltage changes, so will the current.

To make it a little easier, NiMh tend to have a much more level voltage across the discharge cycle.

So you see, it is nigh on impossible to accurately calculate what you are asking even if you did provide all the information you possibly could about your circuit, there are just too many variables affecting it. Ambient temperature even could be a factor, battery life is shortened if it is colder. etc

Hope you got plenty to go away and think about..

Graham

To make matters worse, the amp hour capacity depends on the rate at which it’s being discharged. Faster discharge is less efficient, and different battery chemistries behave differently (in terms of how much of an efficiency loss) at faster discharge rates.

And beyond that, manufacturers, particularly chinese ones, routinely lie about capacity. I’ve gotten “4000mAh” lithium rechargables that we tested at around 600-700mAh. Even the reputable manufacturers are very optimistic with their specs.

A major limiting factor of batteries is their internal resistance, that increases to a point that makes them unusable LONG before they are exhausted. So even when they have half their amp hours still available, the current available to your circuit will be greatly reduced.

Chances are, if you start with a fully charged battery and then note the current used before it becomes unusable, you'll be lucky if you're getting even a half of the rated "amp hours".

ghlawrence2000: But a general explanation is because the voltage regulator burns off the extra voltage as heat..... you want 5V, you are giving it 12V, your circuit is drawing 60mA, you are wasting 7V*60mA = 420mAh, but when you are supplying your circuit with 6V, you are only wasting 1V*60mA=60mAh.

That will have a measurable effect over several hours on the life expectancy of your fully charged battery.

Hopefully grumpy mike is watching. ;) :D

Regards,

Graham

the result of V x A is NOT Ah !! it is W (Watts) you are right about the power lost in heat, and yes, the power lost will in this case be 0.06 x 7 = 0.42 W , but there would be (almost) no difference in the time of use between : - 12 V 4Ah battery + 5v regulator powering a 60mA load and - 5V 4Ah battery powering the same 60mA load

ghlawrence2000: This could actually become unduly complicated if you wanted it to...... You could factor in the type of battery you are using. For example a 12V lead acid battery may easily start off freshly charged at 13.8V rapidly falling to 12.4-12.6 for quite a while, before starting to taper off towards fully flat at approx 10.8V is considered flat for a 12V lead acid battery. Therefor you cannot calculate in an easy linear way the life expectancy of your fully charged battery based on a given fixed load, because as the voltage changes, so will the current.

Yes, and no. That is only true if there is a linear load on the battery.

In this case, an Arduino, the regulator keeps 5V at the Arduino. So the Arduino continues to draw the same current as long as the battery voltage is high enough for the regulator to continue to put out 5V.

Where it gets complicated is if you are using a SMPS regulator. In that case, it will draw a lot less current from the battery, as it is no longer burning up more than 50% of the power as heat in the regulator. So if the Arduino draws 60mA at 5V, that is 300mW. With a 95% efficient SMPS regulator, that is 316mW.

But to draw that from 12V, that is 316mW/12V = 26mA.

As the voltage drops, it will draw more current to supply the same power to the Arduino. But you can see how you can get a -lot- more lifespan from the battery in this way.

alnath: the result of V x A is NOT Ah !! it is W (Watts)

OOPS!!! ;) :stuck_out_tongue: I am blaming it on lack of sleep!! Thank you for pointing out that obvious mistake!

Regards,

Graham

olf2012 wrote:

Your battery has 5 Ah, that is 5000 mAh. Divide 5000 mAh by 60 mA, you get 83 hours

wrong.

ghlawrence2000 wrote:

Apply a nominal factor of say 75% of useable battery capacity would give a reasonable guess at your battery life

Getting close to being correct.

raschemmel wrote:

4 Ah = 4000 mA / 50 mA => 80 hrs. Is that so hard to understand ?

not correct and same as olf2012 wrote.

ghlawrence2000 wrote:

This could actually become unduly complicated if you wanted it to...... You could factor in the type of battery you are using

Very good point.

A battery capacity as stamped on the battery is not the capacity you can actually draw from the battery in a ongoing cycled based usage. The stamped capacity is a figure that tells you the 100% capacity of the battery.

But of course you can not use all this 100% of the battery or you wil have a dead weight with no use in a short time.

Battery cycle life is measured as a depth of discharge and typically DOD will be between 50% and 80% depending on the battery chemistry.

Depending on the battery chemistry you use, will suggest how much energy you can take from a given capacity battery.

For example, if you are using lead acid, you will only want to take 50% of it's rate capacity, and no more than 70%.

With LiFePO4, you can take up to 80% of its rated capacity.

So, if you have a 5Amp/hr lead acid battery, you should be basing your calculations on only using 50% of that rated capacity, and that is 50% of 5Amp/hr which is 2.5Amp/hr.

There are many other good points made in this thread that also need to be considered.


Paul

So what your saying is that it should be 0.8 * 80 hours = 64 hours ?

if (you_mean == me) { yes; }

Like I said, the capacity rating of a battery as stamped is its 100% capacity, and you simply can not not all that 100% capacity or you will have a nice paper weight in no time.

Like I also said, the available capacity you will want to take as a maximum will depend on battery chemistry. Lead acid batteries, the figure is around 50% typically. LiFePO4 batteries, the figure is around 80% typically.


Paul

Venki: Hi, I have another one doubt, I have made some gadget, It draws 60 mA, 5 volts, I'm using 5AH battery so How to calculate the ampere hour for this. And also, give me exactly battery working time.

Exactly? Impossible! Not only for the reasons previously discussed but manufacturers blatantly lie about the performance of their batteries. There are so many variables that a guestimate is just as good as a calculation.

LIPO batteries are labeled with their discharge rate which typically STARTS at !00% and can be up to 4500 % depending on the brand and model battery. That spec is called the "C" rating. 1 "C" is 1 charging rate which is assumed to be the rating (in Ah) of the LIPO battery. These days the cheaper ones are 25C, meaning they can be discharged at 25 x 1C. For a 2200 mAh battery , 1 "C" is 2.2 A, and 25C means a discharge rate of 55A. In addition, it is not uncommon for the LIPO battery to be labeled with the BURST discharge rate, (10 seconds or less) which can be 1.5 x the C rating so a 25C battery COULD be rated for 55A continuous *1.5 = approx 90A BURST. If it the burst rating is not clearly labeled it cannot be assumed. All of this is info is readily available on Hobbyking, or Thunder Power websites.

LIPO BATTERY SPECS

To say that a Lipo battery cannot be discharged at 1C (as opposed to it's clearly labeled rate of 25C) just doesn't make sense. Do you have any experience using LIPO batteries ? (what you said applies to OTHER technologies)

FYI, Lithium Iron Phosphate ( LiFePO4) is not Lithium Polymer (LIPO)

raschemmel wrote:

Do you have any experience using LIPO batteries ?

who is who you refer to, as I'm confused when you don't say who is who or who mentions what? :(

If it is me, then I can say I know more about LiFePO4 style chemistry as I have lived on lead acid for the past 18 years and am about to dish out a rather large wad of folding stuff :money_mouth_face: for a new LiFePO4 battery bank in few months to continue to work from renewable energy. Yep, all this is brought to you by free and off grid electrons :P

All the OP wants is to know how to calculate the capacity of a battery and solar panal to power his very small gadget, as he says:

I have made some gadget, It draws 60 mA, 5 volts

I'm guessing the OP is likely not that interested in knowing it can do 90Amps in burst mode. :o


Paul