Figuring your Battery Capacity (Li-Ion)

[jayrob]

I field quite a few questions along this subject...

I'm sure there are several posts that cover this. But I don't know where they are. So I'm making a brief post that I can just refer people to.

Please post your input for good information to be learned by others!






Li-Ion Batteries:

Most Li-Ion batteries have the claimed capacity stamped on them. But the only way to know for sure is to run them and time it.

(as far as I know)

Of course, it will depend on what current draw they are at.

FlexDrives are a boost driver. So it's tricky to get completely accurate estimates. Because the current draw on the battery will normally change as the battery drains and voltage drops.

It's because the driver uses extra current to boost diode voltage demand.

For instance, if you have a blu-ray diode, which is asking for 5 volts, then your FlexDrive will boost the needed voltage from a single 4.2 volt Li-Ion by using extra current. The driver does this very efficiently.

Now the tricky part. As the battery drains, and the voltage starts to drop. The driver will use more and more current to keep the diode at what ever voltage it asks for.

So as your battery drains, it must keep supplying more and more current as it drains.

That is why you need a high quality battery like AW brand. Or an IMR Li-Ion... (with a high current build - low current builds can get away with lesser quality batteries)

For 445 builds. The voltage of the diode is closer to the battery voltage. (Li-Ion) Around 4 volts. So the difference is not as bad as with a blu-ray build.

But still, the battery is probably at maybe 3 volts by the time it needs to be charged. So the 'boosting' is still needed even with a 445 build...

Anyway, you can just kind of estimate what is going to be the run time for your battery.

You can get a pretty good idea if you understand what is going on with the way it works.

If you have a linear driver, such as a DDL, or the FlexModP3 that I use in my Maglite builds, it is a different story.

The current to the diode is the exact same as the battery current draw when using a linear driver.

So your math is easy to figure. Example:

If your battery is exaclty 2900mAh's capacity, and your current is 1450mA's with a linear driver, then your battery will last for 2 hours. (if it is truly 2900mAh's capacity)




Batteries 'in parallel' vs 'in series':

This is also kind of basic, but good to be mentioned for those who do not know. Or perhaps never really thought about it.

When you have more than one battery, and they are connected in parallel (negatives together, and positives together), this will be a battery pack with the same voltage as a single battery. But multiply the capacity.

On the other hand, if you have more than one battery, and they are connected in series (positive to negative), this will be a battery pack that will have the same capacity as a single cell. But multiply the voltage.

 

[Cyparagon]

Now the tricky part. As the battery drains, and the voltage starts to drop...

Calculate using Wh instead of Ah. If the average voltage is 3.7, and the capacity is 2.5Ah, that's just over 9Wh. A buck or boost driver will draw roughly the same power - about 4-5W for a 1A blue.

 

[HIMNL9]

Calculate using Wh instead of Ah. If the average voltage is 3.7, and the capacity is 2.5Ah, that's just over 9Wh. A buck or boost driver will draw roughly the same power - about 4-5W for a 1A blue.

Ehm ..... sorry, but remember that "Ah" indication is a relative value, not an absolute value ..... as example, "2.5Ah" ..... it just mean that the battery can give you "at least" 2.5A "for a whole hour" ..... but, depending from his internal resistance and health condition, it can give you 5A for half hour, or 10A for 15 minutes, or also more A for less time, if the battery is in good condition and high quality / low internal resistance.

Calculating how much current you can get safely and constantly from your battery for a certain amount of time, is not just as calculate Wh (also cause increasing the current and proportionally reducing the time, you still get the same Wh ).

 

[Cyparagon]

...depending from his internal resistance and health condition, it can give you 5A for half hour, or 10A for 15 minutes...

Maybe you misunderstand what a watt-hour is. it is a unit of energy. If your power company charges you for 20,000 Wh, it is NOT saying you used 20,000 watts for one hour during the month.

5A at 3.7V average for half hour is 18.5W for half hour. This is equal to 9.3Wh.

10A at 3.7V for 15 minutes is 37W for 1/4 hour. This is equal to 9.3Wh.

37W/0.25h = 37/.25 Wh = 9.3Wh

 

[BShanahan14rulz]

So, can I use a linear driver to drive ~2.6V worth of resistive load at a constant current to drain battery versus time?

 

[LORDJET]

In any event , thanks for giving this post. It contains good basic information that many people should find useful. Batteries are something that may seem so basic and simple to most. But like many things, there is much more going on than at first seems apparent!

 

[Sparkles]

Maybe you misunderstand what a watt-hour is. it is a unit of energy. If your power company charges you for 20,000 Wh, it is NOT saying you used 20,000 watts for one hour during the month.

5A at 3.7V average for half hour is 18.5W for half hour. This is equal to 9.3Wh.

10A at 3.7V for 15 minutes is 37W for 1/4 hour. This is equal to 9.3Wh.

37W/0.25h = 37/.25 Wh = 9.3Wh

18650 battery test with capacity curves for many cells

I frequently reference this awesomely done page for information.
As you can see in the graphs of current and voltage versus time, the Wh varies with the resistance/current required. So the first problem of giving a simple Wh energy stat is that it does not give a perfectly accurate picture of the batteries capacity.

The second problem, as jayrob was saying, also depends on the type of driver you are using. The buck driver will use the Wh capacity more closely because it uses the full voltage available and varying current to match the diodes needs.
The linear driver on the other hand will NOT use the Wh figure because it uses a set amperage/current draw at all times. The linear driver "bleeds" off the excess voltage as heat, and in regards to the FlexMod P3 driver brings the voltage down to 1.5 V above what the diode requires. Again as jayrob was saying, the math is simple at this point and you only use the mah capacity stat.

As HIMNL9 was saying the capacity of the cells are "1C". Which is how much mah the battery would have if it was discharged in 1 hour. The useful mah goes down as the current goes up. The webpage I referenced shows that excellently.

Hope that clears up any confusion.

 

[lazeerer]

I always had a hard time figuring this out and never really thought about. Bu now i have a better idea on how its done.

Thanks Jay :beer:

+1

 

[LtKernelPanic]

If you want more info then you ever wanted to know about battery capacity Dave over at EEVblog did two episodes on the subject a couple weeks ago.