Does anyone know of a way to test the mah rating of a battery ??? Say an EGO battery , say some one tells you it's 1100 mah , and in reality , it's only a 650 mah . Is there a way to test them , to find out ????
Well, there's no easy way - that's for sure.
The rating is an expression of the amount of time it takes to drain a battery by using a known load. It should be do-able but finicky.
Do a thread search - I seem to recall some time back there was a suggestion that one could get a rough idea by firing an atty of known resistance in 5 second bursts and counting these 'pulses' until the battery worked no more.
That is pretty much it, operate it under load over a period of time to see if the math works.
But I thought that was some of the Mystery to mAh.
What load is it being tested under and does Everyone use the Same Load when they come up with their mAh Rating?
... Having a known resistance in this circuit doesn't help you, because the voltage will drop as the battery discharges, thus any calculation you make with ohm's law will not be accurate. ...
Hope that helps.
To test the mah rating you'd have to put a fully charged battery under a known load until it is completely drained. An eGo battery, or a protected battery of any type will cut off when the voltage drops to a pre-set level. Therefore, you'd have to remove the protection circuit in order to test this type of cell. That wouldn't be adviseable with a li-ion cell.
As an example, if you start with a 1000 mah battery at full charge and put a 10 mah drain on it, it should take 100 hours to completely discharge. If it takes 80 hours, well you've got an 800 mah battery instead of a 1000 mah.
That's the theory anyway. Other factors can influence your reading such as temperature, or age of the cell.
There is the aspect of the chemical power source dropping off the E as it discharges but other wise it is straight math with a known load and time.
If I'm understanding this correctly.
The Straight math part is put a Constant Load on the Battery and then Time how long it takes the Battery to go from Full Charge to Full Discharge.
Then Multiply the mAh times the number of Hours
Correct?
Yes, this is true. As a Battery is used more and more, it's mAh will decline.
BTW - I didn't want to imply that the Math was Difficult in Determining the Total mAh a Battery may have. Multiplying a mAh times a given amount of time seems pretty Straight Forward.
But what if you would like to Determine when a Battery Reaches 75% of its Maximum Capacity? Or it output Voltage at Any given time between Fully Charged and Fully Discharged?
I am going to know the battery is in decline when I have to swap them more but my juice use is the same.
Thanks, it does.
Yeah, that’s where those God Awful Differential Equations come in.
Since the Voltage Rate of Change from say 4.2v ~ 3.3 volts is Non-Linear, the Dif-Eq will probably be Nasty at Best.
So If the Same 1,000 mAh Battery was tested at 20 mAh, would the Discharge time be Exactly one half the time of what it was when tested at 10 mAh?
If it does then I think I'm getting this.
I guess I should have paid more Attention in those EE classes instead of checking out the Undergrad next to me in the Short Skirts back in the 70’s.
...
Me too. I'd be able to explain it all better if I could remember all the math.
And then there's still the problem of Ego type batteries not fully discharging.
When voltage gets to a certain point they stop working and the light flashes.