Calculating ''C'' rating
 

I ordered this thing here to test my lipos

http://www.westmountainradio.com/CBA.htm

I have never been good at Math so I have to ask if I got it right. With the help of this battery analyzer, I want to record the specs of my lipos and monitor how well they perform and for how long. This thing will display voltage, current, amp, watts and temp throughout a discharge of up to 150 watts or 40 amps. One thing I want to calculate with the data that I will get from this device is if the C rating of a lipo is rated accurately by the manufacturer, so I want to make sure I have the formula right. It should be:

Amp/hrs X 1000 = C rating
lipo's Mah

Am I correct?

Nope. That formula will always give you 1 as the answer. (Ah*1000)/mAh=1.

There is no way to calculate C really. The C is a capacity multiplier to get a total discharge rating. And, you'll need to determine an end voltage. As the current load increases, the battery voltage will drop. If you are happy with 3.2v/cell at load, then just measure the current at that voltage. Then, divide that current by the Ah (or mAh*1000) to get the C rating.

I really thought I had it right you bubble burster :lol:

I figured that to get the output in amps it is: Mah capacity x C rating / 1000 = output in amps. So if I have the output in amps and I want to get the C rating I would go (amps X 1000) / lipo's mah = C rating.
I don't get why I would get 1 as an answer. For example, if I get an output of 98 amps, I would go: (98 X 1000) / 5000 mah = 19.6 C.
You're confusing my simple little brain Brian.

mistercrash, I hope you measure those Turnigy packs you bought...

I need 20-30C

Haven't used em yet though :]

lol, sorry. :oops:

From the initial formula, it looked like you were taking Ah, multiplying by 1000, then dividing my mAh. Which would always be 1.

I don't really get why you are trying to calculate the C rating. For one thing, the manufacturer doesn't state the voltage that will be present at the advertised C rating. The only benefit I can see to this test is the verify that the C rating is approximately accurate, but again, since you don't know what the manufacturer considers an acceptable voltage at that C value, verification will be kinda pointless. Or are you just checking to be sure that the pack will deliver the current you need in your setup without tripping the LVC?

Regardless, testing is difficult. So, you get 98A, what does that tell you? Yeah, you can then assume the C rating of a 5000mAh battery is 19.6C. But again, what battery voltage is the pack holding at that value? Unless you set an arbitrary minimum acceptable voltage, you aren't really getting what you want IMO.

And for another thing, 98A is far beyond the 40A and 150w rating of your device. Even a 1s pack (one cell) running at 98A is pushing the device almost 2.5 times its rating, and the power developed at 98A (around 360w for a single cell at 3.7v) is 2.4 times the device rating.

Sorry, I'm not trying to be a "bubble buster".

I ordered this device mostly to record and store some data of the lipos I use and monitor how they perform. That is the main reason but I was hoping that switching the numbers around could tell me the actual C rating of a lipo. You clearly show me that it's more complicated than I thought. Thanks Brian, you're input is always appreciated.

Ray

Again, sorry for raining on your parade. :(

I guess I forgot to answer your initial question: C=max_current/Ah

So, for a 98A discharge for a 5000mAh pack, it would be 98/5, or 19.6C.

Do you still have the receipt for the CBA?... :oops: :lol:

yes.


Edit: And I don't wish to return it, I want to use it.

I wonder if there is a way to use part of that CBA and mod it as necessary? Hmm...

now you're talking. Me like modifications, me like a lot. :smile:

Just add one or two amps from cba or camlight. It works very slick.

When i dont want to use brain power i use this http://www.cheapbatterypacks.com/lipocalc.asp

I can't believe they are using ASP to do those calculations. Perfect use of client-side scripting IMO...

Is that a good or bad thing? :neutral: