brushless

I recently purchased an Integy Indi Zero-Equalizer and used it on my batteries for the first time. I let the batteries sit on it for about 24 hours. I then charged them and they took a 3800 Mah charge. They are GP3300s. How can a 3300 battery take a 3800 charge? I only had the peak set on 15 mV.

Still Bill

iTrader: (16)

Had the Indi Octane 2 but have upgraded to a Novak Smart Tray.

I usually get 3800-3900 mahr numbers on my better batteries after discharging to .9V on the Smart Tray. Otherwise, the very first charge on a new pack usually is around 4000-4500 mahr depending on the overall quality of the pack.

Read somewhere about how the GP batteries are rated as being 3300 mahr; but, cannot quite remember the explanation for such. Try looking on the SMC Batteries string or ask "Korgae" (he is with SMC) who knows such things.

All my experience is with SMC and Promatch batteries with higher voltages (1.175+) and lower IR figures (below 2.5).

Pleskit

How did you like your octane 2? Did it work well, help batts, etc?

brushless

So are you saying that battery companies rate their batteries lower than what their actual capacity is?

ALBERTO

iTrader: (23)

hey i thought it was bad to let NIMH discharge below .9?? Or maybe i miss understood you,

crimson eagle

Hi ,

the rating is on discharge, you always put more in than you get out and 3300 is the minimum the cells will provide, generally we use the better cells.

Chris

Kennedy

GP3300's should have Peak Detect@ 5mv p/cell or 30mv as a whole pack. 15mv is only suited for NiCads. Your cells would have got really hot i bet

Still Bill

iTrader: (16)

The Octane 2 worked fine.

My only problem with it was I occasionally left my packs discharging on it too long (to somewhere below .9 V). That is never an issue with the Novak Smart Tray.

Still Bill

iTrader: (16)

Hi Chris,

Yep. That's what I couldn't remember.

Have you been racing at Ripon lately? Don't see any recent posts. Stockton (carpet) is getting pretty close. Hope to see you and Reuben soon.

crimson eagle

Hi Bill,

No, Ripon racing has been conspicuous in it's absence. Apparently there is two more weeks of Modesto racing before the rug season. I'll be at Stockton every Saturday as usual, maybe one or two Wednesdays.

Chris

crimson eagle

Back on topic...

I always equalise my cells down to 0.1v. Any given cell may have more or less energy left in it when you only discharge to 0.9v, so saying at 0.9v the cells not really equalised. My friend in England did a test and consistantly when a pack is equalised (in this case to 0.1v) you get an extra 20 seconds on discharge at 30A.

Chris.

brushless

If I set my peak detect at 30, my pack would melt. It gets too hot. I set it at 15 and it is just warm when it completes charging.

I still don't understand why my batteries can hold 3800 if they are just 3300. Can I actually get 3800 mah out of them if that's what charge they are taking?

galway

I peak detect at 30mv max. At 30 they get up to about 145 external temp. I don't want them getting any hotter then that.

When charging you often put a lot more into a battery then it is rated at, don't really know why this is though. When discharging you usually are right at the rated capacity.

brushless

I don't want my batteries getting that hot though.

Shouldn't you get out of a battery whatever charge you put into it?

crimson eagle

Using the theory that energy is never created only changed from one form to another in this case we are converting from electrical energy to chemical energy and then back again. The question is where are we losing this energy that we put in but does not come out? The answer of course for the most part is heat. We do notice that as the pack reaches full charge it gets hotter, this heats does not come out of no where. Therefore we can assume that part of the energy put into the pack goes to heating the cells, this energy is then dissapated to the air and is now therefore not available during discharge. Also the heat generated in the pack during discharge is not available as electricity to the motor in the race - more losses.

The next logical questions would be - why and how is this happening. My assumption is resistance in the cell. We all know that resistance causes heat in wires (this is how incandescent light bulbs work). And if you also assume that the more resistance in the cell, the more voltage is required to push the same amperage into the cell and the higher the volts the more resistance, heat etc. And therefore less you will get out for what you put in. To digress a little, this is why I personally like to have a lower peak voltage during charge than the stated IR numbers (which I have no way of verifying) to assess my packs.

Chris.