AFR CMD VS ACTUAL AFR

2012 chrysler 300C V8 5.7L .

Im running open loop and have tune NN but have hard time to get 12.5 on WOT.

My NN at WOT is like -13. I remove -13 and log again and surprise nothinh change !

I change value on FA Enrich MAX au cat to 0.1628 because this table is fraction so 1.0000+0.1628= 1.1628 14.68÷1.1628=12.5 AFR

So NOW i Stiller have my protection and when apply nothing go can go wrong so i log and do a WOT get me 11-10 AFR !

I go gear shift and do a pull only in second gears and right on 12.5 all the WOT !!!

Try in third WOT instantly 10-11 AFR :(

Try in 4 WOT instantly 10-11AFR :(

Try in 5 WOT instantly 10-11 AFR :(

Im losing my head, i need help

Your abbreviations aren't helping us understand what you are doing. I get WOT and AFR, but what is "NN" and "FA Enrich". The 1.168 value appears to be an Equivalence Ratio (the inverse of Lambda Ratio). So 16.8% rich or .856 Lambda.

Is this the factory ECU or a stand-alone?

What software are you using to tune your ECU?

Perhaps all of the tables for your ECU aren't defined. Perhaps there are separate tables for each gear as the load (and time spent at load) will change with the gearing.

In FUEL-TEMPERATURE CONTROL, FA Enrich is responsible for protecting the catalyst by adding PE. FA (Fuel Adder) is a fractional value, not a raw multiplier. So for a ratio of 14.7, calculate this as follows: lambda 1.000 + 0.1628 (FA PE) = 1.1628.

14.7/1.1628 = 12.68 AFR. Now, catalyst protection is enabled, but it can't distort my AFR since power enrichment can only give an AFR of 12.5.

After that, when I do a WOT in 3, 4, or 5 gears, my actual AFR is immediately between 10 and 11, but during a WOT in 2 gears, it gives me 12.5 all the way.

I want a real AFR of 12.5 during my WOTs, but I'm confused as to why I only get it in the second gear.