Leaves the mainloop???

[Acetronics2]

. If I used a car battery would the result be the same? I need to try that.

After what you have seen ... I wouldn't even try !!!

that means your electronic load draws lots of current @ start ( called an overshoot ) and you are not sure it's internal circuits can withstand, say, 40, 50 or more Amps the car battery could easily give ...
" magic smoke " probably in sight ...

Glad you got it in the end ...

Alain

[Ioannis]

My personality is hurry up and ask questions and get an answer. I need to slow down and think of my questions first and organize them. I want the asnwer in miliseconds - lol.

no no no - asking me if I had a scope was not rude. I was trying to edit my post and the 30 minute limit hit and my post didnt mean to infer that - sorry for that. I thought the rude remarks came from someone else and if you read the thread you can see from who. On top of all that I'm asked if I had a scope and I took it the wrong way. Email and texts and forums can suck sometimes because you cant hear tone.

currently I still do not know whats up. I brought my circuit to my EE friends at work and they think the load is a resistive load where it really should be an inductive load (service vehicle lights etc) - they are thinking the for the first turn on its shorting my supply for a very short time, resetting the PIC. The supply I use has 15 amps capability, so IMO its beefy enough. I aslo put a larger cap across the 5V with no luck.

The circuit works under 1 amp so that might be true of load dumping the supply > 1 amp. The scope I have should catch this, I agree. My next idea was to change the load from resistive to inductive.

And read more ideas here on this forum.

No hard feelings. We are here to help each other and not to argue or fight. More info you give the better the answer you 'll get.

It the first time that I see you use electronic load to test the circuit. So, what is that load? DIY or a ready made device? How is that set up? Can you adjust it?

Ioannis