Dreamer Mech

[EbbiVapes]

Alright guys ty so much for your Help !!

 

[EbbiVapes]

Worst best case scenario:

2 coils @ 0.13Ω
Total resistance - 0.13Ω/2 = 0.065Ω
1 x 30T, theoretical maximum current draw = 35A
1 x 30T, fully charged (in peak condition) = 3.6V

Current draw according to Ohms law (excluding switch, battery and contact resistance)
3.6V/0.065Ω = 55.4A

Power dissipated = 55.4A x 3.6V = 199W

You want to be drawing a maximum of 30A (80% of capacity to be safe, even less if the battery is old or not genuine)

So in reality, you are running your battery at ~ 60% over its rated maximum.

No wonder you have thermal issues. Please take the advice others have given above, under these conditions you are an accident waiting to happen.

These figures are an estimated best case, if your battery has an internal short, not only will the figures be greater, but thermal runaway will make it worse.

mate dual coils come out to 0.13 not 0.065 dual

 

[Old Greybeard]

mate dual coils come out to 0.13 not 0.065 dual

No worries, sorry, misunderstood.

Revised figures:

2 Coils @ 0.26Ω
Total resistance - 0.26Ω/2 = 0.13Ω
1 x 30T, theoretical maximum current draw = 35A
1 x 30T, fully charged (in peak condition) = 3.6V

Current draw according to Ohms law (excluding switch, battery and contact resistance)
3.6V/0.13Ω = 27.7A

Power dissipated = 27.7A x 3.6V = 99.72W

Thats a lot healthier .

 

[Topwater Elvis]

The cells we use are 4.2v fully charged.

When using a mechanical you should always use 4.2v when calculating battery amp demand.

4.2v /.13Ω = 32.3a

 

[greek mule]

@EbbiVapes if you forget,i made a question at post #17