Hi just had a quick question.. I needed to know if a fused clapton coil running at .14 ohms is safe to run.. I normally do 6 wraps each that ohm out to like .3 but the deck is much too small and I could only do 5 wraps each. Any help on this is appreciated.
Coil ohm
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Hello @Spyroklown, is it a regulated mod and could you tell us what batteries and wattage you'll possibly be using?
Yea it is a regulated mod.. I'll be using 2500 mah 3.7v 20amp batteries ..and I really dont know I didnt want to put cotton in it till I knew for sure it was gonna work
Hi,
IMO:
If the mod will fire it, you should be fine.
With an average 20A battery, you are good for about 60 watts.
You will have a VERY short battery life at 60 watts or above.
The easiest solution is to source thinner wire for that RDA or use a dual battery device.
Cheers
I
IMO:
If the mod will fire it, you should be fine.
With an average 20A battery, you are good for about 60 watts.
You will have a VERY short battery life at 60 watts or above.
The easiest solution is to source thinner wire for that RDA or use a dual battery device.
Cheers
I
@Spyroklown, battery voltage is 4.2 when you do any calculations.
And what make and model battery or batteries please?
And can you clarify around what possible wattage range might you use?
But at least with a regulated mod you do have a margin of safety.
And what make and model battery or batteries please?
And can you clarify around what possible wattage range might you use?
But at least with a regulated mod you do have a margin of safety.
As long as your resistance falls within the devices operating range you are good to go. Should you approach that operating ceiling, the device should give you a resistance or atomizer error if it feels the resistance falls out of range.
With a regulated device the battery does not see the coil resistance, it only sees the requested power from the regulated circuit. As the battery voltage depletes with use, current draw from the battery increases to meet that power demand. Simple Current = Power/Voltage. Saying that, to determine the current draw on the batteries with a regulated device we need to use the operating wattage, the devices low voltage cut off and factor in the devices efficiency. To calculate that, take your power level, divide by the amount of batteries, divide by your devices low voltage cut off, and then divide that by the devices efficiency. 3.2 volts is the average low voltage cut off (some devices are even lower, some a little higher), and an average efficiency to use is 90%. So with a single battery device operating at 50 watts would be …
50 watts/1 battery/3.2 volts/0.9 = 17.3 amps
A dual battery device using the same parameters would be half that current draw 8.65 amps.
3.7 volts spec is the batteries Nominal Voltage, not the fully charge voltage.
@Spyroklown, Mad Dog Mooch is one of our battery experts. See his videos here - HookTube search
View this - Mooch's Recommended Batteries
Read through his whole blog here - Mooch's blog
View this - Mooch's Recommended Batteries
Read through his whole blog here - Mooch's blog
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