Is it safe to vape the rx200 at 250 watts (latest update) with only .1 ohms resistance? I know that if you use the formula I=P/V with I = 250 watts / 9 volts then the maximum amp draw would be 27.78 but I do have some concern because when you use an ohm's law calculator with .1 as the ohms and 250 as the watts (power) it says the current (amps) is 50 even though the volts are only 5. Due to the RX200 maximum's output voltage the maximum ohm's you can have to reach 250 watts is .324. I guess I just don't understand why the regulated mods are not supposed to factor in the ohms for ohms law safety. I also know that the highest 18650 amperage rating for continuous amps is 30. I also know that some can go past the rating but I'm more comfortable sticking with the rating of the battery. Thanks to anyone who can help in this matter. Also does anyone know the low voltage cutoff point for the rx200? Thanks again.
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Because the mod is regulating the voltage to the coil. If i am drawing 100 watts from the batteries, it doesnt matter if i am applying it to a .1 ohm coil or a 100 ohm coil. 100 watts is 100 watts. The mod is going to pull however many amps at the current battery voltage to get the 100 watts. It is then going to step up or down that voltage and apply that new voltage to the coil.
On a regulated device the resistance of the coil is irrelevant. Regulated mods separate the input and output voltage, in other words they separate the battery from the atomizer. The only relevant value is the wattage, and the remaining voltage in the battery. The wattage is generated by the mod by multiplying the volts by the amps. As the voltage falls, the mod will increase the amp draw to maintain the selected wattage from the remaining voltage level. You need to know the amp draw at full charge, and when the battery is discharged as this value will be the highest. Most regulated mods are about 90% efficient, so you will also need to factor this loss into your calculations as it will marginally increase the amount of amperage pulled from the battery.
To find the amp draw use I=P/V (-10%)
Eg.
50w divided by 4.2v equals 11.9 divided by 0.9 = 13.22 amps
50w divided by 3.2v equals 15.6 divided by 0.9 = 17.33 amps
If go to this page and click on the 'how it works' link in the bottom left there is a slightly more detailed explanation.
To find the amp draw use I=P/V (-10%)
Eg.
50w divided by 4.2v equals 11.9 divided by 0.9 = 13.22 amps
50w divided by 3.2v equals 15.6 divided by 0.9 = 17.33 amps
If go to this page and click on the 'how it works' link in the bottom left there is a slightly more detailed explanation.
Thanks for explaining this better, I believe I get it now. Basically I have the rx200 and it uses 3 cells. Since it is a series mod it shares the voltage and my minimum voltage would be 9.3 since the low battery warnings keep the device from operating when the cells hit 3.1 volts. So if I am operating at 250 watts the most amps my device would ever draw before it stops working would be 29.87 (this includes the -10%). Although if my volts go that low that would leave absolutely no headroom but I always recharge my batteries with my nitecore i4 charger when I'm done vaping for the day so that's not much of an issue.
250w divided by 9.3v (3.1v per cell with three cells) equals 26.88 divided by 0.9 = 29.87 amps
Thanks again sonicbomb for explaining it better. Please correct me if I'm wrong on my above calculations.
I plugged the numbers into steam-engine and got 22 amps at 4.2 volts and 29 amps at 3.2 volts for each battery.
Mooch wrote a blog entry about the this subject specifically with the Reuleaux in mind. Have a read, then read the rest of his blog entries while you're at it.
Calculating battery current draw for a regulated mod | E-Cigarette Forum
Mooch wrote a blog entry about the this subject specifically with the Reuleaux in mind. Have a read, then read the rest of his blog entries while you're at it.
Calculating battery current draw for a regulated mod | E-Cigarette Forum
Yeah I read that article a few times but for some unknown reason I didn't quite understand it until I read your post. I understood what he meant, I just didn't understand why it was like that lol. Thanks again.
Is it possible to know the remaining voltage of the battery being used? I mean, would the mod have an indicator showing the remaining voltage, or would we have to calculate it on our own based on the remaining battery power? (i don't own a box mod)
Easiest way to do it is just base it off the cutoff voltage. Most mods cut off at about 3.2v per cell. If you have a single cell mod, use 3.2v. If you have a 2 battery mod, use 6.4v. If you have a 3 battery mod, use 9.6v.
That tells you the highest amp draw you should encounter throughout the entire discharge cycle. When calculating, I like to add in 10% to account for the inefficiency of the mod.
Examples:
You have a two battery mod and run at 110 watts. Add 10% to that and you get 121 watts. Divide that by 6.4 and you get 18.9 amps drawn from each battery at the cutoff voltage. If you assumed the batteries are at about half charge(3.7v each) and did the same calculation, you would get 16.4 amps(121/7.4=16.351 amps)
For a three battery mod running at 110 watts, add in the same 10% putting you at 121 watts. Divide that by 9.6 and you get 12.6 amps being drawn from each battery at the cutoff voltage. If you did the same calculations at approximately half charge(3.7v per cell), you would get 10.9 amps(121/11.1=10.9009 amps)
As you can see, using the cutoff voltage gives you the highest possible amp draw, and lets you be sure you are always in the safe range.
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