So confused. I dont wana die.

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AzPlumber

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AzPlumber

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The board does the calculations. I'm not really following what you mean by that, to be honest.

The board determines what goes to the coil NOT what is being drawn from the battery(s). Your wattage setting determines the demand on the battery(s)
 

Kiyle the Manlet

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AzPlumber

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The only difference between what I am saying and what he is saying is that he is solving for amperage given wattage and voltage.

which instead of I=v/r and IV=W, you get I = w/v.

It's still ohms law. I don't see how that refutes my point.

Your statement above says a 20 amp battery is good for 84 watts, that is NOT the case in a regulated device.
 
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Kiyle the Manlet

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Your statement above says a 20 amp battery is good for 84 watts, that is NOT the case in a regulated device.

if it is safe in a mech mod, it's applicable to a regulated mod because of the internal resistance on the device is higher than a mechanical mod. I'm talking about purely theoretical limits.

In practice, it's actually going to be lower at the same resistance, because of the internal resistance.

I'm not saying that it's the exact same thing, but it's not irrelevant with regulated mods.

In all cases, power is defined as:

P - Work done per unit time (joule per second)
Q - Electric Charge (coloumbs)
V - Voltage
and t in time.

P = (VQ)/t

which is equivalent to P = V/I.

1 coulomb is equal to 1 amp per second. Which essentially simplifies to N amps over 1 second in this application.

Edit: I forgot to define a joule.

One joule is equal to the amount of energy dissipated when something moves a current of 1 amp through a resistor of 1 ohm for 1 second.
 
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AzPlumber

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if it is safe in a mech mod, it's applicable to a regulated mod because of the internal resistance on the device is higher than a mechanical mod. I'm talking about purely theoretical limits.

In practice, it's actually going to be lower at the same resistance, because of the internal resistance.

I'm not saying that it's the exact same thing, but it's not irrelevant with regulated mods.

Dude, the highest demand on a battery in a mechanical device is at a full charge, the highest demand on a battery in a regulated device is at low battery cutoff, then add the inefficiency of the regulator. THEY ARE NOT THE SAME.
 

Kiyle the Manlet

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Dude, the highest demand on a battery in a mechanical device is at a full charge, the highest demand on a battery in a regulated device is at low battery cutoff, then add the inefficiency of the regulator. THEY ARE NOT THE SAME.

Okay, if that's true, it's possible that I'm missing an equation for the right calculations.

I'd like to see it if you could share it.
 
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Kiyle the Manlet

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Dude, the highest demand on a battery in a mechanical device is at a full charge, the highest demand on a battery in a regulated device is at low battery cutoff, then add the inefficiency of the regulator. THEY ARE NOT THE SAME.

Okay, I read what you posted, and all he's doing is changing the constant value by which you're dividing your resistance.

It doesn't make ohms law irrelevant for regulated mods.

It's in fact, the exact same thing as ohms law. The only change is the constant in the numerator.

It IS the same thing.
 

Walee

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Here is an example of real world unregulated mod operation. The build is .47Ω and the batteries are VTC5A charged to a full 4.2V.
Across Coil.jpg

Across the coil 7.8V is being applied.

Across Batteries.jpg

Across the batteries we see the same 7.8V being provided (+/_ 1% approx) You can see the batteries come under load in the circled area.

The pair of series VTC5A's are running at 7.8V under a load of .47Ω providing approx 16A continuous and developing just over 124W.

This is where Mooch's recommendations are essential to most of us who don't do actual measurements. We can still be safe.

As far as published specs, give me a break. Two battery mods running at 225W and 18650 batteries putting out 40A? Not real world.
 

Kiyle the Manlet

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Feb 12, 2018
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Here is an example of real world unregulated mod operation. The build is .47Ω and the batteries are VTC5A charged to a full 4.2V.
View attachment 722665
Across the coil 7.8V is being applied.

View attachment 722667
Across the batteries we see the same 7.8V being provided (+/_ 1% approx) You can see the batteries come under load in the circled area.

The pair of series VTC5A's are running at 7.8V under a load of .47Ω providing approx 16A continuous and developing just over 124W.

This is where Mooch's recommended are essential to most of us who don't do actual measurements. We can still be safe.

As far as published specs, give me a break. Two battery mods running at 225W and 18650 batteries putting out 40A? Not real world.

This is a fascinating post. Thank you for sharing this.

What model oscilloscope are you using? I was looking at getting one.
 
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Kirill2525

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I've been vaping this build for a few days actually. The dipper is a Vandy Vape Pulse so the airflow is on the side and its actually working very well. after i primed the coil and cotton the airflow holes show up. thanks for all the advise everyone. i will let you know if i blow up before i can install a new battery. i will say, the flavor on this thing is pretty good i will attach some more pics
 
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