Ohm's law...

[Topwater Elvis]

There is also the Samsung INR18650-29e & newer f.
All the 2900mah IMR/hybrid batteries I am aware of fall into the 2.75a ~ 10a CDR range.

 

[coilburner]

This implies to me that you're using a regulated VW or "variable-wattage" device. Ohm's law is more important on an unregulated, mechanical mod. Ohm's law states that the current (amperage) through an electrical conductor between two points is directly proportional to the voltage, and inversely proportional to the resistance (ohms). The equation is as follows:

I = V/R, where I = current, V = voltage, and R = resistance.

This says that Amps = Volts / Ohms. The power equation (sometimes referred to as "Watt's Law") is a related concept, which states that power (wattage) is directly proportional to the square of the voltage, and still, like with the amps, inversely proportional to the resistance:

P = V^2 / R, or Watts = Volts * Volts / Ohms.

As these are mathematical equations, you can rearrange the terms and they will still hold true. For example, Ohm's law (I=V/R) could also be stated as V=IR or R=V/I. It just depends on which are the knowns, and which are the unknowns.

Ohm's law is important in a mechanical mod because it tells you how many amps you are drawing from the battery. Try to pull too many amps, and the battery goes *BANG!* (Or, more likely, it goes "pfffFFFFFFFTSSSSSSSSSSSSSSSHHHHHHHH" as it vents hot gases and flames -- it's not likely to actually explode unless it's confined inside a rigid container like a metal tube without adequate venting, aka certain mechanical mods that do not have vent holes. Regulated devices too, I guess. Anything with a battery in it should have adequate venting, even your laptop and cell phone use the same kind of batteries we use and can vent if abused.) Ohm's law calculations should always be done assuming 4.2v, as with a fully-charged battery.

With a mechanical mod, the battery voltage is applied directly to the atomizer, and so the wattage (and amperage) will vary along with the battery charge state. With a regulated device, however, the wattage is held constant by the circuitry inside. Let's take another look at Watt's Law:

Watts = Volts * Volts / Ohms

Note that "volts / ohms" makes up a part of the equation. But "volts / ohms" is also equal to another value we have already discussed, the current (amps). So you could also state it as Watts = Volts * Amps. If you know the watts (determined by user setting) and volts (battery charge state), but not the amps, just divide both sides by volts to end up with Amps = Watts / Volts. If your device is set at 60 watts, for example, and the battery has 3.7v charge remaining, then you are drawing 60 / 3.7 = 16.216 amps. Note that I didn't mention ohms. That's because ohms don't matter on a wattage-regulated device. All that matters are the wattage setting, and the battery charge state. If your battery can safely push 60 watts, then it doesn't matter if it's pushing them through two ohms, or two-tenths of an ohm -- the amp draw is the same in either case. Since the amp draw is highest on a weak battery (as opposed to a mechanical mod, where it is highest when the battery is at full charge), Watt's law calculations should always be done assuming 3.2v or whatever the low-voltage cutoff is on your particular device, since that's when the battery will be worked the hardest at a given wattage setting.

Note that we are talking about the output voltage (charge state) of the battery which is often represented only by a "fuel gauge"-type icon on the device's interface, and not the applied voltage delivered to the atomizer, which may be displayed as a numerical value. These are not the same value (not usually, anyway, but even a broken clock is right twice a day.) If you want to know the output voltage more precisely than "well it looks to be about two-thirds full, or so" then you will need to check it with a multimeter. Some chargers, and a few (but not very many) devices also show the battery voltage as a number, but usually it's just the not-very-precise battery bar on the device, or a blinking red/green light on the charger.

so with a regulated mod the only thing that effects battery drain is the Watts your using and the volts of the battery? The resistance never comes into play?

 

[bwh79]

so with a regulated mod the only thing that effects battery drain is the Watts your using and the volts of the battery? The resistance never comes into play?

That's correct.

 

[Baditude]

so with a regulated mod the only thing that effects battery drain is the Watts your using and the volts of the battery? The resistance never comes into play?

It all depends upon the processor's amp limit. If the resistance of the coil is too low for the processor's amp limit, the processor will not fire the coil.

 

[David Wolf]

so with a regulated mod the only thing that effects battery drain is the Watts your using and the volts of the battery? The resistance never comes into play?

Yes resistance comes into play on the output of the regulated mod. And ohms law applies there as well:
I = V/R.
However the regulated output voltage can be lower (some mods) or higher than the input battery voltage, depending on the users watt or voltage setting, therefore current will be different from the battery than it will be to the coil. Steam engine addresses both sides of a regulated mod (battery input side and regulated output atomizer side), I will find the link and update it here. One thing affecting current on the battery side is the efficiency of the regulated mod, calculations that don't consider this will be inaccurate.
Update:
Here's the link to steam engine calculator for regulated mods (you must click "regulated (APV) button under "Device Type" in the upper left of the page:
Battery drain | Steam Engine | free vaping calculators

 

[Douggro]

Yes resistance comes into play on the output of the regulated mod. And ohms law applies there as well:
I = V/R.
However the regulated output voltage can be lower (some mods) or higher than the input battery voltage, depending on the users watt or voltage setting, therefore current will be different from the battery than it will be to the coil. Steam engine addresses both sides of a regulated mod (battery input side and regulated output atomizer side), I will find the link and update it here. One thing affecting current on the battery side is the efficiency of the regulated mod, calculations that don't consider this will be inaccurate.

To deepen this out a little more, volts and amps can be exchanged but it's a one-way deal, right? I know it's done with AC via transformers but my quick search for application on DC yielded nothing. My reasoning says it should be the same in principle, as AC and DC differ only in the phasing of the current.

 

[Rabbit Slayer]

Does this apply to the coil builds in heads for tanks as well? Like how a .3 coil is recommended used between 70-80W, would you be able to crank it to, say 100w on a vv/vw mod?

It would depend on the coils, some you can go a little higher than the recommended range and some you can go much higher.

 

[David Wolf]

To deepen this out a little more, volts and amps can be exchanged but it's a one-way deal, right? I know it's done with AC via transformers but my quick search for application on DC yielded nothing. My reasoning says it should be the same in principle, as AC and DC differ only in the phasing of the current.

Yes, if running Wattage mode, say 20 watts, then the voltage on the regulated output goes up and the current goes down, Power must be conserved - Power on the input = Power on the output (+ efficiency loss), P = I x V.

 

[Douggro]

Yes, if running Wattage mode, say 20 watts, then the voltage on the regulated output goes up and the current goes down, Power must be conserved - Power on the input = Power on the output (+ efficiency loss), P - I x V.

And this is where the CDR of the battery (or batteries) becomes important, correct? Because we are exchanging amperage draw to generate the voltage needed, yes?

 

[David Wolf]

And this is where the CDR of the battery (or batteries) becomes important, correct? Because we are exchanging amperage draw to generate the voltage needed, yes?

Yes. The current from the battery goes up to provide the higher voltage on the output of the regulated mod. You are right to look at it working like a transformer does with AC power as far as voltage and current transformation goes, though the DC input circuitry on a regulated mod uses switching circuit designs to vary the voltage on the output from what I've read. I added the link to the steam engine calculations for regulated mods to my earlier post above you should find interesting. I had a typo I fixed in the earlier post, P= I x V.

 

[Douggro]

Yes. The current from the battery goes up to provide the higher voltage on the output of the regulated mod. You are right to look at it working like a transformer does with AC power as far as voltage and current transformation goes, though the DC input circuitry on a regulated mod uses switching circuit designs to vary the voltage on the output from what I've read. I added the link to the steam engine calculations for regulated mods to my earlier post above you should find interesting.

Thanks for this. I thought I had a decent grasp of the concept but it's nice to know I wasn't too far off.

 

[David Wolf]

Thanks for this. I thought I had a decent grasp of the concept but it's nice to know I wasn't too far off.

I thought you had an excellent grasp of it.