Ohms Law question - kind of ...

[EasyGoing1]

As most of you are well aware, the whole vaping thing has become fairly popular lately... I was having a discussion with someone in that realm who builds coils and the issue of current flowing through a coil came up and so I did a quick calc for the current going through a coil whose resistance is .15 ohms with an applied voltage of 8.4 volts, and I came up with 56 amps ... which is a hell of a lot of amps ... but doesn't seem inconsistent given the effects it has on the coils ...

But then I started to think about it and realized that 18650 batteries are not capable of delivering 56 amps ... so naturally, something has to give in the equation and that something has to be voltage because the resistance is static...

However, two fully charged 18650 batteries in series, will be 4.2 * 2 = 8.4 volts ... yet they aren't capable of delivering 56 amps to a .15 ohm load ...

So ... how does THAT work? Does the voltage somehow drop when the batter is being taxed that heavily?

[petehall347]

highest rating you can get is 3.6AH as far as i know .

[sam_sam_sam]

Do you actually have one of those coils to do this measurement correctly I am just asking the question not wanting to counter dictating your results with ohms law math but actually testing the coil circuit working would be interesting to know what this circuit does and how it does it

No I am not a nicotine user at all but have wondered how this technology actually works

[sam_sam_sam]

Originally Posted by petehall347

highest rating you can get is 3.6AH as far as i know .

If you short a 18650 monetary you can get upwards of 6 to 8 amps this however is not recommended to do and things can go horribly wrong quickly if done to long ( this depends on the series of battery that we are talking about because most of them are limited to 4 amp according to the data sheets )

[petehall347]

if i calculated right about 50 amps for 5 minutes

[eccerr0r]

ohm's law doesn't work here only because the circuit isn't fully modeled.
ESR is being improperly ignored.

[PeteS in CA]

Originally Posted by eccerr0r

ohm's law doesn't work here only because the circuit isn't fully modeled.
ESR is being improperly ignored.

IOW, the battery has an internal resistance which would drop some of the battery's voltage if connected to a .15 ohm load (ignoring the inductance and charge time of the inductance).

[EasyGoing1]

Originally Posted by sam_sam_sam

Do you actually have one of those coils to do this measurement correctly I am just asking the question not wanting to counter dictating your results with ohms law math but actually testing the coil circuit working would be interesting to know what this circuit does and how it does it

No I am not a nicotine user at all but have wondered how this technology actually works

I vape with a "controlled mod" - which is basically the circuit I described only it has a microcontroller and it varys the current through mosfets based on parameters you set.

HOWEVER, I absolutely do have multiple 18650's - the high drain rated ones and I have a ton of raw coil material so this would not be a difficult experiment to set up at all.

My only concern is that this new Fluke meter has fuses in it for the current settings ... my other Fluke does not ... but I'll be damned if I know where it is. But I think the Fluke is able to read up to 50 amps maybe?... I already blew a friggin $20 fuse but that was before I even realized it had fuses that you could blow ... which I think was on the microamp setting or something ... either way I fucked it up and had shell out $20 for a fuse and now I'm paranoid about it ... but I shouldn't be, I should just know its limits and make sure I don't exceed them ... lol

[EasyGoing1]

Originally Posted by eccerr0r

ohm's law doesn't work here only because the circuit isn't fully modeled.
Esr is being improperly ignored.

esr?

And ohms law HAS to work here because ... being a LAW and all ... just need to figure out what I'm missing ... and Petes in CA mentioned internal resistance of the battery ... that might be something to consider as it would certainly have a real causal affect on the circuit...

[sam_sam_sam]

Originally Posted by EasyGoing1

My only concern is that this new Fluke meter has fuses in it for the current settings ... my other Fluke does not ... but I'll be damned if I know where it is. But I think the Fluke is able to read up to 50 amps maybe?... I already blew a friggin $20 fuse but that was before I even realized it had fuses that you could blow ... which I think was on the microamp setting or something ... either way I fucked it up and had shell out $20 for a fuse and now I'm paranoid about it ... but I shouldn't be, I should just know its limits and make sure I don't exceed them ... lol

You could use a meter that has a external shut something like 100 to 300 amp they are not very expensive I have bought couple of them and have used them in the past and seem to be fairly accurate

[EasyGoing1]

Originally Posted by sam_sam_sam

If you short a 18650 monetary you can get upwards of 6 to 8 amps this however is not recommended to do and things can go horribly wrong quickly if done to long ( this depends on the series of battery that we are talking about because most of them are limited to 4 amp according to the data sheets )

Look at the amp rating of the batteries I use for my vape...

[eccerr0r]

See you neglected that 0.16Ω printed right on the battery (which is the "ESR" or "internal resistance", and this value changes as the battery ages, per lot, state of charge, etc., etc.)! That's why Ohm's law is not working for you.

Actually looking at that again, probably need to rotate that battery some. 160mΩ seems very very high for a battery designed to pass 22A continuous. Should be much less than 160mΩ, probably on the order of 10 to 20 mΩ.

[redwire]

Life 4 battery datasheet. The "0.16Ω" spec is the 'ohm limitation' to achieve "non-stop CDR power". CDR is the continuous discharge-rate where the cell apparently does not overheat. You can draw more current- but the battery will quickly overheat. It naturally does make it's own heat due to the battery's internal resistance, ESR.

A way to look at is everything has resistance. The battery (internal resistance), the switch+wiring+battery holder contacts, the heater resistor.
So all of them will make heat, the question is who can't take it, what is their limit.

A 0.16Ω coil with 8.4V is 52.5A/441W which is not realistic. In real life the battery's voltage will sag under overload well below 8.4V, they can't put out more than 31.5A before blowing up anyway.
The heater coil will likely just blow up as they are maybe 80W at most for a few seconds. It will also overheat and melt the wire.

Do you see as you increase the power, the heating element runs hotter, the batteries run hotter, the switch/wiring run hotter - losses go up. Which is fine unless it's too much and the part roasts and burns up.

[eccerr0r]

Indeed 0.16Ω is declared to be the minimum resistance of the load for one cell that's printed but couldn't read because the label needs to be rotated a bit...

However if 2 cells are in series, then the minimum load resistance should be 0.32Ω. So either way yes the batteries are being abused. With the two batteries in series and a 0.16Ω load, assuming 0.02Ω ESR per cell, the cells would droop over a volt a piece, and burn 26 watts for each cell (though with the voltage droop, not really this much but still) ... quite hot! Not healthy at all and you wouldn't get anywhere near full power on the 0.16Ω load.