Sub Ohm Vaping Chart

[danwldr]

Interesting, it seems 0.7ohms seems to be the best way to go for the duration of your battery

[TopCookie]

That makes perfect sense in many ways really...  The current draw from these 0.5 ohm and below builds is absolutely tonking the battery, good style...!!!

[danwldr]

Yeah, I'm just ignoring it

[TopCookie]

 

[rhay80]

Is that chart for single coils or duel coils, as i would imagine the results would be different for duel coils.

[TopCookie]

The maths side of it would be pretty much identical, just the end results that would differ really, I expect...

[CraigHB]

It depends if you're looking at the overall resistance or the resistance of each coil.  For example, two .7 Ohm coils would .35 Ohms and two 1.4 Ohm coils would be .7 Ohms.  Electrically, it only matters what resistance the battery is seeing.  Whether that's coming from one coil or ten coils is immaterial.

One thing to keep in mind with sub-ohm coils is that internal battery resistance and resistance from connections and contacts becomes significant.  The lower the atomizer resistance, the higher the currents and the more significant small resistances become.

For example, a battery may have 50 mOhms internal resistance (like an AW IMR 1600mAh) plus there might be 150 mOhms encountered in the electrical path to the atomizer.  The battery sees the total resistance of the circuit.  That would be the battery internal resistance plus the resistance of the electrical path (contacts and connections) plus the resistance of the atomizer.  The battery may actually be seeing .9 Ohms from that .7 Ohm coil.  Amperage would be lower since that's battery voltage divided by *total* resistance.  Power would be lower since that's amperage squared times atomizer resistance.

What I'm saying in so many words is that you get a lot of power loss on the way to the atomizer with a sub-ohm coil.  For the example I made, only about 80% of  the power delivered by the battery is actually getting to the coil.  For dual .7 Ohms coils, the loss doubles.  In that case, the battery sees .55 Ohms, .35 Ohms from the atomizer plus .2 ohms from the rest of the circuit.  Only about 60% of the battery power is getting to the atomizer.  The battery and electrical paths are consuming almost as much as the atomizer!!!

These numbers can be greatly improved by reducing extraneous resistance.  Using a battery with lower internal resistance and using materials with better conductivity can reduce those losses quite a bit.

So, the power figures in that table are only valid for voltage at the atomizer itself.  To get an accurate idea of atomizer power consumption, you need to measure voltage at the atomizer while it's powered.  That can be tricky since an inline voltmeter may add resistance to the circuit introducing error into the measurement.   You also need an accurate measurement of the coil resistance which can be hard to get with such low values.

[gaffer]

So you're saying it's liek quantum physics where you change the result by looking at it? See I knew electricity was some crazy voodoo stuff 

[Lordy]

Don't suppose anyone has a table showing amperage output at sub ohm levels do they?

[danwldr]

Don't suppose anyone has a table showing amperage output at sub ohm levels do they?

watts/volts

[Lordy]

Handy! I didn't know that formula... thanks

So that means... with a Panasonic CGR18650CH cell, you can safely run 0.4ohm setups if you so wish 

[TopCookie]

Don't suppose anyone has a table showing amperage output at sub ohm levels do they?

Not a chart, but extremely useful calculator:  http://www.the12volt.com/ohm/ohmslawcalculators.asp

[Lordy]

Bookmarked! Thanks TC

[TopCookie]

I now have the short cut icon on my pc desktop, I use it that often....  It's so easy punching numbers in for instant results, and being able to swiftly change them etc... 

[BigT]

 , very handy indeed

[vapingfool]

Don't suppose anyone has a table showing amperage output at sub ohm levels do they?

watts/volts

The following may be a bit complicated for some - and not absolutely necessary to know. But when playing with low resistance coils and high battery currents, it is nice to be able to verify that everything is fine before risking blowing up a battery.

To expand on what Dan wrote a bit:

You want to vape at a maximum of 25 watts. What is the circuit resistance you need?

A fully charged battery is 4.2 volts. Dividing 25 by 4.2 gives 5.95 amps.

You can now find the necessary resistance - since Resistance = voltage divided by amps.

R = 4.2 / 5.95 or approximately 0.7 ohms

Hopefully, the coil will be most of this resistance - with the internal battery and connectors contributing a negligible amount.

If you have an accurate voltage meter, you can fire the coil with a fully charged battery and see what the voltage measurement is across the coil. It should be very close to 4.2 volts. If it isn't, you can figure how many watts you are losing in the battery/connectors.

Assume that the voltage across the coil is 4.0 volts. You know the coil is 0.7 ohms, since you carefully wound it and measured it at the binding posts. 

power dissipated by the coil = voltage(across the coil) squared divided by the coil resistance

So power (watts) = 16/0.7 = 22.9 watts

This means you are losing 2.1 watts in the battery and connectors.

You don't have 5.95 amps flowing through the coil as was calculated for an ideal circuit - you only have 4/0.7 = 5.7 amps.

There are other ways to get to the same answer, especially if you know all the relationships between watts, volts, ohms and amps.

[DavidP1970]

Thank you again!!