Mech Mod Voltage drop - why it is important :) UPDATED

[robby]

With many vapers now using mech mods, I thought I would give you my take on voltage drop and hope it helps people without electrical knowledge to understand what is going on.

In the example above, the resistance of the coil is 0.8 ohms.
The resistance of the other bits and pieces (tubes, switch, 510 connector and spring) is 0.05 ohms.
 
The important factor is the numerical relationship of the two resistances because the voltage available from the battery will be split between the two resistances in the ratio of those resistances ie. - 0.8 to 0.05, which is 16 to 1.

So even though the battery is 4.2 volts, there will only be 3.95Volts available to heat the coil. In this case, the voltage drop across the mech mod is only 0.247Volts. That isnt a bad figure in this case. There is always some voltage drop across the tube, switch, 510connector etc. 





Now if you have a mech mod that has a higher resistance or you have dirty contacts, maybe a poor spring etc it can have a major effect on the voltage available to the coil. The effect of a small increase in resistance of 0.1 ohms (which is still a tiny value) is significant and shown below.




So, we still have 4.2Volts in our battery but only 3.53Volts is available for the coil. 0.66Volts is being lost because of the resistance of the tube, switch spring etc.
So if sub ohming, the effect of this voltage drop is significant because of the relationship between the coil resistance and the Tuberes resistance. If we were using a 0.5 ohm coil in the second circuit we would drop 0.97Volts and only have 3.23Volts available for the coil.

That effect is reversed with a higher resistance coil. If for example we use a 2.4 ohm coil in the second example we only lose 0.247V.


Voltage drop is unavoidable, there will always be a small amount. There is another reason for voltage drop and that is that we lose some because of the battery internal resistance, but we dont really need to go into that now.


So it is very important to keep all threads clean and all contacts clean, spring, end cap, switch contacts and 510 connector contacts.



I hope this makes some sense, it is not an easy thing to explain well. Any questions, please ask

[Ripshod]

Brilliant. I'd like to be able to explain this myself, but I find it hard to explain even tho I have the electrical knowledge. To look at it as a simple ratio is a stroke of genius.

[robby]

The mech mod circuit in it`s simplest form.

[markm]

Great work Robby,
I slippery subject explained in easily digestible terms.

[Super-Shiny]

Nice one robby

[VapingBad]

Nice job robby, if I fell a mech is not firing well the first thing I do is a quick voltage drop test with a tank-o-meter, clean it and test again. If you haven't done this you would be surprised how much dirty (even when they don't look dirty) threads and the internals of the switch make.

I also just got a Stingray clone and the magnets in the switch stuck out slightly so they were acting as electrical contacts and not the silver plated screws that should be. I carefully pushed them in with a vice and the voltage drop (4.08 V Sony VTC4 with 1 ohm coil) went from 250 mV to 100 mV, the small parts are particularly important. Chrome plating is also a rubbish thing to have on small parts like contact pins (510 or switch), some times I replace the 510 pin with a brass screw to avoid having to alter the stock one.

[ronaldo]

also dont buy painted aluminium tube mods the current cant flow properly through paint so makes it all worse sorry for butting in robby

[Die5el]

Great work Robby

[djs]

Steady on robby.

[robby]

May 30, 2014 9:24:35 GMT ronaldo said:

also dont buy painted aluminium tube mods the current cant flow properly through paint so makes it all worse sorry for butting in robby

I`m sure any advice would be welcomed by newbies ronaldo

[halight]

Thanks robby

Still trying to get my head around all that's been written. Might have to read a few more times yet to fully understand, but I have more knowledge about it now I did before.
So many thanks

[dave]

Very useful robby - a good clear explanation. It is surprising how much effect the small resistance of the mod and the battery can have.

Let me add an extension to this for those who want to be even more adventurous

If the battery voltage is 4 volts, the coil resistance is 1 ohm and the resistance of everything else is 0.1 ohm then using Robby's method the voltage is split in a ratio of 1 to 0.1, or 10:1. Putting this a different way the voltage drop is 1/11 (about 9%) of 4 volts so what is actually going to the coil is 10/11 of 4 volts (about 3.64 volts)

The vape is most closely related to the power we are experiencing though, so what effect does this have on that?

If we just had 4 volts going into a 1 ohm coil we would get 16 Watts of power. If we add in the effect of the additional 0.1 ohm resistance what we actually get is a proportion of that which is actually the square of the proportion worked out above - 10/11 squared. In other words 10/11 multiplied by 10/11 which is just under 83%, so the 16 Watts is reduced to 13.2 Watts.

I've played around with these sorts of figures a bit in the past and my guess is that with a decent mod and battery the other bits of resistance come out at around 0.05 to 0.1 ohms at a rough estimate. The practical significance of this for me is that I like to vape at about 9 Watts so if I am just using a mech mod and want a reasonable amount of life out of a battery (say down to 3.7 volts or so) I need to use a coil which is at most about 1.5 ohms.

[robby]

A little more information.

Another major factor in assessing Voltage Drop is the internal resistance of the battery

I have just tested two batteries, the Panasonic NCR18650B 3400mah battery(green) and the Panasonic NCR18650 2900mah (grey)

The no load voltage of the 3400mah battery is 4.15V. The voltage with a .8 ohm coil firing is 3.6V. It has an internal resistance of <100 milliohms which is .1 ohms
The no load voltage of the 2900mah battery is 4.12V. The voltage with a .8 ohm coil firing is 3.8V. It has an internal resistance of <60 milliohms which is .06 ohms

So in all the drawings above there is another factor. All batteries have a resistance value. It is tiny but significant when sub ohming because of the increased current draw (more amps) when sub ohming.

So with the two batteries above it means that you will not have 4.15V and 4.12V volts with a coil firing, you will only have 3.6V and 3.8V available for the coil. So internal resistance of a lion battery is a very important factor often ignored. The lower it is the better. Lower quality batteries than the batteries I have used will give you a lower voltage available to your coil even though they may read 4.2V with no load. This has been my take on this, please correct me if I have made an error

So, you can have two different batteries reading 4.2V off load which give you a different vape experience.

I used my all copper Nemmy for the test.

The Sony US18650VTC3 18650 has a much better internal resistance of only 12 milliohms which is .012 Ohms. This will probably give you over 4V available to your coil (.8 ohms) when connected.

I have these from Fasttech,

 www.fasttech.com/products/0/10004182/1659701-lg-icr18650he2-3-6v-2500mah-rechargeable-li-ion

 They are only .016 Ohms. They are on charge at the moment, I will post the voltage figures when using one of these to see the difference.

[tomj777]

That "new" button on the forum menu bar is a boon (as is Timeline within Tapatalk).
Without which I wouldn't have found this.
robby, many thanks - this is beautifully presented and very clear.

[hijack]

Well written robby and worth bearing in mind when using a mech with a rebuildable that performs best within a certain window.
So does this mean that when working out the drain on your battery when sub ohming, this effect of the added resistance reduces the Amp drain on the battery?