Mflb dual battery adapter(looking for electrical engineers to help)

Hi everyone! I just got my mflb a little over a month ago. I had smoked a few times before that and hated the smoke part. I'm fairly new to smoking overall but I'm a research addict so I feel like I know my stuff pretty good.

So basically I love the mflb but hate how long it takes to take a good hit at a fairly higher temperature. I know the power adapter helps with that but I don't want to spend 60 bucks and I'm usually not by a plug anyways.

So what im thinking is a dual battery adapter. More power is more heat and I can take quicker draws at the same temperature.

Here's where it gets kinda technical. I'm going into electrical engineering and was hoping there are some other electronics guys on the site.

So basically when the battery is connected now it has a 1.2v source with an internal resistance of 30 milliohms. This means there is a .75v drop across the mflb and 15 amps of current or about 11.25 watts of power.

If I took two batteries in series, this would still have a 1.2v source, but the total batteries internal resistance would only be 15 milliohms. This would cause a .9v drop on the mflb and 18 amps of current or about 17 watts of power or about 1.5x the power.

If I had two batteries in parrallel, then the new voltage would be 2.4 volts, but with an internal resistance of 60 milliohms. This would lead to a 1.1v drop, 22 amps of current and 24 watts of power of about double the power.

So basically my question is will the extra power damage the mflb and or the light inside due to the extra voltage and current, how much voltage, current and power does the power adapter put out(at max setting) and also which would be better if this does work because when the batteries are setup in series the batteries would last longer for more bowls than one battery, but in parrallel more power would be lost internally in the batteries so the batteries would die quicker.

Any responses about anything i said would be awesome! And mainly I really want to compare this to how much the power adapter outputs so if someone could tell me that, that would be great!

only the A123Systems batteries are capable of putting out that kind of current - up to 120 amps. Those "AA" cells are only good for an amp or two (maybe).

Actually the nimh batteries have such a low internal resistance that they can creat currents that high. It's also on the mflb website that it runs at 15 amps...

Sorry, my bad ... i stopped looking at batteries when i found the A123Systems for my design.

Aside from your having 'series' and 'parallel' backwards, you're sorta on the right trail.

Some of the puzzle is internal resistance (makes the battery hot) but there are also losses in the rest of the circuit at that level. Miliohms count. So I'd expect two in parallel to provide more current, but not really a lot more.

A bigger factor is internal resistance growing with discharge. I suspect what your dual battery idea would be more effective at is delaying that decay. That it it wouldn't be all that much hotter, but it would be 'sweet' twice as long.

A better solution might be the c or even 'sub c' NiCd cells used in power tool packs?

And small, 15 Amp switches don't grow on trees.....at least not around here.

Good luck.

OF

Oh yes you're correct that I had series and parrallel backwards. Thanks.

And ya I'm guessing that 2 in parrallel would give more current. Basically I just want to make sure I'm not gonna blow up the mflb with the extra current before I try this haha.

And thanks again for the help.

the resistance element (heater screen) of the mflb doesn't change, so the current draw will be the same ... unless you put the two cells in series, doubling the voltage across the resistance.

ohms law: v = i * r

Yeah, sorta. In practice it's the total resistance (screen, wiring, contact and internal resistance) that determines the current. Of that total, we can frig a fraction (the internal resistance part) at the possible expense of more wiring and contact resistance issues being added. That part (internal resistance) is something on the order of 20% IIRC (you can probe the free end of the contact wire through the hole in the back and the body of the battery). Even if we managed to cut that in half we'd get something like 10% more voltage at the load (less 'sag' under load) and therefore an additional 10% boost in current for a total of 21% boost in power. Useful, but not earth shaking IMO.

Pak does point to a real problem, push it too far at that lovely balance we use to control the vaping action for us gets out of kilter and the box can turn from best friend to something that needs watching.

Fun stuff, but them's the rules as I understand 'em. The battery is a pure voltage source with a series resistance inside. The more current we pull, the larger the drop across that internal resistor (generating heat by Ohm's law) and the lower the voltage we see on the outside terminals as the voltage 'sags under load'. The rub is not only doesn't the power show up in the load where we want it, it shows up heating our poor battery up, which we definitely don't want. Life is like that sometimes.

OF

Thanks for the info. So what I'm getting from you guys is basically it will easily combust and almost be unusable which kinda sucks. As long is wont break from the extra power, which seems to be the consensus, I think I'm gonna go ahead and try it.

Thanks for you help guys and I'll let you know what happens!