DIY E-Cig 200W with regulated wattage

[Oczek]

Hi guys. I’m interested in making my very first e-cig based on atmega (probably Arduino). It would be used to monitor cells and regulate wattage/voltage. What do I know so far:
- gonna use atmega328p
- 0.96” 128x64 OLED I2C
- 2xLi-Ion 18650 in series
- of course 510 connector

I want to reach 200W using 0.3ohm heater. Maximum cells discharge current is 30A.

And I have few questions:
1) what’s the best way to regulate voltage/wattage? I was thinking about MOSFET controlled by PWM but maybe there are better solutions.
2) I need to calculate and show wattage using atmega. I was thinking about ohm’s law. I can measure voltage without any problems, but I also need resistance/current. Which one would be better and easier to measure and how?
3) How should I charge cells using USB port? I want to be able to use cig while charging. I was thinking about IC with BMS but can’t find any. Any solutions?
4) do you have some good and tested circuits to look on?

Any advices will be strongly appreciated. Thanks in advance

 

[Izan]

Hi and welcome,

What is your budget?
Why not just buy something OTS?
Where in the vaping world are you located? US/UK/EU/AUS/?
What other devices have you constructed?

Cheers
I

 

[bombastinator]

you wont reach 200 irl with two cells.

 

[Oczek]

Hi,
Budget shouldn’t be a problem, I hope 30€ for electronic components is fine (not including case). I’d like to do something on my own. I live in EU, Poland to be precise. I’m more like programmer rather then engineer and stuff, but I designed few circuits and pcbs (mainly keyboards).

 

[Oczek]

you wont reach 200 irl with two cells.

I saw Sony cells to have discharge rate of 30A. I can reach 200W using 0.35ohm and when fully charged - 8.4v. It takes 24A. Correct me if I’m wrong.

 

[stols001]

The ratings are a bit artificially inflated, even as authentic cells. If you want a more in depth understanding on that, I suggest this blog, Mooch tests batteries including batteries from the "big three" manufacturers and gives a recommended "highest load" based on real life testing. Even if your batteries were accurate, you still don't want to push the cells at their highest rating at all times, really, it's bad for battery life and can grow increasingly dangerous as the cells get used and stressed.

Mooch's blog | E-Cigarette Forum

Best of luck

Anna

 

[Oczek]

The ratings are a bit artificially inflated, even as authentic cells. If you want a more in depth understanding on that, I suggest this blog, Mooch tests batteries including batteries from the "big three" manufacturers and gives a recommended "highest load" based on real life testing. Even if your batteries were accurate, you still don't want to push the cells at their highest rating at all times, really, it's bad for battery life and can grow increasingly dangerous as the cells get used and stressed.

Mooch's blog | E-Cigarette Forum

Best of luck

Anna

Thank you for info. I will decrease wattage then. Still didn’t get answer for questions stated in post. Anyone?

 

[Izan]

Thank you for info. I will decrease wattage then. Still didn’t get answer for questions stated in post. Anyone?

Yeah, the road is steep,
Check here for supplies and such: ModMaker

Google: "arduino based E cigarette" for additional information.


What e cigarettes do you have/ have you used? What topper(s) are you using now? Power level/nicotine level?
Singles, dual/ exotic?
Do you have a model or brand in mind you are trying to emulate? Evolv/dicodes/Mark bugs/Provari/?


cheers
I

 

[Oczek]

Yeah, the road is steep,
Check here for supplies and such: ModMaker

Google: "arduino based E cigarette" for additional information.


What e cigarettes do you have/ have you used? What topper(s) are you using now? Power level/nicotine level?
Singles, dual/ exotic?
Do you have a model or brand in mind you are trying to emulate? Evolv/dicodes/Mark bugs/Provari/?


cheers
I

I was googling and couldn’t find anything good and useful. I haven’t really used many cigs and I don’t know to which one I can compare it. Just a simple one with lcd and very basic variable wattage.

 

[Izan]

Battery Mods

Wiring Diagrams
Mod DIY
DIY Modder Parts and Supplies

Building my first lipo PWM mod

 

[Oczek]

Thank you for the links. Thanks to Battery Mods, Now I know that I don’t want PWM anymore, more risky and I can’t really control wattage. I think I’m gonna use DC step down. Can you recommend any with 200W output? Other links didn’t really help me, in the PWM mod he decided to take ready board, no circuits and solutions there. And I still don’t know anything about measuring resistance

 

[untar]

I saw Sony cells to have discharge rate of 30A. I can reach 200W using 0.35ohm and when fully charged - 8.4v. It takes 24A. Correct me if I’m wrong.

In theory you could reach 200W with that. For real world vaping purposes - nope. If you draw that much current your batteries will drop voltage like it's a burning coal and your mod will have to draw even more current.
You might get near 200W for a single hit and then that's it. In addition you'll probably have to buy new batteries more often then you like. What Sony batteries are you using?

Your calculation should be more in the direction of: max power divided by lowest battery voltage (the number you want your mod to allow to discharge the batteries to). Let's say that's 200W and 3.2V. That's 62.5 Amps.
No way in hell you're going to find an 18650 that does that (safely).

I'd start with a more realistic setup - go to 100W max. And if I were you I'd ditch the homebrew USB balance charge as well.

 

[Oczek]

In theory you could reach 200W with that. For real world vaping purposes - nope. If you draw that much current your batteries will drop voltage like it's a burning coal and your mod will have to draw even more current.
You might get near 200W for a single hit and then that's it. In addition you'll probably have to buy new batteries more often then you like. What Sony batteries are you using?

Your calculation should be more in the direction of: max power divided by lowest battery voltage (the number you want your mod to allow to discharge the batteries to). Let's say that's 200W and 3.2V. That's 62.5 Amps.
No way in hell you're going to find an 18650 that does that (safely).

I'd start with a more realistic setup - go to 100W max. And if I were you I'd ditch the homebrew USB balance charge as well.

Minimum voltage would be 6V (3V for cell). Wattage is the least important thing right now. The maximum wattage will be calculated based on resistance and actual battery voltage by micro controller. I don’t know how to change voltage and measure resistance

 

[untar]

You'll need to look up arduino (or atmega) and generic voltage converter (for vaping usually inductor based ones, you'll need a buck-boost one) and resistance meter circuits, that's not the problem, there's quite a few online for different purposes.
The problem with very high current applications is to find the proper components and heat sinks for the job, as well as designing appropriate security measures (eg your mod should not fire a battery with a certain voltage sag no matter if it's above 3.2V because then the battery is too weak for the power you want to draw), while accounting for the losses. You won't get good power if your circuit only has 70% efficiency.

What you're trying to do by yourself in your first vape circuit is something that manufacturers figured out over years while collecting experience with lower powered devices. That's why I suggested setting the aim a bit lower, 100W is still a buttload of power for such a small device to handle and it's still stretching what can be done from scratch without much experience.

I'm no electronics expert and I only scratched the surface in my own research on how to create such a board, I don't claim to be correct with everything I said but that's about the gist of how far I got.
I started a similar project (I'm also more of a computer science guy) but got sidetracked into learning CAD and creating a 3d printed mod body for a DNA board first

Edit: thanks to azplumber for correcting the mistake I made in the last post

 

[AzPlumber]

Calculating battery current draw for a regulated mod | E-Cigarette Forum

 

[Oczek]

You'll need to look up arduino (or atmega) and generic voltage converter (for vaping usually inductor based ones, you'll need a buck-boost one) and resistance meter circuits, that's not the problem, there's quite a few online for different purposes.
The problem with very high current applications is to find the proper components and heat sinks for the job, as well as designing appropriate security measures (eg your mod should not fire a battery with a certain voltage sag no matter if it's above 3.2V because then the battery is too weak for the power you want to draw), while accounting for the losses. You won't get good power if your circuit only has 70% efficiency.

What you're trying to do by yourself in your first vape circuit is something that manufacturers figured out over years while collecting experience with lower powered devices. That's why I suggested setting the aim a bit lower, 100W is still a buttload of power for such a small device to handle and it's still stretching what can be done from scratch without much experience.

I'm no electronics expert and I only scratched the surface in my own research on how to create such a board, I don't claim to be correct with everything I said but that's about the gist of how far I got.
I started a similar project (I'm also more of a computer science guy) but got sidetracked into learning CAD and creating a 3d printed mod body for a DNA board first

Edit: thanks to azplumber for correcting the mistake I made in the last post

Calculating battery current draw for a regulated mod | E-Cigarette Forum

Thank you for your replies. So buck converter is better than PWM? I have read that PWM is kinda dangerous because it doesn’t lower wattage but only deliver it for less time (eg. with fully charged cells and 0.2ohm resistance it would give 350W which would kill the battery - more than 40A current draw, am I right?). I found smth like this: https://power.murata.com/data/power/okr-t10-w12.pdf

Is it good? Is it able to deliver this maximum of 110W without much heating? What if I exceed it? Can you recommend something better with higher wattage? Size doesn’t really matter

Edit: from the link from AzPlumber, I’ve read that the connection doesn’t matter. I thought that batteries connected in series don’t sum up discharge rate. So how is it?

 

[untar]

Batteries connected in series don't sum up capacity, that has little to do with discharge rate. 2*2000mAh in series is still 2000mAh (but twice the voltage (2*3.7V)), 2*2000mAh in parallel is 4000mAh (but still the voltage of a single battery (3.7V)). The overall usable energy (measured in Wh= Watt hours) stays the same in theory
2000mAh*(2*3.7V) = 14.8Wh = (2000mAh*2)*3.7V
in practical applications one may be slightly better than the other due to specific circuit properties.

The calculation from the link is used to determine the max amp load on a single battery in a pack (series or parallel) and has nothing to do with capacity. I made an error in an earlier post that you also picked up on, I forgot to divide by 2, the link has the correct formula shown.

The component you linked requires a 20A fuse at the input and has a maximum of 10A at the output and an overall absolute maximum power output of 50W, so I doubt this would be suitable for a regulated mod that aims to deliver 110W

Your calculation regarding what happens on the firing side and on the battery side is not correct for a regulated mod. In a regulated mod the two sides are strictly separated by your circuit. You dial in a regulation (that can be voltage or watts) and the circuit will draw the requested power from the batteries (in case of voltage dial you need to calculate the power in watts). This means that the lower the battery voltage gets over time, the more current (amps) needs to be drawn from the batteries, that's why we have that calculation from the link that works with the lowest battery voltage to select the proper batteries.

I'll give a concrete example with fictional batteries that don't sag (don't lose voltage) under load, in series, as well as a fictional circuit with 100% efficiency:
let's say you have a 0.2Ω resistor and you dial in 6V into your regulator circuit. This means the circuit needs to draw 6V*30A = 180W from the batteries.
If your batteries sit at 3.7V that means your circuit needs to draw 180W/(2*3.7V)= 24.32A from your batteries.
Now imagine you used your mod for a while and your batteries sit at 3.3V, now your circuit needs to draw
180W/(2*3.3V)= 27.27A from your batteries.
The more the batteries are discharged the more current needs to be drawn to match the power requirement on the firing side of the circuit.

In the real world it is a little more complicated, the batteries will sag, eg if they sit at 4.2V and you draw 180W from them they may instantly jump down to 3.7V and your circuit will need to adapt constantly.
Then your circuit will never have 100% efficiency, some power will get lost to heat. In very good circuits that's 5-10%, in less good circuits that can be far lower.
If you have 80% efficiency that means 20% of your power gets lost to heat, so to meet a 180W requirement on the firing side you need to provide 225W from the battery side (180 + 45, 45 = 20% of 225).
You see it gets more complicated real quick.

I doubt you will find a ready-made voltage/current regulator that you can use, you most likely will have to design your own (this is what companies like DNA or Yihi do).


I sure hope I didn't put my foot in it again somewhere in that great wall of text

 

[AzPlumber]

Edit: from the link from AzPlumber, I’ve read that the connection doesn’t matter. I thought that batteries connected in series don’t sum up discharge rate. So how is it?

I'm not too sure what you are asking but yes in series voltage increases not capacity or CDR. It is just calculating max amp draw per battery with a simple mathematical equation.

Max wattage divided by min voltage will give you max amps per battery. You can see the total current at 200 Watts is the same for all three examples but by changing the min voltage it gives you a per battery answer.
200W/3V=66.6666 Max amps (one battery)
200W/6V=33.3333 Max amps (two batteries) 33.3333x2=66.6666
200W/9V=22.2222 Max amps (three batteries) 22.2222x3=66.6666

 

[Oczek]

Oh Thank you, everything is clear now. I’ve mistaken links, wanted to send this one https://www.digikey.com/product-det...s-inc/OKL2-T-20-W12N2-C/811-2961-1-ND/4738836

So I need to take converter efficiency and with that in mind calculate maximum wattage? Let’s say that efficiency is 90% and cells max safe discharge rate at 20A. Let’s assume that cells are fully charged at 8.4v. It gives 168W from batteries. So the efficient wattage would be 168W*90% = 151.2W. Having 0.3ohm resistance it would give me a voltage range of 0-6.73v without exceeding max wattage from converter. Having 0.2ohm resistance it would give 0-5.5v. Now let’s assume that each cell is at 3.5v. It gives 7v in total and 140W. On converter 126W.
0.3ohm - max 6.14v
0.2ohm - max 5v.

Is that all correct now? (assume that converter can go up to these values, no power limit here)

 

[untar]

It's almost correct. To have a better approximation you should calculate with the nominal battery voltage, which is 3.7V for our cells. So "full cells" means you have ca. 7.4V under load (until the battery gets used a while, then less). As soon as you hit the fire button your 4.2V batteries won't have 4.2V any more.
So the fully charged calculation should be with 148W, so on the firing side with 90% efficiency you'd have 133.2W left as max. power if you don't want to draw more that 20A from your batteries.

Now let's check what happens at the lowest voltage, let's say 3.2V cutoff again (on battery side):
148W/2=74W per battery
74W/3.2V = 23.125A <- maybe a problem for your component, you will need to regulate power down

That component you linked looks good, but it can only go up to 110W. Now we're almost at the 100W I suggested in my 1st post
Also keep in mind that your batteries will need to power your atmega + your voltmeter (+ your Ω-meter) and lose heat there as well, so the efficiency will go down further.