Firstly, let me say thank you for all the responses. I appreciate it.

Further to my last, I today tried a little experiment - I ran my 3Ah 12v lithium polymer battery to a heating element through a variable voltage tester I made (basically a potentiometer with a voltage gauge attached). I also connected my multimeter to measure the amperage.

I found if I kept the voltage to around 5v, this kept the power consumption to about 2.8A and the element worked, although it didn't get very hot.

If I went to 6v, the power went to 3A and the battery overload protection kicked in.

Tomorrow I'm going to try using a regulator to output 5v and see if I can make it work that way. If it does, then in theory I could attach any battery to the system (over 3Ah) and not have to worry about the overload protection.

Thank you all again for the help - I'm certainly heading in the right direction.


First off, let me start by saying my speciality was automotive engineering and as such I am a hopeless duffer when it comes to the magical dancing pixies of electricity, so go easy on me.

Since I retired, I've been building all kinds of weird stuff in my workshop in the garden, one of which is smoke generators; specifically small ones, from a few inches on a side down to my smallest which is less than an inch square.

I've built loads of the things for various projects and it's all been fine. I power them from sealed 12v 1.2Ah lead acid batteries, the kind you get for alarm systems, and they work great, to a point. The trouble is, I want to improve the electrical side more. I'd like to use 12v Lithium batteries, add timers, Arduino controls and so forth, but the problem is whenever I try to do this, the inbuilt overload protection in the battery kicks in and shuts it all down.

I think I know why this is happening - the heating element in the smoke generator is basically shorting out the battery - I'm using resistance to turn the electrical pixies into heat (the element wire comes from a hair dryer), so what I need is a way to limit the current going into the generator (I think).

Now, as I mentioned, I'm useless with this kind of stuff. I can build basic circuits, solder and such but other than that I'm at a bit of a loss. I put a multimeter on it and it looks like it's drawing about 6v and 10A, which is kicking the snot out of it as it's a 1.2Ah battery, so I'd also like to do this to help save the battery as they have a very short lifespan because I'm abusing the heck out of them. I can use higher capacity batteries, but they're obviously bigger and heavier which kind of defeats the object of making tiny smoke generators.

Basically, I want to limit the output from the battery to avoid tripping the overload protection on the electronics. I tried using a 10ohm 10w resistor to limit the draw, which works on the multimeter (output is 11v 1.1A) but the resistor soaks up all the power so the element doesn't heat up.

Any assistance would be greatly appreciated.

For info, I've added a link to a short video showing one of the smoke systems in operation to give an idea of what I'm rambling on about.


  • \$\begingroup\$ You have your heating element that is 60 Watts, 10A at 6V. The load is 0.6 Ohm. If you really need that much 60W evaporator, then the only solution is to have a bigger battery. Again, if 60W is ok, you might need to increase the resistance of heater, because from a bigger 12V battery it will suck about 240 Watts and burn out. Or you need a more sophisticated DC-DC switching downconverter, from 12V to 6V, it will take only about 5-6A from the battery while delivering same 60W, so you will still need a bigger battery. \$\endgroup\$ Nov 4, 2016 at 21:14
  • \$\begingroup\$ 6A at 12V is nothing, and a 3-cell (11.1V nominal) lipo battery from an RC-hobby store such as Hobby King can handle pulsed and continuous loads in the 10's and 20's of amps. Something like this would work fine. It does NOT have an integrated protection circuit either, so it literally will never trip. hobbyking.com/en_us/… \$\endgroup\$
    – KyranF
    Nov 4, 2016 at 22:49
  • \$\begingroup\$ If your numbers are correct, 6V, 10A, your load is around 0.6 Ohms. You could just use two Lithium batteries in series. Then you might not need any additional circuitry. Find high-drain batteries that are rated to supply, say, 15A. \$\endgroup\$
    – user57037
    Nov 5, 2016 at 3:59

2 Answers 2


If you continue to use a 12V battery, you could limit the current by using a heating element with more resistance, or by placing multiple heating elements in series. If you place 2 of your elements in series, that should cut the current in half.

Another way to lower the current would be to use a battery with a lower voltage. You can buy 3.7V lithium batteries online (one option). The 12V lithium batteries are probably 4 lithium cells in series, so using one of these 3.7V batteries should cut the current running through your circuit by 75%.

If you use a lower voltage battery, you may still need to raise the resistance of the heating element by using a different element, or putting multiple in series, in order to stay within the current rating of the battery.


Do you know what Ohm's Law is. Or the Power formular?

If not then you really MUST find out how this affects what you are doing.

The biggest issue is HOW MUCH POWER DO I NEED?

And as you already discovered, adding resistors to limit current wastes a lot of power, there are better ways although they are more complex, PWM switching is one such way which is quite efficient.

You said it was "drawing about 6 volts and 10 Amps" {one doesn't 'draw' voltage, we don't say that}, from a 12 volt 1.2Ahr battery, and you said the heating element is shorting out the battery, - you are not wrong. Remember it is a 12v battery, at full charge it will be nearer 13volts. Any load you put on it that causes it to fall below 12 volts is what we call serious abuse for these small batteries. I know car batteries are pulled much lower during starting but thats another story.

I suggest you make a much larger element and see if it still works. Use a longer length of the same hairdryer heater wire. If it still works then make it larger again until you find out what does not work. You will Draw less current with the longer wire and therefore less power, and as a consequence you won't pull the battery towards death's door as quickly. Of course the wire won't heat up so much and the size might be a problem. If it is coiled then you need to be sure adjacent turns don't contact each other and create a short cicuit, which will negate all the forgoing.

Once you have an answer about how much power is needed then you can start to do something more around controlling it. You said you want to use Arduino to implement control. Fortunately Arduino has some simple to use PWM functions.- see analogWrite() in arduino reference. This is Pulse Width Modulation. Basically if you have 12 volts and turn it on and off fast enough {depending on the response time of the system} you can control the "effective" voltage you apply to something. If it is on half the time and off equally e.g. 50:50 {percent} you will be effectively applying 6 volts i.e. half the voltage. During the on time however, the current will be the same maximum value as if it were on permanently so you will have to deal with this. Your meter will only show average voltage or average current.

Once you have some of this worked out you are going to need a way to turn the power on and off, I suggest using Power MOSFETs. There are some with really low on resistance for use in low voltage high current applications. You will have to be sure the Arduino can turn it on with the 5 volts signal it can output, so make sure the MOSFET Gate threshold voltage is less then that.

You have your homework assignment, good luck. You'll probably need it. Be prepard to convert components into miniature single use smoke generators.


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