How to reduce amperage from an 18650 li-ion cell to 3 amp 1.2 Volts?

Project is to adapt an 18650 cell to power an RC Glow plug. I just need to get the battery to power the glow plug to a working safe level.

From my multi meter I read around 5 Amps out of the battery and around 4.3 Volts when fully charged.

I have already tried a 2.7 and 3.3 Ohm resistor and its not worked. These look to have reduced the Amperage far to low.

By plugging the battery directly to the glow plug it burns the plug out.

From my understanding I need to reduce the supply down to 3 Amps and 1.2 Volts.

Anybody have a suggestions?

Further reading on the glow plug:

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luxstar4 years ago

Efficient dc to dc converter. If the wattage or amperage are not high enough Murata makes higher rated devices. This one can be set for an output of one half volt up to 6 volts at 3 amps.

It’s here:



Harb247 (author) 4 years ago
Thanks for all the great answers.

Having a read I think I'm better off with a voltage regulator.

The reason for the 18650 cells is mostly because I have lots off them, their capacity is excellent, easy to charge and the cells are easy to swap when it starts to fail.

Can anybody point me to a suitable regulator I can use? Maplin sell them but there's always better value from eBay.

I have come across one particular regulator which specifically mentions giving a guaranteed value of 3A and within the Voltage range I need. Do I need to purchase something like this?


For the adjustable regulators am I correct in thinking I need to use only 2 resistors to make it work and add 2 capacitors for stability? How do I work out the values of the resistors to give me what I need?

Heat could be an issue as I am enclosing the entire thing inside a torch housing which fits perfectly onto the fitting from a dead glow plug starter I had already.

I came across an excellent tutorial on using Voltage regulators but it talks mostly about fixed voltage regulators instead of the more common adjustable that I have found.


I could use the body as a heat sink but I expect it to get very hot over time especially as its very long but not very thick. the cars can be difficult when cold and normally results in the battery's dying on these glow starters.

Toby Robb4 years ago
I am using an lm 350k linear regulator with a heat sink. Inefficient as heck but runs on a 3s or 2s lipo and only has 4 parts. i only use it for a brief period anywayz. gets VERY HOT.
Re-design4 years ago
Don't worry about the amps. You need at least 3 amps to light up the glow plug if that's what the spec's are.

Voltage is what's burning out your plugs.  You should use a voltage regulator to get the voltage down to 1.5 volts.  Click here for examples.  I used to fly RC planes and ran my plugs off a 12 volt gel cell using a voltage regulator,
I think the most satisfying way to do this would be using a switching regulator,or PWM, rather than a linear regulator, or some fixed resistance in series, just because you'd waste less power that way.

By the way, I estimate the amount of fixed resistance you'd want in series with this thing to be about 0.8 ohm.  That number came from assuming that your battery voltage is 3.6 volts, a current of 3.0 amperes flows through it and the resistor and the glow plug all in series, and 2.4 volts drops across the resistor, and 1.2 volts drops across
glow plug.

The glow plug itself, with a current of 3.0 A through it, and a drop of 1.2 V across it, kind of "looks like" a 0.4 ohm resistor, since (3.0 A)*(0.4 ohm) = 1.2 V

In terms of power, that's 3.6 watts dissipated by the glow plug, and 7.2 W dissipated by the resistors, and an efficiency of only 33%.

The power losses using a linear regulator look the same, but the advantage of using a linear regulator is that it sort of like a resistor that "adjusts" itself,  in such a way as to keep the output voltage constant.  Coincidentally the regulation scheme for the the popular LM317 wants tries to keep the voltage between its Vout pin, and its Vadj pin, equal to 1.25 V, which is a number that is close to your desired voltage of 1.2 V, so you could probably use the LM317 without any external resistors.  Just tie Vadj to ground, and that'd be a neat trick

For the PWM (pulse width modulation), the way that works is you put your glow plug in series with a transistor that is quickly turning on and off, but is only on for fraction of each period. E.g. if the glow plug truly looked like a 0.4 ohm resistor, you could 9 A pulses at it (since 3.6V/0.4ohm = 9A), but only have these pulses be on 1/3 of the time, so that the time averaged current would be only 3A.

A switching regulator looks a lot like PWM, in that there is a transistor quickly switching things on and off.  The difference is that a switching regulator uses filtering, to sort of smooth out the pulses. Also the switching regulator uses feedback to make some characteristic of its output be constant. Usually it's constant a output voltage, or a constant output current, the regulator is trying to achieve.

Constant current switching regulators used to be rare, but they are more common now, because this is the kind of power LEDs want.  So high powered LED drivers are typically switching regulators designed to deliver constant current.

The reason that information is relevant to you, is that I think such a regulator intended for driving an LED would work for your glow plug, provided you can find one that regulates to a constant current of around 3A, and that will accept input voltage of around 3.6 V or so, provided by your Li-ion battery.

Also a constant voltage regulator would probably work too, and for that the spec you are looking for is a constant output voltage of around 1.2 V, again with an input that is willing to accept the 3.6 V or so provided by your battery.  

It makes me wonder why they do not build regulator/drivers specifically intended for driving a glow plug from a single Li-ion battery.
frollard4 years ago

This circuit will limit the current to whatever you need;

You don't need to limit the voltage to the glow plug, you need to limit the current - and since it's (I presume) a boring old resistive load, this will work well.

By my calculations they say current is 0.5/R3 in the diagram
so R3 needs to be .5/3 ohms = 0.1667 ohms.
Why not use an AA battery instead?

You have a couple of options for reducing the voltage. You can use a voltage divider or you can use a voltage regulator like the LM317.

With the right resistor you can set the output of the regulator where you want it. Fortunatly 1.2V is about as low as it will go but your limited to about 1A. But you can connect several of them in Parallel to get more amperage. Pluss you will need a good heat sync for them. 

For the voltage divider you'll have to use some high power resistors. Especially with the amount of current you want to pull through them.