I have a fan RF remote which runs using the a small 12v A23 battery. Note this is not 23 Amps but the battery model. They can provide something like 50mAh and are quite small.

I am hacking the remote for use with a home automation system however have been unable to run it free of the battery. I have tried connecting the battery terminals up to a 12V 2A power supply however it blew up the remote (capacitors I think) and made a burning smell.

I had a few of them so assuming I must has used the incorrect polarity I exploded another remote.

I am totally at a loss why I cannot run this from a power back, and the only working theory I have is a minimum power draw for my 12v 2Amp supply?

Does anyone have any other ideas?

Should I risk my last remote on a smaller power pack, say a 12v 600mAh?

  • 1
    \$\begingroup\$ Use an LM317. Connect a R = V/I = 1.25/50 mA =~~ 22 Ohm resistor from Vout to Vadj. Connect 12V to Vin, output from Vadj (NOT from Vout). This is now a ~~ 50 mA max current source. Try and draw more an Vout will fall. Adjust R as above to vary Imax. Note that Vout is max of about 1.5V - 2V less than Vin. Probably no problem here. \$\endgroup\$
    – Russell McMahon
    Jul 30, 2013 at 11:30

3 Answers 3


Older transformer based power supplies only have the stated voltage close to the stated current. If you measure the 12v 2 Amp or the 0.6 A supply with nothing connected, is it still 12V? Also, even newer switching supplies have a minimum current for stable operation (Typically 10%). That means the 2A supply might need atleast 200mA draw to have a regulated 12v.

The 23A is essentially 8 coin cell batteries in series. It is 55maH capacity, and it's internal resistance is unknown. This internal resistance acts like a resistor in series with any connected load, so current would be limited. This is why you can connect an typical LED to a 3v coin cell without a resistor, and not blow the LED. It simply can't provide too much current without it's voltage sagging (voltage droop). A generic datasheet for a generic 23A shows that with a 100Ω load (for 0.3 seconds testing), the battery voltage drops down ~4v from 12v to 8V. That's only 12v / 100 =, or 120mA draw.

Basically, what happens is that the battery itself acts like a resistor, preventing high current draw and dropping it's voltage as current draw increases. A remote could have been designed around this fact, as a way to protect the remote. Or your power supply is a unregulated/badly regulated supply. Since the caps blew, this might be the case. Caps normally blow because of high voltage. Test them first, if the 12v 600 mA is 12v without a load, try connecting it to a remote with a 100Ω resistor in series. This should limit it somewhere between 120mA to 10mA draw (depending on how much voltage the resistor drops).


The 12V supply requirement for your remote seems to imply that this may be an RF remote. (Most infrared remote controls run on a pair of AA or AAA cells and so this helps support this conclusion). I've also seen multiple instances of garage door remotes and wireless security system door sensors that operated in RF mode and used these very small 12V batteries.

RF remote transmitters often times use the series resistance of the battery to limit the current into the final output stage of the transmitter instead of using some other circuit design to do this limiting. There are advantages to this including:

  1. The technique reduces component count and cost.
  2. Less components can lead to smaller size.
  3. The circuit works over a wider range of battery voltage as the battery discharges.

In your case you may need to put a current source circuit between your power supply and the remote module to limit the total current to the transmitter.

One way to evaluate the voltage versus current draw for your transmitter is to use a variable lab bench power supply that supports adjustable current limit. Set it initially to a very low setting. Then slowly increase the current a step at a time and note the current and voltage at the inputs of your module.

  • \$\begingroup\$ I just assumed the internal resistance was designed into the remote, glad to have confirmation of that. \$\endgroup\$
    – Passerby
    Jul 30, 2013 at 22:02

As an alternative, try a 9v battery. I just put my garage door remote on a benchtop supply and tried it on various voltages. It did just fine at 9.6 and 9v. All the way down to 7.0v just fine. That was enough to make me happy and I'm about to solder a 9v battery connector in parallel to the A23 (in case I want to go back to 12v A23) and will mount an external 9v low-self-discharge NiMh battery to the remote. No doubt, a 7.2v NiCad or NiMH pack would also work or two LiIon or LiPo cells for 8v.


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