I have a small hobby project for which I wired 4 AA batteries in parallel to achieve the desired result, which comes to 1.5V, as the batteries are in parallel, which is what works for my project.

My project works without the indicator light but the problem is that I have wired in a standard rocker switch to turn it on and off, but I would like an indicator light, and the light integrated to the rocker switch is a standard rocker switch led requiring assumedly 1.8V+ to power the integrated LED light.

Obviously it works fine without the light and I can visually see the I and O to know if it is on or off (O is off), but as this is a hobby, I want to have an indicator light.

For example when I wire 2 batteries in series, which makes 3V, then the rocker lights up.

I'm looking for a way to either increase the voltage only for the rocker switch but not out to what it is powering which needs to remain at 1.5V; or, to wire a second indicator LED (simply) to have an indicator light when it is on.

One way I thought to do it would be to wire each pair of 2 AA batteries in series, and the two packs in parallel, so that I get 3V output, which lights up the rocker - but then after the rocker, use a voltage regulator to step down the voltage from 3V back to 1.5V -- only problem is when I did a search, the devices seem to be rated at 3A, and 4 AA batteries are running at about 10A.

Note that this rocker switch is rated to 250V and lights up at 3V, but not 1.5V.

How to get an indicator light or light the rocker without increasing the voltage to the output device?


  -  +
  |  |
  ROCKER SWITCH [indicator?]
  |  |  |
  -  +  +(out)
  • 1
    \$\begingroup\$ Assuming the input voltage doesn't exceed the LED's voltage, you might consider using a basic joule thief circuit. However, keep in mind that this approach consumes some power in switching the transistor and other components which aren't 100% efficient. So the overall efficiency of the circuit goes down a bit. It might consume 8 - 10mA or more depending on the inductor as well. Additionally, it will continue operating until the battery voltage is too low to switch the transistor any more. \$\endgroup\$
    – 15 Volts
    Commented May 10 at 15:50
  • \$\begingroup\$ electrical, if you are willing to invest some time, I'd recommend looking into high efficiency LEDs to use together with something like the LM3909 (don't know if they are made anymore.) It can be done with a discrete circuit, too. About 2 mW. Just one 1.5 V battery, I'm thinking. (I'm zeroed in on your diagram there showing 1.5 V.) \$\endgroup\$ Commented May 10 at 16:33
  • \$\begingroup\$ LM3909 was discontinued over 10 years ago. \$\endgroup\$ Commented May 10 at 21:23
  • \$\begingroup\$ @JasenСлаваУкраїні I was worried. But a discrete circuit can be made that works almost alike. So it can be handled that way. \$\endgroup\$ Commented May 11 at 7:36

3 Answers 3


It's not easy or cheap to do with the internal LED in your stock inexpensive rocker switch, because you would have to generate a negative voltage due to the internal connections in a typical such switch. That, in itself, is not so hard, but the very low voltage input is an issue, especially if we take into account the minimum voltage of an alkaline cell under load.

If you are willing to use an external LED, you could use a small boost converter such as the QX2303 series. One of them is even available in hobbyist-friendly TO-92 as well as SOT-23. They will start up at 800mV minimum (and continue down to 600mV) at low output currents, and provide 5V out (say at 10mA) for your LED. For 10mA at 5V (50mW), assuming 80% efficiency, the draw from your batteries at 1.5V would be about 40mA.

enter image description here

Just connect Vin to your load and Vout through a resistor suitable for 5v to your external LED.

If you are insistent on using the internal LED, tack on an ICL7660 and a few ceramic capacitors to generate -5V and connect that through a resistor suitable to run the LED (it will get about 6V with 10mA draw). The battery current draw will be 10mA (or whatever LED current you choose) higher because of the LED plus a bit more to operate the 7660's internals.

  • 2
    \$\begingroup\$ Yep, either that or a double LVCMOS inverter configured as a voltage doubler and/or inverter+doubler. They have pretty low output impedance of say 10Ω or less - good enough to drive a LED at a couple mA. \$\endgroup\$ Commented May 10 at 19:02
  • 1
    \$\begingroup\$ @Kubahasn'tforgottenMonica LVC CMOS parts are spec'd down to 1.65V, the ultra-low power 74AUP down to 0.8V (though output current is uA at that voltage). I don't see typical figures. I imagine the 74AUP series would work, for some reasonable definition of work. They sometimes put really garbage LEDs into those switches, so the minimum LED current and the minimum operating voltage would be the TBDs. \$\endgroup\$ Commented May 10 at 19:35

With having 1.5V for input, the easiest solution is to use a 1.5V lamp.

The second easiest solution is to use a switched mode converter like the "joule thief", which has its own tag here, and is also covered on Wikipedia. Here's the circuit:

Joule Thief (wikipedia)


Here is another possibility, building on @Jasen's answer:

I've flipped the standard circuit to accommodate the internal connections of the rocker switch and suggested magnetics that worked for me typically.


simulate this circuit – Schematic created using CircuitLab

The transformer is an off-the-shelf 68uH toroidal inductor with an equal number of turns (25 or 30) added for the base drive using AWG30 wire-wrap wire.

This circuit typically starts at around 700mV and puts a bit less than 20mA through the LED at 1.6V in.

The LED current is not well controlled with this simple circuit so some experimentation and testing may be required to keep from overdriving the LED.

  • \$\begingroup\$ I've had my eyes on the LL series for a while. Did it give decent efficiency? Love how your design came out! \$\endgroup\$ Commented May 17 at 14:14
  • \$\begingroup\$ @MicroservicesOnDDD I didn't attempt to measure efficiency, but I glanced at the power supply current meter and I think efficiency was in the 60% range at 1.5V in. \$\endgroup\$ Commented May 17 at 14:18

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