I have a design that runs from a lithium coin cell battery with a nominal voltage of 3V but running down to ~2.2V. I'd like to be able to momentarily (~250-500ms) pulse a small Blue LED with a Vf of 3.3V

I am aware of conventional options like voltage converters, the joule thief, etc. but I am looking for a very simple, low cost, and power efficient method to drive this LED with the battery I have. A microcontroller with various peripherals is available on the board.

Any clever suggestions? Is a voltage doubler via microcontroller PWM a reasonable method to drive a few mA?

  • 3
    \$\begingroup\$ "very simple, low cost, and power efficient " are mutually exclusive. \$\endgroup\$ – PlasmaHH Nov 16 '17 at 20:05
  • \$\begingroup\$ How many if "a few" \$\endgroup\$ – Trevor_G Nov 16 '17 at 20:15
  • \$\begingroup\$ see this post electronics.stackexchange.com/questions/340226/… \$\endgroup\$ – Trevor_G Nov 16 '17 at 20:18
  • \$\begingroup\$ Stay completely away from the Joule Thief for this application. It's highly inefficient. It's fantastic for what it achieves (when designed and built right, which it almost never actually is.) But efficiency isn't even in the same universe with it. Also, coin cells are targeted at hundreds of microamps, not dozens of milliamps (or still more, if using the Joule Thief.) How often will this be pulsed? Are you only looking for a few mA through the LED during this pulse? Or? Is this MCU triggered? Or are you looking for a blinking LED, only? \$\endgroup\$ – jonk Nov 16 '17 at 21:26

If you have a microcontroller, then from the top of my head:


simulate this circuit – Schematic created using CircuitLab

To light the LED, set the "DRV" pin to output a "0" then pulse the "CP" pin with a square wave. This makes a crude voltage doubler / charge pump.

The LED current limit comes from the fact that the cap will transfer a certain amount of charge per pulse, and charge multiplied by frequency (ie charge per second) is current. Also your micro's IO pins will have 20-50 ohm output resistance. So you don't need a resistor... you can add one if you want.

That's gotta be the cheapest way to do it. Efficiency won't be that good, but it will light your LED. If you only want to light it once in a while as an indicator, not keep it lit for a long time to make a flashlight, it'll be fine. You can also dim it by adjusting the frequency.

Not bad for 2 parts ;)

  • \$\begingroup\$ This might be the best solution even if you do not already have a uControlers since adding one might be easier and cost less than other alternatives and make it possible to do fun things like have the LED "breathe" or blink. BTW, on many uControllers, D1 may even be unnecessary as you can use an IO pin to drive the top leg (shown as Vcc here)and float that pin in software during the boost phase. I've done exactly this on AVRs and it works great with 3V coin cells! \$\endgroup\$ – bigjosh Nov 17 '17 at 18:16

A joule thief together with an oscillator sounds like one solution for you, especially an inverter ring oscillator. And to make that ring oscillator oscillate slower, in the Hz region, put one RC low-pass filter between the logical inverters.

This was a comment, and I knew that Pipe would've come by and said that this was an answer in the comment section. Therefor it's now as an answer.

To make this more of an answer, this is the oscillator I have in mind:

enter image description here

Here's the link for it in case you want to play around with it.

The 5 V supply would of course be connected to the output of the joule thief. As you can see, it reaches 3.07 Hz which is roughly 330 ms period, so if you want to make it oscillate faster then reduce the RC values.

  • \$\begingroup\$ Wait, where's the 5V supply come from? Isn't OP's supply ~2.2V? \$\endgroup\$ – calcium3000 Nov 16 '17 at 20:26
  • \$\begingroup\$ @calcium3000 Re read the last paragraph. I'm assuming the joule thief will make some voltage in the 5-10 V regions. Also re read my first paragraph that states that a joule thief together with this oscillator above = one solution. And re read this comment stating everything again. \$\endgroup\$ – Harry Svensson Nov 16 '17 at 20:29
  • \$\begingroup\$ Ah okay, got it. I thought the circuit above was the whole shebang. Maybe add in an image of the joule thief circuit? \$\endgroup\$ – calcium3000 Nov 16 '17 at 20:37
  • \$\begingroup\$ @calcium3000 Making that particular circuit is difficult to make in circuitJS because of the windings of the core, so instead I'll just make it more clear with text in the image as well. Well actually.. I could make that schematic but no one would be able to tell "Ah! That's a Joule thief!". \$\endgroup\$ – Harry Svensson Nov 16 '17 at 20:58
  • \$\begingroup\$ @calcium3000 The joule thief is very inefficient. \$\endgroup\$ – jonk Nov 16 '17 at 21:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.