I am working on an ultra low power infrared wake-up circuit for a microcontroller (Atmel XMEGA). I have IRDA comms working but the receiver uses too much current to keep on all the time. I need to run for five years on a single battery, so <3uA is the goal. Cost is always a factor, as usual.

In fact my company already produces other products with such a wake-up circuit but they use somewhat expensive op-amps and take up quite a bit of board space. I'm thinking it can be done with just a single comparator, something like an MCP6546.

(can't post images, look here)


The resistors will set the comparison voltage based on what photodiode we pick, obviously. The reason for using it in photovoltaic mode like this is to reduce power consumption.

Will this work? Can it be improved upon? I found this design which is similar but uses a comparator with built in reference voltage.


I could do that too, but it would need to be something like an MCP65R41 that costs a lot less and is available.

  • \$\begingroup\$ What research have you done into the sepc of the photodiode in your circuit - have you managed to find out what signal it will produce when the correct amount of light hits it? \$\endgroup\$
    – Andy aka
    Commented Apr 25, 2013 at 17:09
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    \$\begingroup\$ -1 for directing us not to a schematic or image, but to something that tried to pop up another window and then started playing a advertisement. No thanks. \$\endgroup\$ Commented Apr 25, 2013 at 17:27
  • \$\begingroup\$ I have been using ImageShack for years and didn't get any complaints, but then again I use AdBlock so don't get any pop-ups or ads anyway. I'll avoid them in future. BTW, I have been using StackExchange for a while, strange that my reputation doesn't carry over. \$\endgroup\$
    – user
    Commented Apr 25, 2013 at 22:35
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    \$\begingroup\$ Consider the presence of light in the environment when using a simple comparator. Your circuit could stay awake constantly in outside applications or well-(sun)lit rooms. \$\endgroup\$
    – Mels
    Commented Jul 25, 2013 at 12:12

1 Answer 1


Okay, we managed to get down to about 3uA using an MCP65R41 in photoconductive mode. The MCP65R41 is so low power it gave us near 2uA to play with so using a large resistor to create a voltage for comparison became possible. It isn't quite as good as I was hoping for but meets our battery life requirement.

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    \$\begingroup\$ Would it be possible to share the design that worked for you - only the relevant portion as relates to the question, of course, not proprietary information. Also, what photodiode did you use? \$\endgroup\$ Commented Jun 25, 2013 at 11:27

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