So you think there's no way I could get an IR receiver that draws 0.35mA to draw less and still work?
@Nick: a quick comment - when you measure the voltage drop on a 1Meg resistor pls do consider a standard meter has ~10Meg internal resistance. So, 1Meg || 10Meg = 0.909Meg, thus your nA values will be 9% off..
I'm between Sketch C and D in the link and reading 0.40mA after by turning ADC off. I should expect a huge power reduction according to his method, but I'm not. I don't know if I'm missing any steps leading up to that reduction, but if you could help, that'll be fantastic!
Okay, so I tried your method and measured the voltage across the circuit which comes to about 4.8V. Putting a 1M Ohmz resistor in series and measured the drop across it comes out to 4.1V. Did Ohmz law and got 4100 nanoAmps.
You know it sleeps properly when you are able to measure the current dissipation across the chip to be less than .1mA. The chip can go as low as .1uA if you set the fuses and clock correctly.
set_sleep_mode(SLEEP_MODE_PWR_DOWN); sleep_enable(); sleep_mode();
So I'm as confused as ever. Going into sleep mode definitely uses less power (I can measure the difference!) but consumption doesn't seem to be anywhere near as low as it should be.