Just trying to understand the difference between the battery usage between both of my EmonTH’s. One has a reed switch connected and the other is the optical sensor sold in the store.
The EmonTH with the optical sensor far outperforms battery life compared to the one with the reed switch. By way of comparison the EmonTH with the reed switch has already had it’s batteries replaced and they only lasted 2.5 months (full to empty). In the same timeframe the EmonTH with the optical sensor has dropped 0.1V.
Not that I’m in any way electrically minded but why would one use far more power than the other? The reed switch doesn’t have an LED so surely the only thing it would be counting is pulses and not necessarily powering the sensor. I have a capactior across the wires of the reed switch to smooth out any bounce.
Both are using the same batteries - Ikea rechargables and are in an outdoor meter cupboard. The optical sensor is measuring electricity and the reed sensor a gas meter. Given the time of year gas usage is much lower than electricity so the pules are far less on the gas side. This has left me completely stumped.
The question is, how long is the reed switch closed? If you stop using gas while the magnet is over the switch, then it’s drawing a small but measurable current for the whole time until you use some more gas and move the magnet away from the switch, in addition to the current needed to run the emonTH.
The resistor that limits the current is part of the processor, and its value is anywhere between 20 kΩ and 50 kΩ, meaning a current of between 66 µA and 165 µA.
In the case of the electricity meter, we know the sensor draws almost no current (the ‘dark’ current of the photodiode) except for the duration to the light pulse.
Thanks Robert, you’re a wealth of knowledge. Having a constant draw of power when the magnet is activating the switch is something that never occured to me.
I don’t suppose there is anything that I can do to extend the life of the batteries?
It’s an interesting thought: maybe you could turn off the internal pull-up resistor and replace it with an external component of higher (and more closely defined) value. The first thought is to try a 47 kΩ external pull-up. It’s within the range of the internal resistor, but close to the high end. If your pulse input happened to have a pull-up resistor that was at the low end of the range, it represents a quite significant current drain.