I have a device that needs to detect the presence of voltages from 12V DC up to 240V AC RMS (Absolute maximum of 9V DC to 250V AC RMS) This device is going to have up to 8 channels that should be able to work independently with complete isolation between them.
I started working on a circuit that could accomplish this but can't seem to find a fit that wouldn't burn up with high heat dissipation or fail to trigger with low current.
The 2 options that I was thinking of were:
Use a small value capacitor or high-value resistor to limit current to an optocoupler. A capacitor wouldn't work with DC voltages, (if I am not mistaken) and a high-value resistor would not be able to drive enough current to pull down a 100k resistor on the transistor side of the optocoupler at low DC voltages. If I lowered the resistance there would be enough current at low DC voltages but my calculations show it would be in the order of 2 watts of dissipation at the resistor.
Use a high-value resistor with a Zener to limit voltage to around 9V on the sensing side. This should have the added benefit of triggering at a set voltage rather than varying based on the current transfer of the optocoupler. My issue here is that I then need to get the signal back over the optocoupler, there is not enough current left in the single to drive the Opto without some buffer that would require an additional isolated power supply. This could work but having an isolated supply per channel is quite of bit of wasted space and value. The stepping nature of the supplies would add a lot of noise too.
Possible solutions:
Is there a better circuit or IC that could accomplish this task?
Is there a low cost, low current (< 100mA), small footprint isolated DC-DC power supply available? I found the RFM-0505S, it an isolated 5V -> 5V supply at 200mA, much more than required and relatively large but could work.
Would there be enough current available after running 12v DC through a 300k+ resistor to drive the base of a transistor or would I need an op-amp to buffer the voltage?
Thanks in advance for your help!