Electronic – Leach circuit to power microcontroller

current-sourcelow-powermains

Like many, I'm trying to design small circuits to do things. One objective is a device to report home power use. Such a device is relatively easy. Powering such a device for the very long term is hard. Batteries die, solar panels advertise that there's something in the meter box worth stealing, etc.

On the other hand, why can't I just connect a current clamp (http://dlnmh9ip6v2uc.cloudfront.net/datasheets/Sensors/Current/ECS1030-L72-SPEC.pdf) to the mains supply and run the monitor from the leached power?

Use a blocking diode and a fairly large(?) capacitor across the terminals of the clamp and add a 3.3V voltage regulator, with ground and supply to the respective sides of the capacitor.

The capacitor won't be "perfect" and should add enough resistance to limit the voltage, I'd have thought.

Anyone tried anything like this? Would it work? Suggestions, comments and crazy to add? (Note that I wouldn't take it amiss if you said "Go home, you're drunk". This would, right now, be good advice!)

Thanks

Edit to add: I can't plug in an adapter. I have the mains cables, but this is at the meter box outside. No sockets are present.

Edit 2: Target power "generation" would be ~1ma at >5V, feeding a 3.3V LDO regulator. Power use to be ~0.1mA, with a <1s pulse at ~40ma every ~2 minutes. Capacitor to support pulse.

Best Answer

It's a valid plan and I've seen others implement it before. Here's a link to a evaluation board that does AC lines monitoring and powers itself off of the AC Mains. Look at the schematic on page 10. The regulator is U2 and the rectification diode is Z1. You might be interested in the board anyways, it already does the voltage and current measurements and outputs the value over UART/I2C/SPI.

Back to your idea though. I would suggest the following modifications. We'll talk about the clamp in the next paragraph. Instead of a shunt capacitor make it a series capacitor. This provides AC coupling and blocks any DC current from forcing the diode on. The diode is a cheap form of rectification. Half wave rectification is probably good enough for your purposes. See the wiki page for some details on rectification.

Make sure that the regulator can support the output voltage of the rectifier. If you bring in the voltage directly from the line, the rectified voltage will be very large. This is one of the reasons I would use a transformer to step down the voltage. Additionally, placing a shunt zener diode can help decrease the rectifier output voltage. You might want this if the rectifier output is too larger for your regulator. See what they do in the schematic linked above. U7 is the shunt zener. Make sure that you add appropriate resistance to prevent excessive current flow if the zener conducts.

One issue I see with the plan is the current transformer that you are planning to use. Current transformers are typically used for current measurements. If you look at the I-V curve in the datasheet that you've linked, it provides mV of output voltage for tens of Amps on the primary side. This cannot supply the rectifier and regulator. Alternatives would be to use a transformer to reduce the AC voltage to something manageable and use the same rectification scheme. Another method would be to attach directly to the AC line as in the board I linked to. Be very sure that all of your parts are rated for the expected voltage or current.