>From: Bob Drzyzgula >Subject: Current sensing again >What I'd most like to do is monitor the current draw from >each 20A circuit. 20 cheap analog AC ammeters, little bits of black cardboard glued onto the meter needles, and LEDs and phototransistors to detect when the meter gets above a certain point? No? :) How about this, run the power through several relays in parallel. In series with each relay is a fuse...20A for the first relay, 15A for the second, 10A for the third...etc. Your PIC monitors the input power (after all the relays) and drives the relay coils. First you turn on the 20A relay...then the 15A relay, then turn off the 20A...then turn on the 10A, turn off the 15A, etc...eventually you will blow a fuse and lose power, at which point you know the draw is somewhere between the last two ratings you used. Sort of like an R/2R resistor ladder from hell. >There are probably other good transducers >that don't need all those volts, and some which may not >require any power supply at all. I'd love to hear any >suggestions along these lines. It looks like you're using an industrial-control type 4-20 mA output sensor of some kind. That will work, but it's sort of like a 10k linear pot with an RS-232 interface. :) I always thought that current transformers just produced an AC voltage all by themselves. Essentially, you are wiring a transformer with a very-small-voltage primary in series with the power to the device of interest, and looking at the voltage generated on the secondary. You might want to rectify it before shoving it at a PIC, but I don't think you'd have to do much more. I have an All Electronics ( http://www.allcorp.com/ ) catalog that lists a current sensor. It's rated for 1-100 KHz, so it may not be as good at powerline frequencies, but it claims: "can read currents as high as 35A with a sensitivity of 10mA to 1A per volt, depending on the terminating resistor (200 ohms nominal)." It appears to just have two pins, and a hole in the middle that your wire would be routed through. The model given is "Falco #CS200". Price is $2 each. I don't work for them, but I have ordered from them in the past and been happy with the service. You may want to try building a "splitter" for the hot and neutral wires as you mentioned, then just trying a few dozen turns of small-gauge (24 or less?) insulated wire around the hot wire, and see what that gets you. If you won't want to build several splitters, you *may* be able to locate your current transformers in a handy junction box, or at the main breaker panel...check code/an electrician on this, though. I know I saw a PIC-based power meter project recently... I think it may have even been in a Microchip app note. Anyway, they were measuring the voltage drop across a "resistor" - a known length and gauge of wire whose (low) resistance was known. I think they were measuring amps and fudging a little by saying P = V x I which is not always strictly true for AC. But you might be able to adapt the same sensing technique. If you used an 0.01 ohm resistor (yes, you can get them this low - check Mouser), it would drop 0.2V at 20A and dissipate 4 watts doing so. You'd probably want to isolate this from your circuit though...maybe a small 1:1 transformer or even an optocoupler would do it. It would be fairly linear with current... 0V at 0A, 0.1V at 10A, 0.2V at 20A. You can get a better range by going to a bigger resistor, but then you "lose" more voltage as heat. (0.1 ohm resistor gives you 2V at 20A but also must dissipate 40 W). I hope this helps! Matt Roberds mroberds@worldnet.att.net