4

I want to use a clamp-on ammeter (like this one) to detect whether a table saw (230V) is switched on or not. The clamp-on ammeter is meant to be used with a volt meter and has an 1 mVAC / A output signal.

I'm looking for a way to hook the clamp-on ammeter to a raspberry pi. All I need to know is whether the saw is switched on or off. Do I need some sort of ad converter and/or rectifier?

  • You're missing the current (or the power) of the table saw - without that there is no way to tell how much voltage can be obtained. – Mister Mystère Jan 2 '16 at 21:21
  • Grüezi, BetaRide. The question of Mister Mystere is still valid. What's the power rating / the current drawn of the saw? How fast does the detection have to be? I think that an opamp based rectifier could be sufficient here (no ADC needed). Such rectifiers could be simpler if just half-wave instead of full-wave. – Ghanima Jan 2 '16 at 21:49
  • The saw uses up to 2200W. But if I get this working I might usw this for other machines which use only somme 100W. – BetaRide Jan 2 '16 at 21:58
  • If the detection takes a 1 second, that's ok. Faster would be plus. – BetaRide Jan 2 '16 at 22:00
4

The first thing I would do is plug it into your multimeter and see what readings you get.

Note that current clamps (with very few exceptions) have to be clamped round the live wire only. Clamping round the whole cable will not work.

You may be able to get a larger (and thus easier to work with) reading by looping the live wire through the clamp more than once.

Then you will need to build an amplifier circuit, probablly based on a low noise op-amp. I would probablly use an AC coupled inverting amplifier design to minimise DC offsets on the output.

This would then be followed by an envelope detector to convert the AC to DC.

And finally a comparator to compare the DC voltage with a reference and provide a signal to the Pi.

| improve this answer | |
3

As @Peter Green suggests, I would definitely recommend wrapping the wire around the current clamp - indeed I would wrap both Live and Neutral in opposite directions so this will act as a multiplier!

Given the quoted maximum of 200A (for a single core) and the OP's Switzerland location where single-phase plugs are rated at 10 or 16 Amprms at 250 Vrms then 10 / 6 loops of both wires respectively means that a 2.2KW load (divided by a nominal 250Vrms) would produce an 8.8rms Amp current which if multiplied as suggested by 2x10 or 2x6 respectively would appear to be 176 or 105.6 Arms .

Using the 1mV/A conversion factor of the clamp this would yield a representative voltage of 176 or 105.6 mVrms. One thing that is not immediately obvious is that the Digital-Multi-Meters that the clamp is designed to be connected to is a High Impedance (probably of the order of 10 MOhms) to achieve this you are probably going to want to use an Op-Amp circuit - however you also want a rectified and smoothed DC voltage to feed into the Pi - whereas the voltage you are getting is an AC one.

In this situation however the alternating voltage is in you favour - you can capacitively couple the AC voltage into the Op-Amp with a gain of somewhere between 10 to 20 times (very easy to achieve) and capacitively couple the output to a half wave rectifier (diode) feeding a low-pass filter, then you can configure things so that you get a voltage that is of a magnitude (but clamped to a maximum of 3.3V !) that the Pi can monitor directly to confirm that a certain current level has been reached.

The only proviso is that if you are splitting a mains lead so that you can wrap the individual (live and neutral) cores around the clamp is that that will remove one of the two layers of insulation originally in the main cable, either you will need to enclose those wires and the clamp in a suitable box or you will need to make an in-line "adapter".

My suggestion for this adapter is a "loop" of main cable sufficient to wrap around the clamp the required 6 or 10 or whatever number of times and two more short lengths of the same cable with a plug on one length and a socket on the other and a suitable terminal block and an enclosure with five or six terminals. Wire the "plug" lead to three terminals and the "socket" lead to the other three (or two and use a common (the middle) terminal earth for a five terminal block case). Then you wire the neutral and earth wires from one end of the "loop" lead and the live from the other end to one set of terminals and the other wires to the other set of terminals. The idea here is to get the current flowing in the live wire in the "loop" to be flowing in the same direction as that in the neutral wire - in a normal cable they will be in opposite directions. Now we have that reversed we can simply coil the whole "loop" wire around the measuring clamp with the properly doubly insulated cable safely in the clamp, and provided the terminal block is properly put together with clamping/strain relieves on all four cables (the "plug" one in, the "socket" one out and the two ends of the "loop" cable) and securely closed to enclose the individual cable wires you will have a "safe" construction - just do not leave any magneticly sensitive things like credit cards in the vicinity of that loop when it is powered up and current is flowing!

| improve this answer | |
2

With the assumption of 10 A and a sensitivity of 1 mV/A of the clamp the resulting output will be a alternating voltage with an amplitude of 10 mV - that's not much. (If the current to operate the saw is significantly different, adjust the numbers accordingly but the idea is the same).

I would suggest a opamp based precision rectifier which is also used to amplify the voltage to detectable levels (see here). While precision is not needed here, the idea of an active rectifier is to rectify voltages smaller than the forward voltage of diodes. Note that this circuit is a half-wave rectifier only, so every second half-wave will not be detected (which would lead to an additional detection latency of 10 ms).

Selection of the opamp should mainly consider two aspects: single supply - to run from +5V only and low offset - since we're working on a few millivolts only here.

| improve this answer | |
  • Thanks for your answer. I do understand the idea but have bot enough knowledge to find the exact parts. Any chance to get more details? – BetaRide Jan 2 '16 at 22:15
  • I was afraid you'd ask ;) (that's why they should have kept the question at electronics...) The thing is that I am really not sure whether the offset is going to be of trouble here. The diodes are afaik of little interest they could be any cheap type (say 1N4148), the resistors are to be selected to provide the suitable amplification (see linked page). – Ghanima Jan 2 '16 at 22:31
2

plug it into a USB sound card.

| improve this answer | |
  • This isn't as silly as it sounds - it might work under the right circumstances! – SlySven Jan 3 '16 at 3:49
0

Not knowing your end goal, this might not help but.. If you only want to record whether the saw is operating or not, why not just take a feed from the switched side of the saw, measure the voltage there and then use a small transformer to make that feed 3.3v DC so you will get a HIGH or LOW reading on your Pi?

| improve this answer | |
  • A transformer would certainly work for stepping the 230V down to a lower voltage. However, you can't connect the output of the transformer directly to the GPIO. This is because, for one, the Pi can't handle AC inputs, and secondly the current would also need to be stepped down. These would both require using some kind of voltage/current regulation circuit. But I don't think it is the easiest solution to implement. – Darth Vader Feb 23 '17 at 12:57
  • I see what you are saying. My technical language skills need brushing up. I see now that I wrote transformer when what I actually meant was the inside of a AC/DC wall adapter like you might use to power a Pi or charge a phone, with a transformer, regulator, rectifier etc all ready to go. Then use a fuse or MCB to protect from over-current. With that said, I see that there are other issues with that and this would turn out to be more work than I originally thought. – P3Z8 Feb 28 '17 at 9:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.