292Views10Replies

Author Options:

Inductor Usage Answered

I have an 10mH inductor pressed against the active line of an extension cord connected to a load (a washing machine) that produces 9 amps. How do I measure the voltage produced by the inductor? Does it need to have a current already in order to react to the magnetic field? Does the negative lead produce a negative AC and the positive frequency such that if I attached a voltometer I would get a 0 reading? That's what I am getting even on my voltometer, even on the finest setting. I should be reading 0.09V, I believe.

10 Replies

user
NachoMahma (author)2007-08-02

> How do I measure the voltage produced by the inductor?
. With a voltmeter. Open circuit voltage will usually be higher than if there is a load attached.
.
> Does it need to have a current already in order to react to the magnetic field?
. Yes. An inductor is just half of a transformer - no current flow = no changing field = no output.
.
> Does the negative lead produce a negative AC and the positive frequency such that if I attached a voltometer I would get a 0 reading?
. No. As LV pointed out, AC is always changing polarity.
.
. As LV also pointed out, you will need a current tap (just a coil of wire) around _one_ of your power lines. Make sure R1-R4 are set for the appropriate values (based on the output of your CT). You will probably need a load/dropping resistor across the CT.
.
. All that said, wouldn't it be easier to hook up a relay to the washing machine timer?

Select as Best AnswerUndo Best Answer

user

> How do I measure the voltage produced by the inductor?
. With a voltmeter. Open circuit voltage will usually be higher than if there is a load attached.
So attach another voltage source and measure the difference?

> Does it need to have a current already in order to react to the magnetic field?
. Yes. An inductor is just half of a transformer - no current flow = no changing field = no output.
Ok so then it definitely must have a voltage source. If I use DC the AC voltometer will still read the new higher limit AC correctly I hope. Right?

> Does the negative lead produce a negative AC and the positive frequency such that if I attached a voltometer I would get a 0 reading?
. No. As LV pointed out, AC is always changing polarity

I don't know why I keep getting the properties of DC and AC confused. Inexperience I guess. I should know better because I've studied a beginning passive circuit course last Spring.

As LV also pointed out, you will need a current tap (just a coil of wire) around _one_ of your power lines. Make sure R1-R4 are set for the appropriate values (based on the output of your CT). You will probably need a load/dropping resistor across the CT.
This is a different recommendation than using a premade inductor I think. If I understand correctly, I wind some wire around the hot wire (with a load dropping resistor somewhere) then I can math out the inductance (*shivers*) and work from there.

All that said, wouldn't it be easier to hook up a relay to the washing machine timer?
Yes. It would be so much easier. I don't know why I had decided that non-invasive was THAT important for this exhibit (a Maker Faire exhibit). I could always prototype a non-invasive version if I had some actually subsequent commercial interest. However the point was really the remote indicators and not the sensor so yes. It may well be worth it to attach a relay instead, and then move on.

Thanks for all of your feedback by the way. It helps to not be utterly alone with this voodoo science.

Select as Best AnswerUndo Best Answer

user

> So attach another voltage source and measure the difference?
. No. No external V source is needed for the inductor. A current flow is induced in the inductor by a changing field.
.
> I wind some wire around the hot wire (with a load dropping resistor
somewhere) then I can math out the inductance (*shivers*) and work from there.
. The neutral wire will work just as well. Since you are just looking for an on/not on signal, construction won't be critical as long as it puts out enough for the opamp, but not so much to blow it out.
.
> I don't know why I had decided that non-invasive was THAT important for this exhibit
. I don't blame ya - it does have a high "cool factor" for a project. I'm just afraid that you haven't learned enough (yet) to pull it off.
.
> Thanks for all of your feedback by the way. It helps to not be utterly alone with this voodoo science.
. No problem. Capacitors have always been voodoo to me - I can do the calcs, but it doesn't make sense to me.

Select as Best AnswerUndo Best Answer

user
watchingmachine (author)2007-08-02

Thanks alot for the reply, LV. Fact is I am not trying to salvage electricity so much as merely trying to create a tiny amount of AC to get an op amp to illuminate an LED. If I get there, in the next phase I will have it send a simple "on" or "off" signal wirelessly to another device instead of illuminating the LED. The circuit plan I am using is the simple on in the upper leftmost corner.

Select as Best AnswerUndo Best Answer

user
gmoon (author)watchingmachine2007-08-03

Have you tried the circuit, or just tested the inductor? Also, have you thought about using a sensitive Hall effect sensor, instead of the inductor? They even make 'split ring' versions for non-invasive sensing (what you're attempting.) I'd build the thing, then remove R4--make the op amp a comparator and see what happens. When (if) the LED lights, it will be pulsed, but that shouldn't matter. The advice about sensing off a single AC line is a good one. And I assume you're going to power the op amp with a battery....

Select as Best AnswerUndo Best Answer

user
westfw (author)2007-08-03

An inductor designed for use in an electronics circuit is probably designed NOT to pick up voltages from external magnetic fields (or the other way around - not to pollute the nearby environment with magnetic fields), and is not the best choice for sensing current as you've described. Usually you'd want something called a "current transformer" to do this sort of measurement. They're usually torroidal things with lots of windings, carefully made so output voltages is proportional to th e current on a single wire strung THROUGH the torroid. Here's a picture of a smart current-monitoring outlet-switching power strip (broken, alas...)

Select as Best AnswerUndo Best Answer

user
lemonie (author)2007-08-02

You are using an AC setting on the voltmeter, and taking the reading while the appliance is drawing it's 9A (on a heating or spin cycle for example)?

L

Select as Best AnswerUndo Best Answer

user
watchingmachine (author)lemonie2007-08-02

Now that I think of I am not sure I was using the AC setting a voltometer. However I was using it while the machine was running. I have a clamp ammeter and it reads the current in the extension cord perfectly.

Select as Best AnswerUndo Best Answer

user
watchingmachine (author)2007-08-02

I forgot to mention that I have in fact built this. I am trying to troubleshoot it now (assuming that it works).

Select as Best AnswerUndo Best Answer

user
LasVegas (author)2007-08-02

You're trying to construct a transformer... AC is AC. It doesn't have polarity, but Alternates the Current between positive and negative. Now... You're placing the inductor (coil) next to a power cord containing a hot line, a return line and a ground twisted around each other. The magnetic field emanating from that cord is almost nil. The twisting of the power and return lines effectively cancel each other's magnetic fields.

Even if you were to seperate the hot line alone, you would have to wrap it into a coil around the other coil or through a magnetic cunductor shared by the other coil. The resulting voltage would be the ratio of windings between the two after the voltage drop of the washing machine. You wouldn't be gaining any free electricity but taking it away from the washing machine.

Select as Best AnswerUndo Best Answer