Hi -

I'd like to find a way to remotely tell when my air conditioner's compressor is running.

My first attempt at this was to use a clamp-on ammeter with an external winding around its iron core as a current transformer, and light a pair of low-wattage lamps when the higher compressor current is being drawn. I used two lamps in parallel for safety, so if one burns out the other will prevent the CT secondary from opening up.

It didn't work, and here is a diagram showing the setup and describing the results:

http://www.randomcollection.info/curr-trans-setup.jp

I've found cold air blowing out of the vents is a very reliable indicator.

THe compressor outside is closely tied to the blower fan inside, unless you have yours set to run constantly.

I'm trying to have the compressor indicator show in places away from the air conditioner.

The circulator fan does run continuously, but the current is far lower than the compressor - doesn't even budge the needle on the ammeter set to the 6 amp scale.

Still, the big puzzle is why when the CT secondary is open, there isn't even a tiny vestige of voltage at the lamp sockets.

Eleanor White

You should have the CT's around the motor T leads, not the meter. I hope you know the safety issues involved with an open secondary on a CT.

A suggestion that might be better for you. If you can find a small 24 VAC contactor with an auxiliary contact on the side, you could use a circuit through the aux. contact to turn a light on. It won't tell you the motor is drawing current, but it will tell you when it should be on.

I am familiar with the hazards of open CTs.

The meter's iron core loop is being driven by the motor leads' current, and the meter core's flux is proportional to the motor's current. I can understand if the meter's iron core loop isn't hefty enough to drive the lamps, and I may have to try a larger iron core, eventually.

But that doesn't explain ZERO voltage across the OPEN secondary, a hazardous condition, but I tried it with high voltage probes, when the lamps are both disconnected.

Even if the meter's core is too skimpy to drive the lamps, at least there should be some voltage when the secondary is open.

I think I'll hunt up a heftier core and see what happens.

Eleanor White

Eleanor,
I kind of get the impression that you are more interested in making the CT work than you are in the result. That is all fine. Take the challenge if that is your thinking.
If you are not, then there are probably better and much safer ways to get an indicator that will work. You just let us know please.

If you are set on using the CT, put it around the motor lead so that you have a better idea of how to rate the CT. Example 30amps = 50 millivolts for the CT. Even though the clamp on meter is proportional, it isn't probably anywhere near the flux of the wire. ( My opinion only for what it's worth). If I designed CT's and meters I could help more. I used them throughout my career, but always purchased the CT for the application. If you are trying to design CT's for a class project or something, then better do some heavier reading from a more in depth source. Always glad to try if you have questions. There are some engineers who float in out of this site who may have more background in design of the CT if they respond.

What I'm trying to avoid is having to disassemble the air conditioner.

The only difference between the line cord and the motor leads is the very small current which is used to drive the circulator fan. The circulator fan is very small - it doesn't even move the meter needle on the 6 amp scale. That means that the small circulating fan current can be ignored for practical purposes.

As to safety, I live alone, and with two lamps, if one burns out I would notice it right away and replace it. The turns ratio is designed to power one lamp at its normal rated current, 0.5 amp so a single lamp would not flare up and burn out if its partner burnt out. Nobody else around to disturb the setup.

I may end up using a sensitive electronic circuit and relay to operate the indicator lamps, instead of attempting to light them directly from a CT, though the CT setup would have been simple.

I still don't understand why the AC voltmeter reads absolutely ZERO when one lamp is removed, and finally when both lamps are removed, creating an open circuit. If that is a true reading, I couldn't even use the CT idea to trigger a sensitive detector circuit.

Eleanor White

You show a green wire carrying current. That is not permitted in the US. Is there no black or red? Sure you are not on the low voltage side?

The most common design of CT consists of a length of wire wrapped many times around a silicon steel ring passed 'around' the circuit being measured. The CT's primary circuit therefore consists of a single 'turn' of conductor, with a secondary of many tens or hundreds of turns.

http://upload.wikimedia.org/wikipedi...chnung.svg.png

Sorry, I had an some formulas for you to use to help you figure out your question, but when I submitted them they disappeared. I have had that happen too often on this site. To make a long story shorter, try taking the power lead and making an extra loop through your clamp on. The more loops you make the closer the primary ratio is to the secondary ratio of the CT you are trying to create. In the previous picture "n" represents the number of turns in the secondary of the CT. The current in the secondary winding(yours) times the "burden" resistance should give the voltage of the CT at the burden point. Try this and get back to me. It is late and I don't want to assemble everything all over again tonight.

Primary Current
Secondary Current
Number of turns in primary coil
Number of turns in secondary coil
Resistance of Load
Voltage drop across load





Keep in mind as resistance of the load increases so does voltage drop across the load.

Example of Primary Coil Looping.
Attachment 46164

Thanks to all who have answered. I'm answering two replies here:

"You show a green wire carrying current. That is not permitted in the US. Is there no black or red? Sure you are not on the low voltage side?"

EW: I just used green for the diagram. The actual wire is the "hot" (not ground, not neutral) wire powering the air conditioner. It is a short ribbon-style extension cord with the conductors separated for the clamp-on meter.

"The most common design of CT consists of a length of wire wrapped many times around a silicon steel ring passed 'around' the circuit being measured. The CT's primary circuit therefore consists of a single 'turn' of conductor, with a secondary of many tens or hundreds of turns."

EW: I understand. All I need for this particular low current setup is a turns ratio of 1:6, to reduce the single "turn" carrying 3 amps to the half-amp lamp rating.

The meter does have some type of ferrous ring, and no doubt many turns around the ring to reduce the current to whatever the meter movement requires. My setup has nothing to do with the internal works of the meter - I just wrapped wire around the same ferrous ring, trying to make the ferrous ring do two jobs.

I can understand if the small ferrous ring doesn't carry enough flux to light the lamps, but what puzzles me is why, even with an open circuit, the 6 turns shows ZERO voltage. That doesn't make sense.

By the way, I did test the (new) lamp sockets before attaching them to the 6 turns for internal shorts. The resistance was infinity - no shorts there.

One reason for wanting a setup like this is that it could be used on any air conditioner or other appliance with the same current range. No need to "do surgery" on other appliances.

Another reason is I'm in my 70s and would have to get someone to pull the air conditioner out, do a lot of tinkering, then get someone to put it back in the window. Not something I'd like to do, if a simple external circuit would do the job.

I may try a non-ferrous, non-metallic core form with a gap so it will slip over the hot power lead, and wind that "air" core with some number of turns, and see if I can at least get a little voltage to show. With that "air" core, it's very loosely coupled and relatively safe compared with an iron core. (Still I will observe high voltage safety.)

I could then feed that small voltage into an op-amp, say, and drive a relay.

But regardless, I would like to know why the original setup provides zero voltage with an open circuit, and the meter reading 3 amps.

Eleanor White

Replying to Studs ad:

"Sorry, I had an some formulas for you to use to help you figure out your question, but when I submitted them they disappeared. I have had that happen too often on this site. To make a long story shorter, try taking the power lead and making an extra loop through your clamp on. The more loops you make the closer the primary ratio is to the secondary ratio of the CT you are trying to create. In the previous picture "n" represents the number of turns in the secondary of the CT. The current in the secondary winding(yours) times the "burden" resistance should give the voltage of the CT at the burden point. Try this and get back to me. It is late and I don't want to assemble everything all over again tonight."

Same result - lamps don't light, and zero voltage even though the 6 turns are open-circuited (lamps removed from sockets.)

I do understand the theory of how CTs work - that's not the issue.

The issue is that even with the lamps removed, open circuit in other words, why is there ZERO voltage across the 6 turns? The meter shows 3 amps, so clearly, the ferrous ring of the meter is carrying flux.

Eleanor White

Eleanor,
When answering questions on this site, we seldom know the expertise of the people we are dealing with, so we have to generally dumb down our answers so to speak to communicate. You obviously have more back ground and I apologize if we treated you with less respect than you deserve. I can't tell you the makeup of your meter flux, but it is quite apparent that it isn't sufficient to induce current into the secondary winding or you would have some voltage at the open circuit.

formula: V (secondary) = I (secondary) x R (secondary). Since you have a very high resistance on the secondary, any current at all should give a secondary voltage. Your meter impedance is probably in the mega ohms, so for example 1,000,000 ohms x .0001 amps would equal 100 volts which you could read. Based on what I can see you are not inducing anything into your secondary CT coil.

Studs ad wrote:

"formula: V (secondary) = I (secondary) x R (secondary). Since you have a very high resistance on the secondary, any current at all should give a secondary voltage. Your meter impedance is probably in the mega ohms, so for example 1,000,000 ohms x .0001 amps would equal 100 volts which you could read. Based on what I can see you are not inducing anything into your secondary CT coil."

That's it, exactly. That's why I posted here.

I can't tell how massive the ferrous ring in this particular meter is, however, I've seen rings in other clamp-on meters and their rings are fairly substantial, at least to the cross section of small audio transformers. I don't want to split open the case in my meter to find out.

I'm tempted to try a heftier iron core next, rather than building an amplifier to drive a relay.

Eleanor White

MORE: I took a magnet, 1" dia. x 2" long, and put it against the plastic case which houses the ferrous ring on my clamp-on meter.

There was a substantial attraction there - I'd estimate maybe 20 lbs. force needed to pull the magnet off. This suggests to me that the meter's ferrous ring is substantial.

Eleanor White

Interesting. Does the magnet have that much pull on other metal or only against the meter ferrous ring on the meter? Either way it is not inducing current into your homemade secondary. There are some differences between magnetic flux and electromagnetic flux. I can't remember what they are, but it had something to do with Faraday, Lenz and the boys trying to make sense of it all. It may be that the difference has nothing to do with what we are trying to do, but it might. Your pushing my memory back a lot of years on this one.

Most Ct's have a lot of windings in the secondary, but adding more would make our situation worse in ways. The current should get smaller and the voltage higher. If you have one winding , it should equal the current of the source on the primary. You have 6, so you would be seeing 1/6th the current- see earlier diagram.

It would be interesting to know how many turns the meter uses for its winding around the ferrous clamp. For curiosity, have you watched the meter needle when you are doing this? It might be a possibility that your winding is canceling the field of the other secondary winding. If you are familiar with mag amps, you can add or subtract from the mag amp core by how you wind your other feedback windings. It is a long shot, but another possibility. What do you think? Are we having fun yet?

I have used and seen many CT devices set ups and they all were connected to a ground and power source of the device or system to be monitored. Could be you are seeing a ground default symptom that needs correcting.