Breaker panel current transformer?

Your question implies that you are passing those turns through something, but you haven't said what that something is.

A current transformer is just a transformer. The ration of the current in one winding is related the current in another winding by the inverse of the turns ratio. What current transformer do you have in hand or in mind?

Then you need a circuit that converts AC current (out of the current transformer) to a proportional DC voltage. A full wave rectifier and RC filter may serve. you will also probably want to put a zener across that filter to prevent an over voltage on a big current surge. All this has to be scaled to match the voltage range of the A/D converter you have in hand or in mind.

Thanks for the reply John... lets assume 5V a/d... I guess If I want

10A in the mains to be 100ma in the secondary, I need 100 turns of the PRIMARY mains wire, not 100s of turns on the small secondary, right? Guess I got it updide down? The run the secondary thru a resistor to give me the 5V peak I want.

For a current transformer, you don't wrap wire around the wire being measured. All you get that way is a little bit of capacitive coupling. You need to thread your primary wire through a core (typically a toroid), and wind turns around it, in typical toroid fashion, and then the current in the secondary (the wire you added) is simply 1/N of the current of the primary (the sense wire), assuming that your toroid has good permeability at 50/60 Hz.

Good Luck! Rich

I say we should boycott google kiddies. This dingus obviously doesn't even know what a current transformer _is_, let alone how to use one.

Maybe we should suggest that he measure it by firmly grasping the conductor in question, while standing in a bucket of salt water....

--
The Pig Bladder from Uranus

Use your google skills and look up "transformer" first.

Good Luck, Rich

Sorry for getting off on the wrong subject. If you guys wanted to measure juice in your breaker box down to a couple ma, whats the easierst, simplest, safest thing to hookup? Somekind of hall sensor taped to the black wire? OK, BP... you first... want to hear how a nondingus thinks.

Go to the hardware store and buy an Amprobe.

Cheers! Rich

Thanks Rich. At least you were a little more polite than ol' PB, Bet he aint got all his teeth if he talks like that within about arms reach of most of the good old boys I know.

You need to pass the power wire through a transformer core that has lots of turns already wound on it. Lets say you get one of these from Digikey:

It has 1000 turns already wound around the toroid core, so 10 amperes passing through the hole will cause 10 mA (10 amperes divided by 1000) to pass through the 1000 turns if the voltage is keep low enough that the core dies not saturate.

You connect the CT output to a bridge rectifier and then to a parallel resistor (10 mA at 5 volts would require 500 ohms) and a capacitor (say, 1000 uF for a .5 second time constant) to smooth the current so you can digitize it. This will actually give something close to an average of the absolute value, instead of an RMS reading, but if you have some other amp meter to check the calibration, you should be able to correct that.

And as I said before, you should parallel the resistor with a 5.1 volt zener diode to clamp the over voltage if there is an overload on the line, till the breaker pops.

Better to terminate the CT with a low value, such as 50 to 100 ohms, then run the CT output through an amplifier before rectification. Otherwise, you lose the low end of the current range due to diode drop in the rectifier. Also, notice the output voltage when the CT is lightly loaded.. it gets pretty nonlinear for terminations above 500 ohms. Don't know if these curves are similar across the mfgs, but I suspect that they are. CT's are made to be loaded in the 10-100 ohm range.. and the output amplified accordingly.

Cheers!!!

--
Dave M
MasonDG44 at comcast dot net  (Just subsitute the appropriate characters in 
the address)

Never take a laxative and a sleeping pill at the same time!!

The lower resistor is a good idea to keep the volt seconds per half cycle low for the core, but as long as the core has enough flux swing capability, the diodes cause almost no error, because it is a "current" transformer. The current into the diodes is almost exactly the same as the current out of them. The exact voltage (or the wave shape) across the secondary is not very important as long as it is low enough.

For the one I mentioned, 500 ohms causes a high enough voltage across the secondary that it begins to loose some of its accuracy (it is a pretty small core), so using a 100 ohm resistor and a gain of 5 amplifier would improve the accuracy a bit. But the diodes still go between the transformer and the resistor.

Put the terminating resistor _after_ a bridge rectifier. That will put the diodes in the constant current part of the circuit, so their voltage-drop won't matter.

Then use an RC network to get an average value.

--
~ Adrian Tuddenham ~
(Remove the ".invalid"s and add ".co.uk" to reply)
www.poppyrecords.co.uk

Well, Duhhhh Now that my head is out in open air, of course... For some reason, I couldn't get past the thinking that the load always had to be connected directly to the CT.

--
Dave M
MasonDG44 at comcast dot net  (Just subsitute the appropriate characters in 
the address)

Never take a laxative and a sleeping pill at the same time!!

This is something that's always scared me about current transformers. If the load resistor is, say, 1 Meg, then it will develop A THOUSAND VOLTS to make a milliamp flow.

Thanks, Rich

What I used to do, back in the days when I designed switchers, was to AC-couple the loaded CT thru some OpAmp gain.

Then I DC-restored the signal, clamping the bottom of the waveform to zero. Then a peak detector (comparator) could give quick trip of the current limit.

...Jim Thompson

-- | James E.Thompson, P.E. | mens | | Analog Innovations, Inc. | et | | Analog/Mixed-Signal ASIC's and Discrete Systems | manus | | Phoenix, Arizona Voice:(480)460-2350 | | | E-mail Address at Website Fax:(480)460-2142 | Brass Rat | |

| 1962 | I love to cook with wine. Sometimes I even put it in the food.

If

VOLTS

But not for very long. Remember that the core saturates and the device effectively ceases to be a transformer after the core flux reaches saturation after some number of volt seconds is applied to it, each half cycle. It produces a pulse of voltage and saturates early each half cycle. Not much of a hazard unless it is a big core. And, in any case, it can't produce more volts per turn than the line voltage passing through it on a one turn winding. Because that is the most volts line can apply to that turn.

If you want to see what is involved in the design of an energy monitor then take a look at

...But until it has enough flux to saturate, it _will_ develop higher and higher voltage trying to get the flux-cancelling current through the secondary. Hundreds, even thousands of volt are possible, at least with larger CT's.

That's why the commercial manufacturers warn you to NEVER put current through the primary of an improperly loaded current transformer. It can break down the insulation and start fires.

John Perry

things can get exciting if the secondary wire falls off, too :)

Cheers Terry