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Owl Energy Monitor

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echase
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Post Number: 382
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Posted on Thursday, 02 December, 2010 - 11:54 am:   Edit Post Delete Post Print Post    Move Post (Moderator/Admin Only)

Anyone got a Owl Energy Monitor? Do you know what the output of the sensor is? Looks like hall effect or current transformer to me. This plugs into a battery powered electronics unit that radio transits the current value to the display unit. I am hoping to tap off the sensor output and use it for another purpose. If hall effect it is probably easy. But looks like a 2 wire connection so more likely a transformer. What simple circuit can I use to translate this into an analogue value to input to a PIC? Op amp true RMS converter? Tapping into their electronic unit is a possibility but I suspect itís a custom chip so I could not access the signal before its digitised and transmitted as a serial data stream.

http://www.theowl.com/shop/index.php?target=products&product_id=7

http://www.theowl.com/shop/index.php?target=products&product_id=41

Pics do not show the transmitter box.
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johnmosborneuk
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Posted on Friday, 03 December, 2010 - 02:14 pm:   Edit Post Delete Post Print Post    Move Post (Moderator/Admin Only)

I don't know about the OWL energy monitors but the Efergy one I have uses a split core current transformer.
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ianjoh
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Posted on Saturday, 04 December, 2010 - 07:51 pm:   Edit Post Delete Post Print Post    Move Post (Moderator/Admin Only)

This is just a theory as I have yet to investigate it, but my digital electric meter has an led that flashes for about 500mS with each fraction of KWH that is used. The greater the energy that is used the faster it flashes.
If a light sensor can be used to pick up these flases then surely with some clever maths via a pic the KWH useage can be calculated.
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azayles
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Posted on Saturday, 04 December, 2010 - 08:09 pm:   Edit Post Delete Post Print Post    Move Post (Moderator/Admin Only)

My electricity meter (an EDF key meter) has the same flashy light. There are words next to it, "800 lmp/Kwh" which I *think* means the light flashes 800 times for every kilowatt hour used.
I was thinking of making some circuit to measure energy consumption and display it in real time, but didn't even think to use this light.
Cheers, Ian!
Keeping the Magic Smoke in since 1978
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johnmosborneuk
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Posted on Sunday, 05 December, 2010 - 11:35 pm:   Edit Post Delete Post Print Post    Move Post (Moderator/Admin Only)

My meter has 100Imp/Kwh next to the LED, you would think the imp/kwh would be standard but I guess that'd be too much to ask.

I had been thinking about a similar pulse counting idea to display real time gas and water usage but sadly I don't think we're allowed to plug into the pulse output on the gas meter
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echase
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Posted on Monday, 06 December, 2010 - 02:47 pm:   Edit Post Delete Post Print Post    Move Post (Moderator/Admin Only)

I was thinking of putting a photodiode over the LED on my solar electric meter. It has 2 LEDs 10mm apart. One red one pulses at 1000 per kWhr and the other I am not sure about. So need to ensure the photodiode reads only one. It is not that simple to mount a photodiode accurately over the right spot without gluing it to the meter face. But I guess silicone mastic would work and could be scraped off if anyone came round to look at the meter Ė the utility company might be suspicious that one was tampering with the reading.

http://www.leiderdorpinstruments.nl/International%20web/Info%20displays/index.html

is an Owl like device. On this the sender unit (probably using a series resistor in neutral line) issues pulses per kWhr hour and I think itís settable to how many per kWhr. There was a new unwanted one for sale on eBay recently for £60 but no one bought it, I guess because the cheaper Owl pretty much does the same thing, although maybe less accurately.
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echase
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Posted on Monday, 06 December, 2010 - 03:42 pm:   Edit Post Delete Post Print Post    Move Post (Moderator/Admin Only)

A sucker could be ideal for holding the photodiode but not very easy to adapt a sucker to take a diode, whilst still being airtight. Tip of a childís arrow with the diode siliconed in instead of the arrow shaft? Needs an opaque sucker. Many are transparent.

To answer my own question there are ICs that do the true RMS to DC conversion for metering. The AD736/7 seem to be the most common and lowest cost but still costs at least £5 and much more if you get the more accurate versions. I need to look back at the EPE electricity meter from 2-3 years ago and see how they did it.

I suspect a dual/quad op amp with rectifier may do it if accuracy and temperature drift are not important but such analogue design is beyond me.


(Message edited by echase on 06 December, 2010)

(Message edited by echase on 06 December, 2010)
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azayles
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Posted on Monday, 06 December, 2010 - 03:49 pm:   Edit Post Delete Post Print Post    Move Post (Moderator/Admin Only)

Maybe a lens system installed the other side of the meter cupboard which focuses the light from the meter onto a photo diode.
This assumes, however, the meter is installed on a side wall, not the back wall, and the cupboard door is kept mainly closed.

Failing that, just bluetack a photodiode to the meter face.
Keeping the Magic Smoke in since 1978
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echase
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Posted on Monday, 06 December, 2010 - 04:01 pm:   Edit Post Delete Post Print Post    Move Post (Moderator/Admin Only)

Ah bluetack, that is a good idea.
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mark
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Posted on Thursday, 16 December, 2010 - 03:34 pm:   Edit Post Delete Post Print Post    Move Post (Moderator/Admin Only)

Tracking the kWh led seems like a very good idea, as it should agree perfectly with the bills. Just measuring the current and doing rms conversion is not very useful in the case of high power factor loads where there is a phase angle to consider as well as the actual current. I think (hope) the meter takes account of this in its power measurements. Most of the add on devices sense only current and so can't measure true power. Probably accounts for some of the silly over estimated power readings that standby things are said to give. If the telly really did take 20 watts on standby it would get quite warm -- imagine a 20 Watt bulb in there!
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echase
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Posted on Wednesday, 29 December, 2010 - 11:46 pm:   Edit Post Delete Post Print Post    Move Post (Moderator/Admin Only)

I have now bought an Owl unit and the sensor is a current transformer with two C cores to make a split ďcircleĒ around the wire to be sensed.

I measured it with scope at 70mV output per amp in the sensed wire.

If I wind more than one turn through the core will the output voltage be a pretty exact multiple of the number of turns? As itís a 70A sensor and I only am measuring about 15A a few extra turns would increase the output voltage.

What would be a good circuit to convert these few hundreds of mV into a 0-5V DV signal for the PIC to read? Strictly it does not have to be a true RMS circuit as the waveform out of the solar panelís inverter is pretty much a constant waveform and power factor. So I can adjust the calibration to take out any non pure sinewave effects.

Far as I can tell the AD736 only outputs the same voltage in DC as the AC input so Iíd need a fixed gain op amp after it to amplify it, unless I can squeezeze about 4 primary turns into what is quite a small hole in the core.
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terrym
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Posted on Thursday, 30 December, 2010 - 01:55 am:   Edit Post Delete Post Print Post    Move Post (Moderator/Admin Only)

This (or variations of) is what I have used on many occassions.
R1 is the burden or load resistor specific to your CT. Resistors R2/R3 set the gain of the first stage. This first stage with the low pass filter (R4/C1) may be all you need. If you want a DC output, fit diode D2 between pin 1 of ic2a and R4. The second stage is actually a comparator stage (which you dont need at this point), the trip level set by R8. You could modify this to be a second amplifier stage as per stage 1.

You will need to play with the values of R2/R3 to get the gain you need and maybe the values of R4/C1 to adjust the low pass filter frequency (depends on output harmonics etc of the inverter)

CTIface

TM
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echase
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Posted on Thursday, 30 December, 2010 - 11:07 pm:   Edit Post Delete Post Print Post    Move Post (Moderator/Admin Only)

Thanks Terry. How does the circuit get around the non linearity in the diode to stop low inputs causing excessively low outputs, due to the 0.6V drop?

And how do you cleverly insert a diagram into your post?
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terrym
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Posted on Thursday, 30 December, 2010 - 11:37 pm:   Edit Post Delete Post Print Post    Move Post (Moderator/Admin Only)

Most items I use this in has a micro after it, so the micro programming takes care of most lineraity problems. Unfortunately, I can't post any code, as it is used in commercial products.

You can also connect the end of R3 that is connected to pin1, to the junction of the diode and R4. This way, the diode is within the feedback loop of the op amp (same as half wave precision rectifier). The gain of the op amp can be adjusted such that minimum input gives an output above .6v also. Very much depends on how much post processing power is available and how much you want or are able to do.

The circuit is drawn in sPlan7 from Abacom. I export it as a GIF and use a graphics program to reduce it to the proper size (max 800 x 800). Then use the upload attachment box next to the 'Preview/Post message' button below.

TM
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terrym
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Posted on Thursday, 30 December, 2010 - 11:53 pm:   Edit Post Delete Post Print Post    Move Post (Moderator/Admin Only)

Here's another circuit based on a precision rectifier, that I've also used. The first stage is a unity gain buffer with low pass filter in front. Second stage has a gain of ten.
Adjust the gain of the second stage to suit your needs by adjusting the ratio of R5/R6.

CT/Iface 2

TM
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echase
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Posted on Friday, 31 December, 2010 - 11:57 am:   Edit Post Delete Post Print Post    Move Post (Moderator/Admin Only)

Excellent. I can put in an offset in software if needed to take out any non linearity.

Fortunately the inverter in my loft gives the exact amps reading so I calibrate against that. I did read though that circuits like this are very temperature prone e.g. because the diode changes. So I am not sure how accurate it will be, but Owl only claim 5% at FSD and much worse at low currents.

On an alternative tack I see that RS do very cheap (apart from min order numbers and handling charge) current transformers http://uk.rs-online.com/web/search/searchBrowseAction.html?method=getProduct&R=2509299877
that are just like a conventional transformer where the sensed current passes through the primary. No data sheet though. It occurs to me that if I just take a normal mains transformer driven backwards and replace what was the secondary with a few thick turns then the old primary will give tens or even hundreds of volts. This is very easy to rectify and convert to 0-5VDC with just a diode and passive components. I able to insert something into the circuit rather than rely on the non interference toroidal option.

But is there something special about this RS current transformer that keeps the unit more linear than my DIY solution?
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echase
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Posted on Friday, 31 December, 2010 - 05:02 pm:   Edit Post Delete Post Print Post    Move Post (Moderator/Admin Only)

Hundreds of volts may be an exaggeration, judging by the datasheet of the RS one,s which I have now found on the manufacturerís site.

As I donít want to drop more than a volt in the primary and the turns ratio might be no more than 50 to 1 then 50V is a maximum.
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terrym
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Posted on Saturday, 01 January, 2011 - 12:46 am:   Edit Post Delete Post Print Post    Move Post (Moderator/Admin Only)

Temperature can be a nuisance depending on how accurate you want to get. As you are using a micro, you can of course add a temperature measurement device to calculate it all out.

A backwards mains transformer will work, just make sure the secondary can carry, at a minimum, twice the current you want to measure (for overheating and safety reasons).

You can get 1000's of volts out of a CT if you don't have load across it, very nasty. Out of a backward mains transformer, as long as it is a low voltage secondary, you "should" only get a few volts if properly loaded. There is an art to calculating the load value,but I can't find the article at the moment. A Google search for info re CT's should find it.

TM
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ianjoh
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Posted on Sunday, 02 January, 2011 - 11:04 am:   Edit Post Delete Post Print Post    Move Post (Moderator/Admin Only)

I have got current transformer up for grabs if anyone wants it. It is a 200/5A made by RITZ model ASTW 6, with a 5va rating. I think it originally came in a RS components box but that has long since gone. It measures 3"x2" with tags for the primary current.
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echase
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Posted on Sunday, 02 January, 2011 - 01:24 pm:   Edit Post Delete Post Print Post    Move Post (Moderator/Admin Only)

Yes I could be interested thanks. But if the 5A means it only does 5Amp on pirmary then it's not enough for my 15Amp requirement.

Am trying to get my head around the theory of CTs to see find one that gives enough output for me.
E.g. http://www.butlerwinding.com/store.asp?pid=28351
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echase
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Posted on Sunday, 02 January, 2011 - 02:11 pm:   Edit Post Delete Post Print Post    Move Post (Moderator/Admin Only)

To paraphrase Wikipedia a 200:5 CT would provide an output current of 5 amperes when the primary was passing 200 amperes. On that basis a 200/5A at 5VA would produce 1V on secondary as 5VA= 5A x 1V. That is at 200A primary. So at my 15A it would only give a few tens of mV which is not enough.

I see from websites that a secondary load resistor must be connected or else massive overvoltageís can occur, causing insulation breakdown. But not clear how to calculate the right resistor value. Presumably the one that gives the largest load power, which could be derived by experimentation. But I guess your load resistor is moulded into the unit and so not changeable.

To be clear ianjoh does yours have 4 terminals? 2 for the sensed current and 2 secondary. Or is it the type with a hole in middle for the sensed wire?
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alec_t
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Posted on Sunday, 02 January, 2011 - 03:55 pm:   Edit Post Delete Post Print Post    Move Post (Moderator/Admin Only)

As I understand it the 200:5 CT outputs 5A for a 200A input current. Hence it outputs 375mA for 15A input. So if, for example, you pass the 375mA through a 10 Ohm load resistor you will get 3.75V across the resistor, with a dissipation of 1.4VA. Won't that be ok?

Alec

PS If you use a 4 Ohm load resistor with this CT you should get a convenient 0.1V output per Amp input. The maximum load resistor value is about 35 Ohms, to keep within the 5VA rating.

(Message edited by alec_t on 02 January, 2011)
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ianjoh
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Posted on Sunday, 02 January, 2011 - 09:56 pm:   Edit Post Delete Post Print Post    Move Post (Moderator/Admin Only)

@ echase, Yes it does have four terminals and looking at it should easily handle the 15A through the primary. I think alec_t has got his maths right, the secondary voltage must be adjustable with the load resistor.
pm me your address & I will post it to you.
Ian
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echase
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Posted on Sunday, 02 January, 2011 - 11:18 pm:   Edit Post Delete Post Print Post    Move Post (Moderator/Admin Only)

Alec the theory sounds good unless the CT has a lower value resistor already moulded into it. The Owl one has a semi integral resistor.

I happened to blow up a 70W halogen spotlight transformer today (switched mode PSU with 12VAC output. That has a 25mm diameter torroid with 8 turn secondary in it that might work as a CT. But maybe no more than 10A judging from wire thickness. And not sure whether a ferrite one like this will work well at 50Hz as it presumably switches at a much higher frequency. But the Owl one has a ferrite core too.

I shall need more than 1 CTs to measure all the parameters I have in mind.
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terrym
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Posted on Monday, 03 January, 2011 - 12:39 am:   Edit Post Delete Post Print Post    Move Post (Moderator/Admin Only)

One of the ways CT's are spec'd is by number of turns. The commonly available ones start at 50 turns and go up to 1000 turns. So, as the primry turn is only one turn, you can class them as 1:50, 1:100 and so on turns ratio.

As Alec alluded to, the load resistor value will be calculated such that at full load, the VA rating of the CT is not exceeded. If you have one with an inbuilt resistor, you can add an extra turn or so on the primary, but again, make sure you don't excceed the VA rating.

TM
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terrym
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Posted on Monday, 03 January, 2011 - 12:50 am:   Edit Post Delete Post Print Post    Move Post (Moderator/Admin Only)

Here's a quickie app note on CT's

http://www.hurricaneelectronics.com/appNote108.php

TM
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echase
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Posted on Monday, 03 January, 2011 - 10:11 am:   Edit Post Delete Post Print Post    Move Post (Moderator/Admin Only)

Slightly worrying is that the saturation voltages (on secondary) quoted in the linked datasheet http://www.hurricaneelectronics.com/currentTransformers.php are very low like <<1V for the small ones. But

1) ianjohnís one is towards the larger end of that datasheet so presumably might have a saturation voltage nearer the 5V I need.

2) As I am operating it at 7.5% of its FSD is it less likely to saturate? Afraid my knowledge of magnetics is limited.
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alec_t
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Posted on Monday, 03 January, 2011 - 11:20 am:   Edit Post Delete Post Print Post    Move Post (Moderator/Admin Only)

Out of interest, in CT world does simply passing a wire through the core in a straight line count as 1 turn or a half turn?

Alec
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dselec
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Posted on Monday, 03 January, 2011 - 08:18 pm:   Edit Post Delete Post Print Post    Move Post (Moderator/Admin Only)

What counts is how many times the wire passes thru the core so pass once can be called turn, even if the wire turns and comes back to the opening of the core but not pass thru it is still 1 turn and the magnetic field around the wire in the outside has no effect

dselec

(Message edited by dselec on 03 January, 2011)
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terrym
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Posted on Monday, 03 January, 2011 - 11:16 pm:   Edit Post Delete Post Print Post    Move Post (Moderator/Admin Only)

Echase, the link I gave was just for the app note data rather than their CT products.
Operating your CT at 7.5% of fsd will keep it within spec.

The only way you are going to expand your knowledge of magnetics is:

1/ Play with the devices and see what happens.

2/ Google for the info.

3/ Ask questions as you have been doing.

TM
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dave_g
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Posted on Sunday, 09 January, 2011 - 07:44 am:   Edit Post Delete Post Print Post    Move Post (Moderator/Admin Only)

I've been working on a project measuring mains current and calculating power in the load using a PIC.

So far I've avoided CT's using the Sparkfun ACS712 breakout board - in my case, the low current version.

http://www.allegromicro.com/en/Products/Part_Numbers/0712/

http://www.sparkfun.com/search/results?term=acs712&what=products


I'm having trouble with drift and varying gain, but it uses a hall effect sensor. Worth a look, it's a neat solution if only it would settle down. I think the trimpots on the low current version could be the problem.

There is a version of the sensor chip suitable for 15A.
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armadillo
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Posted on Sunday, 09 January, 2011 - 09:31 am:   Edit Post Delete Post Print Post    Move Post (Moderator/Admin Only)

dave_g

I hope you are aware that the isolation of the ACS712 is only classed as 'basic'?

i.e. If the sensed current is at mains voltage, the secondary is not considered as safe to touch.

Armadillo
There's no such thing as gravity..........
The earth sucks!
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dave_g
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Posted on Sunday, 09 January, 2011 - 12:16 pm:   Edit Post Delete Post Print Post    Move Post (Moderator/Admin Only)

Armadillo - you are correct if on 240v mains, which I am. This design is completely enclosed (i.e.) no external connections.

Thanks for the reminder, however.

http://www.allegromicro.com/en/Products/Part_Numbers/0712/0712-13_faq.asp#Q1
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azayles
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Posted on Sunday, 09 January, 2011 - 01:53 pm:   Edit Post Delete Post Print Post    Move Post (Moderator/Admin Only)

This from Sparkfun might also be worth a look.
http://www.sparkfun.com/products/9028
Measures up to almost 90 Amps.
"If you can't open it, you don't own it"
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echase
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Posted on Monday, 10 January, 2011 - 01:35 pm:   Edit Post Delete Post Print Post    Move Post (Moderator/Admin Only)

Jaycar do a Hall Effect clip on (so mains isolated) AC/DC sensor for about £10. Not sure how accurate.

Many thanks for the CT ianjoh. Not tried it yet but looks just the ticket. Plenty of room to add primary turns unlike the standard Owl one, although I think Owl do a different sort as well for fatter cables.
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terrym
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Posted on Tuesday, 11 January, 2011 - 01:00 am:   Edit Post Delete Post Print Post    Move Post (Moderator/Admin Only)

From memory , the Hall Effect clip-on is not good at low currents.

I remember we tried it for one of the products we were prototyping and dismissed it.

Think it is only linear within a certain range too.

TM
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armadillo
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Posted on Tuesday, 11 January, 2011 - 08:45 pm:   Edit Post Delete Post Print Post    Move Post (Moderator/Admin Only)

To improve accuracy of hall effect devices at low currents pass the wire multiple times (in the same direction) through the clip on.

E.G. make a winding of say 10 turns and put the clip around all 10. This gives a transformer effect (even if it's DC). Obviously the reading has to be divided by 10.

Armadillo

(Message edited by Armadillo on 11 January, 2011)
There's no such thing as gravity..........
The earth sucks!
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echase
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Posted on Thursday, 13 January, 2011 - 11:21 am:   Edit Post Delete Post Print Post    Move Post (Moderator/Admin Only)

Have now tried ianjohís CT. Itís the type with a hole in the middle, so I wound some turns on it and connected a Ďscope. The output waveform is grossly distorted much of time so I am wondering why.

I started gingerly with one primary turn, a 10-40 ohm secondary load and 5amps into a fan heater; the waveform is nearly sinusoidal at around 1Vrms with slight Class B amplifier type crossover distortion.

But when increasing the input flux at 10 amp there is more crossover distortion and it gets worse if I add more turns or open circuit the secondary. There is a positive and negative spike in the middle of the peak input wave such that on 90% of the cycle itís outputting 0V and 10% of the time it has these sharp peaks, like a very narrow half sine wave. Here my magetics knowledge fails me as I would have thought that if it were a saturation problem then it would clip the sine wave to make it look like a square wave. But itís doing the opposite in that it seems to take a certain level of instantaneous input current before any output happens, Class B style. But then this conflicts with the first test where low levels of input are giving the best waveform.

Another puzzling thing it that even with 10A though 10 turns, which is equivalent to 100A through one turn (I think), this 200A CT only gives 2V peak output when o/c. This is nowhere near the hundreds and thousands of voltage predicted in postings above.

(Message edited by echase on 13 January, 2011)
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echase
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Posted on Thursday, 13 January, 2011 - 11:28 am:   Edit Post Delete Post Print Post    Move Post (Moderator/Admin Only)

http://www.electronics-tutorials.ws/amplifier/amp_7.html shows my waveform but imagine the flat part at 0V taking up 90% of the width.
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ianjoh
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Posted on Thursday, 13 January, 2011 - 08:03 pm:   Edit Post Delete Post Print Post    Move Post (Moderator/Admin Only)

I am glad you are putting it to use .
Have you wound your turns and connected them in series with the secondary?
It could be possible that they are working against the existing windings because they are reverse connected (or wound).
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picnut
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Posted on Thursday, 13 January, 2011 - 09:48 pm:   Edit Post Delete Post Print Post    Move Post (Moderator/Admin Only)

I was somewhat pleasantly surprised to see the Andover (Hants) library had the OWL energy monitor available on ticket loan.

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