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jim hardy

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www05.abb.com/global/scot/scot235.nsf/.../ittechinfoappguide.pdf

Probably the transformer has a shorted turn and you'll have to clear the short before it can induce any sigal into your new secondary.

Can you drive it with a step-down transformer and variac to secondary windiing? Plot Currrent vs Volts, current should stay quite low until it saturates at several volts making a distinct "knee".. if it has shorted turns current will rise quickly, no knee, and you may feel it getting warm.

I'd say strip the core and rewind it. It's an opportunity to count turns. If you find only nine we learned something.

If you're lucky you'll find the short at a terminal and can fix it with epoxy.

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The 2 parallel wires are each 0.038" in diameter, like maybe 19 gauge? I'm wondering if I could just run 9 or 10 turns of a single, 16 gauge wire, instead of paralleling 2 smaller ones like they did. But I'm wondering if they had a good reason for paralleling those 2 smaller wires. I mean it does seem like you'd be spreading out the winding a bit more by doing that. That parallel setup spanned about 3/4 of the core circumference.

The laminated core is merely a thin strip of laminated steel, rolled up into a coil and spot welded on both ends to keep it from unraveling. And it has a 2-halved protective plastic cover that goes over it. Do you think I could get away with using insulated wires if I got rid of the plastic cover?

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jim hardy

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here it is through google

http://www.google.com/url?sa=t&rct=...ytj-AQ&usg=AFQjCNGKy8swuzdp-t8dpWyhgMX3vsrpTg

hope that works

read section on current transformers..

Shorted turns, you found?

That's how current transformers fail. They try to be an ideal current source and can develop quite high voltage if operated into an open circuit. They'll pierce their own insulation.

I cant say i understand the nine/ten turn bifilar winding.

Initial reaction is that gave them nine and a half turns, leaving a half turn of primary current to account for magnetizing current of core. But i want to do some reading.

Okay - what to do next?

Best thing would be repair the burnt spots on old wire and put them back.

A coat of good varnish or enamel paint on core should insulate it.

After winding, it could be wrapped with electrical tape. There's a white fiberglass variety that looks quite nice, Scotch 27.

Or get creative with epoxy .

Next best would be get new wires same diameter and replicate their winding.

late entry - check old wires - are they copper? see later on

But - if you don't need high accuracy, just go with ten turns of #20. That'd be a quick test you could do with duct tape.

You can bet they did, and bigger mystery is unequal numbers of turns.But I'm wondering if they had a good reason for paralleling those 2 smaller wires.

Thinking aloud - parallel windings of unequal turns will circulate current

and the circulating current will oppose flux in one winding but aid it in the other.

Amount of circulating current will be (volts mismatch)/ohms

and (volts mismatch) is in proportion to flux so will be a measure of the error in transformer's current ratio.

perhaps they got clever and linearized the transformer? Or just accounted for magnetizing amp- turns? I suspect the latter.

Those windings were both wrapped same direction around core, right?

Would you try to measure their resistance ? Can you tell if they're copper?

0.038 copper should be 7 milliohms/ft maybe you can put an amp DC through it and measure?

Anyhow dont throw them away.

What do you know - that transformer was

Is there a transformer doctor in the house?

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jim hardy

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PS - Insulated wires will be fine.

Here's a good writeup on CT theory of operation..

Could that have been two nine turn windings? It's easy to lose track of one at the terminals..

Here's a good writeup on CT theory of operation..

Could that have been two nine turn windings? It's easy to lose track of one at the terminals..

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The voltage difference is like 10%.

When the regular CT is open circuited, very high Voltage can get induced.

But in this CT, when the secondary is open circuited, Let the voltage induced in the coils be 10V and 9V. Now, the circulating current I = (10V-9V)/R

Now the circuilating current flows in both windings, but the Amp-turns cancel each-other.

Hence net Amp-turns = (10-9)*I = V/R

This amp-turns must equal primary Amp-turns = I_primary

hence, I_primary = V/R

hence, V = I_primary*R

Hence, there is limit to what the open-ckt terminal voltage of the CT will be.

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jim hardy

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the workings of it are still percolating in my alleged brain.

There's the current that the CT supplies to its load flowing too...

and that has me stumped at the moment.

But i have always been too dumb to give up.

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jim hardy

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And i see i forgot to attach link on CT theory.

http://www.butlerwinding.com/store.asp?pid=28351

and from their design hints

http://www.butlerwinding.com/store.asp?pid=28351

and from their design hints

[/The Desired Current Ratio – This is simply the desired secondary current value (at a specified value of primary current) divided by the primary current value that generates said value of secondary current. Alternatively, a turns ratio could be specified.but don’t expect the current ratio to exactly equal the turns ratio.http://www.butlerwinding.com/store.asp?pid=28353 [Broken]

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Full scale reading on the volts/amps/watts meter is 25A.

It uses the same meter movement for all 3 functions, i.e. volts, amps and watts.

As I mentioned before, the core is made of thin, laminated steel, rolled into a coil, a with spot weld at the beginning (inner most part) and another at the end (outermost part) of the coil. But I noticed that the spot weld at the innermost part isn't actually welded. There's just a hole burned through it, like maybe it was a bad weld that burned through the steel and melted the middle of the top layer without fusing it to the next layer. The outermost weld has a good, solid weld in it, and the coil is immovable there. But you can move the innermost part of the coil around.

Could that be the problem?

And what would you think the secondary voltage would be if the core was working properly?

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jim hardy

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see thread "power in an inductor" in classical physics, last few posts.

https://www.physicsforums.com/showthread.php?t=585779

A CT should not be asked to deliver much voltage. I was expecting more like 1/4 volt.

First thought is this:

CT's have a trait that seems incredible. They can get permanently magnetized andthat may well have happened when yours failed.

The phenomenon is described here.

www05.abb.com/global/scot/scot235.nsf/.../ittechinfoappguide.pdf

page 27 0f report, 29 of pdf.

but for some reason google has interposed themselves as middleman and i had to get there by this route:

http://www.google.com/url?sa=t&rct=...nZi3Dw&usg=AFQjCNGKy8swuzdp-t8dpWyhgMX3vsrpTg

or you can search on that first link.

Anyhow we neeed to de-magnetize your CT for starters to remove that as possiblity.

Can you put five turns of your shop-vac cord through CT and run it, with your 10 turn secondary attached to your wattmeter? If it is permanently magbetized that should walk it down the hysteresis curve and make it better.

I have been playing with some algebra regarding the two secondaries abd it's very interesting. Will scan and post some more tonight.

Now - you got 1VAC with ten amps thru donut and 1 amp by clamp on, 1.5 by meter in your instrument.

Make sure nothind else is in series with winding and instrument. CT's are accurate only at low output voltage.

Try demagnetizing it by instructions on that ABB link, page 27 & 28 of report page 29 & 30 on PDF page counter. Then take same readings again.

I'll be back as soon as have worked some more algebra - wish me luck !

I'm glad you have some test equipment - this is going to be alearning experience. I did not know CT's as well as i thought i did. The textbooks never mention thet 9turns-10turns trick.

old jim

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jim hardy

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In other words, what impedance does CT see looking into wattmeter?

Correct term for that is "Burden" and you'll see it in that ABB literature.

I'd expect less than an ohm.

If wattmeter has an internal resistor failed open placing high burden on CT, that could be big part of trouble

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Unfortunately, I don't have an accurate Ohm meter, i.e. none that measures a decimal point of an ohm. But with both Ohm meters I have, the resistance was the same as the internal resistance of those meters, i.e. the Ohm's reading with the probes shorted together (about 2 ohms) which seems to indicate less that it's less than 1 Ohm.

In this configuration - where the load current carried by the primary winding is always there as long as something is plugged into the load receptacle in the meter, it's my understanding that a "shorting" block (something fairly conductive) is typically used in the meter to keep you from putting the CT secondary into a "no load" scenario, where the voltage might become excessively high when the test instrument is not actively reading amperage. And that would be the case here, i.e. if you're measuring voltage, because the unit only has one meter in it. So that single meter is used to measure, amps, volts and wattage.

So I wasn't surprised to see close to 0 ohms across the secondary. And I decided (after first making sure no voltage was flowing back through the secondary wires) to check the resistance of the secondary windings with them disconnected from the meter, but with the meter powered up and that 10 amp primary current load going through the donut hole.

I checked the resistance first with the system in the amps position. And next with it in the "volts position". Surprisingly, both tests showed less than 1 ohm. But I assumed that the resistance in the meter circuit, i.e. with the switch in the amps mode, would probably be quite low as well.

Any guess about how many volts the CT secondary would output with the secondary wires disconnected from the unit, with a 10 amp primary current? I had thought that would be fairly high with no secondary load there, which is supposedly why they have those shorting blocks in them to prevent the no load scenario.

In other words, what impedance does CT see looking into wattmeter?

Correct term for that is "Burden" and you'll see it in that ABB literature.

I'd expect less than an ohm.

If wattmeter has an internal resistor failed open placing high burden on CT, that could be big part of trouble

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And I also determined how that solder joint likely got cracked. The CT is an Sht model, i.e. with no mounting tabs on it, and it was merely lying in the bottom of the case under the circuit board. So it would come flying up and strike the wiring and possibly the bottom of the circuit board when the unit got turned upside down. :(

And when I saw the arced appearance of the bottom portion of the core case, I figured that it had been hit by some fairly high voltage with that bad secondary solder joint creating a "no load" secondary scenario.

- #14

jim hardy

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Very good ! And THANKS!

Your broken solder joint might be what wrecked the CT windings.

I think your gizmo is probably close to working.

We need to get confidence in the CT then the meter.

Have you a Variac ? if not that's okay.

One experiment would be place your DMM on ten amp scale and connect your ten turn CT secondary directly to it, then measure running current of shopvac. That should give of course 1/10th shopvac current,

That'll confirm CT works at very low output voltage, for your DMM is probably only ~ 0.2 volt drop on amps scale.

Then loop your shopvac cord through twice, curent should double.

Then loop it again... if triples we probably have an operable CT operating as CURRENT transformer, that is it's operating at low flux.

Reason flux is low is the secondary amp-turns cancel the primary ones. Ideally they'd cancel perfectly but we know there has to be a couple left over to make that little bit of flux flow round the donut.

.................................................. ............

Now let's figure out how to test the core itself.

Can you rig an ordinary lamp with a split core so we can excite that core with lower current?

A 60 watt light bulb should be around 0.5amp and we can measure that with DMM.

One turn through core would be 0.5 amp turn

two turns would be 1.0 and so on.

Just a few amp turns i'd think would not saturate the core.

So we could measure open circuit volts on yur secondary and plot open - circuit votage versus amp-turns on graph paper,,

and see if it looks like the magnetization curve - straight line until we approach saturation where it levels off.

In industry we test CT's by forcing them into saturation with a Variac adjustable transformer applying AC to secondary, measure current and raise voltage until it forces saturation and current shoots up.

If it accepts current at too low voltage it has shorted turns or has lost inductance perhaps through an air gap in core. A big CT might take 200 volts to saturate, real small ones just a few volts. The sharp current increase at saturation is the key.

That test also removes any permanent magnetization. ...........................

Now, keep this though in back of your mind... " the open circuit voltage your CT makes is a direct measure of the flux in the core__so long as flux doesn't depart from a sinewave__. . "

...............................................

Can you measure cross section of core?

Your measured 1 volt ac infers 1/377 square meter at 10,000 gauss, or 1/565 at 15,000.

1/565 m^2 would be ~4 cm square cross section which seems too big.

Something else is going on - is your meter a true RMS? It could be reporting effects of saturation.

.................................................. ........................

But there's a twist to that thought, and it's the key to understanding CT's. I can remember when they were a mystery to me - they're not intuitive..

Please indulge another of my simplified thought experiments, this is how i think and figure things out.

Ten amp-turns might well be enough to drive your core into saturation on the current peaks. Let's walk through one sinewave cycle starting at zero.

At zero current there's zero flux .

As current increases toward positive peak, flux follows current up until the core can hold no more of it. That's saturation.

When that happens, flux stops increasing and stays "flat" for some time.

The core might even get some permanent magnetization if the current peak is high enough.

Next,

When current passes peak and starts decreasing, flux stays flat until amp-turns let flux start decreasing, and as current heads toward negative peak flux gets pushed toward negative saturation.

After the negative peak, current starts positive again and this repetitive cycle continues.

Now here's what causes the trouble.

Instead of a nice smooth sinewave flux, we have a more flat-topped square-ish looking flux wave that**snaps back and forth **between positive and negative saturation. That snap is very quick compared to a sinewave cycle.

That snap-action is what kills insulation.

Recall that voltage is rate of change of flux, e = n_{turns} X d[itex]\Phi[/itex] /dt, and the edges of our square-ish flux wave have very high rates of change.

So induced voltage waveform is no longer a nice smooth sine wave but a series of very high, narrow sharp "needles" .

Since the needles are narrow they dont have much average or RMS value and a voltmeter doesn't even notice them, but they'll poke right through insulation just like the needles they are.

I have more but this is plenty for one session.

To Do:

1. You mentioned "looseness" in the core. Try to get rid of that, a CT core cannot tolerate air-gap it needs high permeability. (Well there's special CT's that can handle DC component in their current but i do not know how they work.)

So find out what's up with that. Perhaps you can lace the core with some sort of twine to tighten it up and get rid of air gaps. If that's a tape wound core it'll need to be tight. i use braided tournament grade fishing line myself. Epoxy invites creativity....The "burnt spot" you spotted on core could be acting like a shorted turn. maybe you can get some varnish or epoxy in there.

2, Try that open circuit voltage versus amp-turns plot.

If it won't saturate there's something wrong with it, iMHO.

A soldering gun might be a useful de-gausser, hold it near then draw away while holding trigger.

I'm back to what i mentioned last night trying to figure out the dual secondaries in parallel.

Your broken solder joint might be what wrecked the CT windings.

I think your gizmo is probably close to working.

We need to get confidence in the CT then the meter.

Have you a Variac ? if not that's okay.

One experiment would be place your DMM on ten amp scale and connect your ten turn CT secondary directly to it, then measure running current of shopvac. That should give of course 1/10th shopvac current,

That'll confirm CT works at very low output voltage, for your DMM is probably only ~ 0.2 volt drop on amps scale.

Then loop your shopvac cord through twice, curent should double.

Then loop it again... if triples we probably have an operable CT operating as CURRENT transformer, that is it's operating at low flux.

Reason flux is low is the secondary amp-turns cancel the primary ones. Ideally they'd cancel perfectly but we know there has to be a couple left over to make that little bit of flux flow round the donut.

.................................................. ............

Now let's figure out how to test the core itself.

Can you rig an ordinary lamp with a split core so we can excite that core with lower current?

A 60 watt light bulb should be around 0.5amp and we can measure that with DMM.

One turn through core would be 0.5 amp turn

two turns would be 1.0 and so on.

Just a few amp turns i'd think would not saturate the core.

So we could measure open circuit volts on yur secondary and plot open - circuit votage versus amp-turns on graph paper,,

and see if it looks like the magnetization curve - straight line until we approach saturation where it levels off.

In industry we test CT's by forcing them into saturation with a Variac adjustable transformer applying AC to secondary, measure current and raise voltage until it forces saturation and current shoots up.

If it accepts current at too low voltage it has shorted turns or has lost inductance perhaps through an air gap in core. A big CT might take 200 volts to saturate, real small ones just a few volts. The sharp current increase at saturation is the key.

That test also removes any permanent magnetization. ...........................

Now, keep this though in back of your mind... " the open circuit voltage your CT makes is a direct measure of the flux in the core

...............................................

Quick answer: I think ~377 volts per square meter of core material for each turn so long as it doesn't saturate !.Any guess about how many volts the CT secondary would output with the secondary wires disconnected from the unit, with a 10 amp primary current? I had thought that would be fairly high with no secondary load there, which is supposedly why they have those shorting blocks in them to prevent the no load scenario.

Can you measure cross section of core?

Your measured 1 volt ac infers 1/377 square meter at 10,000 gauss, or 1/565 at 15,000.

1/565 m^2 would be ~4 cm square cross section which seems too big.

Something else is going on - is your meter a true RMS? It could be reporting effects of saturation.

.................................................. ........................

You are quite right that's exactly what the shorting block is for.those shorting blocks

But there's a twist to that thought, and it's the key to understanding CT's. I can remember when they were a mystery to me - they're not intuitive..

Please indulge another of my simplified thought experiments, this is how i think and figure things out.

Ten amp-turns might well be enough to drive your core into saturation on the current peaks. Let's walk through one sinewave cycle starting at zero.

At zero current there's zero flux .

As current increases toward positive peak, flux follows current up until the core can hold no more of it. That's saturation.

When that happens, flux stops increasing and stays "flat" for some time.

The core might even get some permanent magnetization if the current peak is high enough.

Next,

When current passes peak and starts decreasing, flux stays flat until amp-turns let flux start decreasing, and as current heads toward negative peak flux gets pushed toward negative saturation.

After the negative peak, current starts positive again and this repetitive cycle continues.

Now here's what causes the trouble.

Instead of a nice smooth sinewave flux, we have a more flat-topped square-ish looking flux wave that

That snap-action is what kills insulation.

Recall that voltage is rate of change of flux, e = n

So induced voltage waveform is no longer a nice smooth sine wave but a series of very high, narrow sharp "needles" .

Since the needles are narrow they dont have much average or RMS value and a voltmeter doesn't even notice them, but they'll poke right through insulation just like the needles they are.

I have more but this is plenty for one session.

To Do:

1. You mentioned "looseness" in the core. Try to get rid of that, a CT core cannot tolerate air-gap it needs high permeability. (Well there's special CT's that can handle DC component in their current but i do not know how they work.)

So find out what's up with that. Perhaps you can lace the core with some sort of twine to tighten it up and get rid of air gaps. If that's a tape wound core it'll need to be tight. i use braided tournament grade fishing line myself. Epoxy invites creativity....The "burnt spot" you spotted on core could be acting like a shorted turn. maybe you can get some varnish or epoxy in there.

2, Try that open circuit voltage versus amp-turns plot.

If it won't saturate there's something wrong with it, iMHO.

A soldering gun might be a useful de-gausser, hold it near then draw away while holding trigger.

I'm back to what i mentioned last night trying to figure out the dual secondaries in parallel.

Last edited:

- #15

jim hardy

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Darn the editing wiped out all my bolds and underlines and quotes.

Read it again in a few minutes.

...........................

Edit -

Got 'er done !

Read it again in a few minutes.

...........................

Edit -

Got 'er done !

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Start with Maxwell's equation in differential form

[tex] curl\space E = - dB/dt [/tex]

Integrate to get Faraday's Law:

[tex] \oint E\space d\ell= - \frac{d}{dt}\int_{A}^{}B\cdot n \space dA [/tex]

The induced voltage in an N-turn loop around the perimeter of an area A is equal to minus the time derivative of the perpendicular magnetic field integrated over the area inside the loop. If B(t) = B

[tex] V= \frac{\omega NA_0 B_0}{\surd 2}[/tex]

Using B

So if the current transformer secondary current is 15 amps/N =1.5 amps, and the core cross section is 0.02 x 0.02 meters

Bob S

- #17

jim hardy

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Well - i finally worked through my arithmetic. Will try to scan and post it in the morning when fresh.

To summarize -

Indeed since a transformer requires some magnetizing current,

the ratio of currents in primary and secondary will differ slightly from what the turns ratio says they should be.

That's because I primary includes magnetizing current.

Remember we usualy ignore that for power transformers.

SO what they did on yours is let two parallel secondaries share the load current.

Were the core perfectly permeable, ie no magnetizing current, a ten turn secondary would work fine.

But since there IS magnetizing current they adjusted for it.

They could have put a 9 turn secondary, that's give 11.1% more current but that's too much. Magnetizing current isn't that high, 11%, if core is good.

And you can't make 9.5 turns or (whatever non-integer you'd need..)

So they made two secondaries and let them share the load..

The formula i came up with needs three numbers :

Resistance of nine-turn winding let's call it R_{9}

Resistance of ten- turn winding,,, let it be R_{10}

and Volts per Amp-Turn of the core from the 60 watt lamp test described earlier. Call it V/AT

Let us also define

Resistance of transformer load = R_{bur} , for burden

and current through nine-turn secondary = I_{9}

and current out of transformer to its load (burden) = I_{bur}

The nine turn secondary's share of output should be

I_{9} / I_{bur} = R10 / ( R9 + R10 + V/AT )

if my arithmetic was right.

So we need to find that wire you unwrapped and see which they made those windings from - ordinary copper or some higher resistance wire.

And plot a line of volts per amp-turn, get its slope .

You said your Fluke isn't so great on low ohms,

Maybe we can make a wheatstone bridge or something? Do you have a source of steady DC at an amp or two??

Does Fluke have a millivolt scale? Have you an old analog mcroamp-meter movement?

You guys can check my work tomorrow .

I have sure learned something from this and i never saw it in a textbook.

Thanks for letting me play in your sandbox !!

old jim

To summarize -

Indeed since a transformer requires some magnetizing current,

the ratio of currents in primary and secondary will differ slightly from what the turns ratio says they should be.

That's because I primary includes magnetizing current.

Remember we usualy ignore that for power transformers.

SO what they did on yours is let two parallel secondaries share the load current.

Were the core perfectly permeable, ie no magnetizing current, a ten turn secondary would work fine.

But since there IS magnetizing current they adjusted for it.

They could have put a 9 turn secondary, that's give 11.1% more current but that's too much. Magnetizing current isn't that high, 11%, if core is good.

And you can't make 9.5 turns or (whatever non-integer you'd need..)

So they made two secondaries and let them share the load..

The formula i came up with needs three numbers :

Resistance of nine-turn winding let's call it R

Resistance of ten- turn winding,,, let it be R

and Volts per Amp-Turn of the core from the 60 watt lamp test described earlier. Call it V/AT

Let us also define

Resistance of transformer load = R

and current through nine-turn secondary = I

and current out of transformer to its load (burden) = I

The nine turn secondary's share of output should be

I

if my arithmetic was right.

So we need to find that wire you unwrapped and see which they made those windings from - ordinary copper or some higher resistance wire.

And plot a line of volts per amp-turn, get its slope .

You said your Fluke isn't so great on low ohms,

Maybe we can make a wheatstone bridge or something? Do you have a source of steady DC at an amp or two??

Does Fluke have a millivolt scale? Have you an old analog mcroamp-meter movement?

You guys can check my work tomorrow .

I have sure learned something from this and i never saw it in a textbook.

Thanks for letting me play in your sandbox !!

old jim

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- #18

jim hardy

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I tried to make those neat equations but was unable to get subscripted text aove and below the line.

How'd you do that?

Perhaps you'll help us more with the math ?

old jim

- #19

dlgoff

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I tried to make those neat equations but was unable to get subscripted text aove and below the line.

How'd you do that?

Perhaps you'll help us more with the math ?

old jim

You can right-click on the equations then "Show Math As"="TeX Commands" to see the code. Redbelly has posted a nice guide Using LaTeX at Physics Forums.

- #20

jim hardy

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Thanks

and please bear with me.. Since DCCAD no longer runs under windows i am without drawing program.

I did my best to figure out that current transformer without going into heavy duty math. It'll probably take me three posts. I type as slow as i think.

Here goes:

Here's a sketch of current transformer.

Let ## flux = ## [itex]\Phi[/itex]

## \Phi = \frac {\sum amp-turns } {Reluctance} ##

note with assumed directions, primary amp-turns are aided by upper secondary and opposed by lower one.

## \Phi = \frac{I1 + 9I9 -10(I9 + Ibur) }{Reluctance} ## (oops i shortened Iburden to Ibur )

## \Phi = \frac{I1 -I9 -10Ibur}{Reluctance} ## ; EQ 1.

We will look again at reluctance soon enough.

Now let's try a loop current method on the electrical circuit and see if anything fortuitous happens.

Here it is.

Each secondary is represented as an ideal voltage source in series with the resistance of its winding.

Value of voltage source is K X [itex]\Phi[/itex]

I know, it has to be d[itex]\Phi[/itex]/dt but remember we're dealing with sine waves.

So there's a 60 hz sinewave flux of RMS amplitude [itex]\Phi[/itex] that produces 377 volts/weber in every turn.

For this little core we should probably be talking microwebers and millivolts.

For now let constant K include all the multipliers for core area, permeability, and the 377 that results when we differentiate a 60 hz sinewave ( 377 = 2∏X60 ) .

So, for flux ## \Phi Sin(wt)## we get voltage ## K\Phi Cos(wt) ## in each turn.

## Ibur Rbur = 10K\Phi - R10(I9+Ibur) ##

and

## Ibur Rbur = 9K\Phi + I9R9 ##

Subtracting,

## 0 = K\Phi - I9(R10+R9) - R10Ibur ##

## K\Phi = I9(R10+R9) + R10Ibur ##

## \Phi = \frac{I9(R10+R9) + R10Ibur} {K}## ; EQ 2

---------------------------------------------------------------------------------------------

I think we're almost there, fellows.

Now we have two different equations for flux.

Set ## EQ 1 = EQ 2 ##

## \frac{I1 -I9 -10Ibur}{Reluctance} = \frac{I9(R10+R9) + R10Ibur} {K}##

Now let us set ## I1 = 10Ibur ##

because that's the objective of this whole experiment (well, really for me it's to understand this current transformer better).

## \frac{10Ibur -I9 -10Ibur}{Reluctance} = \frac{I9(R10+R9) + R10Ibur} {K}##

## \frac{ -I9 }{Reluctance} = \frac{I9(R10+R9) + R10Ibur} {K}##

## \frac{ -I9 }{Reluctance} = \frac{I9 R9}{K} + \frac{I9 R10}{K} +\frac{Ibur R10}{K} ##

## \frac{Ibur R10}{K} = -\frac{ -I9 }{Reluctance} -\frac{I9 R9}{K} - \frac{I9 R10}{K} ##

## \frac{Ibur R10}{K} = -I9 (\frac{ 1 }{Reluctance} +\frac{ R9 +R10}{K} ) ##

## Ibur R10 = -I9 (\frac{ K }{Reluctance} + R9 +R10 ) ##

## -I9 = \frac{Ibur R10} { R9 +R10+\frac{ K }{Reluctance} } ## ;; EQ 3

------------------------------------------------------------------------------------------

My negative I9 means i assumed it in wrong direction which you probably remember from early posts.

R9 and R10 we can measure if Mr Tinkeringone still has some of the wire.

##\frac {K}{Reluctance} ## is going to be (milli)volts per turn per amp-turn of magnetizing current, which we can measure and plot from the sixty watt lightbulb test.

Then i think we'll have this transformer characterized.

We can easily calculate how the load current is shared between secondaries.

I'll be iterested to see how sensitive it is to the resistance of the wire in secondaries.

I feel inadequate that i cant get here via Maxwell's equations.

BUT to have got all this typed is a milestone for me - first time Latex experimenter.

If you find any errors of thought or of transcription, please advise.

I'm pooped. See ya later

Improvements and correct are welcome,

respectfully submitted,

old jim

and please bear with me.. Since DCCAD no longer runs under windows i am without drawing program.

I did my best to figure out that current transformer without going into heavy duty math. It'll probably take me three posts. I type as slow as i think.

Here goes:

Here's a sketch of current transformer.

Let ## flux = ## [itex]\Phi[/itex]

## \Phi = \frac {\sum amp-turns } {Reluctance} ##

note with assumed directions, primary amp-turns are aided by upper secondary and opposed by lower one.

## \Phi = \frac{I1 + 9I9 -10(I9 + Ibur) }{Reluctance} ## (oops i shortened Iburden to Ibur )

## \Phi = \frac{I1 -I9 -10Ibur}{Reluctance} ## ; EQ 1.

We will look again at reluctance soon enough.

Now let's try a loop current method on the electrical circuit and see if anything fortuitous happens.

Here it is.

Each secondary is represented as an ideal voltage source in series with the resistance of its winding.

Value of voltage source is K X [itex]\Phi[/itex]

I know, it has to be d[itex]\Phi[/itex]/dt but remember we're dealing with sine waves.

So there's a 60 hz sinewave flux of RMS amplitude [itex]\Phi[/itex] that produces 377 volts/weber in every turn.

For this little core we should probably be talking microwebers and millivolts.

For now let constant K include all the multipliers for core area, permeability, and the 377 that results when we differentiate a 60 hz sinewave ( 377 = 2∏X60 ) .

So, for flux ## \Phi Sin(wt)## we get voltage ## K\Phi Cos(wt) ## in each turn.

## Ibur Rbur = 10K\Phi - R10(I9+Ibur) ##

and

## Ibur Rbur = 9K\Phi + I9R9 ##

Subtracting,

## 0 = K\Phi - I9(R10+R9) - R10Ibur ##

## K\Phi = I9(R10+R9) + R10Ibur ##

## \Phi = \frac{I9(R10+R9) + R10Ibur} {K}## ; EQ 2

---------------------------------------------------------------------------------------------

I think we're almost there, fellows.

Now we have two different equations for flux.

Set ## EQ 1 = EQ 2 ##

## \frac{I1 -I9 -10Ibur}{Reluctance} = \frac{I9(R10+R9) + R10Ibur} {K}##

Now let us set ## I1 = 10Ibur ##

because that's the objective of this whole experiment (well, really for me it's to understand this current transformer better).

## \frac{10Ibur -I9 -10Ibur}{Reluctance} = \frac{I9(R10+R9) + R10Ibur} {K}##

## \frac{ -I9 }{Reluctance} = \frac{I9(R10+R9) + R10Ibur} {K}##

## \frac{ -I9 }{Reluctance} = \frac{I9 R9}{K} + \frac{I9 R10}{K} +\frac{Ibur R10}{K} ##

## \frac{Ibur R10}{K} = -\frac{ -I9 }{Reluctance} -\frac{I9 R9}{K} - \frac{I9 R10}{K} ##

## \frac{Ibur R10}{K} = -I9 (\frac{ 1 }{Reluctance} +\frac{ R9 +R10}{K} ) ##

## Ibur R10 = -I9 (\frac{ K }{Reluctance} + R9 +R10 ) ##

## -I9 = \frac{Ibur R10} { R9 +R10+\frac{ K }{Reluctance} } ## ;; EQ 3

------------------------------------------------------------------------------------------

My negative I9 means i assumed it in wrong direction which you probably remember from early posts.

R9 and R10 we can measure if Mr Tinkeringone still has some of the wire.

##\frac {K}{Reluctance} ## is going to be (milli)volts per turn per amp-turn of magnetizing current, which we can measure and plot from the sixty watt lightbulb test.

Then i think we'll have this transformer characterized.

We can easily calculate how the load current is shared between secondaries.

I'll be iterested to see how sensitive it is to the resistance of the wire in secondaries.

I feel inadequate that i cant get here via Maxwell's equations.

BUT to have got all this typed is a milestone for me - first time Latex experimenter.

If you find any errors of thought or of transcription, please advise.

I'm pooped. See ya later

Improvements and correct are welcome,

respectfully submitted,

old jim

Last edited:

- #21

jim hardy

Science Advisor

Gold Member

Dearly Missed

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Getting back to that ## \frac{K}{Reluctance} ## term:

## \Phi = \frac {\sum amp-turns } {Reluctance} ## ;

SO

##Reluctance = \frac {\sum amp-turns } {\Phi } ## ; units are ## \frac{amp-turns} {weber} ##

AND K was constant of proportionality between volts and flux, 377 volts/weber

so ## \frac{K(\frac{volts}{weber})} {Reluctance({\frac{amp-turns}{weber })}} = \frac{K}{Reluctance} ## in units of ## \frac{volts}{amp-turn} ## How about that - looks almost like ohms !

Now , the transformer can be tested with an open secondary and exciting surrent set by a 60 watt lamp to plot volts/ampturn*for that particular core*,.

Result gives that core's ## \frac{ K}{Reluctance} ## term for EQ 3.

Perhaps we'll rearrange EQ3 :

EQ 3: ## -I9 = \frac{Ibur R10} { R9 +R10+\frac{ K }{Reluctance} } ##

Allow me to correct the sign and divide by Ibur

##\frac{ I9}{Ibur} = \frac{ R10} { R9 +R10+\frac{ K }{Reluctance} }##

The nine turn secondary contributes that share of output current, ten turn secondary contributes the remainder.

It will be interesting to plug in some reasonable guesses at winding resistance and volts/ampturn for the little core. Or maybe tinkeringone has some measurements?

I'd start with his measured lengths at ~7 milliohms per foot. ten miilivolts per turn per ampturn. That gives a 39/61 % current split.

## \Phi = \frac {\sum amp-turns } {Reluctance} ## ;

SO

##Reluctance = \frac {\sum amp-turns } {\Phi } ## ; units are ## \frac{amp-turns} {weber} ##

AND K was constant of proportionality between volts and flux, 377 volts/weber

so ## \frac{K(\frac{volts}{weber})} {Reluctance({\frac{amp-turns}{weber })}} = \frac{K}{Reluctance} ## in units of ## \frac{volts}{amp-turn} ## How about that - looks almost like ohms !

Now , the transformer can be tested with an open secondary and exciting surrent set by a 60 watt lamp to plot volts/ampturn

Result gives that core's ## \frac{ K}{Reluctance} ## term for EQ 3.

Perhaps we'll rearrange EQ3 :

EQ 3: ## -I9 = \frac{Ibur R10} { R9 +R10+\frac{ K }{Reluctance} } ##

Allow me to correct the sign and divide by Ibur

##\frac{ I9}{Ibur} = \frac{ R10} { R9 +R10+\frac{ K }{Reluctance} }##

The nine turn secondary contributes that share of output current, ten turn secondary contributes the remainder.

It will be interesting to plug in some reasonable guesses at winding resistance and volts/ampturn for the little core. Or maybe tinkeringone has some measurements?

I'd start with his measured lengths at ~7 milliohms per foot. ten miilivolts per turn per ampturn. That gives a 39/61 % current split.

Last edited:

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