How to calculate Toroidal Core Maximum VA capacity

[SanPhysics]

Hello hardy sir
Can it is compulsory to add thin copper strip on core between primary and secondary winding to avoid loss of magnatic flux. Because I can't find thin copper strip in my area .please help

 

[jim hardy]

So my core is not saturate in 400 va

it's probably not quite saturated at your 230 volts/755 turns.

The core has no idea what is amp load, just voltage.

I think theas charts not for toroidal core (m4 crgo)

no, they were a guess at what steel you might be usungm4 crgo hmmm
http://www.kryfs.com/knowledge-center/technical-papers/pdfs/article1.pdf

looks like you're about up to the knee
you should see some distortion from sine in the magnetizing current wave
from http://www.vikarsh.com/toroidal-cores.html

Comparative Chart

Toroidal core & C.T manufacturers need to know the B-H values of CRGO Materials after annealing over a wide range of Flux densities ranging from 100 Gauss to 19000 Gauss. We give below the actual values obtained on a core of size 150 mm OD x 100 mm ID x 25 mm HT with a stacking factor of 95% and a frequency of 50Hz.

....Tesla...M4..at/cm...23MOHgrade at/cm.. two other specialty cores...

 

[jim hardy]

Can it is compulsory to add thin copper strip on core between primary and secondary winding to avoid loss of magnatic flux.

no, that shield's purpose is not to contain magnetic flux, it's to remove effect of interwinding capacitance .
http://electrical-engineering-portal.com/purpose-of-shielded-isolation-transformer

I probably would forego the shield.
if you do add a shield be sure it doesn't form a closed loop or it will be a shorted secondary.
You could use very thin brass shim stock from an auto parts or machine tool store.
Even aluminum foil if you can figure how to connect to it.

 

[SanPhysics]

Thanks for reply

 

[SanPhysics]

Thanks for reply

Hardy sir can you suggest any method to test output current of transformer so I can confirm that transformer is working properly .

 

[jim hardy]

Hardy sir can you suggest any method to test output current of transformer so I can confirm that transformer is working properly .

What test equipment do you have available ?
if you can observe primary current waveform with an oscilloscope you should see effect of BH curve nonlinearity . The closer you are to the knee the more distorted will be the sine wave current.

If you have only a voltmeter and ammeter -
Check open circuit voltage to confirm turns ratio
measure output voltage with a resistive load to measure "regulation" Δvolts/Δamps , perhaps some large lamps, or a heating element perhaps a kitchen toaster or small electric heater.

 

[SanPhysics]

Hello hardy sir

I am completed all secondary winding
As follows 35 25 0 25 35 .When I measure 35 volt winding respective 0 it showing both 35 volt is OK but when I measure 25 volt winding respective to 0 volt one 25 volt tap shows 10 volt why this happening it shows 25 0 10. attach photo of completed transformer https://drive.google.com/file/d/0B5ROBprWe44vNDIyUlFoVmctLUk/view?usp=sharing

 

[jim hardy]

hmmm i can't quite tell how you wound it

looks like you wound two wires ?
Here's your "start point"
do both wires go up over top and continue wrapping
or does left one go underneath ?

Next, verify
do you have two completely separate windings?
Place labels on your wires like this and find with ohmmeter which three are with each winding.
AND
that the two windings are indeed separate.

Then re-arrange your labels so that one winding has odd numbers 1(start), 3(25 tap), 5(35 finish)
and other winding has even 2(start), 4(25 tap), 6(35 finish) .

Verify again that even labels are on one winding and odd on other

Now energize your transformer and measure the voltages
1-5 (expected = 35V) measured = ? ____
1-3 (expected = 25V) measured = ? ___
3-5 (expected = 10V) measured = ? ___

2-6 (expected = 35V) measured = ? ____
2-4 (expected = 25V) measured = ? ____
4-6 (expected = 10V) measured = ? ____

Next, connect your two start wires 1 and 2 together
measure voltage 3 to 4 , I expect zero, what do you measure ? ____
measure voltage 5 to 6, I expect zero, what do you measure ? ____

If you get those readings try this

disconnect 1 from 2
connect 1 to 6 and use that for 0volt

voltage 1&6 to 5 ? (expected 35) __
voltage 1&6 to 2 ? (expected 35) __
voltage 1&6 to 3 ? (expected 25) __
voltage 1&6 to 4 ? (expected 10) __
voltage 5 to 2 ?... (expected 70) __

Might be we'll have to unwrap some turns and relocate one of the 25 volt taps...

old jim

 

[SanPhysics]

hmmm i can't quite tell how you wound it

looks like you wound two wires ?
Here's your "start point"
do both wires go up over top and continue wrapping
or does left one go underneath ?
View attachment 104463
Thanks hardy sir
Yes both wires go up over top and continue wrapping

Next, verify
do you have two completely separate windings?
Place labels on your wires like this and find with ohmmeter which three are with each winding.
AND
that the two windings are indeed separate.
View attachment 104464
yes two winding is separate

Might be we'll have to unwrap some turns and relocate one of the 25 volt taps...

old jim

All your suggestion is correct.And all result as you expected.
I am using only one tap at a time means 35-0-35 or 25-0-25 .so I don't have to unwrap turn.

 

[jim hardy]

Yes both wires go up over top and continue wrapping

All your suggestion is correct.And all result as you expected.

Okay, there's our trouble
i didn't think about having to place the 25 volt tap 10 volts away from opposite ends of the two windings if they're wrapped same direction
That's one morelittle detail to be aware of, we will both do a better job of helping the next guy on that point...
If you want to you could add more turns to that 10 volt tap to get 25 volts, it'd work, just a minor "ugly" add-on.

On the positive side
you successfully wound that transformer without creating any shorted turns
and it behaves exactly as it should for how it was wound.
I call that a SUCCESS !
If you want to start a few more turns at your 10 volt tap and bring out a new 25 , you'll have a transformer that is functionally just what you wanted.

And - look how much we learned !

Congratulations on a nice job.

old jim

 

[SanPhysics]

Hello Hardy sir

What are you doing
I got new toroidal core from my friend. He want wound power transformer for him.
Core size is OD-140mm, ID-90mm, Ht-50mm (M4 0.27mm CRGO) core wet about 3.3Kg.With my little knowledge an experience in transformer wounding I think
This core is more than 1000VA He want use this transformer of drive 500w audio amplifier. Is my gassing is right please replay soon.I will post core photo soon.

 

[jim hardy]

Core size is OD-140mm, ID-90mm, Ht-50mm

Try that link Anorlunda gave ?
http://www.brighthubengineering.com/diy-electronics-devices/96783-designing-your-own-transformer/

Your core cross section is (7.0-4.5) X 5.0 = 12.5 cm², 1/800 m²

at 1 Tesla 50 hz it'd carry 100π X 1/800 = 0.39 volts per turn
so you'd need 230/0.39 = 589 turns at 230 volts
current for 1kva at 230 volts = 1000 /230 = a little over 4 amps, call it 5
#18 would carry that easily, #20 would probably work fine

window area is π X 4.5² = 63.6 cm²
and per anorlunda's link here's how many turns per square cm you can get with careful technique

at 60.8 turns per cm² 589 turns of #18 would take up 9.7 cm², less than 1/6 of your window

looks to me like you have plenty of core there for 1 kva .

Which isn't surprising, the core has a window 1.5X bigger than your 418 VA core, and 418 X 1.5². = 940,
since it's also got 25% more cross section multiply by 1.25, 940 X 1.25 = 1175 .

Try your spreadsheet ? It seemed to work okay.

This time we'll remember to put the voltage taps close to the dotted end of their respective windings...

old jim

 

[SanPhysics]

Hello dear hardy sir
Thanks for your useful thought

Your core cross section is (7.0-4.5) X 5.0 = 12.5 cm2, 1/800 m2

Can I don't have to consider stacking factor of core.

and per anorlunda's link here's how many turns per square cm you can get with careful technique

This table an my swg table on this link https://en.wikibooks.org/w/index.php?title=Engineering_Tables/Standard_Wire_Gauge&printable=yes
have very much different which table is right

current for 1kva at 230 volts = 1000 /230 = a little over 4 amps, call it 5
#18 would carry that easily, #20 would probably work fine

#18 swg carry 3.7A with circular mil (750 kcmil/A) and 4.61A with (500kcmil/A) which circular mil have to consider.
And can I wound for more than 1.3 Tesla on this core.

 

[jim hardy]

#18 swg carry 3.7A with circular mil (750 kcmil/A) and 4.61A with (500kcmil/A) which circular mil have to consider.
.

hmmm after looking i find recommendations from 350 to 1000. I think i'd use 500, though 750 will run cooler.
Magnetics uses 500 for their inductor design example.
http://www.mag-inc.com/design/design-guides/Inductor-Design-with-Magnetics-Ferrite-Cores

4.61A with (500kcmil/A)

4.61amps X 500 cmil/amp = 2305 circular mils which is closer to #16 per this table
http://www.mwswire.com/pdf_files/mws_tech_book/page4_5_6_33_34.pdf
so i stand corrected

This table an my swg table on this link https://en.wikibooks.org/w/index.php?title=Engineering_Tables/Standard_Wire_Gauge&printable=yes
have very much different which table is right

your table gives turns per cm, how many adjacent turns will fit in a centimeter or inch, observe it's 1/diameter.
Turns per square cm or inch would be the square of that number

And can I wound for more than 1.3 Tesla on this core. (M4 0.27mm CRGO)

Can I don't have to consider stacking factor of core.

What does the datasheet for the core say?

 

[SanPhysics]

observe it's 1/diameter.
Turns per square cm or inch would be the square of that number

Suppose I use #17 gauge wire as per my table for primary winding so wire diameter is 1.42mm. so turn per cm2 is 10/1.42= 7.04, so 7.04*7.04=49 turn per cm2.
nearly 600 turn for primary. so 600/49=12.24cm2 window are required is it right sir.

What does the datasheet for the core say?

no datasheet for core but manufacture test core with 10 turn and volt is 3.95.

 

[jim hardy]

Suppose I use #17 gauge wire as per my table for primary winding so wire diameter is 1.42mm. so turn per cm2 is 10/1.42= 7.04, so 7.04*7.04=49 turn per cm2.
nearly 600 turn for primary. so 600/49=12.24cm2 window are required is it right sir.

i think so. It looks correct, with condition you won't get perfect lay on the wires so it'll usea kittke nore that 12.24 cm²
Have you tried your spreadsheet , the one you used for your transformer ?

no datasheet for core but manufacture test core with 10 turn and volt is 3.95.

0.395 volts per turn equates to how much flux ? 0.395 /100pi = 1.26milliwebers, in cross section of 1/800 m² = only 1.006 Tesla RMS = 1.42 T peak.

See if you can find a recommended operating flux for that core, or stick with 1T rms..

 

[SanPhysics]

Have you tried your spreadsheet , the one you used for your transformer ?

Spreadsheet result have a look. Please Zoom in
https://drive.google.com/file/d/0B5ROBprWe44vaXA3RmkwczU0Mjg/view?usp=sharing

 

[jim hardy]

Wow that'll be a challenging wind, manhandling 300+ turns of #12.
For 230 to 115 I'd look into buying one. Try a search on THG step down transformer to see what's available in your part of the world.

From your spreadshseet

great regulation,,
2.6 turns per volt = 0.385 volts/turn
0.385 / 100π = 1.22 milliwebers , in core area of 12.5 square cm
1.22 X10^-3 W / 12.5 X 10^-4m² = 0.979 Tesla, should be safe for most transformer cores i'd think , it's 1.39T peak

Sanity check looks okay

losses at full load show around 35 watts primary , 75 secondary ,,, so at 2kva it'll run warm..
But i think it'll be a better transformer than you could buy . And you'll have fun.

keep us posted ?

 

[SanPhysics]

Hello hardy sir
Thanks for your thought.
Sir how to calculate quickly both copper winding use how much mm redious of ID.Is that any formula.

 

[jim hardy]

how to calculate quickly both copper winding use how much

What ? To figure power I used volts drop (from your spreadsheet) X amps at 2000VA, 2000/230 for primary and 2000/115 for secondary

how much mm redious of ID.Is that any formula.

i don't know what you are asking.

 

[SanPhysics]

i don't know what you are asking.

I mean How to find both primary and secondary magnet wire use how much Inner diameter (not area in cm2)of the core?

 

[SanPhysics]

Hello hardy sir

Which adhesive i should use on bare core before start winding because core is old and have some rust on core.I have standard Epoxy adhesive pack which contain 2 small tubes one is resin and other is hardener.with mixing two same quantity both tube.I got jelly type transparent liquid.Can i apply this liquid on bare core its got hard in 40 to 60 minit.But I don't know it's uses for electrical purpose.
And can you suggest any insulating electrical tape for insulating primary and secondary winding like professional transformer use for their transformer.

 

[jim hardy]

And can you suggest any insulating electrical tape for insulating primary and secondary winding like professional transformer use for their transformer.

here's 3M's selection.
http://multimedia.3m.com/mws/media/103938O/3m-oem-insulating-and-conductive-tapes-brochure.pdf

myself i use mostly Scotch 27 glass cloth tape but only because i had several rolls of it . I'd like to try their polyester.

I mean How to find both primary and secondary magnet wire use how much Inner diameter (not area in cm2)of the core?

How many layers of winding will you have ? What is thickness of wire ? Each layer will subtract that thickness from radius.

 

[jim hardy]

some rust on core.I have standard Epoxy adhesive pack

Rust on core ? You want to arrest that before you cover it. I would apply a phosphoric acid treatment .

Rust removal
Phosphoric acid may be used to remove rust by direct application to rusted iron, steel tools, or other surfaces. The phosphoric acid changes the reddish-brown iron(III) oxide, Fe2O3 (rust) to ferric phosphate, FePO4. An empirical formula for this reaction is:

2 H3PO4 + Fe2O3 → 2 FePO4 + 3 H2O
Liquid phosphoric acid may be used for dipping, but phosphoric acid for rust removal is more often formulated as a gel. As a thick gel, it may be applied to sloping, vertical, or even overhead surfaces. Different phosphoric acid gel formulations are sold as "rust removers" or "rust killers". Multiple applications of phosphoric acid may be required to remove all rust. Rust may also be removed via phosphate conversion coating. This process can leave a black phosphate coating that provides moderate corrosion resistance (such protection is also provided by the superficially similar Parkerizing and blued electrochemical conversion coating processes).

It's in hardware stores as Naval Jelly or Ospho. It turns the rust black . Then apply your epoxy. I've never had rust eat through paint when i Ospho'd the metal before painting.
Of course, wire brush or steel wool the loose rust away, and get it thoroughly dry before applying paint or epoxy.. I put my treated parts in the oven on "warm".

 

[Jimmy Lalani]

Hello San,
If I already have the transformer ready then how can I calculate the effective area Ac of the core?
My core Dimension are OD=80mm, ID=46mm H=30mm, 100VA
VA=5.0*J*Bm*f*Ac min*ID²*10-7

 

[SanPhysics]

Hello San,
If I already have the transformer ready then how can I calculate the effective area Ac of the core?
My core Dimension are OD=80mm, ID=46mm H=30mm, 100VA
VA=5.0*J*Bm*f*Ac min*ID²*10-7

Hello Jimmy Lalani
Actual bare core dimension required for calculation, is your provided sizes are bare core or after transformer winding completed.please clarify.or proved primary and secondary VA rating.

 

[Jimmy Lalani]

Hello San,
The transformer winding is already completed.

 

[Jimmy Lalani]

try this approach

volts per turn is a good measure

e = n dΦ/dt
if Φ = A sin(100πt)
dΦ/dt = 100πAcos(100t)

so e = n_turns X 100πA cos(100t)
you know n = 755
and e = 230√2 cos(100t)

so A = 230√2cos(100t) / (755 X 100πcos(100t))
A = .00137 Webers

did i make any arithmetic mistakes? Took me several tries to get same answer twice in a row.. darn that Windows calculatordivide Webers by area of core in square meters to get Teslas

volts per turn divided by ω gets magnitude of flux,

0..305 / 100π = .000971 Weber RMS flux, which is .00137 peak, divide them by area of core to get flux density

volts per turn at your line frequency is a handy thing to know about a core.

Now - what do you get for your flux density ?

Hello Hardy Sir,
I was just going through the conversation for calculating Maximum VA capacity. I found this conversation. In this you have taken e=230sqrt(2) cos(100t). I am a bit confused. Normally we take e=Em sinwt. Then why is it so here? Sorry , if it seems a stupid question to ask.

Jimmy Lalani

 

[jim hardy]

. I found this conversation. In this you have taken e=230sqrt(2) cos(100t). I am a bit confused. Normally we take e=Em sinwt. Then why is it so here?

That was just an example of starting with an assumed flux instead of an assumed voltage.
Flux and voltage have a derivative relationship, do they not ? One can ssume any flux one wants and voltage will be its derivative . I started with sine because its derivative has same sign. And san said he has 230 volts RMS.(not peak) where he is.

 

[Jimmy Lalani]

Hello Hardy Sir,
Thank you very much. Now it is crystal clear in my mind.
However, if I assume Bm in order to calculate various parameters of transformer at the very begining, then later do I again have to find the new value? I asked few people and they said that my be I have to iterate until I get a suitable or final value? I could not understand what they meant? Can you please help me.

 

[jim hardy]

However, if I assume Bm in order to calculate various parameters of transformer at the very begining, then later do I again have to find the new value? I asked few people and they said that my be I have to iterate until I get a suitable or final value? I could not understand what they meant?

I don't know why they said that.

If you make a good guess at how much flux density your core will happily accommodate, then the magnetizing current required to produce it will be reasonable. For that reason you need to be aware of you core's flux capability when you choose your operating flux density.
Since cores are available in discrete sizes and number of turns is an integer, you may have to adjust your Bm slightly to get your desired voltage with a standard size core .

Here's an example from the internet, worked out by what appears to be a methodical student with a lot of common sense and a good "Do It Yourself" mindset:
:
:http://engineerexperiences.com/design-calculations.html

1. Core Calculations:
Calculate area of core (central limb) by using following formula:

Ai= area of core
F= operating frequency
Bm= magnetic flux
Te= turns per volts
(for derivation of this formula Click Here)
Assumptions:
So, we know the frequency of the power system. We need magnetic flux and turns per volts. For designing a small transformer magnetic flux is averagely taken as 1 to 1.2.
By putting values we will get the area of core.
Current density of copper wire is taken as 2.2 A / mm2 to 2.4 A/ mm2 (approximately).
So, putting values
F= 50 hz
Bm = 1.2 wb/m2
Te = 4 (turns per volts)

As, we are going to design a practical transformer so we must consider the core available in market. The standard Bobbins available in market practically is 1”x1”, 1.25”x1.5”, 1.5”x1.5” and so on. We took nearest core area available to our calculation. We took bobbin of 2.25 inch2 (1.5”x1.5”) or 0.00145161 meter square. We have the core area. We can calculate turns per volts using this area by following:
Putting f=50 hz; Bm = 1.2 wb/(m^2); Ai= 0.001451 m^2, we got:

here he shows pictures of the transformer he built and it's a pretty nice job...
http://engineerexperiences.com/hardware-design-.html

 

[Daniman]

I don't know why they said that.

If you make a good guess at how much flux density your core will happily accommodate, then the magnetizing current required to produce it will be reasonable. For that reason you need to be aware of you core's flux capability when you choose your operating flux density.
Since cores are available in discrete sizes and number of turns is an integer, you may have to adjust your Bm slightly to get your desired voltage with a standard size core .

Here's an example from the internet, worked out by what appears to be a methodical student with a lot of common sense and a good "Do It Yourself" mindset:
:
:http://engineerexperiences.com/design-calculations.html
here he shows pictures of the transformer he built and it's a pretty nice job...
http://engineerexperiences.com/hardware-design-.html

Why did he take the value of turns per volts=4.

 

[Tom.G]

Unfortunately Jim Hardy is now deceased, so he won't be answering.

I haven't been following the thread, but just from reading the last post it seems the calculation used an iterative approach; that is, take a guess at the unknowns, work the formulas, and see if all the constraints are met.

The 4 Turns-per-Volt ( T_e) was an initial guess (likely based on experience as being 'safe').
This was immediately followed by the calcualtion for the actual Turns-per-Volt which was found to be T_e = 2.6 for that particular core.

Cheers,
Tom