How to calculate Toroidal Core Maximum VA capacity

In summary, The conversation is about the technical details of a toroidal core OD-10cm, ID-6cm, height-5cm and weight of 1.845 kg (M4 grade CRGO). The person wants to make a toroidal transformer using this core and is asking for help in calculating the core area and maximum VA capacity. They also mention attaching a test certificate of the core and provide a link for designing a transformer. The conversation also discusses the use of a toroidal core for an AC Mains step-down transformer and the importance of safety considerations. There is also a discussion on how to determine the required number of turns and the maximum size wire that can be used for the transformer. The conversation ends with a question about
  • #1
SanPhysics
31
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  • #3
SanPhysics said:
What are the specs of the transformer you want to wind on this core? What frequency? What input and output characteristics? What gauge wire(s) do you plan to use for Primary and Secondary?

Are you using the toroidal core for noise reasons? How will you wind the Primary and Secondary? Will they be sector wound, or full-circumfurential (bifilar?)?
 
  • #4
berkeman said:
What are the specs of the transformer you want to wind on this core? What frequency? What input and output characteristics? What gauge wire(s) do you plan to use for Primary and Secondary?

Are you using the toroidal core for noise reasons? How will you wind the Primary and Secondary? Will they be sector wound, or full-circumfurential (bifilar?)?

My line voltage is 230VAC/50Hz I wanted single phase step down transformer to make from this core. But how to determine Core maximum VA capacity. is it 100VA or 200VA how to calculate that. Is Core wet is important in calculating VA .This core is M4 CRGO.Or I have to first wind Primary winding but how to test core maximum VA through that please help.
 
  • #6
SanPhysics said:
My line voltage is 230VAC/50Hz I wanted single phase step down transformer to make from this core. But how to determine Core maximum VA capacity. is it 100VA or 200VA how to calculate that. Is Core wet is important in calculating VA .This core is M4 CRGO.Or I have to first wind Primary winding but how to test core maximum VA through that please help.
Why are you wanting to use a toroidal core for an AC Mains step-down transformer? And what experience do you have working with AC Mains projects? What safety considerations are you keeping in mind in this project? What will the transformer be used for?
 
  • #7
SanPhysics said:
is it 100VA or 200VA how to calculate that.

anorlunda's link looks to me like a good one.

As you read through it
here's the direction you need to be heading -

1. How many turns of wire must you have to in order for your core to support the voltage you want?
That depends on the cross sectional area of the core. I cannot read that test you posted, too blurry and small.
Look for a volts per turn rating.

2. Your toroid has area of X in which you can place those turns of wire.
They call it 'Window Area 'in transformer circles.
What is maximum size wire of which you can physically fit your required number of turns into that area ?

Ampacity of that wire X voltage your core can support is your maximum VA.
 
  • #8
berkeman said:
Why are you wanting to use a toroidal core for an AC Mains step-down transformer? And what experience do you have working with AC Mains projects? What safety considerations are you keeping in mind in this project? What will the transformer be used for?

  • Because I wanted to build toroidal transformer.If toroidal core is used Transformer wet is much lower than same VA EI transformer And low humming high efficiency.I make few mains EI transformer and they working good And I lived without any accident. want to use toroidal transformer for amplifier.Can you help me now.
 
  • #10
jim hardy said:
anorlunda's link looks to me like a good one.

As you read through it
here's the direction you need to be heading -

1. How many turns of wire must you have to in order for your core to support the voltage you want?
That depends on the cross sectional area of the core. I cannot read that test you posted, too blurry and small.
Look for a volts per turn rating.

2. Your toroid has area of X in which you can place those turns of wire.
They call it 'Window Area 'in transformer circles.
What is maximum size wire of which you can physically fit your required number of turns into that area ?

Ampacity of that wire X voltage your core can support is your maximum VA.

Dear hardy sir
  • You have to zoom that image . This report says 3.30 Turn per volt for 1.5T. suppose we assume 3 turn per volt so 230*3=690 primary turn is required but how to calculate window area of toroidal core and how to chose wire diameter for primary if I chose 0.80A(dimeter-0.67mm,used-750 kcmil/A) supported wire size for primary. Than input VA of transformer is 184VA.Can I use this wire if it fits in Half Window Area of toroidal core without considerations of Core VA capacity.
 
  • #11
SanPhysics said:
Core VA capacity.

Why do you think the core cares how many VA it is transporting ?
Flux is proportional to magnetizing current not load current.

Its VA capacity is limited by how much wire you can fit through the window. The bigger the core cross section the fewer turns you need.

SanPhysics said:
Can I use this wire if it fits in Half Window Area of toroidal core without considerations of Core VA capacity.

That's your constraint - does it fit ? Will the winding run cool enough ?I'd go a little under 1.5T not over. Look at the magnetizing currrent !

230 X 3.3 = 759
and do you really have 230 V or closer to 240 ?
 
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  • #12
jim hardy said:
Why do you think the core cares how many VA it is transporting ?
Flux is proportional to magnetizing current not load current.

Its VA capacity is limited by how much wire you can fit through the window. The bigger the core cross section the fewer turns you need.
That's your constraint - does it fit ? Will the winding run cool enough ?I'd go a little under 1.5T not over. Look at the magnetizing currrent !

230 X 3.3 = 759
and do you really have 230 V or closer to 240 ?

Dear Hardy sir

I found following link.They calculate Maximum power handling capacity of toroidal core.as bellow

VA=5.0*J*Bm*f*Ac min*ID²*10-7

where:
VA =Power handling capacity (VA)
J =Current density (A/ mm2)
Bm =Maximum induction ( Tesla)
f =Frequency ( Hz)
ID =Inside diameter (mm)
Ac min =Effective cross sectional area (mm2)
If I enter My core values in that equation result is as follows.

VA= (5.0*3.1*1.5*50*1000*3600)*0.0000001=418.5
Is that means I can make toroidal transformer up to 418 VA.this website has all information about toroidal core calculation.

http://www.eilor.co.il/page_13250
 
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  • #13
SanPhysics said:
Is that means I can make toroidal transformer up to 418 VA.

There you go - Well Done !
That's a practical page !

I do a reasonableness check on formulas before i trust them
VA=5.0*J*Bm*f*Ac min*ID²*10-7

VA =Power handling capacity (VA)
J =Current density (A/ mm2)
Bm =Maximum induction ( Tesla)
f =Frequency ( Hz)
ID =Inside diameter (mm)
Ac min =Effective cross sectional area (mm2)

Let's see here
jim hardy said:
Its VA capacity is limited by how much wire you can fit through the window.

J X ID2 tells how many amps i can get through the window for a given current density, which will define amps X turns
Bm X Acmin tells how much flux i can get through the core , which when multiplied by frequency defines the volts per turn

amps X turns X volts/turn = VA
so the form of the equation passes my sanity check.
They collected their constants in those two numbers, 5.0 & 10-7
and i didnt check them. This is just a sanity check after all ...

SanPhysics said:
VA= (5.0*3.1*1.5*50*1000*3600)*0.0000001=418.5
I see where you got every number except the 3.1 for J
and i got the same result as you when i multiplied

so yes i think your 418.5 VA is the capacity of that core at 3.1 amps/mm2
418 VA / 230V = 1.82 amps / 3.1 = 0.59mm2 wire , approximately #19 which seems reasonable
Be aware it might take considerable patience to get enough turns on it when winding by hand instead of using a machine that lays them in neatly
but it's really instructive to wind a core and run a saturation curve on it - plot magnetizing current versus volts per turn.

I'm very much impressed with that Eilor tutorial you found. NIce Job !

Have Fun . We learn 10X better by actually doing things than by just reading about doing them.

Keep us posted ?

old jim
 
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  • #14
jim hardy said:
There you go - Well Done !
That's a practical page !

I do a reasonableness check on formulas before i trust them

Let's see hereJ X ID2 tells how many amps i can get through the window for a given current density, which will define amps X turns
Bm X Acmin tells how much flux i can get through the core , which when multiplied by frequency defines the volts per turn

amps X turns X volts/turn = VA
so the form of the equation passes my sanity check.
They collected their constants in those two numbers, 5.0 & 10-7
and i didnt check them. This is just a sanity check after all ...I see where you got every number except the 3.1 for J
and i got the same result as you when i multiplied

so yes i think your 418.5 VA is the capacity of that core at 3.1 amps/mm2
418 VA / 230V = 1.82 amps / 3.1 = 0.59mm2 wire , approximately #19 which seems reasonable
Be aware it might take considerable patience to get enough turns on it when winding by hand instead of using a machine that lays them in neatly
but it's really instructive to wind a core and run a saturation curve on it - plot magnetizing current versus volts per turn.

I'm very much impressed with that Eilor tutorial you found. NIce Job !

Have Fun . We learn 10X better by actually doing things than by just reading about doing them.

Keep us posted ?

old jim

Hello hardy sir
  • Thanks for your replay and all your suggestion. This page have one table under power handling capacity in that table current density A/mm2 depend on toroidal core wet. my core wet is 1842 grams according to that table if core wet is under 3500 grams current density should be 3.1A/mm2.
  • I found one excel sheet of toroidal transformer winding calculation when i searching Toroidal core on ebay. Can you tell me is this excel sheet is right for calculation of toroidal transformer. https://drive.google.com/file/d/0B5ROBprWe44vTDVfZGFWQjBkMUk/view?usp=sharing
 
  • #15
SanPhysics said:
Can you tell me is this excel sheet is right for calculation of toroidal transformer. https://drive.google.com/file/d/0B5ROBprWe44vTDVfZGFWQjBkMUk/view?usp=sharing
How would i know? All i can do is sanity/reasonableness checks on some of the entries

Most of the them look reasonable to me
#19 wire is 0.9mm diameter plus insulation
~1mm wire wound around a 6 cm ID circle should be 60π turns, 188 and he got 191 very reasonable
next layer should be a few less turns because ID is smaller and he shows that,
...
1 amp would be 230 watts
i figured about 3 ohms of primary resistance (791 turns at 5 + 5 + 2 + 2 cm per turn = 11 meters X )0.0264 ohm/meter = 2.9Ω )
so i guess his entry for primary length must be centimeters not feet not meters
and the primary resistance enty must be milli-ohms not ohms
that'd be 3 volts of primary drop out of 230 applied, 3/230 = maybe 1.3% which falls between his Excel entries for 100 and 200 wattsSo i'd say it passes the sanity/reasonableness check
though it sure would have instilled more initial confidence had he said what were his units
and i wonder what is "final window" ?

Try changing things like wire gage , core thickness, and see if all the numbers move in right direction

That's all i can say from that screenshot.

I'd have wound it based on pencil&paper calculations.
That spreadsheet would give me confidence in them because somebody else got similar result.

old jim
 
  • #16
jim hardy said:
How would i know? All i can do is sanity/reasonableness checks on some of the entries

Most of the them look reasonable to me
#19 wire is 0.9mm diameter plus insulation
~1mm wire wound around a 6 cm ID circle should be 60π turns, 188 and he got 191 very reasonable
next layer should be a few less turns because ID is smaller and he shows that,
...
1 amp would be 230 watts
i figured about 3 ohms of primary resistance (791 turns at 5 + 5 + 2 + 2 cm per turn = 11 meters X )0.0264 ohm/meter = 2.9Ω )
so i guess his entry for primary length must be centimeters not feet not meters
and the primary resistance enty must be milli-ohms not ohms
that'd be 3 volts of primary drop out of 230 applied, 3/230 = maybe 1.3% which falls between his Excel entries for 100 and 200 wattsSo i'd say it passes the sanity/reasonableness check
though it sure would have instilled more initial confidence had he said what were his units
and i wonder what is "final window" ?

Try changing things like wire gage , core thickness, and see if all the numbers move in right direction

That's all i can say from that screenshot.

I'd have wound it based on pencil&paper calculations.
That spreadsheet would give me confidence in them because somebody else got similar result.

old jim
Hello sir

In that sheet instruction tap have all information about using worksheet "final window" is diameter of the remaining hole in the core center, before any final insulation layer is applied. in millimeters. Is wind transformer about 350VA using that calculation is good in real life.I attach instruction tab information in pdf form bellow.
 

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  • #17
Curious, the guy mixes inches and millimeters

but the calculations are straightforward enough
i see you have to tell the program turns per volt
so it's not doing any magnetic calculations just physical ones to fit your turns into the space available.
It doesn't do any heat calculations either
calculate your I^2R losses at rated power
if they are less than 1 watt per square inch of surface i'd say you're in fine shape(old rule of thumb)
otherwise it might be prudent to up wire by one AWG .

i thought that was a screenshot but i see it's an active spreadsheet out there someplace

old jim
 
  • #18
https://drive.google.com/file/d/0B5ROBprWe44vTDVfZGFWQjBkMUk/view?usp=sharing
jim hardy said:
Curious, the guy mixes inches and millimeters

but the calculations are straightforward enough
i see you have to tell the program turns per volt
so it's not doing any magnetic calculations just physical ones to fit your turns into the space available.
It doesn't do any heat calculations either
calculate your I^2R losses at rated power
if they are less than 1 watt per square inch of surface i'd say you're in fine shape(old rule of thumb)
otherwise it might be prudent to up wire by one AWG .

i thought that was a screenshot but i see it's an active spreadsheet out there someplace

old jim

Hello Hardy sir
No sir Turn per volt calculate spreadsheet automatically try changing core size value. And Can you tell me what is used of "Power load,W" column. Is for how mush regulation of transformer is given in particularly used VA rating.How to calculate I^2R losses.Tell me if you want this active spreadsheet.
 
  • #19
SanPhysics said:
No sir Turn per volt calculate spreadsheet automatically try changing core size value.
i don't know how to make it go. But that is to me amazing ... Handy program to know about!

SanPhysics said:
And Can you tell me what is used of "Power load,W" column.
Hmm it's green, so you had to type them in ?
I'd guess the program figures regulation at those points. ame as you proposed


Is for how mush regulation of transformer i think so

is given in particularly used VA rating. i don't know what you mean by that

How to calculate I^2R losses. i estimated simply by resistance of the winding X square of current at selected VA. At the time i didnt know he'd given resistance in milliohms so estimated from a wire table and my own estimated length so i probably differed slightly from his numbers.


Will you be winding that transformer soon ? Keep us posted !

old jim
 
  • #20
hello hardy sir

Finally Done primary winding today of my toroidal core.Attach photo of transformer below.
755 turn of 0.914 mm magnet wire.
core size before winding OD-100mm, ID-60mm, Ht-50mm (M4,CRGO) as we discuss early.
Sir when I test for secondary voltage I get 2.7 volt on 10 turn of magnet wire . why I get less volt per turn. Because before full primary winding when I wind 10 turn for testing perpose on core I get 3.2 volt.Please replay soon
 
  • #21
SanPhysics said:
Finally Done primary winding today of my toroidal core.Attach photo of transformer below.
755 turn of 0.914 mm magnet wire.
230 volts across 755 turn primary = 0.305 volts per turn
That's how much flux is in your core and that's what it will induce into secondary.

Do you have a slider transformer(Variac) ?

The .27 volts per turn you measured sounds about right
try swapping your volt meter leads, the one turn they complete might explain the 10% difference between .27 and .305

old jim
 
  • #22
SanPhysics said:
Because before full primary winding when I wind 10 turn for testing perpose on core I get 3.2 volt.Please replay soon
Ahh i missed that part , you tested it with a partial primary ...

How many turns were on your primary when you made that partial test ? Sounds like you had a lot fewer.

Volts per turn is the key to understanding what goes on in a core, it's a measure of induction (amount of flux)

The fewer primary turns, the more volts per turn

so when you tested with a partial primary, what was the turns ratio?

Very nice looking job in the pictures, by the way... Will you tape it when through? Or dip in varnish ?
 
  • #23
jim hardy said:
230 volts across 755 turn primary = 0.305 volts per turn
That's how much flux is in your core and that's what it will induce into secondary.

Do you have a slider transformer(Variac) ?

The .27 volts per turn you measured sounds about right
try swapping your volt meter leads, the one turn they complete might explain the 10% difference between .27 and .305

old jim

Thanks Hardy sir
What is slider transformer its sound veritable voltage transformer please clear for me. why in secondary drooping volt per turn
 
  • #24
jim hardy said:
Ahh i missed that part , you tested it with a partial primary ...

How many turns were on your primary when you made that partial test ? Sounds like you had a lot fewer.

Volts per turn is the key to understanding what goes on in a core, it's a measure of induction (amount of flux)

The fewer primary turns, the more volts per turn

so when you tested with a partial primary, what was the turns ratio?

Very nice looking job in the pictures, by the way... Will you tape it when through? Or dip in varnish ?

hardy sir
I tested with my other transformer who have 12 volt secondary make input supply for this partial test 37 primary turn and 20 secondary turn i get 6.4 to 6.5 volt in output.
I dint use any tape or varnish but simple cotton tape without any adhesive. wrap cardboard first on bare core and apply cotton tapes tightly over on cardboard Than wind primary turn. Can it is OK to use cotton tapes.How to measure of induction (amount flux) of core.
 
  • #25
SanPhysics said:
What is slider transformer its sound veritable voltage transformer
Yes, they're also known as "Variac" but hat's a trademark
Staco makes them too
SanPhysics said:
clear for me. why in secondary drooping volt per turn
A transformer has a turns ratio
Vsecondary / Vprimary = Turns secondary / Turns primary

so Vsecondary/Turns secondary = Vprimary /Turns primary

volts per turn is same both sides

when you add turns to primary and appply same voltage(230?) you reduce volts per turn on both

SanPhysics said:
I tested with my other transformer who have 12 volt secondary make input supply for this partial test 37 primary turn and 20 secondary turn i get 6.4 to 6.5 volt in output.
12 X 20 / 37 = 6.49, it's working just right !
 
  • #26
SanPhysics said:
How to measure of induction (amount flux) of core.

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 = nturns 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 ?
 
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  • #27
jim hardy said:
Yes, they're also known as "Variac" but hat's a trademark
Staco makes them too
A transformer has a turns ratio
Vsecondary / Vprimary = Turns secondary / Turns primary

so Vsecondary/Turns secondary = Vprimary /Turns primary

volts per turn is same both sides

when you add turns to primary and appply same voltage(230?) you reduce volts per turn on both12 X 20 / 37 = 6.49, it's working just right !

So hardy sir
to get 70 volt 230*232/755=70.67 volt 232 turn for secondary.
But Actually after I finish primary winding. try finding secondary turn per volt wind 10 turn for testing i get 2.7 volt means 0.27 volt per turn so to get 70 volt 70/0.27=259 turn. so how many turn for secondary 232 or 259 please help.
 
  • #28
place 1 turn around the core, make it from just your meter leads
see wnether you get 0.27 or 0.30 volts, write down result

then see which is closer - 232 or 259

then wind 232 turns but don't cut your wire, leave enough for 27 more turns (i'd leave enough for 40 or 50 more and measure voltage
if you come up short of volts keep adding turns until it's where you want it

what flux density did you find ?
 
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  • #29
jim hardy said:
place 1 turn around the core, make it from just your meter leads
see wnether you get 0.27 or 0.30 volts, write down result

theb see which is closer - 232 or 259

then wind 232 turns but don't cut your wire, leave enough for 27 more turns (i'd leave enough for 40 or 50 more and measure voltage
if you come up short of volts keep adding turns until it's where you want it

what flux density did you find ?

Hello sir
0.00137*0. 001=1.37 is it right
Can I get multi pal voltage on same secondary winding how to do it
 
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  • #30
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  • #31
jim hardy said:
multiple ?

tapped windings are done all the time

Just atop winding for moment, bring out a small loop in your secondary wire at however many turns you want, then continue winding
That loop becomes a tap.

http://www.bing.com/videos/search?q...8EEA92C1EEA1ECB69D6B8EEA92C1EEA1E&FORM=VRDGAR

hello hardy sir
I am winding center tap trafo of 35 0 35 using duel winding method(2 wire at same time ) ,if i want lower voltage from same winding can it possible ?
e.g for 25 0 25
 
  • #32
SanPhysics said:
I am winding center tap trafo of 35 0 35 using duel winding method(2 wire at same time ) ,if i want lower voltage from same winding can it possible ?
e.g for 25 0 25
sure, bring out a tap at that number of turns
remember your turns ratio equation ?

230/755 = 25/X
X about 78 turns ?

did you ever figure out what is your flux density ?
 
  • #33
  • #34
SanPhysics said:
divide Webers by area of core in square meters to get Teslas
0.00137*0.001=1.37 Teslas.Is it right
sorry, i missed your earlier reply

from your first post
SanPhysics said:
Technical details of core OD-10cm, ID-6cm, height- 5cm
let's see
that gives for cross section 2cm difference in radius X 5 cm height = 10 square cm = 0.001 square meters

looks right to me !You must have a pretty good core there.

http://www.mag-inc.com/products/tape-wound-cores/typical-hysteresis-loops-for-tape-wound-cores
Orthonol.jpg


magnesil.jpg



old jim
 
  • #35
jim hardy said:
sorry, i missed your earlier reply

from your first post

let's see
that gives for cross section 2cm difference in radius X 5 cm height = 10 square cm = 0.001 square meters

looks right to me !You must have a pretty good core there.

http://www.mag-inc.com/products/tape-wound-cores/typical-hysteresis-loops-for-tape-wound-cores
View attachment 104284

View attachment 104285old jim

Hello hardy sir

So my core is not saturate in 400 va use what's your thought .I think theas charts not for toroidal core (m4 crgo)
 
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