How to calculate Toroidal Core Maximum VA capacity

  • #51
jim hardy
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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..
 
  • #53
jim hardy
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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

upload_2016-8-31_9-36-7.png


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 ?
 
  • #54
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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.
 
  • #55
jim hardy
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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 dont know what you are asking.
 
  • #56
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i dont 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?
 
  • #57
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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.
 
  • #58
jim hardy
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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.
 
  • #59
jim hardy
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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 .
.wikipedia.org/wiki/Phosphoric_acid said:
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".
 
Last edited:
  • #60
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
 
  • #61
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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.
 
  • #62
Jimmy Lalani
Hello San,
The transformer winding is already completed.
 
  • #63
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 = 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 calculator


divide 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
 
  • #64
jim hardy
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. 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.
 
  • #65
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.
 
  • #66
jim hardy
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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:

R8K922p6WSLZ8AgcqDwgG2yX7r98pZ6Z6lxEoxiLV9BTNoUc6OW8IW_zbCElF3mTxCN_BM9tI-tf4WvvNk_J=w223-h57-no.png


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)
8ad6246daa744c16b31f400d50fce49b?AccessKeyId=59DEEF39E23371504A83&disposition=0&alloworigin=1.png

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:
te.png
te1.png

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

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