What's minimum turns for 220v 50hz isolation transformer ?
Is only 12 turns in primary and secondary enough ?, wire is 4.0mm diameter

 

Winding some wire around a random core does not an isolation transformer make.

There is a lot that goes into designing a transformer.

 

Winding some wire around a random core does not an isolation transformer make.

There is a lot that goes into designing a transformer.

What do you mean ?, I want to make it with 1 : 1 ratio, fully separated wire

 

I don’t know the answer, but I am pretty sure 12 turns is not even close.

The actual answer will depend strongly on the geometry and material of the core.

Bob

 

I don’t know the answer, but I am pretty sure 12 turns is not even close.

The actual answer will depend strongly on the geometry and material of the core.

Bob

Ok... Is I need only to fill up bobbins ?

 

Ok... Is I need only to fill up bobbins ?

Not even close.

You would need to calculate the correct number of turns and wire gauge based on the magnetic properties of the core and the power levels involved.
Your core may not even be suitable for the transformer you have in mind.

What kind of core do you have?

 

Step 1: measure the core cross section A and apply (turns to volt)=50/A (cm2). Example, core is 4.2x3 cm thus the A=25 cm2 and t/V=12, then 220 V demands 440 turns.
Step 2: measure the window area Q. Divide it to half and immagine that You have wound in the one solely alone turn filling all that Q/2. Then make a correction for the wire is round (0.784) and correction that wire have insulation (thick wire 0.8, thin wire 0.5) and correction that wires sometimes crosses (0.85) thus the total corr factor for 0.2 mm will be around 0.2 and for 0.8 mm be around 0.6. So, now multiply Q/2*k(all corrections)*J(max)*(Volts per turn) and get the gabarite-power. Note that Volts per turn=1 divide the turns per Volt). And note that current density 3-3.5 is realistic maximum for long run, or if without of human observation then better 2.5 and only for short work with long interruptions (ca 5 min/2 hr) J may near to 4.5 A/mm2.
Step 3: When power is at the hands, choose the wire size from (turns*pi()*d^2/4)*k(all corrections)<Q/2. I mean, d here is what You are looking for. Then make a winding and condemn the ill hate if last 1% of turns have no a proper place . Means, throw it away and take one size smaller wire . But better, just take bit more in the safer side those correction factors for "filling the window with the copper"

 

...Or just buy one.

 

Basically you need to wind enough turns so the magnetic core does not saturate with the voltage and frequency you are applying to the primary.
That depends upon the core size, shape, and material.

 

@MisterBill2
Is this formula true ?
CA = 1.152 ×√ (Output Voltage × Output Current)
TPV = 1 / (4.44 × 10-4 × CA × Flux Density × AC frequency)

Example :
Pri : 220v 50hz 6a
Sec : 220v 50hz 6a
CA = 1.152 x√(220 x 6) = 41.9
TPV = 1 / (4.44 × 10^(−4) x 41.9 × 1 × 50)
TPV = 1.08

NP = 1.08 x 220
N = NP:NS = VP:VS
N = 220 : 220
N = 1
NP = 238
NS = 238

N, Turn Ratio
NP, Primary Turn
NS, Secondary Turn
VP, Primary Voltage
VS, Secondary Voltage

Is that right ?

 

@MisterBill2
Is this formula true ?
CA = 1.152 ×√ (Output Voltage × Output Current)
TPV = 1 / (4.44 × 10-4 × CA × Flux Density × AC frequency)

Example :
Pri : 220v 50hz 6a
Sec : 220v 50hz 6a
CA = 1.152 x√(220 x 6) = 41.9
TPV = 1 / (4.44 × 10^(−4) x 41.9 × 1 × 50)
TPV = 1.08

NP = 1.08 x 220
N = NP:NS = VP:VS
N = 220 : 220
N = 1
NP = 238
NS = 238

N, Turn Ratio
NP, Primary Turn
NS, Secondary Turn
VP, Primary Voltage
VS, Secondary Voltage

Is that right ?

How to know flux density ?

 

No, 1:1 means there will be more turns in the secondary.

Why ?, to prevent losses ?
Is this formula true ?
NS = 1.04 x (TPV x VS)
NS = 1.04 x (1.08 x 220)
NS = 1.04 x 238
NS = 248
Is that right ?
Also why ?, to prevent losses or what ?

How to know flux density ?

Also it is good idea to use parallel magnetic wire ?, example : 2x 2.0mm in parallel or 4x 0.2mm in parallel, is this good idea ,

 

What advantage would it bring? I never saw an isolation transformer use this scheme.

I need at least 25kva and 50kva isolation transformer, MUST BE DIY, COST ISN'T PROBLEM, even buy is cheaper I still choose DIY, winding 4.0mm or 6.0mm enameled wire isn't easy, I need about 6.0mm wire, equipment is also not problem I can buy almost every tools, no budget limit /$100k limit

 

A lossless transformer does not exist.

Yes... But you was say 1 : 1 secondary turn needs more than primary turns, is my formula true ?, and WHY secondary turns need more than primary turns even in 1 : 1 ratio ?, It's my question, not lossless transformer, I WAS KNOW THERE IS NO 100% EFFICIENT TRANSFORMER

 

It's easy if cost is not a limiting factor.

Yes... and that's for research purpose...

 

So? What part don't you understand?

Yes... But why secondary need more turns than primary ?

 

Yes... But why secondary need more turns than primary ?

I don't ask to make a lossless transformer, I think only solid state transformer can do that, even maybe limited to about 99.9% efficiency

 

About all possible research has already been done on transformers. Perpetua energy doesn't exist.

Yes... I want to make isolation transformer just that research...

 

To make a 1:1 transformer.

Yes... but why ? 1:1 need's more turns in secondary ?

 

What is a solid state transformer?!?

https://en.wikipedia.org/wiki/Solid-state_transformer