Thickness of wire in SMPS transformer?

Can you tell me if my primary wire is too thin on my offline full-bridge smps transformer?


The maximum primary current is 2.4Amps RMS.

Originally , i specified 84 turns of (2*0.5)mm enamelled copper wire.....-in 4 layers

(-thats two strands of 0.5mm enamelled copper wire acting like "one" wire)


However, the transformer manufacturer came back to me and said that i should just use 84 Turns of 0.7mm TEX-E triple insulated wire.....this has an inner conductor diameter of 0.5mm.....and could give the 84 turns in just two layers

Switching frequency = 84KHz.

Do you think that 0.5mm wire is too thin for 2.4 Amps rms?
-Also, TEX-E triple insulated wire has a insulative coating and surely would mean the inner conductor overheating whilst carrying 2.4 Amps RMS?

the ferrite core is an ETD44.
Max power is 335W at the output.

It's really hard to say without knowing all the details of the transformer construction. Using heavy insulation sounds like a dubious proposal, unless you actually need it for isolation reasons. Having less layers though can be a good thing, though, since the heat from the inner layers will escape easier. It can result in cooler operation even if total losses are higher due to higher primary resistance.

Yes the trafo needs 3500vrms isol from pri to sec

yes, I think it would lead to overheating. if I got my calculation right transformer would operate around 12Amp/mm2 current density, which if very high for even bare wire, forget about insulation. if you plan to use some kind of cooling you want to see what would be steady state temperature. Without forced cooling, I would not go for it.

Right, now that I look at the numbers, 2.4A is quite a bit to stuff through 0.5mm diameter wire at that frequency. Even two 0.5mm wires in parallel may not be enough. If I were you I would wind one myself and just run DC current through it and see how hot it gets.

Hi,

an accepted rule is:

".... in a copper wire, a current density J = 4 A / mm^2, will not heat the wire..."

Example
--------
- Let's say in primary circuit you have an RMS current of 0.4 A
- The area of the wire is equal to I / J = 0.4 / 4 = 0.1 mm ^2
- From the area you can then calculate the diameter of the wire

Always use the RMS value of the current.

Enrico Migliore

thanks but is that not the rule for 50Hz transformers.....as opposed to smps transformers?

Current density numbers like 4 A/mm² are useless for SMPS transformers. Usually they are designed for a specific overtemperature based on surface area and losses, power losses are e.g. equally shared between core and copper. But skin and proximity effect are severe for 84 kHz and 0.5 mm wire gauge, expectable AC/DC resistance factors are 5 or more. I like the EPCOS ferrite tool for an overview calculation.

Hi,

> Current density numbers like 4 A/mm² are useless for SMPS transformers.
It's a starting point for designing the diameter of the primary and the secondary wires.

The most important things are the measurements taken on the prototypes.

> power losses are e.g. equally shared between core and copper.
That's true.


> thanks but is that not the rule for 50Hz transformers.....as opposed to smps transformers?
The rule applies also to 50 Hz transformers of course.
You just need to know the RMS value of your current.

Enrico Migliore

Current density numbers like 4 A/mm² are useless for SMPS transformers.

Click to expand...

...i quite agree......but page 313 of "switching power supply design" by Pressman, Billings and Morey tells that 1 Amp per 500 circular mils is a good current density

(thats 4 Amps per mm^2)

This page also says that one should never go above 1 Amp per 300 circular mils

Conversion circular mils to mm^2:
Circular Mils to Square Millimeters Conversion Calculator


....so basically its saying NEVER go more dense than 6.7 Amps per mm^2.

Do you agree with this?

The 1A per 500 circular mills rule is something I've heard applied to DC bus bar, not transformers. Could be mistaken though.

I think you can apply numbers like that to high frequency transformers if you calculate the effective conducting area of the wire and use that instead of the total copper area. So your area isn't pi*r^2, but rather 2*pi*r*δ.