I´m currently designing power transformers in a pretty basic way, using AWG tables for the section of the copper wire, and not knowing what are the real losses, just overdesigning.

  • Three phase transformers: Secondary power between 3-15 kW, star-delta or delta-delta configurations. Input voltage: 380 Vac. These start to give me troubles when the power is more than 5 kW approximate.

  • Single phase transformers: Secondary power between 100-3000 W. Input voltage: 220 Vac. These start to give me troubles between 1800-3000 W.

Specially three phase ones, I've searched a lot of these ones, and not being able to understand the calculations. In my case, they are working in saturation 100% of the time

Let's say that 200 A 24 Vdc are required at output. If the design is delta-delta, it will be:

V = 24 Vdc / sqrt(3) = 13.85 Vac

It = 200 A

I ( tower ) = 200A / sqrt(3) = 116 A

P = 1600 W per tower

(added information⬇⬇⬇⬇)

Core selected:

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Maximum power of core: (5.1 cm x 9 cm) x (5.1 cm x 9 cm) = 2106 W

I primary = 1600 W / 380 V = 4.2 A

Turns x Volt = 45 / core area = 45 / 45.9 = 0.98 Turns x volt

Primary side turns: 380 V * 0.98 = 372 Turns

Secondary side turns: 13.85 V x 0.98 = 14 Turns

(added information⭡⭡⭡⭡)

Then, I would search the section of wire in an AWG table.

Are the calculations ok? I'm missing losses somewhere? Why when I make the bigger ones, it can't reach the calculated current?

Thank you!

enter image description here

(example of calculations for a single phase transformer)

  • 1
    \$\begingroup\$ Could you explain what "troubles" you are having? If your problem is temperature rise, what are you using to estimate power dissipation and thermal resistance? If saturation, what are you using to calculate core area and turns count? \$\endgroup\$ Commented May 11 at 22:58
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    \$\begingroup\$ The problems are saturation related. I use the transformers for charging lead acid batteries, they are connected to a rectifier and then directly to the battery ( controlling the primary side with a relay ). So they are working all the time in maximum power. I added a photo of the way that I use for calculations. \$\endgroup\$ Commented May 11 at 23:02
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    \$\begingroup\$ There are numerous textbooks on transformer design. But if you don't like books, this slide presentation from IEEE gives a good overview. Transformer Design and Design Parameters \$\endgroup\$ Commented May 12 at 13:54
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    \$\begingroup\$ Did you read the datasheet of the cores to find the maximum power at the mains frequency for each core? \$\endgroup\$
    – Uwe
    Commented May 12 at 15:56
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    \$\begingroup\$ The problems are … please don't comment comments asking for clarification or complementary information: edit your post. then directly to the battery gives an awful conversion_ratio/Crest factor. \$\endgroup\$
    – greybeard
    Commented May 12 at 16:54

2 Answers 2


You need to have a minimum number of turns to prevent saturation. The flux density equation is: $$ \hat B_e = {\hat E \over {\omega \, N\, A_e}} $$ Where:
\$ \hat B_e \$ = peak flux density [T]
\$ \hat E \$ = peak emf for sine wave. For square wave, multiply the peak voltage by 1.57.
\$ \omega \$ = angular frequency = \$ 2 \, \pi \, f \$
\$ A_e \$ = core area [\$ m^2\$]
\$ N \$ = number of turns

You also need a minimum number of turns to set the magnetizing inductance which is a parameter you select for your application. Chances are, you'll have enough magnetizing inductance once you meet the saturation requirements.

There are also wire diameter limitations which affect the DC and AC losses. Proximity effect will be something to watch out for, however, you're probably at a frequency low enough where this won't be a problem.

  • 2
    \$\begingroup\$ In case it wasn’t clear to the reader: the magnetization inductance is what the power source “sees” without load. It needs to be sufficiently high to keep magnetization current manageable. A transformer without a load is an inductor after all. It needs to have enough reactance to idle at a reasonable current when fed with mains voltage. \$\endgroup\$ Commented May 13 at 17:11
  • \$\begingroup\$ So you are saying that probably the problem with the transformers that can´t reach the current desired, are because of lack of turns? What information will give me the peak flux density? Is this one the formula for sinewave peak emf?: E = Turns * B ( magnetic field strenght ) * A (area of coil) * ω( angular frequency )? \$\endgroup\$ Commented May 14 at 23:51
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    \$\begingroup\$ @samuelmattio You stated that the transformer was going in to saturation which means more turns are necessary on the primary. When a transformer goes in to saturation, excessive current will be drawn on the primary. If you add details like core area (you list some sort of area without units), saturation flux density (see data sheet or infer from core material), type of core material, bobbin dimensions or winding window, and other relevant information; perhaps a better suited answer awaits. \$\endgroup\$
    – qrk
    Commented May 15 at 2:24
  • \$\begingroup\$ @samuelmattio I did some quick calculations for a single phase transformer with wild guesses. Assumptions: Ae = 48 cm^2, peak saturation flux density = 10000 G, average length of winding = 140 mm, bobbin width = 90 mm. With that, proximity effect shows up and suggests the primary should be 372 turns #10 AWG which give 0.2 ohms AC+DC loss. The secondary is 14 turns #0, although, you can get away with #2 AWG which gives 1 mohm AC+DC loss. These numbers will change when proper information is given. \$\endgroup\$
    – qrk
    Commented May 15 at 3:01

You didn't state what mains frequency you are using, so I'm going to assume it's 50 Hz.

The saturation limit for a mains transformer is given as

URMS = 4.44 fNAB,

where f is your mains frequency in Hz, N number of primary turns, A is core cross section area in m2 and B the core saturation limit in T.

You posted the bobbin drawing and not the core, but assuming the core is 1 mm less than the bobbin, you have 89*50 mm core area, or 0.004510 m2. For a 1 T saturation limit and 380 V AC, you need 380 turns on the primary in order to not saturate it.

Check the core datasheet for saturation limit, but keep in mind that you need margins in your design for mains voltage variations depending in the grid code in your country.

In general for a battery charger, you'll be in a hard spot between designing for some overvoltage on the secondary to actually transfer power to the battery, due to its EMF simply not forward biasing your diodes other than at the very top of the sine wave (bad crest factor) as greybeard pointed out in the comments versus boiling out the battery if left connected for extended period of time, and at the same time have either plenty of leakage inductance to limit peak current delivered towards a depleted battery in order to not overheat it, unless you have external means to limit the current. Please simulate your circuit with battery counter voltage and battery resistance, transformer leakage inductance and play around with the values and you'll see what I mean.

  • \$\begingroup\$ What does mean "1 T saturation? What value should I use? I added the core data that manufacturer provide, but I think that's not saturation limit... Thanks! \$\endgroup\$ Commented May 24 at 14:45
  • \$\begingroup\$ Each core material has a saturation limit. Check yours. For laminated iron, it’s just north of 1 T. \$\endgroup\$
    – winny
    Commented May 24 at 17:03

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