Power transformer design calculations

Question

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:

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!

(example of calculations for a single phase transformer)

Answer 1

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.

Answer 2

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

U_RMS = 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.