# Help with designing transformer

#### Veldox

##### New Member
Hi, I'm currently working on designing a transformer for a project of mine. I've found a great calculator that does a lot of the maths for me (I've written it out below) but would like to clarify a couple of things.

So my questions are:

Does this “Transformer primary input current (output loaded): 25.575 Ampere RMS” mean that the primary winding will not be able to draw more than 25.5 amps from the power source no matter the size of the load?

And is this “Secondary voltage (output loaded): 39.548 Volt RMS, Secondary voltage (output unloaded): 40.158 Volt RMS” correct as I thought a load on the secondary would significantly drop voltage?

The calculator: http://dicks-website.eu/coilcalculator/#al

here is the design I plan to make, if anything is wrong with this please enlighten me.

Maximum Flux Density in Core: Bmax = 1.1 T (=Tesla)
Effective cross section area of centre core: Ae = 7500 mm2
Effective permeability of the core: ue = 5000
Inductance factor for one core: AL = 392687.5 nH/n2
Number of these cores stacked = 1

Relative Permeability of the core material: ur = 5000 – Estimated
Effective Length of the magnetic path in the core: le = 120mm
Total air gap width: g = 0mm

Total effective cross area of the core: Ae = 0.0075m2
Maximum magnetics flux in core: Magnetic flux max = 0.00825 Wb (=Weber) – Estimated
Effective Permeability of the core: ue = 5000 - Estimated
Total inductance factor of the core: AL = 392687.5 nH/N2

Coil wire size: 5.18922mm = 4 AWG
Average wire length of one turn: 570mm
The coil needs: 126 Turns
Inductance of the coil: 6.234 Henry
Total wire length of the winding: 71.82 metre
DC Resistance of the winding: 0.0594
Total area of the copper for the winding: 2664.79 mm2

Maximum current through coil to keep flux below Bmax: 0.1667 Ampere (DC or AC Peak)
Charging time to maximum current with 230 volts across coil: 0.00451 Seconds
Maximum stored energy in coil: 0.0866 Joules
Maximum AC voltage (sine wave) of 50 Hertz across coil: 230.769 Volts RMS

Primary (input) voltage: Vp = 230 Volts RMS
Secondary (output) voltage: Vs = 41 Volts RMS (not loaded)
Frequency (or lowest frequency): f = 50 Hertz

Number of turns primary winding: Np = 126 turns
Number of turns secondary winding: Ns = 22 turns
Inductance of the primary winding: Lp = 6.234 Henry

Wire size primary winding: 5.18922mm = 4 AWG
wire size secondary winding: 10.4038mm = 000 AWG
Average wire length of one turn on primary winding: 570mm
Average wire length of one turn on secondary winding: 520 mm
Load resistor on secondary winding: 0.27 Ohm

Wire length primary winding: 71.82 Metre
DC resistance of primary winding: Rp = 0.0594 Ohm
Area of the copper for the primary winding: 2664.799 mm2
Wire length secondary winding: 11.44 metre

DC resistance of the secondary winding: Rs = 0.0023 Ohms
Area of the copper for the secondary winding: 1870.23 mm2

Secondary voltage (output loaded): 39.548 Volt RMS
Secondary voltage (output unloaded): 40.158 Volt RMS

Magnetizing current in primary winding (output loaded): 0.1166 Ampere RMS

Transformer primary input current (output loaded): 25.575 Ampere RMS
Power loss in primary winding (output loaded): 38.871 Watts
Power loss in secondary winding (output loaded): 50.526 Watts
Power delivered to the load resistor: 5792.875 Watts

Thanks.

#### Diver300

##### Well-Known Member
Does this “Transformer primary input current (output loaded): 25.575 Ampere RMS” mean that the primary winding will not be able to draw more than 25.5 amps from the power source no matter the size of the load?
No. That is the input current at the maximum load. If you take more than the maximum load, the transformer would overheat, but that would take time. However if you overload the transformer, the input current would exceed 25.575 Amps RMS immediately.

Also a transformer like that will take a huge inrush current, probably over 1000 A, for up to 1/4 of an input cycle, when it is turned on unless something is put in series to limit the inrush.

And is this “Secondary voltage (output loaded): 39.548 Volt RMS, Secondary voltage (output unloaded): 40.158 Volt RMS” correct as I thought a load on the secondary would significantly drop voltage?
That is a serious transformer. The nearest in power that I could easily find is this:- https://mall.industry.siemens.com/mall/en/si/Catalog/Product/4AT3632-5AT10-0FC0 and it's 50 kg. Transformer regulation, the amount by which the voltage increases when the load is removed, is a smaller percentage with larger transformers. Your figures are 1.5 % which seems about right to me, but I've not designed transformers.

I haven't checked all the figures in the calculation but I think that the magnetic core is far too small. Siemens who make the transformer that I linked to would not make a transformer that big if a smaller one would work.

In particular, the primary windings contain 2700 mm2 of copper. The secondary windings contain 1900 mm2 of copper. So ignoring how big the insulation is, and how well you can pack the windings, the hole in the core has to surround 4600 mm2 of copper. That could be done with a round hole 76 mm diameter. That has a circumference of 240 mm.

The magnetic core has a cross section of 7500mm2, so that's going to be at least 50 mm thick (50 x 150 = 7500) so that will add another 25 mm to the diameter of the centre of the magnetic path, making the magnetic path around 100 mm diameter and 320 mm long, and that is with no space for insulation and a strange cross-section of the core, which will be nearer square. So your estimate of 120 mm looks far too small.

In practice, to fit insulation, have a possible packing factor and have a transformer that can actually be made, you are going to end up with something around the size and weight of the Siemens transformer.

What is this for? That is a transformer that takes as much power as a house.