Technical Article

# How to Design a Transformer: Size and Selection

September 27, 2022 by Simon Mugo

## Transformers are key electrical components useful in determining the power reaching the load from the supply. In this article, you will learn how to design a transformer through a well-documented guide that will enable you to select and size transformers.

A transformer is one of the major components when focusing on electricity distribution. It significantly impacts the electrical system's performance, either during normal operations or disturbances. Therefore, smart engineers ensure transformers are correctly selected and perfectly sized for a particular purpose and can deliver sufficient power to the electric loads under certain conditions and regulations.

A transformer helps power systems become scalable, flexible, and cost-efficient by ensuring minimal power loss. This article teaches you how to select and size a transformer.

### Transformer Nameplate

Transformer selection and sizing involve determining the transformer’s basic parameters such as primary and secondary voltages, KVA, winding connection, power factor, cooling methods, winding conductor material, types, mounting arrangement, efficiency, and frequency of operation. All these parameters are written on a plate known as the nameplate, which is fitted on the transformer. When you check on any transformer well, you will see a well-fitted detailed sticker with all the transformer's necessary parameters.

### Parameters to Consider When Sizing a Transformer

#### Voltages Both Primary and Secondary

Primary voltages are all the voltages supplied on the input side of any given transformer, while the secondary voltage is the voltage output by the transformer.

If the supply voltage available is 414 V three-phase and the output required is 240 V single-phase, you must go for a transformer rated 415 V at the input and 240 V at the output.

#### KVA Rating

Calculating the KVA during the selection and sizing of transformers is crucial. This represents the maximum apparent power that a transformer can handle. KVA depends on power factor, voltage, and current. Below are the formulas for KVA calculations:

• Single-phase transformers

• Three-phase transformers

#### Operational Frequency

A transformer operates at a specific frequency. The magnetic current, rated current, and kVA are directly proportional to the transformer frequency. Therefore, the transformer should operate at a rated frequency. The rated frequency of the transformer should always be equal to the input power supply and the load’s operating frequency.

#### Winding Connections

For single-phase transformers, the winding connection is not something to worry about, but the winding connection is a significant parameter for three-phase transformers. Below are several examples of windings configured in a three-phase transformer connection.

##### Figure 1. Winding Connection Configurations for Three-phase Transformers. Image used courtesy of Simon Mugo

Apart from the above-listed parameters, other parameters such as the power factor, transformer type, cooling arrangements, operating temperatures, impulse withstanding capacity, voltage regulation, and mounting arrangements are also considered when selecting a transformer.

### Calculating Transformer Design Parameters

For educational purposes, we are going to design a small transformer. We will be designing a 60 VA step-down transformer of 230 V to 12 V. The reader needs to have basic knowledge about transformers. Detailed calculations are included in this article.

Note that we are designing a small transformer, hence we are going to ignore copper and core losses as they do not affect small transformers.

#### Core Calculations

To design a transformer, we need to have a specific number of turns on both the secondary and primary sides of the transformer to achieve a specific rating. Therefore, finding voltage per turn and turns per voltage design is the first priority. This can be obtained from the existing transformer’s voltage equation below:

$E=4.44fB_{in}A_{i}$

The equation above is a derivation from the basic equations, and we will derive it step by step.

From basic knowledge, we understand that induced EMF is the rate of change in flux:

$e=N\frac{dφ}{dt}$

The Root Mini Square value (RMS) is given as:

$\sqrt{}2$

With peak value

$e=\sqrt{}2E$

Putting all this together, we get:

$E=\frac{N}{\sqrt{}2}\frac{dφ}{dt}$

$φ(t)=ABsin(\omega t)$

Let’s take the derivative with respect to t:

$\frac{dφ(t)}{dt}=\omega ABcos(\omega t)$

$\frac{dφ(t)}{dt}=2\pi fAB_{in}$

From electronics, we know that:

$\omega=2\pi f$

And

$E=N\frac{2\pi FAB_{in}}{\sqrt{}2}$

The formula above gives the total induced EMF.

The main focus is the voltage per turn and hence dividing both sides with total number of turns (N) we get:

$E_{T}=4.44fAB_{in}$

Turns per volt will be given by:

$T_{E}=\frac{1}{4.44fB_{in}A}$

Rearranging the above formula, we get the formula below, which is used to calculate the area of the core:

$A_{i}\frac{1}{4.44fB_{in}T_{e}}$

Where

A= Area of Core, f = Operating frequency, Bin = Magnetic Flux Density and Te = Turns per Volts

• Assumptions

We know the power system frequency. The focus is to have the turns per volt and the magnetic flux. For a smaller transformer design, the magnetic flux density ranges between 1 and 1.2.

Inputting the values in the formula, should result in the area of the core.

Let’s take the copper wire density to be 2.2 to 2.4A/mm2

f = 50Hz

Bin =1.2 Wb/m2

Te = 4 (Turns Per volts)

Ai =1.45 inch2

$T_{E}=\frac{1}{4.44fB_{in}A}$

Inputting the values in the formula, we get that:

Te = 2.6 turns per volts.

For the design, make sure to choose the core that is readily available in the market. From the check, the readily available core is 1.5”x1.5” which gives 2.25 inch2

That is the core area used to calculate turns per volt of 2.6.

### Calculations of the Primary Winding

For the transformer, calculate the primary side parameters then end with the secondary side.

#### Primary Side Current Calculations

Primary voltage = Vp=230 V

Primary Current (I1):

$I_{1}=\frac{VA}{V_{p}}=\frac{50}{230}=0.218A$

But a transformer cannot perform at 100% and hence the efficiency is at 95%, so the primary current becomes

$I_{1}=\frac{VA}{\eta V_{p}}=\frac{50}{0.95\times230}=0.23\,A$

Hence, 0.23 A is the primary current.

#### Number of turns

The total number of turns in the primary side can be calculated as

Total Number of Turns (N1)= Turns per Volts×Primary Side Voltage

$N_{1}=2.6\times230=598\,Turns$

This is approximately 600 turns.

#### Primary Conductor Size

Current density is Current divided by the unit area.

$\delta=\frac{I}{A}$

For copper material, current density is given as 2.3 A/mm2

The copper area on the primary side a1 is given by

$a_{1}=\frac{0.23}{2.3}=0.1mm^{2}$

Use the standard AWG table to determine the size of the 0.1 mm2 where it is 27 AGW.

### Calculations of the Secondary Windings

The secondary Voltage = VS =12 V

#### Calculating the Secondary Current

$I_{S}=\frac{VA}{V_{s}}=\frac{50}{12}=4.2\,A$

#### Calculating Secondary Wire Size for Design of Transformer

$a_{2}=\frac{4.2}{2.3}=1.83mm^{2}$

Checking at the AWG standard table for transformer wire thickness, the thickness will be 15 AWG.

Therefore, secondary wire size = 15 AWG.

#### Number of Turns on The Secondary Side

Calculate this by

Total Number of Turns(N2) = Turns per Volts×Secondary Side Voltages

$N_{2}=2.6\times12=32\,turns$

### Estimating the Weight of the Transformer Winding

This happens when we know:

The density of copper is 8960kg/m3

Approximate length of copper wire = number of turns x perimeter of bobbins.

Cross-sectional area of copper

Volume = approximate length x cross-sectional area

Mass = copper density x copper volume

#### Primary Side calculation of Copper Weight

Perimeter of the bobbin = 1.75 x 4 = 7 inch = 0.1778 m

Therefore, the length of the turn is 0.1778

The primary side total length of the turn = L1

Total length = length of a single turn x total number of turns in the primary side = 0.1778 x 600 = 106 m

The volume of copper wire = area x length = 0.1 x 10-6 = 1.06 x 10-5m3

But we know the density of copper as 8960 m3; hence we can now calculate the weight of copper:

Weight = density x volume = 8960 x 1.06 x 10-5 = 0.095 kg =95 grams.

Therefore, we need 95 grams of 27 gauge wire at the primary side of the transformer.

#### Calculating Winding Weight on the Secondary Side

As calculated, the length of the turn = 0.1778 m

For 32 windings on the secondary side L2 = 0.1778 X 32 = 6 m

The area of the secondary winding wire is 1.83 mm2

Therefore, the volume of the secondary winding = 1.83 x 10-6 x 6 = 1.098 x 10-5 = 0.098 kg

So, we can approximate the weight at 100g of wire size 15 gauge at the secondary side winding.

At this level, all the calculations for the transformer design are done, along with all the transformer-relevant characteristics. Implement the transformer project by building the hardware according to the calculations.

### Key Takeaways of Transformer Sizing and Selection

• A transformer is a common electrical component used in stepping up or stepping down power.
• Every transformer has a transformer nameplate with all the parameters of the transformer.
• We have listed almost all the parameters that every transformer must have.
• We have learned how to calculate transformer parameters which are key to designing, selecting, and sizing transformers.

Featured image used courtesy of Adobe Stock