Calculations for Design Parameters of Transformer

For Quick review and summary of Calculations, see the tables at the end of this article. (Table 1) (Table 2) For better understandings, go through the following steps for calculations

Make sure you have knowledge of Basics of Transfomer

Design Parameters

For designing a transformer, we need:

  1. Power rating
  2. Voltage levels (primary and secondary)
  3. Currents on both sides
  4. Primary and secondary coils wire diameter/size
  5. Iron Core area
  6. Numbers of turns (primary and secondary

We are going to design a 50 VA step down transformer of 230V to 12V. Necessary calculations along with formulae are given below in details:

As we are going to design small transformer (of small power rating) so we are neglecting core and copper losses as they doesn’t matters in small transformers and seriously considered in designing power transformers (high power rating transforms).


1. Core Calculations:

Calculate area of core (central limb) by using following formula:

Area for core central limb

Ai= area of core
F= operating frequency
Bm= magnetic flux
Te= turns per volts
 (for derivation of this formula Click Here)
So, we know the frequency of the power system. We need magnetic flux and turns per volts. For designing a small transformer magnetic flux is averagely taken as 1 to 1.2.
By putting values we will get the area of core.
Current density of copper wire is taken as 2.2 A / mm2 to 2.4 A/ mm2 (approximately).
So, putting values
F= 50 hz
Bm = 1.2 wb/m2
Te = 4 (turns per volts)
As, we are going to design a practical transformer so we must consider the core available in market. The standard Bobbins available in market practically is 1”x1”, 1.25”x1.5”, 1.5”x1.5” and so on. We took nearest core area available to our calculation. We took bobbin of 2.25 inch2 (1.5”x1.5”) or 0.00145161 meter square.  We have the core area. We can calculate turns per volts using this area by following:
Putting f=50 hz;                Bm = 1.2 wb/(m^2);             Ai= 0.001451 m^2,                 we got:Turns per VoltsTe

So, Turns per volts are 2.6 Turns per volts.

Primary Winding Calculations

Primary voltage = Vp = 230 V
Primary current = I1= VA / Vp = 50 / 230 = 0.218

Let transformer we are going to design is 95 % efficient so,
I1 = VA/ (efficiency x Vp) = 0.23A
Primary Current = 0.23 Amp (approx.)

Number of Turns:

Total number of turns = turns per volts x primary side voltage
                      = 2.6 x 230
                 N1   = 600 turns (approximately)

Size of Conductor:

As we know that

Current Density formula

As, for copper the current density is taken as 2.3 A per mm square So, for area of copper conductor

 Primary Conductor size
From the standard American wire gauge table, we can choose wire of the same thickness. It can be seen that it comes out that required primary side wire is of 27 gauge which can conduct required current. (for AWG table Click here)

Selection of wire can also be done by calculating primary current and by cross match the standard table of copper wire according to their current handling capabilities.

Secondary Winding Calculations

Secondary voltage = Vs = 12 V

Secondary current = Is = VA/ Vs = 50 / 12 = 4.2 Amp (approx.)

Secondary Winding Conductor Size

From the standard copper wire, table it can be seen that wire of this thickness is of 15 gauge. So for secondary winding we need 15 gauge wire.


                                     Secondary Wire =15 AWG

Number of Turns:

Number of secondary turns = turns per volts x secondary volts
                       N2 = 2.6 x 12 = 32 turns (approx.)

Weight Estimation of Windings

For weight calculations, following steps will be followed:

  1. Approximate length of copper wire = perimeter of bobbin x number of turns
  2. Cross sectional Area of copper conductor
  3. Volume = Approximate length x Cross sectional area
  4. Mass = density of copper x Volume

Density of copper = 8960 kg/ m3

Primary side:

Perimeter of bobbin = (1.75 x 4) = 7 inch  = 0.1778 m


Length of one turn = 0.1778 m
Total length of all turns of primary = L1

L1 = (length of one turn) x(total number of turns of primary)
L1 = 0.1778 x 600
L1 = 106 m (approx.)

 area of primary conductor = 0.1 mm2
                           = (0.1 x 10 ^ -6) m2
Volume of copper wire = area x length

volume of conductor

And                                        density of copper = 8960 kg/ m3



So we need approx. 100 grams of 27 gauge wire. 

Secondary winding weight:

Perimeter of bobbin = (1.75 x 4) = 7 inch = 0.1778 m


                      Length of one turn = 0.1778 m

Total length of all turns of Secondary 

= L2 = length of one turn x total number of turns of secondary

L2 = 0.1778 x 32

L2 = 6 m (approx.)

area of primary conductor = 1.83 mm2 = (1.83 x 10 ^ -6) m2

Volume of copper wire = area x length

Volume secondary

So, we need approx. 100 grams of 15 gauge wire.

So, from above calculations we can summarize in following table:

Design Summary Primary Side

Design Summary SecondaryAt this point you have done the Calculations and you have the characteristics of transformer components. Now, for making in hard form, see few easiest steps of hardware implementation of calculations:

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Asad Ullah

I am MSc Electrical Scholar under a fellowship program. Working with Modeling and Simulation software related to my field are my activities in leisure.

8 thoughts on “Calculations for Design Parameters of Transformer

    1. As per formula given in start of article, flux density and frequency are inversely related. But, normally in designing flux density is taken constant (as flux required is constant and area is constant). So, if frequency changes the compensation is made on voltage side. i.e. transformer will be designed accordingly with voltages. but in small transformers such parameters are neglected mostly. (but if you want to, Bm will be constant and voltages will be changed.) P.S: sometimes tolerance voltage compensate this change.

  1. Hey just wanted to say thanks you for all of these great articles! They read very very well and you’ve covered all of my questions.

  2. Hi, thank you for your article. On the core calculations section, could you explain where you got the 4 turns per volt from that is needed to plug into the equation to calculate the core area. Also, in the same equation, the constant 4.44 – what does this represent? Thank you

    1. For design of small transformer some assumptions have to made. Turns per volts here are from a standard table of designing transformer. (you can study it in book “Principles of Electric Machine Design by R.K Agarwal” or any transformer design book). This is according to transformer type and rating.
      4.44 is constant in equation of turns per volts comes while deriving this equation with help of fundamental equations.(click link below for detail derivation).
      Hope you get your answer. 🙂

  3. I want calculations for low watts . Ma transformer is pri 220v /sec 1000v and current is 50 ma .
    Second transformer is pri 400v sec 5 v and current is 5 ma …can you help me

What do you think?