Calculations for Design Parameters of Transformer

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).

Calculations:

1. Core Calculations:

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

Ai= area of core
F= operating frequency
Bm= magnetic flux
Te= turns per volts
 (for derivation of this formula Click Here)
Assumptions:
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.

So,

                                     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

So,

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.)

As,
 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,

Weight

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

So,


                      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.)

As,                    
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 (Transformer Design Calculation)
Table 1: Calculation for primary side
---------------------------------------------------------------------------------------------------------------------
Design Summary Secondary (Transformer Design Calculation)
Table 2: Design Summary for Secondary of transformer design

At 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:

For any Queries and further Assistance, feel free to comment below and like our facebook page for new updates.

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.

45 thoughts to “Calculations for Design Parameters of Transformer”

  1. Sir ma transformer banana chatha ho mera pas iron core pare ha his ma 3.8cm5cm ha moja wire no bata dena tarn me tadad be secondary or prembry Dino ma ma an pard ho koch moskil aya gee please ap mer madad kro thanks

  2. Asad, This page is absolutely fantastic. Thank you very much. In the above, "1.75 is the length of one side of the bobbin." In the example on this page, I understood that a standard size of 1.5 inches by 1.5 inches was used for the example calculations. Could you please help me understand where the 1.75 inches is derived from?

  3. Sir,I made a transformer 24V 3A,72VA operated in 220V which heated more without load..and output is accurate, bobbin area is 10.26sqcm , primary turn 937 , turn per volt 4.26, secondary turn 106 ,primary wire gauge 28, secondary wire gauge 19 can anyone please explain why so heated?????thanks

    1. Heating could be due to low quality core material or loose binding of core (air between core stampings) as a lot of power is being lost in it. Use laminated core stamping for core and hard bind them to avoid losses.

  4. By using the same perimeter value for both primary and secondary, you implicitly assume that windings are side by side (split bobbin) rather than one over the other. Standard practice is to use not perimeter but mean length of term (MLT), thus the stated wire lengths are too long. The winding area is not filled 100% because the wire is round and has air included in the winding cross section. Standard practice is to use a fill factor which tends to be around 60% when winding technique and insulation thickness are included. This affects the choice of wire gauge to fit the cross section. From wire length and gauge the next step is to calculate resistances from which the voltage drops can be known, noting that AC resistance will be somewhat greater than DC resistance. Then the design can be iterated to compensate.

    1. Yes you are right. I am aware of 'mean length per turn', 'window effective area', resistance of winding, core geometrical constants and many more factors according to which transformer is designed. Here, we have designed small transformer and with compensation factor, some parameters are neglected and some are included in compensation factor.
      Thank you for your considerations. Design calculations according to all these factor included will be available soon.

  5. as i figured out, all the formulas work only for ac input. what about a rectangular pulse? should i use Fourier expression off voltage and then find EMF equation?

  6. most help full artical or information....welldone
    Asad bhai aap transfomer ka temprature rise find out kr skty hain to kinly mari kuch help kr dain aur L.T ,H.T eddy loss formula k bary ma b bta dain....Thanks

  7. Mr Asad thank you for your explanation. Please I have a problem similar to this...am asked to find the parameters of a Transformer core (shell type) such as height(H), diameter(D), width(W), width and length of the window slot(Ws). Haven being given insulation thickness as 2mm, V=220/18v, P=450W, power factor as 0.8, window factor as 0.3, maximum flux density(Bm=0.42T), F=50Hz, k=1.2.
    Please how can I go about this.

    1. Yes, every material (conductor) have its constant current density as it is property of material. Aluminum has current density of around 0.5 A/mm^2

    1. CT is a transformer with very low voltage drop at primary side and it is stepped up for measurement. Core calculation for small CTs is almost same as other transformers but for very small sized CTs, ferrite core is used which need different set of calculations.

  8. Thank you for explaining the design of transform. I have one confusion that Perimeter of bobbin=1.75 * 4 you have taken but 1.75 where it comes from???

    1. 1.75 is length of one side of bobbin. To calculate length of one complete revolution of wire around bobbin it is multiplied by 4.

  9. Dear Mr. Asad-ullah Sahab. Your Article is very informative and help full. thanks for sharing it. Regards, May Allah live you long..

  10. When you calculate the core area you assumed Te = 4 (turns per volts) but it was not clear where this number 4 comes from. In other calculations the core area is proportional to the squareroot of VA and approximated as Ai = 1.152*sqrt(Output voltage x output current) = 1.152*sqrt(50VA) = 8.146 cm^2=8.146*10^-4 m^2 and then using this Ai value the turns per volt is calculated Te = 1/(4.44AiBf) = 1/(4.44 * 8.146*10^-4m^2 * 1.2T * 50Hz) = 4.6 turns per volts which is close to 4 that you assumed. So actually the turns per volts come from the area and not in the other way because the are comes from the VA. So we don't have to assume anything because everything comes from somewhere.
    Apart from this your work is nice. That was the only thing that was not understandable and i had to look it up. Thanks for your post.

    1. That's good thing you took interest and did research in this. But if you look into literature of transformer design, some parameters are constant for specific core size and specific material. You can find standard tables in many books which give you constant values for calculated core sizes. This 4 turns per volt is from standard table for small transformer design calculation listed in Book by "R.K Agarwal" under the title of " Principles of Electric Machine Design"

    1. For designing a transformer, power factor is not a thing to consider as it depends upon load on transformer. This is one of the important reasons that transformer ratings are in VA (or kVA or MVA).

  11. Thank you so much for explaining the parameters of transformers so precisely. My wife was trying to figure out the area of the core; however, did not understand which formula to use. This information definitely makes it easy for us to just plug in our information -- thank you! We did not know that you could just assume the transformers magnetic flux. Thank you again for the information!

    1. These calculations are for single phase transformer design. For 3 phase transformer, calculations and formulas are different. (this article will be available soon) 🙂
      You can design three phase transformers by connecting three single phase transformers in star. (I personally designed and used it). During connections you must take into considerations phase sequence of all transformers.

  12. محترم اسد اللہ صاحب
    آج کل آئرن کور کی بجائے گریفائیٹ کور کا استعمال عام ہورہا۔ انرجی سیور میں جو کوائل ہوتی ہے اس سے میں نے ایک 12 وولٹ ٹرانسفر مر بنانا ہے ہے۔ کیا آپ میری مدد کر سکتے ہیں؟ الماس شہزاد

    1. Yes! some times graphite core transformers are used in high frequency applications. Its not something you can generalize about "going out" of iron core transformers.
      For that transformers use same calculations for wire of primary and secondary and calculate total area covered by winding according to which you can find out window are and find graphite core of that size.

      1. Asad Ullah! Thank to reply. you can guide me further to make step down transformer using ferrite core or give me any link with will be informative for me?

  13. 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

    1. Calculations for the wire would be same as for others. If you don't find wire of your desire rating, use smallest ( say 30 AWG wire) to wound your transformer. Your first transformer is of course falls in category of high voltage so lamination precautions must be followed. Likewise, you can design second transformer.

    1. Mostly, for Calculation of small transformers core losses are neglected (mostly small transformer are considered up to 1000 VA).
      However, plus minus 5 percent of turns of secondary winding is added to overcome losses or get full voltage in full load conditions.

  14. 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).
      http://engineerexperiences.com/turns-per-volts-transformer-design.html
      Hope you get your answer. 🙂

  15. 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.

    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.

What do you think?