# Design Parameters of Transformer

We are going to learn small Transformer Design Calculation. *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:

- Power rating
- Voltage levels (primary and secondary)
- Currents on both sides
- Primary and secondary coils wire diameter/size
- Iron Core area
- 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 for Transformer Design:

### Core Calculations:

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

$$A_i = \frac{1}{4.44\;f \; B_m \; T_e}$$

Where,

$A_i$= Area of Core

$f$ = Operating Frequency

$B_m$= magnetic flux density

$T_e$ = turns per volts

* (for derivation of this formula Click Here)*

**Assumptions:**

**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 density 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 $2.2 \;to\; 2.4 \;A / mm^2$ (approximately).

So, putting values as;

$f= 50 \; Hz$

$B_m = 1.2\; Wb/m^2$

$T_e = 4 \;(turns \;per\; volts)$

$$A_i = 9.384\times10^{-4}\;m^2 \\ A_i \approx 1.45 \; inch^2$$

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 $0.00145161 \;m^2$. We have the core area. We can calculate turns per volts using this area by following by putting:

$f= 50 \; Hz$

$B_m = 1.2\; Wb/m^2$

$A_i = 0.001451\; m^2$

$$T_e = \frac{1}{4.44\; f \; B_m \; A_i}$$

$$T_e = 2.6 \; (turns\; per \; volts)$$

So, Turns per volts are 2.6 Turns per volts.

## Primary Winding Calculations

For Transformer Design Calculation we first calculate parameters for Primary side then secondary side.

### Primary Current Calculation

Primary voltage = Vp = 230 V

Primary current ($I_1$):

$$I_1= \frac {VA}{ V_p} = \frac{50}{230} \\ I_1 = 0.218 \; A$$

Let transformer we are going to design has 95 % efficiency ($\eta$).

So,

$$ I_1 = \frac{VA}{(\eta \; V_p)} = 0.23 \;A \\ I_1 \approx 0.23 \;Amp$$

Therefore, **Primary Current = 0.23 Amp**

### Number of Turns

Primary side number of turns can be calculated as:

Total number of turns ($N_1$) = turns per volts x primary side voltage

$$N_1 = 2.6 \times 230 \\ N_1 \approx 600 \;turns$$

### Size of Primary Conductor

As the current density is given by current per unit area i.e.

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

As, for copper the current density is taken as $2.3 \;A/mm^2$. So, for area of copper conductor of primary side (**a1**)

$$a_1 = \frac{0.23 \; A}{2.3 \; A\; mm^2} \\ a_1 = 0.1\; mm^2$$

So, from standard AWG table we can find out size of this area. which is **27 AWG.**

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 = $V_s = 12 \; V$

### Secondary Current Calculation

$$Secondary \;Current = I_s = \frac{VA}{V_s} \\ I_s = \frac {50}{12} \approx 4.2 \; A$$

### Size of Secondary Wire

$$a_2 = \frac {4.2 \; A}{ 2.3 \; A\;mm^2}\approx 1.83 \; mm^2$$

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

Hence , **Secondary Wire = 15 AWG**

### Secondary Number of Turns

Secondary side number of turns can be calculated as:

Total number of turns ($N_2$) = turns per volts x Secondary side voltage

$$N_2 = 2.6 \times 12 \\ N_1 \approx 32 \;turns$$

## Weight Estimation of Transformer Windings:

For weight calculations, following steps will be followed:

- Approximate length of copper wire = perimeter of bobbin x number of turns
- Cross sectional Area of copper conductor
- Volume = Approximate length x Cross sectional area
- Mass = density of copper x Volume
- Density of copper = $8960 \;kg/ m^3$

### Primary Side Copper Weight

Perimeter of bobbin = $(1.75 \times 4 )\\=7\;inch \approx 0.1778\;m$

So,

length of one turn = 0.1778 m

and

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 =** 106 m** (approx.)

As,

area of primary conductor =$ 0.1\; mm^2$

$a_1= (0.1 \times 10 ^ {-6})\; m^2$. Also,

Volume of copper wire = area x length

Then volume will be:

$V_1= ((0.1\times 10^{-6})\times 106 \; m^3 \\ V_1 = 1.06 \times 10^{-5}\; m^3$

We also know that density of copper = $8960\; kg/m^3$. So, combining all the dots we can now calculate weight of copper winding

$$weight = density \times volume \\ w_1= 8960 \times 1.06\times 10^{-5} \\ w_1= 0.095 \; kg \\ w_1 \approx 100 \;grms.$$

**So we need approx. 100 grams of 27 gauge wire for primary side. **

### Secondary Winding Weight:

As we know that length of one turn is **0.1778 m**. And for 32 turns of secondary winding there will be $0.1778 \times 32 \approx 6\; m$ of $L_2$.

As, area of secondary winding wire is $1.83 \; mm^2$. So, total volume of secondary winding is:

$$V_2= (( 1.83 \times 10 ^{-6}) \times 6) \; m^3 \\ V_2 = 1.098 \times 10^{-5} \; m^3$$

$$w_2 = 8960 \times 1.098 \times 10^{-5} = 0.098 \; kg \\ w_2 \approx 100 grms.$$

**So, we need approx. 100 grams of 15 gauge wire for secondary winding.**

Finally, from above calculations we can** summarize **in following table:

At this point you have done the Transformer Design Calculation and you have the characteristics of transformer components. Now, for making in hard form, see few easiest steps of __hardware implementation of calculations__:

Here is another article for transformer core optimization using Genetic Algorithm – a heuristic method of optimization.

For any Queries and further Assistance, feel free to comment below and like our facebook page for new updates. Learn projects for your final year project form experienced engineers.

In the original calculation of turns per volt, the imperial measurement for area was used instead of the metric area. Therefore, everything which follows is incorrect. The methodology is correct, so simply work through the calculation with the correct figures.

Sir.the value of current density is not internationally.that u put.

Hi, you are not taking into account the total mass of the core needed due to it’s saturation of magnetic flux Bm

I was looking for formulas to calculate transformar for example i have transformar but i want to rewind it with different input and output ac. i couldn’t get formulas for it please

nicely engineered Design effort

Basant

what is the formula to calculatre te area of auto transformer?

You give formula parameters to use that do NOT make any sense then you solve an equation BUT DO NOT Explain how you derived at simplified variable sub totals and then give a resultant. For example: F= 50 hz

Bm = 1.2 wb/m2

Te = 4 (turns per volts)

Ai= 9.384 × 10-4m 2 ~1.45 inch sq. Really? Are you assuming your viewing audience has taken advanced algebra or is space as a premium you can’t explain how you did this or are you just too damn lazy to explain how you got there?

AS the saying goes —-it’s like listening to one engineer talk another.

There is a formula in “1.Core Calculations.” He gives you values for all variables except one. Then he computes the unknown one. I learned to do that when I was 3rd grade in school. Where do you come from?

hello sir can you make also a calculation for toroidal transformer?

Unfortunately, toroidal transformer calculations are not available currently.

sir, you are put the value of Te=4, please explain this briefly. and next stage u derive this as 2.6 turns. how u adapt the Ai value with this formula Te=1/4.44fABm.

Te=4 is not standard but it is kind of standard for small transformer design. Ai is taken according to the standard bobin sizes available in market. so, by using that size, we can calculate the turns per volts.

or Te=4 is it standard for all type of step down transformer

no mr. g.satheeskumar. tpv wil vary according to BOBBIN SIZE.

Hello sir, can i apply this on SWG magnetic wire? We dont use here in philippines AWG table. There is a big difference on AWG ang SWG

Thank you

You can calculate the current in each winding and according to maximum current, you can choose wire from wire gauge table of your ease.

Hope you got your answer.

If i sent you a nameplate from a transformer do you think you could estimate the copper content of the transformer with much accuracy?

you can check by yourself according to power rating.

I can give it try. you can send on email address. engineerexperiences@gmail.com

No difference inche to mm conversion.

use common sense

Anyone knows why the efficiency effect is only applied on primary current ?? How about the secondary current?

And also what should I do if I require 50% margin for core area? Why is also the perimeter value is using 1.75 instead of 1.5??

primary side current is responsible for induction in secondary side as well as losses in core for magnetization and de-magnetization. That’s why primary side current is considered for efficiency.

Yes, the perimeter should be 1.5 instead of 1.75

1.75 is not wrong (to some level). As we are not taking other parameters into account for winding. So, Doesn’t matter here (to this level of design)

Thank you for your design of this single phase transformer, all your explanations are very clear and straight to the point.

It will be very nice if you could please let us have also a design for a three phase transformer.

Thanks a lot.

Jide T.

thanks a lot sir,

sir, for someone to design a transformer of 10,000 watts, what size of wire will be needed, cause according to your formula the value am getting is excess and I couldn’t see the value i got in the wire Gage data sheet

As mentioned in article, this method is good and suitable for designing small transformers. For high power transformer, other factors affects too.

Anyways, just as a tip: for current handling you can use copper strips for winding and use multiple parallel paths for same winding. (i.e. multiple parallel conductor bars to handle. If a bar handles 10Amps max, you can use two parallel bars for 20Amps.) Again, it is just a tip. use on your own risk.

P.S: But the use of strips are valid.

Hi, Is it ok to use 4A/mm2 current density in a toroidal transformer? I understand your design is for square transformer.

These calculations cum assumptions are for designing of soft silicon steel based core transformers. As far i know, torodial transformers have core of other iron, cobalt, manganese alloy. So, i can’t comment about current density for that currently.

Anyhow, That cores requires less energy to energize with higher current densities to handle… exact figure, i can’t comment at moment.

Most helpful article.I appreciate your work.Great sir

2×5=10″ a\ 60000 fd \ 50hz\ 4.44 gus equatane is t=100000000/4.44*hz*fd*a …. so your t=0.75v

so if you wont 90v taping so 90*0.75=67.5t if you wont 230 v so 230*0.75=172.5t

Sir help me understand how to do calculations for transformer power rating.How do you decide on the size of your transformer in kVA or MVA?

Power rating of transformer depends on the use of transformer. VA, kVA, MVA depends on your application of transformer.

Sir what is the shape of this bobbin and which side are you referring to since we used a 1.5inchx1.5inch bobbin?

http://engineerexperiences.com/hardware-design-.html

hope this link will help you.

its very help full article.

GREAT sir!

Above Calculation based on the Value Te=4 Turn Per Volt. What is benefit/drawback if we choose a big/small size of Bobbin

bobbin size is, somewhat mechanical parameter which decides size of your transformer. Greater size bobbin will only increase cost and weight of transformer material. Smaller bobbin will not be able to hold enough steel core for required power transfer (from primary to secondary).

is it possible to put a 500kva transformer winding in another 500kva transformer core ????

if both are designed to serve the same purpose, there is nothing wrong replacing one’s winding with other’s.

I need to construct a 1.5kVa transformer with 12-0-12 primary and 220V secondary. What is the wire guage needed and number of turns.kindly help

Just follow the formula given above. Nothing will change except the power ratings. Two windings of same ratings with be there on secondary side.

sir, can the size of the copper wire affect the number of turns? if it can, how so? because, as how I understand the solution above, it seems it does not affect. sorry, I’m a newbie. and moreover, we would like to design a 15VA, 220/30 V single phase transformer. we already have 1.25″x1.25″ bobbin and a number 25 gauge wire. how can we determine the number of turns?

size of copper wire is effected by the current. Number of turns on other hand, related to voltage (actually to the generation of magnetic flux and ultimately to voltage generation). So, as in article, number of turns are calculated according to voltage requirement and copper wire gauge is calculated by current requirements. Hope you got your answer.

Sir mai 5kva 8no. Core me(input=90v – 300v) automatic stabilizer banana chahata hu to sir mai kaise calculate kr k turn nikalu.

Calculate turn and wire gauge for each voltage level at secondary. Obviously, you want some steps between these voltage levels so there will be as much number of windings as the number of steps of voltages of secondary. Then by the use of relays, voltage sensor and micro-controller programming, you can make this automatic.

Sir mere paas 7no. core(2″ * 5″) hai to sir mai (input= 90v-300v) tk ka automatic voltage stabilizer ka coil banana chahata hu.

to sir mai kaise calculate kr k turn nikalu.

Plzz bataeye.

where is the EI iron core detail

http://engineerexperiences.com/transformer-basics.html

Hope you can find enough to understand here.

Thanks for this simple explanation. I am an eighty year old man retired and going through these things just to while away the time. I have some engineering background. But have forgotten everything- practically. Thanks once again.

Regards, Guptan

Does this applicable for multi-winding transformers with different voltage and current rating in every windings.

Yes, if you calculate each winding differently.

sir , what is pulse transformer & how to make it .for fence charger input 350-400 volt & output 5000-9000 volt in pulse with 8 mill ampere

i kneed your helper i want to design a step up transformer of 150VA primary voltage is75v and secondary voltage is 220V

Calculations will be same as said in article. just change the voltage levels and you’ll get currents and find number of turns and wire gauges accordingly.

I want to learn about electrical experience more and more.

how to find the bobbin size from power (VA) rating of transformer ?

Actually, VA rating does not have direct affect on bobbin size. It depends on maximum flux density, number of turns and material for core. In this article, we are using laminated stripped iron core and taking standards for maximum flux density and number of volts per turns.

I hope you get your answer.

F= 50 hz

Bm = 1.2 wb/m2

Te = 4 (turns per volts)

How did u get te =4 turns per volt?

The turns per volts are assumptions is constant and almost standardized for designing small transformer.

Actually, magnetic flux density is standard and turns per volt are calculated using that value. You can check derivation on an article linked below:

http://engineerexperiences.com/turns-per-volts-transformer-design.html

Sir kindly answer my question

I made a transformer which have primery winding is 220 turn(18 swg wire 1kg) and secondary turn is 22 turn(10 swg 650gram) now the input voltage is 210 volt and output is 18 volt

bobin size is ( 6inch×3inch) how much current given in output ?????

Plz answer my question

As per calculations, you have current ratio of 1:10 (primary to secondary). Now, as you have 18 SWG so it can have max of 3Amp. So, it can have 3amp to primary and almost 25~30Amp at secondary (at max).

I AM CONFUSED BY THE VERY FIRST FORMULA BECAUSE YOU SAY THAT Bm IS MAGNETIC FLUX_Phi_(Wb) AND YET YOU SUBSTITUTE THE MAGNETIC FLUX DENSITY(Wb/m^2) HOW SO???

It is actually magnetic flux density, no matter the nomenclature in the article it is still magnetic flux density.

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

You want transformer of which ratings ? Wire size depends on power ratings of transformer.

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?

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

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.

pls for large transformer power like 6kva, how will i get the value of flux density (Bm)

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.

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.

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?

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

How to make welding transformer?

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.

What is current density for aluminium? Is the current density value is fixed?

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

How to design k rated transformer

bobbin of 2.25 inch2 (1.5”x1.5”) or 0.00145161 meter square. how dis you get 0,00145161!

2.25 inch square = 0.001455161 meter squares

it’s simple area units conversion.

asslamwalkum sir can i u please explaine me the core calculation of CT

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.

Respected sir,how we find the laminations size if it is NOT C.R.G.O. IT IS C.R.N.G.O

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???

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

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

How to choose flux density? Whether it wil vary for 3 phase transformers?

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.

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”

for design purpose how much power factor will take for distrubution transformer

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).

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!

NEED 3 PHASE LAMINATED TRANSFORMER CORE WEIGHT CALCULATION FORMULA IN SIMPLE METHOD..?

CAN U PLS HELP ME….?

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.

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

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.

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?

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

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.

hi… thank u for this article. upto which power rating core & copper losses r neglected..

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.

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

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. 🙂

4.44 comes from the rms formula derivation.

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.

Great… Thanks.

At 60 hz frequency what will be flux density

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.