*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:**

**Calculations:**

### 1. Core Calculations:

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

F= operating frequency

Bm= magnetic flux

Te= turns per volts

*(for derivation of this formula Click Here)*

**Assumptions:**F= 50 hz Bm = 1.2 wb/m2 Te = 4 (turns per volts)

Putting f=50 hz; Bm = 1.2 wb/(m^2); Ai= 0.001451 m^2, we got:

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) = 0.23APrimary 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

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

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.)

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:

- 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/ 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

And density of copper = 8960 kg/ m3

So,

**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

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

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

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__:

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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.

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.