In this article you will learn how to calculate turns ratio of **ferrite core transformer** for high frequency switch mode power supply inverters. High ferrite core transformers are used in almost every power electronics circuits like **inverters** and** pure sine wave inverters**. They are used to boost up or step up low dc voltage of battery and other dc sources like solar panels. Ferrite core transformers are also used in isolated **dc to dc converters** to step up or step down dc voltage. For example in isolated **buck converter **it is used to step down dc voltage and in isolated **boost converter**, they are used to step up dc voltage. In this article, we will learn how to calculate turns ratio of high frequency ferrite core transformer with examples.

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**Ferrite core turns ratio calculation **

For example in boost up stage we have two options to use from power electronics converters, push pull topology and full bridge. I will explain both methods one by one. Turns ratio calculation formula and concept remains same for both topologies. The only difference between push pull topology and full bridge transformer design is that push pull ferrite core** transformer** requires a center tap in primary winding. In other words, push pull **transformer** have two times primary turn than full bridge transformer.

**Push pull topology ferrite core turns ratio calculation with example **

Let’s start with example. For example we want to design a 250 watt boost up dc to dc converter. We are using push pull topology for this design. We are using 12 volt battery. We want to step up dc voltage from 12 volt 310 volt. Switching frequency of design is 50KHz. We are using ETD39 ferrite core which can handle 250 watt. It is beyond the scope of this topic to tell how to select ferrite core according to power rating. I will try to write separate article on it. The output of ferrite core will be always high frequency square wave of 50 KHz. We need to use full rectifier to convert it into dc of 310 volt. You may also need to use LC filter to harmonics or AC components from output.

### Ferrite transformer primary turns calculation

As you know battery voltage does not remain same all the time. As the load on battery on increases, battery voltage will be less than 12 volt. With no load with fully charged battery, battery voltage will be near to 13.5 volt. Therefore input voltage is not constant, we must consider it while calculating turns ratio of ferrite core transformer. Cut off voltage for battery is usually 10.5 volt. We can take it as smallest possible value of input voltage to boost up dc converter. So we have following parameters now:

Vinput = 10.5 volt

Vout = 310 volt

As we know that formula of turns ratio calculation in transformer is

N = Npri / Nsc = Vin / Vout

Where Npri is number of primary turns and Nsc is number of secondary turns. We have three know variables like turns ratio which can be calculated by above equation, input voltage and output voltage. But we need to calculate primary turns to find secondary turn of ferrite core transformer. Formula to calculate primary turns for ferrite core transformer is given below:

Npri = Vin * 10^8 / 4 * f * Bmax * Ac

But for push pull it will be half the primary number of turns.

- Where Npi is primary number of turn, Vin( nom) is normal input voltage which in our example is 10.5 volt.
- Bmax is maximum flux density. The unit of maximum flux density is Guass. Remember if you are using Tesla unit for maximum flux density, IT = 10^4 Guass. The value of maximum flux density is usually given in data sheet ferrite core. We usually take value of Bmax between 1300G to 2000G. This is usually a acceptable range for all ferrite core transformers. Note : High value of flux density will saturate the core and low value of flux density will lead to core under utilization. For example we will take 1500G for dc to dc converter example.
- f is switching frequency converter. In our example switching frequency of dc to dc converter is 50 KHz.
- Ac is effective cross sectional area of ferrite core. We have to refer data sheet for this value. In this example, we are using ETD39 core. The effective cross sectional area of ETD39 is 125mm^2 or 1.25cm^2.

We have all the values to calculate primary number of turns .i.e.

Vin = 10.5 volt, Bmax = 1500G, f = 50 KHz, Ac = 1.25 cm^2

By putting these parameters in two above formula, we can calculate turns primary number of turns.

Npri = 12 . 10^8 / 4 . 50000 . 1500 . 1.25 = 3.2

Hence Npri = 3.2 But we cannot use fractional turns. So we need to round off primary turns calculated value into nearest whole number 3. The nearest possible whole number is 3. primary number of turns for ferrite core is 3. But before that we need check either for Npri = 3 Bmax is within acceptable range or not. As I have mentioned above the acceptable range for Bmaz is 1300-2000G. But the question is why we need to check the value of Bmax again? Because we adjust the value of primary turns from 3.2 to 3. So let’s calculate value of Bmax for Npri = 3 by using above forumla.

Bmax = Vin * 10^8 / 4 * f * Npri * Ac

Bmax = 12 * 10^8 / 5 * 50000 * 3 * 1.25 = 1600G

So calculated value of Bmax is 1600G which is within acceptable range of maximum flux density. Its mean we can take Npri = 3 for further calculations. Primary number of turns for push pull ferrite center tap transformer is 3 turns + 3 turns. In any design you will need to adjust the value of Npri if it is in fraction. You can easily adjust it. But you need to check value of Bmax every time. We start with assume value of Bmax and calculated Npri. But you can also start with assume value of Npri and check the value of maximum flux density Bmax. For example suppose a value of Npri =1 and check the value of Bmax and keep repeating this process, until it is become in acceptable range.

### Ferrite transformer secondary turns calculation

Now let’s move to secondary turn of ferrite core. In our design the output of dc to dc converter is 310 volt at any input voltage. Input voltage is variable from 10.5 volt to 13.5 volt. We will need to implement

Feedback to get regulated 310 output voltage. So we will take little bit higher value of output voltage so that at minimum possible input we can still get output voltage of 310 volt by changing the duty cycle of PWM. So we should design a ferrite core transformer with secondary rated at 330 volt. Feedback will adjust the value of output voltage by changing the duty cycle of PWM. You should also take care of losses and voltage drops across switching devices and you should take them into account while designing transformer.

So transformer must be able to supply 330 volt output with input of 13.5 volt to 10.5 volt. The maximum duty cycle for PWM is 98% and rest 2% is left for dead time. During minimum possible input voltage duty cycle will be maximum. At maximum duty cycle of 98%, input voltage to transformer is 0.98 * 10.5 = 10.29 volt.

By using voltage ratio formula of transformer = voltage ratio = 330 / 10.29 = 32.1. Voltage ratio and turns ratio in transformer is equal to each other. Hence N = 32.

So we know all values to calculate secondary turns of ferrite core transformer.

N = 32, Npri = 3

Nsec = N * Npri = 32 *3 = 96

So number of primary turns is equal to 3 and number of secondary turns is equal to 96. So it is all about turns ratio calculation for high frequency transformers. If you have any issue, let me know with your comments.

Dear bilal malik

how can we calculate ferrite core inductor for buck converter

i will post a article on it soon

chech this post https://microcontrollerslab.com/output-inductor-design-calculation-smps/

How to get the design frequency

Assala-mu-alaikum .Can you give a full diagram of an inverter circuit using ferrite core transformer. 100w to 300w range will be fine.

you can purchase the design from shop contact me at [email protected]

aslam o alikum sir

dear sir,

i’m confused according ti ferrite core transfomer calculation.

sir i have

Ferrite core selected for Transformer: E 65/32/27

Area of ferrite core Ae = 540 mm2

Max flux density Bm = 0.1 Wb/m2

(for push-pull configuration)

Air gap = 0

but i did not macth result according to your calculation

sir

Vin =12,f=40khz,Ac=5.4cm2,Bmax =1500G but in result just 1 Np turn require but i,m using 3 turn in Np and 106 for Ns

whats happen with me

I’m really grateful for this beautiful piece, pls can u explain how pwm can be controlled in sg3525 chip.. and also how to fix the pin 1 and pin 2 voltage to obtain a nice result as regard to your ferrite feedback compensation calculation. does it mean dat d sg3525 chip must oscillate with 98percent @ all time before the regulation is perfected. I dnt understand these pls. i’d be highly appreciated if u respond quickly sir. thanks

Hi there colleagues, nice article and good arguments commented here,

I am genuinely enjoying by these.

Nice work man…plz,how do we determine the guage of the pry and secondary Windings?

I will try to post a article on it soon or you can find a table for wire gauge selection on google images its not that hard

Hi Bilal God bless you; Good article, congratulations; today I was reading all the article about the calculation of number of turns for primary and secondary in high frequency transformers; yesterday I removed a chopper transformer from a tv switching power supply and removed its windings to rewind it again; But I just wanted to build an easy switching power supply to get familiarized with it; the problems is that I don’t know what kind of ferrite core is; that means its size measurement.

Hi ! I need to a flyback transformer which can convert from 50Vdc to 5Vdc. Also, output will be 5 ampere. How can I chose ferrite core (ETD29,39,49) and calculate primary and secondary turns ?

Hi Bilal,

May I have the link to reliable resource and calculations on how to select wire gauge size for center taped inverter winding?

Dear Sir,

Tell me how to take output current into account.

As i want to calculate turns for 48V, 5A DC to DC converter.

Thank You

Sir this is a great way you have made it clear n simple but i want to ask one question and that is

In the formula for nprim calculation

We are using SI units for voltage flux and bmax but and we are using cm’2 for area rather we should use m’2

Dear Sir,

your article is great and easy to understand for transformer design and i have one of doubt, how to select different types of inductor likes (drum, toroidal core etc.) for circuit designing.

sir can you please help me i have problem on finding flux density of core i have following details

capacity of transformer: 25 KVA

core loss to be mantained at : 70 watt

size of l.t winding = 2.5*6.5

no of turns L.t =100( 25 turns per layer)

size of H.t winding= swg 22( total turns = 5001 and 205 turns per layer)

Dear Sir,

Very nice topics and also helpfull for all.

But I have one question . Pri. & Sec. weir gauge calculation will be as conventional way ,like as iron core transformer ?

dear, how can i calculate inductor for 2 phase 180 degree out of phase buck converter and frequency with duty ratio.

Asalam o alekom.. thanks for the great and wonderful lecture.. i was stuck in it from long time cous i am not a electronic engineering student… 🙂 i am doing by my own.. so thanks for the help ALLAH bless u and one more thing.. plzz share with us calculation of wire gauge # for the P. and S winding

Please I want to know the calculations of the toroidal ferrite core , thank you

Respected sir,

I want to select ferrite core for my high frequency transformer design.

Ratings are:

Vin(primary)=115V

Vout(sec)=230V

frequency= 10khz/20khz

turns ratio=2

Iin=5A

Iout=2.5A

Please help me for designing and selection of ferrite core

Thanks for this article;it was usefull,..sir u said the output of the ferrite transformer will always be 50kHz in the given example, what if the load operates on 50 or 60 Hz wont it be a problem?? If yes, please what are the necessary changes??

thanks a lot for your good article , i enjoyed it

May i use ferrite core for h bridge inverter

I/p V, 7V, o/p V, 220V 80Hz.

Switching frequency 16Khz

hi malik

thanks for the post

I have an issue with the calculation for Npri

my core is EE65 and the Ac is 5.3cm2 by observation.

now, after I calculated for Npri using Vpx10^8/4BmaxAcF which you gave, my result is 0.75471698turns

I feel something is wrong with this calculation

can you please help me with what I am missing?

thanks

please provide all values. Also have you checked and mentioned switching frequency and voltage?

But if you have assumed the Bmax value and Frequency as mentioned in this post then maybe it will make difference with your real number of turns of your transformer.

Hi,

In the beginning, you wrote:

N = Npri / Nsc

And while finding Nsec you wrote:

Nsec = N * Npri = 32 *3 = 96

HOW the formula changed???

Isn’t it should be: Nsec = Npri / N

???

You answer seems right but the formula is confusing me.

Please explain.

(CONT.)

Well, it should be:

Voltage Ratio = Vin / Vout = 10.29 / 330

Voltage Ratio = 0.03118

Voltage ratio and turns ratio in transformer is equal to each other. Hence N = 0.03118.

So we know all values to calculate secondary turns of ferite core transformer

N = 32, Npri = 3

N = Npri / Nsec

Or, Nsec = Npri / N = 3 / 0.03118 = 96.2155

So, Nsec = 96 (after rounding off)

This seems right way to me. But in your case you reciprocaled both formulas.

**Error correction on previous comment:

N = 0.03118, Npri = 3

How can this calculation be applied to EE65 ferrite core transformer.

My calculation seem not to work.

I feel I am missing something

Hi Excellence,

Since you have not provided all the values, let us assume Bmax = 1500 G, Switching Frequency f = 50KHz. Also Effective Area ( Ac ) of EE65 Ferrite core is 540mm^2 or 5.4cm^2 (As per datasheet https://www.tme.eu/es/Document/e9623be7fcdb459866403325c39d6a24/e653227.pdf ). Now put the Vin value in this formula to calculate Npri :

Npri = (Vin * 10^8) / (4 * f * Bmax * Ac).

After calculating Npri you may check Bmax value (if you wish). Then follow those steps and use the formula I have mentioned above in my comment. And if you are facing problems, let me know with more details next time. Have a nice day 🙂

Thanks very much replying me.

And sorry for giving little details.

My concern is this:

After employing the formulae for the calculation, my Np is less than 1.

Here is the breakdown:

12 ×10^8/4×50,000×1500×5.4

My result is 0.74074074 which is less than 1 turn

This looks too little for Primary Turns

Am I still missing something?

Please help me

Its okay 🙂 Happy to help as I am learning too. Are you sure that the Vin you wrote 12V in the formula is actually what you are applying? Is it step-up?

I am certain sir. Its 12v

I intend using it for step up

I actually wanted to use it for Inverter using Switch Mode Topology.

I will be glad if you can give me some practical hints concerning that too.

Its my first attempt.

well, I am not professional in this either. Just leaning as you are. I can’t do practicals but can discuss and maybe sort out the problem. In this case I can’t find out because there can be many cases like:

Case 1 : Your input supply is is greater than 12V

Case 2 : Your switching frequency is lower than 50KHz

Case 3 : Core type is different.

Please counter-check all these things. But unlike all these, Bmax seems ok to me. (You can assume minimum value i.e. Bmax = 1300 too if you like)

Considering Case 1 (which I don’t think should be greater than 12V in your case but let’s try): Let us assume your input supply is 20v. Keeping rest all values untouched. After putting this value in the formula we get Npri = 1.2 ~ 1 [Though this seems wrong to me but it is greater than 0].

Consider Case 2 : Lets us assume little more than minimum switching frequency i.e. f = 25Khz (just half of previous frequency). So, Npri = 2.4 or 2 (after rounding off).

Consider Case 3 : If your core is different then the cross section area may also differ. It could be same too or maybe higher or smaller to 540mm^2. Only in case when it is smaller can result to non-zero primary turns.

Overall, it is important to check all those parameters I suppose because the formula is ok. Hope you find some practical help.

Practically, the more high the frequency, the size of transformer is small.

very good conversation guys (Y)

Okay

If the higher the core size, the lower the frequency, then could it be possible that the core is supposed to operate at the 20kHz to at least get tangible Np using Vp=12v, and maintaining the Bmax=1300

For example:

12×10^8/4×20,000×1300×5.4 = 2.14turns = 2turns approx

But, my question is that what will be the consequence of using 20kHz instead of 50kHz?

Thanks for your responses

**Error correction on previous comment:

N = 0.03118, Npri = 3

I am certain sir. Its 12v

I intend using it for step up

I actually wanted to use it for Inverter using Switch Mode Topology.

I will be glad if you can give me some practical hints concerning that too.

Its my first attempt.

copper wire size calculations

hi.

i have a question. does the size core has any effect on selecting the bmax value?

I’m using a ee85 core and trying to calculate a high output current transformer(about 2-4kA & 2-5 volt).

some websites say the value of bmax should be between 3500G t 4500G.

what should i do??? help plz..

my design:

f=30khz – vin=310vdc – v out- 3- 5volt – IOut max= 4000A – ee85

hi.

i have a question. does the size core has any effect on selecting the bmax value?

I’m using a ee85 core and trying to calculate a high output current transformer(about 2-4kA & 2-5 volt).

some websites say the value of bmax should be between 3500G t 4500G.

what should i do??? help plz..

my design:

f=30khz – vin=310vdc – v out- 3- 5volt – IOut max= 4000A – ee85

SIr,

Can you provide Complete calculations of DC-AC Ballast design T/F and Resonant inductor

Ballast 4W, Input 12Vdc, Push-Pull config.

Hi dear friend

thanks about this post. it’s great

Can you create a post about calculating the number of the wire(wire layer) and the effect of the skin?

with this post’s example.

thanks a lot.

Is it possible to design 4000Watt SMPS inverter from 500VDC input to 220VDC,15A output.

Regards,

Prasada Rao

Hello

I want to design a switching power supply with the following specifications

Topology : Forward or Flyback

Vin (AC) = 220V

Vin (Diode Bridge DC) = 310V

Iout = 13A

EER42/42/20 => Ac = 1.94 cm^2

Bmax = 1500G

Fsw = 45KHz

Thank you for helping me

Very good read. For a full bridge topoplogy it’s the same set of equation for push pull?

You seem to neglect the joule heating losses / coils resistances that will reduce the output voltage at the full charge ?

how to calculate in direct ac-ac high frequency link transformer?