Calculating ferrite transformer is a process in which engineers evaluate the various winding specifications, and core dimension of the transformer, using ferrite as the core material. This helps them to create a perfectly optimized transformer for a given application.

The post presents a detailed explanation regarding how to calculate and design customized ferrite core transformers. The content is easy to understand, and can be very handy for engineers engaged in the field of power electronics, and manufacturing SMPS inverters.

## Why Ferrite Core is used in High Frequency Converters

You might have often wondered the reason behind using ferrite cores in all modern switch mode power supplies or SMPS converters. Right, it is to achieve higher efficiency and compactness compared to iron core power supplies, but it would be interesting to know how ferrite cores allow us to achieve this high degree of efficiency and compactness?

It is because in iron core transformers, the iron material has much inferior magnetic permeability than ferrite material. In contrast, ferrite cores possess very high magnetic permeability.

Meaning, when subjected to a magnetic field, ferrite material is able to achieve a very high degree of magnetization, better than all other forms of magnetic material.

A higher magnetic permeability means, lower amount of eddy current and lower switching losses. A magnetic material normally has a tendency to generate eddy current in response to a rising magnetic frequency.

As the frequency is increased, eddy current also increases causing heating of the material and increase in coil impedance, which leads to further switching losses.

Ferrite cores, due to to their high magnetic permeability are able to work more efficiently with higher frequencies, due to lower eddy currents and lower switching losses.

Now you may think, why not use lower frequency as that would conversely help to reduce eddy currents? It appears valid, however, lower frequency would also mean increasing the number of turns for the same transformer.

Since higher frequencies allow proportionately lower number of turns, results in transformer being smaller, lighter and cheaper. This is why SMPS uses a high frequency.

### Inverter Topology

In switch mode inverters, normally two types of topology exits: push-pull, and Full bridge. The push pull employs a center tap for the primary winding, while the full bridge consists a single winding for both primary and secondary.

Actually, both the topology are push-pull in nature. In both the forms the winding is applied with a continuously switching reverse-forward alternating current by the MOSFETs, oscillating at the specified high frequency, imitating a push-pull action.

The only fundamental difference between the two is, the primary side of the center tap transformer has 2 times more number of turns than the Full bridge transformer.

## How to Calculate Ferrite Core Inverter Transformer

Calculating a ferrite core transformer is actually quite simple, if you have all the specified parameters in hand.

For simplicity, we'll try to solve the formula through an example set up, let's say for a 250 watt transformer.

The power source will be a 12 V battery. The frequency for switching the transformer will be 50 kHz, a typical figure in most SMPS inverters. We'll assume the output to be 310 V, which is normally the peak value of a 220V RMS.

Here,the 310 V will be after rectification through a fast recovery bridge rectifier, and LC filters. We select the core as ETD39.

As we all know, when a 12 V battery is used, it's voltage is never constant. At full charge the value is around 13 V, which keeps dropping as the inverter load consumes power, until finally the battery discharges to its lowest limit, which is typically 10.5 V. So for our calculations we will consider 10.5 V as the supply value for *V*_{in(min)} .

## Primary Turns

The standard formula for calculating the primary number of turns is given below:

*N*_{(prim)} = *V*_{in(nom)} x 10^{8} / 4 x *f* x *B*_{max} x *A*_{c}

Here *N*_{(prim)} refers to the primary turn numbers. Since we have selected a center tap push pull topology in our example, the result obtained will be one-half of the total number of turns required.

*Vin*_{(nom) }= Average Input Voltage. Since our average battery voltage is 12V, let's, take*Vin*_{(nom)}= 12.*f*= 50 kHz, or 50,000 Hz. It is the preferred switching frequency, as selected by us.*B*_{max}= Maximum flux density in Gauss. In this example, we'll assume*B*_{max}to be in the range of 1300G to 2000G. This is the standard value most ferrite based transformer cores. In this example, let’s settle at 1500G. So we have*B*_{max}= 1500. Higher values of*B*_{max}is not recommended as this may result in the transformer reaching saturation point. Conversely, lower values of*B*_{max}may result in the core being underutilized.- A
_{c}= Effective Cross-Sectional Area in cm^{2}. This information can be collected from the datasheets of the ferrite cores. You may also find A_{c}being presented as A_{e}. For the selected core number ETD39, the effective cross-sectional area furnished in the datasheet sheet is 125mm^{2}. That is equal to 1.25cm^{2}. Therefore we have, A_{c}= 1.25 for ETD39.

The above figures give us the values for all the parameters required for calcuating the primary turns of our SMPS inverter transformer. Therefore, substituting the respective values in the above formula, we get:

*N*_{(prim)} = *V*_{in(nom)} x 10^{8} / 4 x *f* x *B*_{max} x *A*_{c}

*N*_{(prim)} = 12 x 10^{8} / 4 x 50000 x 1500 x 1.2

*N*_{(prim)} = 3.2

Since 3.2 is a fractional value and can be difficult to implement practically, we'll round it off to 3 turns. However, before finalizing this value, we have to investigate whether or not the value of *B*_{max} is still compatible and within the acceptable range for this new rounded off value 3.

Because, decreasing the number of turns will cause a proportionate increase in the *B*_{max}, therefore it becomes imperative to check if the increased *B*_{max} is still within acceptable range for our 3 primary turns.

Counter checking *B*_{max} by substituting the following existing values we get:*Vin*_{(nom)} = 12, *f* = 50000, *N*_{pri} = 3, *A*_{c} = 1.25

*B*_{max} = *V*_{in(nom)} x 10^{8} / 4 x *f* x *N*_{(prim)} x *A*_{c}

*B*_{max} = 12 x 10^{8} / 4 x 50000 x 3 x 1.25

*B*_{max} = 1600

As can be seen the new *B*_{max} value for *N*_{(pri)} = 3 turns looks fine and is well within the acceptable range. This also implies that, if anytime you feel like manipulating the number of *N*_{(prim)} turns, you must make sure it complies with the corresponding new *B*_{max} value.

Oppositely, it may be possible to first determine the *B*_{max} for a desired number of primary turns and then adjust the number of turns to this value by suitably modifying the other variables in the formula.

## Secondary Turns

Now we know how to calculate the primary side of an ferrite SMPS inverter transformer, it's time to look into the other side, that is the secondary of the transformer.

Since the peak value has to be 310 V for the secondary, we would want the value to sustain for the entire battery voltage range starting from 13 V to 10.5 V.

No doubt we will have to employ a feedback system for maintaining a constant output voltage level, for countering low battery voltage or rising load current variations.

But for this there has to be some upper margin or headroom for facilitating this automatic control. A +20 V margin looks good enough, therefore we select the maximum output peak voltage as 310 + 20 = 330 V.

This also means that the transformer must be designed to output 310 V at the lowest 10.5 battery voltage.

For feedback control we normally employ a self adjusting PWM circuit, which widens the pulse width during low battery or high load, and narrows it proportionately during no load or optimal battery conditions.

This means, at low battery conditions the PWM must auto adjust to maximum duty cycle, for maintaining the stipulated 310 V output. This maximum PWM can be assumed to be 98% of the total duty cycle.

The 2% gap is left for the dead time. Dead time is the zero voltage gap between each half cycle frequency, during which the MOSFETs or the specific power devices remain completely shut off. This ensures guaranteed safety and prevents shoot through across the MOSFETs during the transition periods of the push pull cycles.

Hence, input supply will be minimum when the battery voltage reaches at its minimum level, that is when *V*_{in} = *V*_{in(min)} = 10.5 V. This will prompt the duty cycle to be at its maximum 98%.

The above data can be used for calculating the average voltage (DC RMS) required for the primary side of the transformer to generate 310 V at the secondary, when battery is at the minimum 10.5 V. For this we multiply 98% with 10.5, as shown below:

0.98 x 10.5 V = 10.29 V, this the voltage rating our transformer primary is supposed to have.

Now, we know the maximum secondary voltage which is 330 V, and we also know the primary voltage which is 10.29 V. This allows us to get the ratio of the two sides as: 330 : 10.29 = 32.1.

Since the ratio of the voltage ratings is 32.1, the turn ratio should be also in the same format.

Meaning, x : 3 = 32.1, where x = secondary turns, 3 = primary turns.

Solving this we can quickly get the secondary number of turns

Therefore secondary turns is = 96.3.

The figure 96.3 is the number of secondary turns that we need for the proposed ferrite inverter transformer that we are designing. As stated earlier since fractional vales are difficult to implement practically, we round it off to 96 turns.

This concludes our calculations and I hope all the readers here must have realized how to simply calculate a ferrite transformer for a specific SMPS inverter circuit.

### Calculating Auxiliary Winding

An auxiliary winding is a supplemental winding that a user may require for some external implementation.

Let's say, along with the 330 V at the secondary, you need another winding for getting 33 V for an LED lamp. We first calculate the **secondary : auxiliary** turn ratio with respect to the secondary winding 310 V rating. The formula is:

N_{A} = V_{sec} / (V_{aux} + V_{d})

N_{A} = secondary : auxiliary ratio, V_{sec} = Secondary regulated rectified voltage, V_{aux} = auxiliary voltage, V_{d} = Diode forward drop value for the rectifier diode. Since we need a high speed diode here we will use a schottky rectifier with a V_{d} = 0.5V

Solving it gives us:

N_{A} = 310 / (33 + 0.5) = 9.25, let's round it off to 9.

Now let's derive the number of turns required for the auxiliary winding, we get this by applying the formula:

N_{aux} = N_{sec} / N_{A}

Where N_{aux} = auxiliary turns, N_{sec} = secondary turns, N_{A} = auxiliary ratio.

From our previous results we have N_{sec} = 96, and N_{A} = 9, substituting these in the above formula we get:

N_{aux} = 96 / 9 = 10.66, round it off gives us 11 turns. So for getting 33 V we will need 11 turns on the secondary side.

So in this way you can dimension an auxiliary winding as per your own preference.

**Wrapping up**

In this post we learned how to calculate and design ferrite core based inverter transformers, using the following steps:

- Calculate primary turns
- Calculate secondary turns
- Determine and Confirm
*B*_{max} - Determine the maximum secondary voltage for PWM feedback control
- Find primary secondary turn ratio
- Calculate secondary number of turns
- Calculate auxiliary winding turns

Using the above mentioned formulas and calculations an interested user can easily design a customized ferrite core based inverter for SMPS application.

For questions and doubts please feel free to use the comment box below, I'll try to solve at an earliest

faith jumbo says

Sir again I want to ask how do I know the area and type of a ferrite transformer salvage from scrab board

Swagatam says

The rule of thumb is that the winding must fit inside the transformer comfortably, then you can assume that the size is correct.

faith jumbo says

Goodday sir please I need help, I rewind a ferrite transformer for a 12v inverter using sg3525, that is high frequency inverter I noticed that when I connect my transformer the frequency will be fluctuating and some times there Will be out while sometimes there’s no output and sometimes both mosfet will be hot whli sometime only one will be hot even if I vary the frequency the problem still happen, but if I remove the the transformer the sg3525 will function normal please what could be the problem and possible solutions

Swagatam says

Good day Faith, sorry without practically checking, the problem can be very difficult to judge.

faith jumbo says

Should I send you the waveform that show when the transformer is connected because when I connect the transformer pulse changes drastically because frequency increase, what causes the frequency to increase

Swagatam says

The problem may be due to inaccurate winding configuration which cannot be diagnosed with waveform images. Ferrite transformers require perfect calculations in order to work correctly.

criss says

For torroidal core transformer how can i calculate nr of turns for primary and secondary?

thank you Swagatam

Swagatam says

Sorry, I do not have the details for torroid coil calculations

SAJI TR says

Copper wire gage is not mentioned

mathieu says

Vd = 0.5V (auxillary winding)

the 0,5V is for one diode rectifier! i have 4 diodes bridge rectifier so i can put 2v to the formula?

thank you

Swagatam says

Sorry did not understand your question? 4 diodes will drop 2V, and you may get 0V at the output

Swagatam says

OK, yes I just missed that Vd is specifically for compensating diode drop…yes that’s right!

mathieu says

Hello Swag i have a problem with the auxiliary winding for the pwm! does it works only in +33V or i can make it work with only +30v? is there a limit for the voltage?

Swagatam says

Hi Mathieu, there’s no limit for the voltage value, you can set it with the help of the given formula.

Mans says

Hello

How can I specify the output current and also the number of the wire used in a SMPS? For example, I want to design a SMPS with 12 volt and 10 amp output, how I can specify the size of the transformer and also the number (diameter) of the wires for two sides?

Thank you

Swagatam says

I am not sure about that, perhaps this table can help

https://www.homemade-circuits.com/56492-2/

It has to be done through some trial and error by making a sample transformer first and checking the current and then modifying it proportionately

Mans says

Dear master

Although the article wasn’t the same respond that I expected (it mostly speaks about the resistance of the copper or aluminum wires) nonetheless thank you for your answer.

Let me I pose my question otherwise:

Dose the number of turns (for an specified output voltage) depend on the amount of the output current (amp), or current is an independent factor to voltage. in the other word, for obtaining a higher current for a specified voltage, does the number of the wiring turns remains changeless but the size of the transformer become larger?

Swagatam says

Mans, once the number of turns are calculated for the specified voltage, the current can be optimized by appropriately dimensioning the thickness of the wire. This must be optimized by winding a bunch fine copper together instead of using a single thick wire, to eliminate skin effect.

Mans says

Thank you Mr. Swagatam

I knew this technique previously, because I not only made a few traditional transformer (with iron core) years ago, but also repaired a few elector motors with burned winding that one of them was a 220 v, 10 amp motor with 1 mm wire. As I hadn’t the necessary equipment and tools I made all its winding with a piece of flat wood and four nails!

At that time, I had a table that guided me how to choose the different double thin wires instead of a single thicker one.

Also I had a formula that offered me the size of the proper transformer and wires for a specified current, but now I forget all, while winding a ferrite transformer is different with an Iron one.

I wanted to know technically, what the ratio of the size of a ferrite transformer and the thickness of the wire (for achieving a higher current with the same voltage) is?

I think, Ac in the formula is the same transformer size (that can be changed for a higher current) but there is no point in the article, about the appropriate diameter of the wires in regard to the transformer size.

Also you spoke about the skin effect (probably the thickness of the isolating resin of wire), what it is?

Swagatam says

Sorry, I don’t have the ratio formula for the ferrite core size to wire thickness…this has to be fixed with some trial and error method so that a benchmark ratio can be determined.

Skin effect happens for thicker single wire winding where the electromagnetic induction does not get absorbed deep into the secondary wire, resulting in inefficient current transfer. To correct this many thin wires are used in parallel for the winding instead of a single thick wire

mathieu says

Hello Swag i have found the teslas value for my etd34 Bmax = (330 mt),(0.33T),(3300Gauss). with your formula,bmax is in tesla,does we calculate in mtesla , or in gauss(3300)?

Thank you

Swagatam says

Sorry Mathieu, I have not yet investigated Tesla value, so can’t suggest much on this.

mathieu says

Hy Swag can i fix the peak value (sec turns) at 115V?

Swagatam says

Hi Mathieu, yes you can….

mathieu says

Hello Swag for the windings of the ferrite transformer do you made it manually or do you send your transformer to a specific factory ?

Thank you!

Swagatam says

Hello Mathieu,

if you have experience with coil winding and have a winding machine then you can do it yourself, otherwise you may have to contact a transformer winder.

mathieu says

Hello swag i have a question on the choice of Bmax between 1300G to 2000g! Is there a precision on the gauss? Do i need to calculate my windings with only 1500G for any core size transformers?

thank you

Swagatam says

hello Mathieu, the value is not critical and can be between 1300 and 2000. It is dependent on the other factors also, and therefore a critical value can never be found. If you try to select Bmax with highest efficiency then that might require the other parameters to be adjusted accordingly, affecting the overall calculations. That is why we have to be satisfied with whatever Bmax comes withing the mentioned range.

Ambrose Fakorede says

Sir, you have a blessing unto us, and you shall increase and multiply sir.

Sir, i need circuit diagram for power bank, that will convert 3.7v to 5v by 1amp. By using 34063 ic and 555timer. Pls sir i need a quick reply sir.

Swagatam says

Thanks Ambrose. the circuit diagram of a boost converter using IC34063 is already given in the datasheet of the IC with formulas….so you can check the datasheet and easily make it from the available information.

Evergreen says

Does that mean using the 10k pot, or is there something else i have to do for the 393 that i will not do with 741? What i mean is, how do u mean configure the output of the ic? Thanks for all ur previous replies, i’m greatful.

Swagatam says

Just connect a 10K from IC output pin to positive line, that’s all rest everything can be same!

Evergreen says

Thanks. Can i use IC LM393 In place of IC 741?

Swagatam says

Yes you can use it, just make sure to configure the output of the IC correctly

Evergreen says

Do u have a link where i can get a visual description(schematic)? I was getting to understand but i got lost when u said, “Used as the feedback with the inverter ic”. i will be using ic cd4047 how do i connect the aux feedback to the ic? (if diagram is not possible, (sorry to say) but theory will work fine also).

Swagatam says

You can check out the diagram just above “How to Setup” in this article:

https://www.homemade-circuits.com/arduino-pure-sine-wave-inverter-circuit/

The bridge input can be connected to the auxiliary winding. Only one transistor is required for your application at the output whose collector could be linked with the R/C junction of the 4047 IC for overvoltage/undervoltage cut off actions

Evergreen says

Sorry for the mutiple questions swagtam, but in the feedback aspect of the article, i was thinking if it means to add the other feedback system found in the feedback link, using 555 timer ic? Simply put should i add the 555 ic feedback system to my ferrite transformer inverter?

Swagatam says

Evergreen, the feedback can be used for shutting down the inverter during a low voltage situation or controlling over voltage etc. For this the auxiliary voltage could be used as the feedback with the inverter IC. In IC 555 for example this can be integrated to pin#5 for shut down.

Evergreen says

My main question is, will it be a center tap or a single? If it is a center tap will it be 9V-0-9V (18V) or 4.5V-0-4.5V (9V) For a 12v battery? And is the aux a must for a ferrite transformer?

Swagatam says

I have already answered. it can be anything between 9 and 11V, center tap or not depends what kind of inverter topology you are using. For the aux winding details please read the auxiliary section of the article.

9 means 9-0-9, 11 means 11-0-11 for center tap.

for full-bridge this will be 0-9V and 0-11V

Evergreen says

Thanks a lot swagtam, i got it now.

Evergreen says

Hi swagtam. If u are to use the transformer for a 12v battery should it not be 9v-0-9v center tap? The diagram shows a centertap transformer but u calculated for single 10.5v.

Swagatam says

Hi Evergreen, the calculation is not critical, because the output voltage range is not critical and can be anything between 210 V to 240 V.

10.5 V is used because the lowest battery level is taken as 10.5 V here. The transformer primary can be anywhere from 9V to 11V

Sam omoniyi says

Hello, am interested in ferrite inverter transformer, pls is there any way I can be getting them?

Swagatam says

Hello, I think you can get a complete ferrite core based converter kit from ebay.

hicham says

Hi thank you for this post i need to know why you said in your post the transformer you had chose has a 250 W but when i read in the datasheet i find that the ETD39 transformer had 450 W

Swagatam says

Hi, the recommended size may have a range between 250 and 450 watts, it cannot be a fixed value. Depends how well the winding is optimized

Matthew Jones says

Not to pick it apart but 125mm ^2 is 12.5cm^2. This same thing is pretty much wrote over here with the same mistake for an ETD39.

Swagatam says

Can you please elaborate a little more on this?

Matthew Jones says

No need, I am wrong. I was reading these articles with idea there are 10mm in a cm, But then I looked it up and there are 100mm^2 per centimeter. Didn’t know that. I would delete this if I could. Sorry to trouble you.

Swagatam says

No problem, thanks for clarifying!

Kamil says

Thanks for this article. Sir, i want to know where 10^8 is gotten from in the formula for calculating d transformer primary winding number of turn. Thanks.

Swagatam says

It’s a part of the formula, a constant.

sheraz says

dear sir! i ‘m using power Esim software for designing smps xformer. i have problem mostly by identifying diameter of secondary turns wire (turns for output voltage let say 16v). they show that N2 = 4 (0.18mm x 7T/W x 9). sir i understand that there are 4 output turns each turn having 9 parallel wire but don’t understand the concept behind (0.18mm x 7T/W)? i don’t know what they mean???

Swagatam says

Dear Sheraz, diameter and number of strands is directly proportional to the current specification of the transformers or the secondary of the transformer. Higher current will require proportionately thicker gauge wire and number of strands

sheraz says

it means that there are 7 strands of wire having diameter 0.18mm and each (0.18mm x 7T/W) must be considered as single wire for higher current and then 9 such wire must be winded in parallel for 4 turns?

Swagatam says

That’s right, 7 strands together must be considered as single wire for the winding.

sheraz says

dear sir! can this formula ” Ns(pri) = ((Vin * 10^8)/(4 * Bm * fosc * Ae)) can be used for winding smps transformer for single power mosfet smps based on uc3842 etc?

Swagatam says

Dear Sheraz, it can be used according to me!

Victory says

1. That’s with the capacitor connected in series with the transformer, does it help to provide inductive reactance?

2. And if so that it helps to provide inductive reactance, removing the cap then, where will the inductive reactance come from since current will only flow in only one direction when the cap is removed because the cap is charged and discharged through the transformer?

Swagatam says

Capacitor works like a series resonant circuit for optimal performance of the inverter

Victory says

Is it safe to remove the capacitor?

And if you did, will the circuit still work?

Swagatam says

If the primary is wound with correct calculation then you can remove the capacitor, otherwise it can burn the MOSFET

Kingsley says

hello Mr. Swagatam!

Thanks for the ferrite core transformer calculation. I have seen this same calculation for ETD39 ferrite core transformer in more than 3 blogs. And when i take the idea of the Bmax of 1300G to 2000G for bigger size ferrite like EE65 ferrite with Ae = 540 millimeter square = 5.4 centimeter square, I ended up getting funny number of turn(s) like 1.7Turn = 2Turns approx for a primary voltage of 24V, a frequency = 50KHz, Bmax = 1300G, and Ae=5.4 centimeter square. And since 1300G is the lowest range of the spec (i.e 1300G – to – 2000G), if i say let me adjust the Bmax to something like 1500G so as to be within the middle region of the Bmax spec, the I got 1Turn approx for primary winding.

You know the funniest thing here? the leftover spaces in the transformer can still accommodate like 5Turns for the primary even after considering the secondary. And being a large size ferrite core transformer, I need to have it well wound by ensuring there is no gap between the cores and the winding to avoid excessive losses.

For this reason therefore, I tried to recalculate for Bmax using N-primary = 5Turns and I got Bmax to be 444.444 = 444G approx. So, my question is can Bmax be reduced to such small value for a large ferrite core transformer?

Thank you my mentor for your expected response.

Swagatam says

Hello Kingsley, If all the parameters are correctly selected then the turn number will be correct, but if one of the parameters is wrongly selected then the turn numbers can go wrong. Are you sure EE65 is correct, and also the thickness of the wire that you may have selected?

Bmax = 444 will not work as per the rules, so you cannot change that, instead try changing the other parameters which could be perhaps wrongly selected.

Kingsley says

Good to here from you my dear motivator.

For the parameters, EE65 is a large ferrite transformer. with all the research i have made, the Ae=540mm square which by conversion is equal to 5.4cm square. For other parameters such as frequency, I chose F=50000Hz, V-in(norm) = 24V, and Bmax = 1500G, yet I got 1Turn for N-primary.

so my calculation was as follows:

N-pri = (24*10^8)/(4*50000*1500*5.4)

Again, I observed that the larger the size of ferrite core transformer, the larger the Ae in mm square. And this is one problem why the Number of turn is drastically reduced. If any parameter spec for the EE65 is wrongly selected, then it should be the Bmax since the datasheet declares Ae to 540mm square.

Hope I am correct saying that 540mm square = 5.4cm square? Or am I wrong? please correct me if i am wrong.

if i should refer your question in the above reply i.e “Are you sure EE65 is correct, and also the thickness of the wire that you may have selected?” I used 10 strands of 20AWG copper coil for the Primary side (i.e 24v battery side) and 5 strands of the same copper coil on the secondary side (330Vac side). Yet 1Turn is too small and enough space still exist in the window area.

Swagatam says

Hello Kingsley,

The formula results are dependent on the turn ratio between the primary and the secondary, and also the E core selection. If your wire thickness and turn calculation are not correct then the core size will also be not correct, and this will ultimately result in the whole calculation going wrong.

How much space is left is not relevant to the number of turns on the primary.

By the way using 20 AWG for the 220V is grossly wrong, how can you have a 330V winding wire having the same thickness as the 24 V primary, that too 5 strands together…? I think this is main problem behind the wrong primary turn results.

Please divide the primary wattage with secondary voltage to get the secondary current and then you can estimate the wire thickness accordingly.

Kingsley says

Ok, I want to ask if the ferrite core transformer will be OK at the frequency of 27KHz since the range of frequency specified is between 20KHz to 500KHz? If that can be fine, then reducing the frequency will take care of the wrong number of turns while still maintaining the Bmax within that same range of 1300G to 2000G; since I can now Obtain up to 2Turns for the primary Turn.

Swagatam says

yes 27kHz will work, but using similar SWG wires for primary and secondary would still be wrong, considering the fact that primary is 24V and secondary 330V, the wire must also proportionately vary in thickness.

Kingsley says

OK since using the same size of AWG but different number of strands won’t work, can you please suggest me the right AWG for the primary and secondary? Again can you also suggest me the quantity of strands for each?

Thank you sir.

Swagatam says

Strands won’t be required for the secondary, since the wire itself will be quite thin. the secondary wire must be 15 times thinner than the primary, considering 24V is 15 times smaller than 310V

Kingsley says

Ok, My boss! I will consider doing it as you have said. Thank you dear.

Reza says

Dear Mr. Swagatam

Thank you for sharing this helpful discussion.But may you discuss about selecting wire tickness (AWG) and wiring transformer(for example first wiring primary then secondary …) due to output current of SMPS?

Thank you

Swagatam says

Dear Reza, wire gauge can be estimated with some trial and error (approximation), or you can refer to the table which is given i this article:

https://www.homemade-circuits.com/56492-2/

Normally the higher voltage winding is wound first and then the lower voltage on top of it, with a layer of insulation.

Victory says

Pls Mr Swagatam, can you help me explain why a capacitor is connected in series with the transformer in the circuit shown from the link below?

http://tinypic.com/view.php?pic=ne9sow&s=9

I understand half bridge transformer connection partially, having read your article about it. But, in the conventional connection used for half bridge transformer connection, there are two capacitors, but in this circuit, only one was used. Why? Does the capacitor act as a power bank, storing power temporarily, to be used for the other half circle when the transistor beneath is switching?

And also, does the capacitor reduce the voltage coming to the transformer?

Swagatam says

The circuit shown in the link is not a push pull topology, therefore only a single capacitor is used. Capacitor will reduce the current, not the voltage, higher value capacitors will produce higher current.

Victory says

Pls can I understand better the what happens in half-bridge, full-bridge and pull-pull connection of ferrite transformers clearly with a bit of diagrams, if possible?

And also, can a ferrite transformer when operated on pulsed DC, still possess reactance to resist voltage and current as would a normal transformer in order to avoid a short circuit when connected to the power supply? Thanks

Swagatam says

Half bridge will pass current in only one direction across the primary winding. A full bridge will pass current alternately across the winding in reverse forward way. The output of a half bridge will be a half wave AC, whereas in full bridge it will a full wave AC.

Half bridge and full bridge both can work with two wire primary.

A center tap push pull topology is also a full bridge topology but it requires two winding at the primary with a center tap.

A pulse Dc operation is another form of half bridge operation,and it is possible but the output will be half wave AC.

Victory says

Please did you mean that you divide into half the value gotten for the primary winding calculation because it’s a push-pull topology?

For example, a primary winding calculated as 20turns will then be divided by 2 to give 10turns because it’s push topology?

Swagatam says

You must double it for a center tap topology. if the result is 20 turns then for a center tap transformer primary it will be 40 turns with center tap at 20th turn

Jonh Smith says

Ferrite cores are used NOT because of their high permeability. Actually, ironcores, and especially permalloy cores, etc could showcase permeability exceeding that of typical transformer ferrite cores.

For the reference, effective permeability of typical transformer core ferrites used in SMPS would usually be about 1000 to 2500 or so, depending on application. Gapped cores, like those in flybacks would have effective permeability much lower than that (though it “coupled inductor” rather than “transformer”). However, transformer-grade steel alloys and especially permalloy-like “low-frequency” cores can showcase permeability well around 20 000, more than 10x higher compared to typical transformer ferrites. Some “special” ferrites also expose permeability about 15 000 or more, but are NOT used for transformers, being high-loss things, so only suitable for ferrites. Who needs high-loss transformer? It would just heat itself up, wasting power on that. Avoiding this scenario is pretty much goal of transformer calculations, especially when it comes to ferrite type selection, etc.

But what makes ferrite transformers small? And why we need ferrite? Its mostly about high frequency operation and eddy currents.

Generally, the higher operating frequency is, the more power could be pushed through same transformer size. So if we want small but powerful supply, we have to go for high frequency. At high frequency one can make far less turns in windings, yet they still would have enough inductance to “resist” incoming voltage. The higher frequency, the less turns we need, the smaller transformer we can afford. To certain extent.

What’s the problem with steel, permalloy, etc? Being metallic they are inherently conductive. At which point, core could act a bit like some kind of (shori-circuited) winding, heating itself by eddie currents. This effect could be put in use, giving a rise to “induction heating” – but we do not want it in transformers. So transformer steels are optimized to have relatively high resistance and also split into separate thin plates to reduce eddie currents. However, the higher frequency is, the more pronounced this “induction heating” gets – and even special steel alloys and very thin plates aren’t enough to prevent core heating. That’s where ferrites come into play. They are ceramic materials, so unlike metals they show rather poor electric conductivity. At which point eddie current loses are mostly mitigated, so we can increase operational frequency a lot compared to steel designs. This implies far smaller windings, so we can push way more power without increasing transformer size. So SMPS would have unusually small ferrite transformer for given power, compared to low-frequency steel transformer designs.

As for topologies, there’re a bit more options. Out of bridge-like, there’re three: half-bridge, full-bridge and push-pull. Half-bridge and push-pull are similar in overall idea, but implementation is different, half-bridge uses just 1 winding and some capacitors to expose winding to both polarities. Push-pull rather uses different halves of winding to achieve same result (in terms of magnetic field behavior). This brings different tradeoffs, and full bridge lacks most shortcomings of both – but being more expensive due to 4 high-voltage transistors reqirement.

Some relatively recent power supply designs could also use e.g. “2-switch forward” approach, which also uses 2 MOSFETs but operates in somewhat different manner. It underuses transformer core, magnetizing it in just 1 polarity, but on other hand it gets favorable MOSFETs voltage and current tradeoffs, pretty much like full bridge, being cheaper solution. So it fairly usual to see 2-switch forward in e.g. ATX power supplies up to about 800-1000W or so. Transformer computed much like “usually”, but with some special considerations in regard of saturation & demagnetizing attached.

There’re also flybacks, that aren’t even transformers by their actions, though turns ratio still haves some meaning. These have their own limitations but are cheapest option up to around ~100W of output power. Above this point different topology could be better consideration though.

Swagatam says

Thank you for the nice explanation!

John Smith says

> being high-loss things, so only suitable for ferrites.

Oh, I meant high-loss high-permeability ferrites are used in EMI filters, etc, where turning noise into heat is rather something good.

Also it worth to mention there is approximately the following trend: the higher intended opreational frequency of ferrite compound is, the less permeability it would get, the higher resistivity it would enjoy, it would have lower core losses but require more turns.

As example, think of N49 vs N67, for example. N49 is tailored up to about ~1MHz or even more, while N67 is barely ok up to maybe 200kHz at very most. N67 would have higher permeability, but would get far higher core losses at high frequencies. Overall it comes to trading copper loses vs core loses, in perfect world we would try to hit minimum of their sum, however numerous secondary considerations make it challenging.

Victory says

Thank you Mr. Swagatam for your reply. Can I understand some better?

Please, what do you mean by Half bridge and full bridge both can work with two wire primary?

Can you explain what this means with the aid of a diagram also?

Thanks

Swagatam says

Hello Victory, please read the last sections of this article:

https://www.homemade-circuits.com/how-to-design-inverter-basic-circuit/

It explains about half bridge and full bridge topology in brief.

Freedom Uyiosa Aisevbo says

How do we determine the swg of the coil used

Swagatam says

SWG is related to current and can be found by estimating how much current a copper wire with particular thickness can carry. Probably this article can help:

https://www.homemade-circuits.com/56492-2/

Ajay Trivedi says

Explained in simple and lucid manner, thank you, sir.

Swagatam says

My pleasure Ajay, Glad you liked it!