## Ferrite Transformer Turns Calculation for Offline SMPS Half-Bridge Converter

On different
forums, I often find people asking for help in calculating the required turns
for a ferrite transformer they are going to use in offline SMPS half-bridge
converters. The half-bridge topology is very popular for offline converters in
the power range 100W to 500W, sometimes going up to even 1000W. In an offline SMPS
half-bridge converter, the line voltage is rectified and filtered and is then
converted to high frequency with 2 MOSFETs – one in high-side configuration and
the other in low-side configuration. This high frequency high voltage AC is fed
to the ferrite transformer to step down the voltage to low voltage high
frequency AC which is then rectified to DC and filtered to provide clean DC
output. A vital thing to remember is that in a half-bridge converter, the 2
MOSFETs work along with 2 capacitors to create the high voltage high frequency
AC. The configuration of the capacitors, MOSFETs and transformer causes the
transformer to be supplied half the voltage of the rectified DC. This means
that, compared to a full-bridge converter, half the number of turns is required for the primary, but the power output
would be half. Thus power/energy
density is halved.

Now let’s move on to the calculation. Calculation of required turns is actually quite simple and I’ll explain this here.

For explanation, I’ll use an example and go through the calculation process.

Now let’s move on to the calculation. Calculation of required turns is actually quite simple and I’ll explain this here.

For explanation, I’ll use an example and go through the calculation process.

Let’s say
the ferrite transformer will be used in a 250W converter that will be used to charge a 12V lead acid battery. The selected topology
is obviously half-bridge. The power source for the converter is the AC mains.
Here I’ll take that to be 220V RMS, 311V peak, 50Hz. So, you must remember that
the mains AC should be rectified to DC first. Output voltage of the DC-DC
converter stage will be 14V. Switching frequency is 50kHz. The selected core is
ETD44. Remember that the output of the transformer will be high frequency AC
(50kHz square wave in this case). When I refer to an output of low voltage DC
(eg 14VDC mentioned above), this is the DC output obtained after rectification
(using schottky, preferably, or ultrafast recovery diodes configured as full-wave
rectifier) and filtration (using LC filter). Since I plan to use full-wave
rectification (with 2 diodes) at the output, the secondary of the ferrite
transformer will be center tapped.

We must take a maximum and minimum input voltage rating for the converter. For our example, these will be a low line voltage of 150V and a high line voltage of 250V. During operation, the output voltage will stay fixed as the converter is expected to have feedback circuitry.

V

We must take a maximum and minimum input voltage rating for the converter. For our example, these will be a low line voltage of 150V and a high line voltage of 250V. During operation, the output voltage will stay fixed as the converter is expected to have feedback circuitry.

V

_{inmin}= 150VAC = (150* √2)VDC = 212VDC
V

_{inmax}= 250VAC = (250* √2)VDC = 354VDC
V

The formula for calculating the number of required primary turns for a forward-mode converter is:

_{innom}= 220VAC = (220* √2)VDC = 311VDCThe formula for calculating the number of required primary turns for a forward-mode converter is:

For our half-bridge transformer, this will be twice the required number of turns, that is, the actual number of primary turns will be half that calculated from the above formula if we use the full voltage, or exactly what is calculated if half the voltage is used. This is because the voltage across the transformer is half the line voltage, as previously mentioned.

So, the actual formula would be:

N

_{pri}means number of primary turns; N

_{sec}means number of secondary turns; N

_{aux}means number of auxiliary turns and so on. But just N (with no subscript) refers to turns ratio.

For calculating the required number of primary turns using the formula, the parameters or variables that need to be considered are:

- Vin
_{(nom) }– Nominal Input Voltage. We’ll take this as 311V. So, Vin_{(nom)}= 311. - f – The operating switching frequency in Hertz. Since our switching frequency is 50kHz, f = 50000.
- B
_{max}– Maximum flux density in Gauss. If you’re accustomed to using Tesla or milliTesla (T or mT) for flux density, just remember that 1T = 10^{4}Gauss. B_{max}really depends on the design and the transformer cores being used. In my designs, I usually take B_{max}to be in the range 1300G to 2000G. This will be acceptable for most transformer cores. In this example, let’s start with 1500G. So B_{max}= 1500. Remember that too high a B_{max}will cause the transformer to saturate. Too low a B_{max}will be under utilizing the core. - A
_{c}– Effective Cross-Sectional Area in cm^{2}. You will get this information from the datasheets of the ferrite cores. A_{c}is also sometimes referred to as A_{e}. For ETD44, the effective cross-sectional area given in the datasheet/specification sheet (I’m referring to TDK E141. You can download it from here:*www.tdk.co.jp/tefe02/e141.pdf*). The effective cross-sectional area (in the specification sheet, it’s referred to as A_{e}but as I’ve said, it’s the same thing as A_{c}) is given as 175mm^{2}. That is equal to 1.75cm^{2}. So, A_{c}= 1.75 for ETD44.

_{pri}– the number of required primary turns.

Vin

Plugging these values into the formula:

_{(nom)}= 311 f = 50000 B_{max}= 1500 A_{c}= 1.75Plugging these values into the formula:

N

_{pri}= 29.6We won’t be using fractional windings, so we’ll round off N

_{pri}to the nearest whole number, in this case, rounded up to 30 turns. Now, before we finalize this and select N

_{pri}= 30, we better make sure that B

_{max}is still within acceptable bounds (it will be since this is such a minor percentage change, but I’ll show this anyways so that you know what to do, just in case). As we’ve increased the number of turns from the calculated figure (up to 30 from 29.6), B

_{max}will decrease very slightly. We’ll now figure out just how much B

_{max}has decreased.

B

_{max}= 1481

The new value of B

_{max}is well within acceptable bounds and so we can proceed with N

_{pri}= 30.

So, we now know that for the primary, our transformer will require 30 turns.

In any design, if you need to adjust the values, you can easily do so. But always remember to check that B

_{max}is acceptable.

- I’ve started off with a set B
_{max}and gone on to calculate N_{pri}from there. You can also assign a value of N_{pri}and then check if B_{max}is okay. If not, you can then increase or decrease N_{pri}as required and then check if B_{max}is okay, and repeat this process until you get a satisfactory result. For example, you may have set N_{pri}= 20 and calculated Bmax and decided that this was too high. So, you set N_{pri}= 30 and calculated Bmax and decided it was okay. Or you may have started with N_{pri}= 40 and calculated Bmax and decided that it was too low. So, you set N_{pri}= 30 and calculated Bmax and decided it was okay.

Naturally, feedback will be implemented to keep the output voltage fixed with line and load variations – changes due to mains voltage change and also due to load change. So, some headroom must be left for feedback to work. So, we’ll design the transformer with secondary rated at 16V. This headroom compensates for voltage drops due to output rectifier diodes. Feedback will just adjust the voltage required by changing the duty cycle of the PWM control signals. Besides that, the headroom also compensates for some of the other losses in the converter and thus compensates for the voltage drops at different stages – for example, in the MOSFETs, in the transformer itself, in the output inductor, etc.

This means that the output must be capable of supplying 14V with input voltage equal to 212VDC and also input voltage equal to 354VDC. For the PWM controller, we’ll take maximum duty cycle to be 98%. The gap allows for dead-time.

At minimum input voltage (when Vin = Vinmin), duty cycle will be maximum. Thus duty cycle will be 98% when Vin = 212VDC = Vinmin. At maximum duty cycle = 98%, average voltage to transformer = 0.98 * 0.5 * 212V = 103.88V.

So, voltage ratio (primary : secondary) = 103.88V : 16V = 6.493

Since voltage ratio (primary : secondary) = 6.493, turns ratio (primary : secondary) must also be 6.493 as turns ratio (primary : secondary) = voltage ratio (primary : secondary). Turns ratio is designated by N. So, in our case, N = 6.493 (I’ve taken N as the ratio primary line voltage : secondary).

N

_{pri}= 30

N

_{sec}= N

_{pri}/ N = 30 / 6.493 = 4.62

Round off to the nearest whole number. N

_{sec}= 5

Now, notice how this rounding up is not an insignificant rounding up. So, let’s try to keep N

_{sec}= 5 and adjust N

_{pri}again.

N

_{pri}= N * N

_{sec}

N

Now let’s check if B

_{pri}= 5 * 6.493 = 32.5 = 33 (rounded off to the nearest integer)Now let’s check if B

_{max}is okay with N_{pri}= 33, ie, if B_{max}is within acceptable bounds.
B

_{max}= 1346B

_{max}= 1346 is okay. So, N

_{pri}= 33 and N

_{sec}= 5. Thus 5 + 5 turns are required for the secondary. With proper implementation of feedback, a constant 12VDC output will be obtained throughout the entire input voltage range of 150VAC to 250VAC.

Of course, notice here that B

_{max}is very small and

*can*be increased to reduce the required turns. So, let’s reduce N

_{sec}from 5 to 4.

N

_{sec}= 4
N

_{pri}= N * N_{sec}= 6.493 * 4 = 25.97 = 26 (rounded off to nearest integer)
Checking B

_{max}again:
B

Here, one thing to note is that even though I took 98% as the maximum duty cycle, maximum duty cycle in practice will be smaller since our transformer was calculated to provide 16V output. In the circuit, the output will be 16V (transformer output will be 14V + Voltage drop of diode), so the duty cycle will be even lower. However, the advantage here is that you can be certain that the output will not drop below 12V even with heavy loads since a large enough headroom is provided for feedback to kick in and maintain the output voltage even at high loads and low line voltages.

If any auxiliary windings are required, the required turns can be easily calculated. Let me show with an example. Let’s say we need an auxiliary winding to provide 17.5V. I know that the output 14V will be regulated, whatever the input voltage may be, within the range initially specified (Vinmin to Vinmax – 150VAC to 250VAC). So, the turns ratio for the auxiliary winding can be calculated with respect to the secondary winding. Let’s call this turns ratio (auxiliary : secondary) N

N

So, N

N

_{max}= 1709Here, one thing to note is that even though I took 98% as the maximum duty cycle, maximum duty cycle in practice will be smaller since our transformer was calculated to provide 16V output. In the circuit, the output will be 16V (transformer output will be 14V + Voltage drop of diode), so the duty cycle will be even lower. However, the advantage here is that you can be certain that the output will not drop below 12V even with heavy loads since a large enough headroom is provided for feedback to kick in and maintain the output voltage even at high loads and low line voltages.

If any auxiliary windings are required, the required turns can be easily calculated. Let me show with an example. Let’s say we need an auxiliary winding to provide 17.5V. I know that the output 14V will be regulated, whatever the input voltage may be, within the range initially specified (Vinmin to Vinmax – 150VAC to 250VAC). So, the turns ratio for the auxiliary winding can be calculated with respect to the secondary winding. Let’s call this turns ratio (auxiliary : secondary) N

_{A}.N

_{A}= N_{aux}/ N_{sec}= (V_{aux}+V_{d})/ (V_{sec}+ V_{dsec}). V_{dsec}is the output diode forward drop (at the secondary). V_{d}is the output diode forward drop at the auxiliary. Let’s assume that in our application, schottky rectifiers with V_{d}= 0.5V is used.So, N

_{A}= 18.0V/14.5V = 1.24N

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

_{aux}= N

_{sec}* N

_{A}= 4 * 1.24 = 4.96

Let’s round off N

_{aux}to 5. Since the rounding up is very small (from 4.96 to 5), the output voltage will be pretty close to the desired voltage, but I'll just show you how to calculate what the output voltage is.

(V

_{aux}+ V_{d}) / (V_{sec}+ V_{dsec}) = N_{A}= Naux / Nsec = 5 / 4 = 1.25
(V

_{aux}+ V_{d}) = (V_{sec}+ V_{dsec}) * N_{A}= 14.5V * 1.25 = 18.13V
V

That is great for an auxiliary supply. If in your designs, you ever find that V

Another option is to recalculate N

So, there we have it. Our transformer has 26 turns for primary, 4 turns + 4 turns for secondary and 5 turns for auxiliary.

Here’s our transformer:

_{aux}= 17.63VThat is great for an auxiliary supply. If in your designs, you ever find that V

_{aux}is far too off the required voltage, a simple voltage regulator (using 78XX for example) should be used to provide the stable auxiliary voltage.Another option is to recalculate N

_{pri}and N_{sec}to accommodate for a near accurate auxiliary voltage but you can just use a voltage regulator to simplify things. After all, the voltage regulator*will*keep the output voltage regulated stable.So, there we have it. Our transformer has 26 turns for primary, 4 turns + 4 turns for secondary and 5 turns for auxiliary.

Here’s our transformer:

Here’s the transformer at work in a circuit (block diagram):

Calculating required number of turns for a transformer for an offline SMPS half-bridge converter is actually a simple task and I hope that I could help you understand how to do this. I hope this tutorial helps you in your ferrite transformer designs for offline SMPS half-bridge converters. Do let me know your comments and feedback.

is the procedure same for a full bridge converter

ReplyDeleteEverything's the same, except that the primary will have twice the number of turns. You should refer to this as well:

Deletehttp://tahmidmc.blogspot.com/2012/12/ferrite-transformer-turns-calculation.html

Regards,

Tahmid.

Can we use the same formula to calculate LLC series resonant converter.

Deletehi tamid in some other articles i have read i usually see 0.5vin(min)*10^8/4*f*b*Ae why r you using vin(norm)

ReplyDeleteI calculate primary turns for nominal voltage but calculate the secondary to allow for minimum input voltage. I'm sure both methods are acceptable - it depends on the implementation.

DeleteRegards,

Tahmid.

hello Tahmid I am waiting anxiously on u for your sugestion for the major problem I am having. I sent the circuit to ur email. I only get the 5v gate voltage on the dc to dc converter fets when I apply a load to the h bridge without load I only get about .19 or so volt I want to know if this is normal. This happens with or without a drive circuit. If i take the b+ from off the fets the voltage also comes back.This is working like a smart circuit and I have read ur blog through and I dont seee u say the sg3525 should work this way. Please help me here I really need ur urgent help

Deletethanks tahmid, your articles are helping me to develop a 12v.10a. s.m.p.s. but the ferrite cores found in kolkata don't have any air gaps. commonly, i do use some insulating material like empire paper or polyester sheet in between " E-E's or 'E-I" s as air gap in flyback s.m.p.s. transformers. should i put some air gap in half bridge power supplies? please clarify. regards, vajanda.

ReplyDeleteYou don't need to use air gap in half bridge power supplies. The glue used to hold the cores together is fine. Don't add any additional air gap.

DeleteRegards,

Tahmid.

Hi Tahmid,

ReplyDeleteI'm planning to make a dc dc converter something in the range 24-200Vdc, I will use the output to feed a pure sine wave ups.

I did allready some tests with stuff I had laying here, the results are acceptable, the results were better with a toroid then with a E core. No problem to light up a 100W bulb at 226V. efficiency about 81% so that could be better? I'm sure my windings can be better.

For the Ic I've used a uc2845 Current Mode Controller, so I only have one primary and one secundary coil. Are 2 primary windings more efficiÃ«nt ?

Wich formula do I have to use for the calculation of my windings?

For the mosfets, if my input is in the 20-30V range, are 55V mosfets to low, and should I use higher voltage fets?

thanks in advance!

nick

I am using sg3525 with pushpull topology to generate 350V DC. Using ETD 49 Core. I am getting 350 V according to calculation but I am unable to draw current from it. As current increases from 100mA waveforms at MOSFET gates are distorted and 350V becomes 26V

ReplyDeleteI am using it at 50KHz

Could you upload a schematic of the circuit you're using?

Deletehello mr.nomi

Deletei am srihari. in your design, please use uf diodes to slove your prob

tahmid sir i need a little help.. in your calculations your havent mentions or use the watt of the transformer.. mean if i calculate turns for 250 watt mean it will be same for 1000 watt if the voltage ratings is same as 250 watt only current will increase .. what will be the difference in calculating turns for 250 watt and 1000 watt ferrite transformer...? material PC40 and EE55

ReplyDeleteand i have bought this core is that suitable for making 1000 watt inverter.?

http://www.aliexpress.com/store/product/large-watt-PC40-material-EE55-ferrite-core-with-bobbin-for-transformer-5sets-lot/406212_528032669.html

The power does not affect the calculation directly. The power affects the core selection which affects the calculation. So, for the same correctly selected core, you can have the same number of turns for 250W and 1000W, provided the wires can handle the current.

DeleteThe core should be okay. Use a high enough frequency. Use good MOSFET driving circuitry.

Regards,

Tahmid.

thank u so much sir i want to ask what will be the Bmax value for this type of core and also effective area Ac .? i searched on net it was written in some document that PC40 core has bmax 500, and also if i use full bridge topology for Dc-DC stage then input current will be calculated by meaning the resistance of the primary winding and if input voltage is 24 then i=v/r am i right.? then according to that current i choose wire and also give attention to skin effect.

Deleteand one thing more sir.. if my output power is 1000watt then input current will be 1000/24=41.6 ampere.. to handle 41.6 amp current wire will be very thick isn't.? which wire gauge i should use .?

DeleteYou should get those values for the core in the datasheet. Just look for them. If you've seen that Bmax can be 500G, then use something lower. Although, I do doubt that Bmax is that low.

DeleteRegarding the current, no. You have to calculate current as I=P/V, taking P as the maximum output power and V as the minimum input voltage. Then, choose the wire for a considerably higher current to compensate for losses (you may estimate the efficiency depending on the parts you are using) and to keep a "safety margin". Yes, give attention to skin effect.

Don't use thick wire. Use loads of thin wires in parallel. Or use Litz wire. Choose the thin wire size depending on frequency. Then, find out how much current that size wire can handle. Then, find the required number of wires. And, of course, use a larger number of wires than the minimum calculated.

Regards,

Tahmid.

thanks sir almost everything is cleared now , but the problem is i haven't get the datasheet of that core from internet . i have also doubt about bmax value. if u have some link where i can get that datasheet then it will be a great help to me thanks :)

Deleteor i have to use spacer in ferrite core .? i read some where that it is necessary to use spacer in between ferrite core.

DeleteI don't think that's necessary. That's required for flyback converters.

DeleteRegards,

Tahmid.

and sir any link for datasheet of PC40 EE55 core .?

Deletein the schematic attached here(and also in the photographs of the half bridge power supplies hand made by you) a polyester capacitor can be seen in series of the transformer. In computer s.m.p.s., a 1 uf /400v. is used.What would be the value of the capacitor in case of a 12v.10A.power supply ? Regards, and thanks in advance! Vajanda.

ReplyDeleteA 1uF capacitor should suffice. 250V is usually ok, but 400V is safer.

DeleteRegards,

Tahmid.

Nice post about SMPS transformers thanks for sharing nice information. We are also Manufacturer & Supplier SMPS transformer in India

ReplyDeletehelo thamid you are real geniues but i had one question if i want to constrcut a 500watt smps using tl 494 which type of core should be used and how the wattage of the core is desided . is their any formula for that and mainly why the primary inductance of the smps transformer is needed

ReplyDeleteFor 500W, I would use ETD44 with a high enough frequency. Or maybe even ETD49.

DeleteRegarding core selection: I'll cover this in a future article and I'll post the link when I'm done.

Regards,

Tahmid.

in computer smps using tl494 the ei33 used is it capable of deliviring 400watt or simply they mention it as a 400watt

ReplyDeleteMost (all I've encountered) ATX power supplies that use the EI33 are actually rated for around 230W or lower. They write 400W or such high powers on the tag specifications, but they can't handle 400W.

DeleteFor 400W, don't use the EI33 cores in the ATX power supplies. Use ETD39 or ETD44 or such cores. Use a high enough frequency.

Regards,

Tahmid.

can you tell me, what is a low side and high side driver ICs and schematic?

ReplyDeleteLow side drive and driver circuits:

DeleteLow-Side MOSFET Drive Circuits and Techniques - 7 Practical Circuits:

http://tahmidmc.blogspot.com/2012/12/low-side-mosfet-drive-circuits-and_23.html

--------------------------

High side drive and driver circuits:

N-Channel MOSFET High-Side Drive: When, Why and How?

http://tahmidmc.blogspot.com/2013/02/n-channel-mosfet-high-side-drive-when.html

Using the high-low side driver IR2110 - explanation and plenty of example circuits:

http://tahmidmc.blogspot.com/2013/01/using-high-low-side-driver-ir2110-with.html

Hope this helps!

Regards,

Tahmid.

hello tahmid in my country the voltage network is 110 volts would you please share with me the formula to develop an SMPS thank you very much

DeleteWhat should be changes in this design for using in PUSH-PULL converter?,

ReplyDeletewhere I get Bmax of a core?

Go through this:

Deletehttp://tahmidmc.blogspot.com/2012/12/ferrite-transformer-turns-calculation.html

If you still don't get your answer, feel free to ask any questions you have after going through the above linked article.

Regards,

Tahmid.

Hi Tahmid, congratulations about your blog. Do you have any spreedsheet to calculate ferrite tranformer?

ReplyDeleteRergards,

Barone

Hi! and my best regards for all friends. I have a question:

ReplyDeleteHow I can selecting the AWG for the primary and secondary?

I working with Half-Bridge topology, i understad the calculation for numbers turns, but

I have problems with the AWG number selection for primary and secundary.

Can you helpme? Thanks..

Hi! and my best regards for all friends. I have a question:

ReplyDeleteHow I can selecting the AWG for the primary and secondary?

I working with Half-Bridge topology, i understad the calculation for numbers turns, but

I have problems with the AWG number selection for primary and secundary.

Can you helpme? Thanks..

Hi, why do you use nominal voltage instead of minimul voltage on input in Nprimary faraday`s law. Whats the diference?

ReplyDeleteI like this simple and straightforward method of calculating transformer turns. I am designing a 48V to 300V DC-DC converter of 800+ watts and about 50 kHz, probably using Epcos E55 N27 Ferrite cores. I made a spreadsheet to simplify the process of wire selection and fill factor and efficiency, but I don't know if it is correct. Please have a look:

ReplyDeletehttp://enginuitysystems.com/files/Ferrite_Transformer.ods

Also, I plan to eliminate the third capacitor in series with the primary and connect just to the center tap between the capacitors. I don't see the need for the third capacitor. I have some 20 uF 100 VAC PP capacitors that should work, according to a simulation:

http://enginuitysystems.com/pix/48V-320V_DCDC_HalfBridge_2Cap.asc

Thanks!

I'll take a look at the spreadsheet.

DeleteThe third capacitor is used as a "DC blocking" capacitor. You should use this capacitor.

Regards,

Tahmid.

I forgot to check the notify box for follow-ups.

ReplyDeletehello tahmid , i wana ask if bmax is lower then the lowest limit then it will decrease in efficiency or it will have some effect on efficiency .? i am forced to use a core for which bmax is very low not in the range. kindly guide me thanks, it is acceptable if core is not fully utilized.

ReplyDeletei will visit it again to check your reply , thanks

dear tahmid, please guide me to make a s.m.p.s. of 600watts ( 60 volts,10 amps. ). input voltage is 220 volts. blessings! vajanda.

ReplyDeleteDear Tahmid,

ReplyDeleteI believe this article is still active (& not to be end anymore). Thanks for your service...

I have very much intrusted to design my first SMPS & this article is helps me to understand the turns ration of the transformer coil easily...

I need to know how to calculate the wire thickness of the primary & secondary winding of the transformer for my required current? Is it the thickness affects the turns ratio?

This question already asked, but I'm still searching for the answer.

Udhay

The turns ratio is affected by the desired voltage ratio. The thickness is important for current handling ability and will also dictate physical winding restrictions as far as spacing goes.

DeleteDeciding the wire based on current depends on the wire's current handling capacity - losses and temperature rise for a given temperature. You also should consider skin effect if you're working at higher frequencies. You can google for tables for ampacity vs diameter or wire gauge.

hello dear.

ReplyDeletefirst i really appreciate ur effort regarding helping peoples. very few can do this.

i also need ur help. wht about primary current? how to decide wire for a specific current? how to set

output current

Deciding the wire based on current depends on the wire's current handling capacity - losses and temperature rise for a given temperature. You also should consider skin effect if you're working at higher frequencies. You can google for tables for ampacity vs diameter or wire gauge.

DeleteRunthala Industries provides different types of transformer for different industries. It provide transformer with digital technique. I appreciate you for your effort regarding helping peoples.

ReplyDeleteTransformer Manufacturers in India

SMPS is an electronic power supply system which transfers power from a source like main power. It regulates the power for different equipment. This blog provides more information about SMPS. Thanks for sharing useful information.

ReplyDeleteTransformers for Industries

hello sir i wold like to know can we calculate turn from given primary inductance and turn ratio.

ReplyDeletei want to design 60 watt 12v@5a output on 85-265ac input. using lnk6766 linkswitch ic.

waiting for your reply

Hi Tahmid

ReplyDeleteThanks for all the explanation related to calculation for number of turns of primary and secondary. With this we can get 12V in output which is our requirement. Now our next requirement was to get output current of 2A. So for this what would be proper calculation for finding wire guage? I mean what wire guage I will have to use if I use core E42/21/15.

Thanks for such a nice and simple tutorial.

ReplyDeleteIn some circuits I have seen a inductor in series with the primary of the transformer. But dont know why it is used. Can you please give some info about that.

Saikat

Thanks for your work,please, I need EDT-5901 DATA

ReplyDeletehello tahmid I need 30V from 230V input. does calculation you provided is same for all type of core because I need to use EE13

ReplyDeletepls reply

Hi Tahmid. I had several transformers saved up, and thanks to your post, using the heat gun, I finally had the courage to dismantle them.

ReplyDeleteI had a strange experience with one of them. Its size very much resembles an EE22 core.

- Before removing the winding, I measured 1.8 mH, then counted 170 turns. If not mistaken, this gives me an Al value of 63. the low value confirmed the air gap. But I could not find any core with such a low Al...

- To be sure, I wound 30 turns on the form, and re-assembled the core. Now I measured 50 uH, giving 63.3 (nH/N^2).

Any idea where this value could come from?

Thanks in advance!

John

is the procedure same for a fly back converter?

ReplyDeleteHi Tahmid, I have been using your method in calculating a ferite transformer which was taken off an ATX 350 computer power supply. This same power supply is to be converted in a dual rail power supply of which the switcher will be a pre regulator and the final portion will be of linear type. I done some reverse engineering and saw what the factory did in terms of winding of the transformer and also measured the frequency prior to dismantling. What I found following your calculations was that the gauss was rather high and was found to be 3300. How ever the power supply was tested under load before i dismantled the transformer and was rather satisfactory. Shall I keep the same gauss as the factory or add more turns to the transformer to bring it down to your recommendations say at least down to 2000 gauss? Will the power of the transformer decrease if I do so now that the magnetic field will be reduced? Your reply will be much appreciated. Thanks.

ReplyDeleteHi...

ReplyDeletewere the ic frÃ©quency set i'm use IR2153 CT0.0022uf and resistor 3Ko or RT 6.2Ko

thank you

Hi ......

ReplyDeleteIt would be great tutorial if you had given real transformer picture...............

hello tahmid,

ReplyDeletethanx for ur detailed explanation, which was simplistic, wonderful and informative to read. there are certain doubts in my mind regarding to wire thickness to be used

plz xplain how to determine which gauge of wire to be used on secondary side and primary side of the transformer, i believe s fixed wire thickness will be say I load(max) current, but this will effect equivalent. inductance provided by the no. turns, so is there ny engineering calculations which covers the primary current, primary voltage, primary inductance and secondary current, secondary voltage and secondary inductance?

Assalamualikum.

ReplyDeletethe following request is required,

FERROXCUBE ETD59/31/22-3C90 FERRITE CORE, ETD, 3C90-from this core can we deliver an output around 2 kilo watt?

(thanks for your kind service for mankind and for easy derivation of formula for we electronic hobbyists and learners awaiting for your reply)

Thank you very much for this article, it has surely helped a lot of people who are interested in switching power supplies.

ReplyDeleteI'm trying to simulate this topology using LTSpice, but I need the inductance value for each coil. How do I calculate it?

Hi Tahmid

ReplyDeleteThanks for an interesting Blog, however you don't mention gapped cores this would help the ferrite not to go into saturation but obviously increase the amount of turns. Do you mention or consider this somewhere on this site, sorry if I missed it. I am in the middle of designing a split rail variable output half bridge and require about 300watts and an output of up to + - 110v this is for testing amplifiers.

Kind Regards Bill

Dear Tahmid, This has been a very helpful topic discussed but I see a missing link as far as enameled copper wire used. Finding quite a few tables of current handling for each gauge and also maximum frequency for each diameter used on the net. But these seem confusing at times. I was reverse engineering some cores from ATX power supplies and measuring operating frequency prior to dismantling and they seem to be off the specs. Regarding power handling and also on Bmax is often being around the 3000 mark and even more at times. Somehow these work satisfactorly

ReplyDelete1) Is the wire gauge regarding frequency is it really that important? At higher frequencies thinner gauge must be used and current handling is reduced. I have seen several designs on the net and everybody seem to have a different idea about this. What is your opinion?

Kind regards

Silvio

Hi Tahmid.

ReplyDeletemy plan was to build 1000w sain wave inverter using without centre tap transformer. After searching some information in the internet, I came across your blog and I found ETD49 transformer.

please I need some advice about witch transformer I can use for my project.

secondly I am really interested to join your blog. Please could you send some information

to my Email cs-topic-max@hotmail.com

kind regards Mohamed sheikh

Can I use the same formula to calculate LLC series resonant converter primary and secondary turns calculation?

ReplyDeleteRegards,

PonGalaban.

why dont you make video of all your project and note

ReplyDeleteHello,

ReplyDeleteThis method is not effective because it implies that you already know Ac .... The professional method consists in deducing the core according to the general characteristics of the supply to be produced, either by the surface products method or by other methods to deduce this fundamental data. There it is called rather of the hack not the state of the art.

Moreover this method does not take into account the skin effect, the current density nor the expansion of the spires in the Aw, the rise temp and so on. Don't use this method for a serious design.

Regards.

Nice Blog Post !

ReplyDeleteHi thamid..!!great job man..you have congratulations from greece..

ReplyDeleteFor a Ferrite toroid core, can these calculations be used??

ReplyDeleteSir iam beginer. (1). I have one doubt.can I use EI ferrite core for high frequency transformer winding instead EE ferrite core.

ReplyDelete(2).Have any problem if I select little more bigger size in EE core (more cross sectional area)instead of actual required cross sectional area?

May i have smps diagram for amp such as 35 0 35,10A and 50 0 50 10A,and how we select guge number of coil?

ReplyDeleteDear Tahmid,

ReplyDeletePlease write about the peak current and RMS current of the primary and secondary for the same design , what is the MOSFET current rating in this design? and how we can choose the mosfet rating for the same.

can this formula "Np = ((Vin(nom) * 10^8)/(4*fosc*Bm*Ae))" be used for single power mosfet based flyback xformer like 10amp-12v???

ReplyDeleteLate answer...no flybacks have their own equations which vary wether you use discontinues or continues conduction design and air gap selection.

DeletePlease can you assist I have built an offline half bridge converter using sg3525as my gate drive after testing individual it works fine but when I power all together there will be no output please what could be wrong

ReplyDeleteInput240v 2.3A/output110v 5A 550watts 53Hhz I need the design of the above mentioned specification. Hoping for a quick reply.

ReplyDeleteVery thanks

ReplyDeleteHi tahmid I hope u are fine...

ReplyDeleteI have done a lot of research on the internet, but your expressions are really clear and beautiful, I don't have any question marks in my head, but I want you to explain it with flyback topology formulas, do you have the opportunity to explain?

Or can you give little information?

I really need help with this...