EI375 steel laminate primary transformer calculation - Page 1
Forum section: electronics / beginners

author: gkmaia
author_rank: Frequent Contributor
timestamp: August 03, 2019, 10:52:08 pm
content: If there are any transformers experts here I would love to learn more about how to wind the primary of my transformer to suit 220v without saturation.The basic thing is it has to be the EI357 laminates/coil former.If I could get some help to fill these values and understand them that would be great.I am using to calculate my primaryhttp://dicks-website.eu/coilcalculator/index.htmlMaximum magnetic flux density in core: Bmax = 1.3.I do not know the BSAT so I am going for an average. Is it ok?Effective cross section area of one core: Ae = 1900mm2The core has 10/10/19mmEffective permeability of the core: μe =Dont know how to get those values bellowμr = relative permeability of the core material.le = effective length of the magnetic path in the coreg = length of the air gap (measured in same unit as le)Inductance factor for one core: AL = 230 nH/n²Value has been collected from TECH11 pdf attached.

author: Jwillis
author_rank: Super Contributor
timestamp: August 04, 2019, 03:45:29 am
content: BSAT and Effective permeability are dependent on core material.Core material greatly determines the maximum flux density of the magnetic field.

author: gkmaia
author_rank: Frequent Contributor
timestamp: August 04, 2019, 03:51:40 am
content: I believe core material is silicon steel. But that is all I know.

author: Jwillis
author_rank: Super Contributor
timestamp: August 04, 2019, 05:07:59 am
content: The relative permeability is around 5000–10,000 for silicon steel and 1.5 Tesla at 50Hz.First off we need the parameters of you transformerPower ratingVoltage levels (primary and secondary)Currents on both sidesPrimary and secondary coils wire diameter/sizeIron Core areaNumbers of turns (primary and secondary)Calculate core area Ai=1/4.44fBmTeAi= area of coreF= operating frequencyBm= magnetic flux densityTe= turns per voltsSo we can say that Te=1/4.44fBmAi assuming its single phase . 3 phase would omit the 4.44Total number of turns = turns per volts x primary side voltageVA rating = (Vs x I)/0.8Primary current = I1= VA / VpSo to prevent over saturation I1 = VA/ (efficiency x Vp) Typical efficiency of EI transformers is around 80-95%Copper has a current density approximately 2.3 A per mm squareCurrent Density= I/A I being the primary amperage and A being the copper area. so Primary conductor size is A1=Primary Amps/Copper Current Density =minimum size of conductor in mm approximately.edit: this may be wrong

author: gkmaia
author_rank: Frequent Contributor
timestamp: August 04, 2019, 06:34:30 am
content: Quote from: Jwillis on August 04, 2019, 05:07:59 amThanks so much Jwillis. I really appreciate your help. Your answers are bellow.Power rating2W @ 110vVoltage levels (primary and secondary)Primary 110VSecondary A 2.9VSecondary B 15VCurrents on both sidesPrimary current 0.018ANot 100% precise about the secondaries so I will base it on primary ratio.Secondary A 0.59ASecondary B 0.13APrimary and secondary coils wire diameter/sizeSecondaries 32AWGPrimary 38 AWGIron Core area10mmx10mmx19mm1900mm2F= operating frequency60HzMaximum magnetic flux density in core: Bmax1.5 TeslaNumbers of turns (primary and secondary)I will have to unwind it to get these values. But before doing that would be good to know where are we heading so I can make sure I understand what our goals are.

author: Jwillis
author_rank: Super Contributor
timestamp: August 04, 2019, 08:01:13 am
content: You have 2 secondary currents so you will need to add those together to get the total power rating required on the primary side.(Vs * I)/0.8S1 =(2.9 * 0.59)/0.8 =1.711VaS2=(15*0.13)/0.8=1.56VaYou should calculate the primary voltage at the standard output voltage of a typical outlet which is 120VThat's a total of 3.271Va so primary current will Va/Vp =3.271/120=0.027AmpsCopper area of 38AWG is 0.0080mm square and have 2.16 ohms per meter maximum current rating is 0.13ACore area is 10 *10 = 100mm squared=0.0001m squared .Height of core isn't important at this point.Te = 1/4.44*60*1.5*0.0001=25 turns /volttotal turns is 120V*25=3000 turnsHeight of core is 19mm .38AWG wire has a diameter of 0.1007mm which allows around 188 turns per winding layer.This gives approximately 15-16 layers of winding.1.5mm thick primary winding.This where it can get complex so lets use an online calculator so find the length of wire.https://coil32.net/online-calculators/rectangular-multilayer-inductor-calculator.htmlI get approximately 128.211 meters of wire with a DC resistance of 432.42Ohms.

author: gkmaia
author_rank: Frequent Contributor
timestamp: August 04, 2019, 08:07:34 am
content: Quote from: Jwillis on August 04, 2019, 08:01:13 amWOW! Thanks a lot for that! I will study your reply as there is heaps going on here and try to learn as much as I can. Will get back you with some more questions later if that is fine.

author: soldar
author_rank: Super Contributor
timestamp: August 04, 2019, 08:41:11 am
content: Quote from: Jwillis on August 04, 2019, 08:01:13 am120? 230?

author: gkmaia
author_rank: Frequent Contributor
timestamp: August 04, 2019, 08:58:44 am
content: Thanks so much again! It is all making much more sense now.In the online calculator please look at the attache image. How did you define de required inductance?On my table, please look attached.A - Secondary Va total. I could not generate the same results as you did by multiplying V*I as you did.B - Max winding layers. where did you get this value from "1.5mm thick primary winding." ?

author: Jwillis
author_rank: Super Contributor
timestamp: August 04, 2019, 09:07:23 am
content: Happy to help.Calculating the total length of wire is done by taking the mean thickness of the winding which include the wire and bobbin. The bobbin is 10mm and the thickness of 7.5 windings.7.5 * diameter is the wire.= 7.5*0.1007mm=0.755mm.Or half the total thickness.So the thickness of the total was about 10mm + (2*0.755mm)=11.51 mm = approximately 46.04mm per turn or 0.04604m per turn. This gives us 0.04604 * 3000 turns Or 138.12 m of wire.Remember these are approximations so hence the difference between the last post and this one.Also the number of windings can be approximate (within reason).A few extra or a few short won't make much difference. A few extra is better.The field will just be slightly stronger.If you don't want invest in a coil winder then mark each layer as you go so if you loose count ,your not completely humped. Tape your primary so to keep it separate from your secondaries.If you want to calculate the impedance of your coil then use XL=2*Pi*f*L where XL is the impedance ,Pi is 3.14. f is frequency and L is the inductance.Some of my calculations may be off for example the total DC Ohms may be closer to 150 ohms.I may have worked in the incorrect values for the gauge of wire.Recheck the wire gauge dimensions and recalculate for turns per layer ,number of layers and thickness of total layers .38 AWG wire should be 0.1524mm not 0.1007mm.

author: Jwillis
author_rank: Super Contributor
timestamp: August 04, 2019, 07:55:35 pm
content: Sorry I was trying to calculate math at 3 in the morning .Not a good practice at all.Please recalculateAnyway the primary thickness is defined as the stacking of the windings. If you know the diameter of the wire and the number of stacks you can work out the thickness of the winding stack.You can calculate the Self Inductance of a Coil asL=μο(N squared*core area/l)L is inductance in Henriesμο is the Permeability of Free Space (4*Pi*10-7)N is the Number of turnsA is the Inner Core Areal is the length of the Coil in metresThe maximum amperage draw a transformer can safely handle is determined by the loads connected to its secondary winding.Since you have to windings on your secondary you need to add the amperage of the 2 secondaries. So you get 3.271Va at the secondary.Power primary is equal to power secondary Vp*Ip= Vs*Is . The current drawn at the secondary is proportional to the current drawn at the primary you get Va/Vp =3.271/120=0.027Amps =3.24VaYour core EI357 should be able to handle up to 3.5 to 4 Va if it indeed 35 mm across the outside.

author: gkmaia
author_rank: Frequent Contributor
timestamp: August 04, 2019, 10:06:02 pm
content: Thanks for that!I just want to narrow down to the logic on primary power requirements so I can get the best wire gauge.The main values I believe we are looking for are:- Primary turns: to align the core specifications to input voltage and avoid the thing to saturate and blow- PA: to define what wire gauge will support the current draw on the primary so we don't get the wire cookedThose are the 2 key things.If you look at the spreadsheet attached I replicated all your calculations on excel and apparently the PA (as per datasheet) matches the PA (based on secondary) perfectly. Just want to make sure I did those right.

author: Jwillis
author_rank: Super Contributor
timestamp: August 04, 2019, 11:56:44 pm
content: Some of your parameters have changed like the primary voltage. This will change all the math. Also recheck your wire gauge diameter. AWG 38 has a diameter 0.1524mm not 0.1007mm. That was my mistake. It wont make any difference to the total number of windings required , but will change the number of windings per stack and the total stack thickness.Also could you post the data sheet of the core you are using.I'm guessing on the dimensions based on the closest example I have.

author: gkmaia
author_rank: Frequent Contributor
timestamp: August 05, 2019, 01:12:13 am
content: Quote from: Jwillis on August 04, 2019, 11:56:44 pmYes!!! I wanted to make a live table I could use with other cores/power/voltages.So if you look at the table bellow you will see that I am using your calculations to generate the results I need to convert my primary from 120v to 240v using the same core.That was the whole idea as I have one of those Fluke 8012 multimeters that is 120 and has not voltage switch. I have to rewind the transformer to get it working on 240.

author: Jwillis
author_rank: Super Contributor
timestamp: August 05, 2019, 08:45:42 am
content: Oh I see. Since New Zealand is 50Hz ,the core area may not be large enough.50Hz core tend to be larger than 60Hz cores for the same power ratings .Does your data sheet mention 50Hz?Transformers rated at 60Hz should not be used on a 50Hz supply due to higher losses and core saturation, and the resultant higher temperature rise.Transformers rated for 50Hz, however, can be operated on a 60Hz supply without any effects.

author: gkmaia
author_rank: Frequent Contributor
timestamp: August 05, 2019, 08:57:02 am
content: The data sheet is not really clear but seems both versions 120 and 240 can handle 50hz or 60hz.But as per calculations, being 50hz means 1000 more turns. Total of 7000 turns instead of 6000By calculations the original primary should have 3000 turns. If it is really 0.127 thick the coil former can hold up to 3400 turns and by the looks of it seems it is nearly there. If I modify it to 235v and use 0.07874 I can fit around 9000 turns, but I only need 7000.How can you define if the size of the core may can handle 50hz?The other thing I find interesting then from what you mentioned. My other 8012A Fluke runs with a step down transformer mounted inside it and it delivers 120@50hz and I have been using it extensively for quite a while.

author: gkmaia
author_rank: Frequent Contributor
timestamp: August 05, 2019, 09:03:53 pm
content: Jwillis I got a question about the laminates.I got to isolate the steel laminates from each other. I've seen people suggesting to use clear spray paint.But in fact any enamel paint will do the job. I think the only issue may be if the paint is metallic it will contain aluminium flakes and they are conductive. I imagine any paint that is not metallic will do.Do you see any problem with that?

author: Jwillis
author_rank: Super Contributor
timestamp: August 05, 2019, 09:39:13 pm
content: Use a clear coat.Then your sure that theirs no other components in the paint that may cause issues.The total turns turns works out to be 7207 240V at 50Hz

author: gkmaia
author_rank: Frequent Contributor
timestamp: August 05, 2019, 09:58:14 pm
content: Yes, will do.Exactly. My excel gave same 7207 240 @ 50hz. Nice to see my excel works!Thanks mate.I've been trying to understand transformer windings for months and in a few posts you made everything easy & clear! Exactly what I needed!If anyone out there need our calculations the final excel is attached.

author: gkmaia
author_rank: Frequent Contributor
timestamp: August 08, 2019, 06:08:12 am
content: Quote from: Jwillis on August 05, 2019, 09:39:13 pmHey JWillis, finally got time to unwind the transformer. I got 4670 turns on the primary.Our initial calculation gave 25 turns per volt.But to run 120v with 4670 turns we would need 38.90 turns per volt.I know more turns is better than less turns. Is there a reason why Fluke did more turns than necessary for the core?

author: The Electrician
author_rank: Frequent Contributor
timestamp: August 08, 2019, 08:01:58 am
content: Quote from: gkmaia on August 05, 2019, 01:12:13 amIt's too bad you didn't say from the beginning that you are simply rewinding an existing transformer to use 220V that uses 120V as is.All you have to do is unwind the 120V primary (which you have apparently done) and count the turns. Then you rewind the primary with twice as many turns of wire having 1/2 the cross-sectional area. For example, if the 120V primary is wound with 36 gauge wire (measure it with a micrometer), then you rewind with twice the turns of 39 gauge wire. Wire cross-sectional area halves with an increase of 3 gauge sizes. Wire gauges that are odd numbers, like 39, can be difficult to find.If the transformer was conservatively designed and doesn't get very hot in use with 120V, you might get away with using a wire gauge 4 times larger--40 gauge in this example.It appears that Fluke designed the transformer to be usable on 50 Hz as well as 60 Hz, and if you just double the primary turns, that will still be the case. That's probably why they used (apparently) more turns than your calculations say would be necessary.You don't need to know the permeability of the laminations, or the turns per volt, etc.Don't try to modify the surface of the laminations by spraying them with clear coat. If you do that, you won't be able to get all the laminations back in place.Transformer steel laminations come from the factory with an insulating coating. That insulation can be as simple as steam oxide, or better, a coating of magnesium silicate (Carlite):https://www.totalmateria.com/page.aspx?ID=CheckArticle&site=kts&NM=101Also:https://www.aksteel.com/sites/default/files/2018-01/carlite201208_1.pdf'If you are careful not to allow the laminations to scrape against one another much when you're disassembling the transformer, the insulating coating, whatever it is, will probably be good enough to just reassemble the laminations without doing anything more to provide an insulating coating. The only effect of a little degradation of the existing coating will be a slight increase in core loss.

author: gkmaia
author_rank: Frequent Contributor
timestamp: August 08, 2019, 08:37:43 am
content: The Electrician, thanks for your answer.Not that bad. I knew that I had to double the windings but I also wanted for a long time to understand better how cores are calculated. Although there is still lots to learn the time JWillis toke to share his knowledge was incredible.Lots os things I did not know before now make much more sense.Also knowing the why is actually better than just doing it.

author: Jwillis
author_rank: Super Contributor
timestamp: August 08, 2019, 08:58:35 pm
content: The Electrician is correct about not really needing to coat the individual laminates. But if you do ,Use the paint sparingly.I sometimes will use a lite wash of epoxy mixed with denatured alcohol as I do a final assembly just to keep the laminates from vibrating. Then a brush over the outside of the core for aesthetics and vibration control. But this can be messy. There is proper transformer epoxy but It's to expensive for a simple hobbyist like me.More turns is not a bad thing because of the increased Magnetic filed .https://en.wikipedia.org/wiki/InductanceExplains " inductance describes the tendency of an electrical conductor, such as coil, to oppose a change in the electric current through it. When an electric current flows through a conductor, it creates a magnetic field around that conductor. A changing current, in turn, creates a changing magnetic field. From Faraday's law of induction, any change in total magnetic field (magnetic flux) through a circuit induces an electromotive force (voltage) across that circuit, a phenomenon known as electromagnetic induction. From Lenz's law, this induced voltage, or "back EMF" in a circuit, will be in a direction so as to oppose the change in current which created it. So changes in current through a conductor will react back on the conductor itself through its magnetic field, creating a reverse voltage which will oppose any change to the current. Inductance,"So having an increased magnetic field will oppose any change in current more.This keeps the transformer from heating up a little better.Learning can be fun and Now you know more than most.I apologize for my crappy math but the formulas are correct. Just plug in the known variables and you can create any transformer you want.

author: The Electrician
author_rank: Frequent Contributor
timestamp: August 09, 2019, 07:11:24 am
content: Quote from: Jwillis on August 08, 2019, 08:58:35 pmMore turns on the primary of a 50/60 Hz transformer does not give more flux in the core for the same applied primary voltage.In another thread:https://www.eevblog.com/forum/beginners/tap-on-secondary-of-transformer/msg2377359/#msg2377359you give the standard formula: N=(10000 * Vp) / (4.44 * f * B * A)Doing some algebra we get: 4.44 * f * B * A = (10000 * Vp)/NWhich becomes B = (10000 * Vp)/(N * 4.44 * f * A), so if N increases, B decreases.This is because the primary of a transformer is typically excited by a voltage source rather than a current source. In that case, we use Faraday's law to determine the flux in the core. More turns means fewer volt-seconds applied to the core, and thus, less flux results. With a constant voltage applied, more turns means higher inductance, which means higher reactance, which means less exciting current. Since the inductance increases as N^2, reactance increases as N^2. So if you double the primary turns, the inductance is 4 times larger and the exciting current is 1/4 as much. Doubling the primary turns would double the ampere-turns if the current were the same as before increasing the primary turns, but the current is only 1/4 as much, so the net result is half the flux in the core.If the primary were excited by a constant current, then more turns would mean more ampere-turns, hence more flux.

author: soldar
author_rank: Super Contributor
timestamp: August 09, 2019, 07:58:48 am
content: I have not calculated and designed a transformer in ages but back in the day I did quite a few and I believe this is the kind of thing that starts out looking quite complex and as you practice you realize it is much simpler than it might look at first sight.I would start by selecting a standard core and bobbin by rough calculation or even by eye.Then I would calculate the wire windings necessary for that core. If they fit in the bobbin I was done. If not then select a slightly larger core and bobbin and start over again calculating the wiring. It was quite simple and mechanical.

author: gkmaia
author_rank: Frequent Contributor
timestamp: August 26, 2019, 01:44:53 am
content: Finally received the wires and put it back together.Good thing is I am off just by 0.2v at the 8v range when compared to the old 110v primary.By our calculations we had to double the 4670 turns to 9340. But that was delivering a bit more than expected. I had to add extra 500 turns to bring it down a bit.The clear coat on just one side of the laminates went fine. Fit not to tight neither loose. Just about right.The only pain is to deal with 42 gauge wire... too thin to manipulate, to remove the enamel and weld.It is now measuring a resistor for more than an hour. Still no magic smoke!!! Transformer is cold and reading stable.I guess that is a winner!

author: gkmaia
author_rank: Frequent Contributor
timestamp: September 24, 2019, 12:48:27 am
content: Quote from: Jwillis on August 08, 2019, 08:58:35 pmHey Jwillis I am working on a new transformer project. Not sure if you have some spare time but I would be great if you could give me some tips.https://www.eevblog.com/forum/repair/tektronix-t910-oscilloscope-high-voltage-transformer-issue/msg2702678/#msg2702678