Help with Pulse Transformer Design - Page 1
Forum section: electronics / projects

author: KurtK
author_rank: Contributor
timestamp: August 22, 2019, 10:09:25 am
content: Hi All.This is my first post so please be gentle.I'm designing a special purpose pulse transformer, and since I really can't find a suitable guides for that on the internet I have to experiment to find find the most efficient design. And I therefore need to know the efficiency of each design so that I can compare them.Here is my first attempt, it's an simple air-core design, how do I calculate the efficiency of this transformer? My idea is to multiply current with voltage for the whole pulse duration to obtain instantaneous power, and from that calculate average power. The divide primary average power with secondary average power to get a measure of efficiency, would that be correct way to do it?CheersKK

author: MagicSmoker
author_rank: Super Contributor
timestamp: August 22, 2019, 11:56:19 am
content: Quote from: KurtK on August 22, 2019, 10:09:25 amA good pulse transformer has very high magnetizing inductance with the lowest possible distributed capacitance, which tend to be conflicting requirements because the former implies lots of turns while the latter implies minimizing the turns. The usual approach, then, is to use a high permeability core material that is much larger than necessary from a flux swing standpoint to get both a high magnetizing inductance with few turns.Neither of which can be accomplished with an air core.

author: KurtK
author_rank: Contributor
timestamp: August 22, 2019, 12:44:11 pm
content: Hi MagicSmoker.Yes I have read exactly that in some technical articles, and I do have some ferrite cores that I would like to try, but unfortunately I don't have any datasheet for them, so I will have to experiment. And if I have to experiment I need a way to measure/calculate efficiency otherwise I will be experimenting in the dark.Can you tell me if my idea of calculating efficiency is correct? Or is there another way?

author: MagicSmoker
author_rank: Super Contributor
timestamp: August 22, 2019, 02:54:14 pm
content: Efficiency is not usually a top design consideration of a pulse transformer as it generally handles very little power, so I am wondering if there might be a "lost in translation" issue here?That said, no, your method won't give you an accurate measurement of efficiency because the energy stored in the magnetizing inductance of the primary is returned back to the supply, minus some losses in the winding resistance and from driving the core (if present), while the actual load current drawn by the secondary is reflected back to the primary and incurs losses in the resistance of both windings. As a result, a pulse transformer without any kind of ferromagnetic core will either have a very high magnetizing current, which incurs a high I^2R loss, or the windings will have an excessive number of turns to minimize magnetizing current, and still incur a high I^2R loss... As us Yanks say, "you can't win for losing."Separating out all of these loss contributions is rather tedious and usually only worth the effort for a transformer that handles significant power which, again, doesn't typically apply to a pulse transformer.

author: KurtK
author_rank: Contributor
timestamp: August 22, 2019, 05:17:46 pm
content: Quote from: MagicSmoker on August 22, 2019, 02:54:14 pmPerhaps "lost in terminology". I would have preferred to just to call it a (step-down) "transformer" but that usually implies 50HZ sine AC. So regardless of the term used I just want to know how to measure/calculate efficiency of the "transformer" at hand. Obviously I want it to be as efficient as possible.Quote from: MagicSmoker on August 22, 2019, 02:54:14 pmI don't quite understand that. In a (50HZ since AC) transformer the efficiency is calculated as primary power / secondary power * 100, and power is calculated as Vrms * Irms * Power factor. So why couldn't I do something similar? I mean in a (50HZ since AC) transformer the energy stored in the magnetizing inductance of the primary is also returned back to the supply etc.? So if I use the entire pulse duration as basis for the calculation, it must include in theory everything including losses you described?Quote from: MagicSmoker on August 22, 2019, 02:54:14 pmYou should come live with us then, here everybody is a winner - even the loosers. ;-)Quote from: MagicSmoker on August 22, 2019, 02:54:14 pmThe application is not a typical pulse transformer, so could we perhaps call it a "power pulse transformer"? In any case, how do I by experimenting with cores and windings and maybe some assistance from general math, obtain a efficient say 90% @ 34KHz "power pulse transformer"?

author: MagicSmoker
author_rank: Super Contributor
timestamp: August 22, 2019, 06:11:16 pm
content: Quote from: KurtK on August 22, 2019, 05:17:46 pmRight, I suspected as much. You are talking about a switchmode power supply transformer. "Pulse transformer" typically refers to small types used to deliver the trigger energy to a MOSFET, thyristor, xenon flash lamp, etc.Quote from: KurtK on August 22, 2019, 05:17:46 pmBut measuring the *real* power is not what you proposed in your first post, rather, it was separately measuring the current and voltage of the primary and multiplying them together, which would give you apparent power. If you have a true RMS / vector wattmeter that can measure both W and VA then you can, indeed, divide the W measured on the secondary by the W measured on the primary to determine efficiency, and this will include all loss mechanisms automatically. Simply measuring the average voltage and current with a DMM or scope won't cut it.Quote from: KurtK on August 22, 2019, 05:17:46 pmToo cold!Quote from: KurtK on August 22, 2019, 05:17:46 pmOh, you should be able to get to 98% at 34kHz fairly easily with almost any decent quality ferrite. This is a rather low frequency by today's standards so even an "old" ferrite formulation like N27 or 3C81 will let you use the entire BH loop (which is to say, you will be saturation limited, rather than core-loss limited) and there won't be too many problems with skin and proximity effect, either. Other tips are to interleave the primary and secondary (e.g.: 1/2 Pri - Sec - 1/2 Pri), use wire of sufficient cross-sectional area to keep current density below 4A/mm^2 and make sure the wire diameter is not too much bigger than about twice the skin depth.

author: KurtK
author_rank: Contributor
timestamp: August 22, 2019, 06:37:20 pm
content: Quote from: MagicSmoker on August 22, 2019, 06:11:16 pmYES, "SMPS transformer" is the magic term then, let's go with that.Quote from: MagicSmoker on August 22, 2019, 06:11:16 pmWhat I thought I proposed was to multiply instantaneous voltage with instantaneous current, and by that I would obtain instantaneous power, and then calculate the average power by summing instantaneous power and divide that by number of samples and then I would obtain average power? Wouldn't that be the correct way to do it?Quote from: MagicSmoker on August 22, 2019, 06:11:16 pmThanks for the tips, but this one I don't think that would fare well, due to the high voltages the windings would arc over.Ohh - I see i forgot to mention that in the graph in my post #1 the X-axis is time in nS

author: Conrad Hoffman
author_rank: Super Contributor
timestamp: August 22, 2019, 07:08:55 pm
content: Never waste time with unknown ferrite (or other) cores. If you can ID them by some tests with a trial winding and looking up any color codes or part numbers, great, but usually you can't. Cores are cheap so buy what you need.

author: KurtK
author_rank: Contributor
timestamp: August 22, 2019, 08:43:40 pm
content: Quote from: Conrad Hoffman on August 22, 2019, 07:08:55 pmI would be happy if I could do that, but I really can't find any ferrite cores that can handle 50KW average power. Are there some manufacturers that specialize in this size?

author: MagicSmoker
author_rank: Super Contributor
timestamp: August 22, 2019, 09:21:14 pm
content: Quote from: KurtK on August 22, 2019, 06:37:20 pmNo, that gives you apparent power as I keep saying. You need to adjust for any phase angle displacement of current with respect to voltage to get true power. This applies whether the transformer is driven by a sine wave from the mains or a square wave from an H-bridge.Quote from: KurtK on August 22, 2019, 06:37:20 pmIt is immensely frustrating when the OP does not include important details... like that the secondary is high voltage or the average power rating is 50kW.No offense - seriously - but if you need a 50kW high voltage transformer then there is no shame in getting it designed for you by a company that specializes in magnetic component design and manufacturing like, say, Spang, which is a division of the US business, Magnetics, Inc.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: August 22, 2019, 09:31:09 pm
content: Okay, what are youreallydoing first of all? Because you literally gave no numbers in your post, nor a description of the figures; and we can design transformers all day and all night but only if we have numbers to put in!Tim

author: tautech
author_rank: Super Contributor
timestamp: August 22, 2019, 10:08:14 pm
content: Quote from: T3sl4co1l on August 22, 2019, 09:31:09 pmYep.For the OP, the first 2 hits from Google on 'smps transformer design'https://wcmagnetics.com/wp-content/uploads/2015/02/appnote11.28.10.pdfhttps://talema.com/smps-transformer-design

author: KurtK
author_rank: Contributor
timestamp: August 23, 2019, 08:51:09 am
content: Quote from: T3sl4co1l on August 22, 2019, 09:31:09 pmLet me address your question first, so there is no doubt about what Ireallywant to do:I want to step-down what present it self on the primary side in the most efficient way possible, that is within my means.Quote from: T3sl4co1l on August 22, 2019, 09:31:09 pmThe first figure is the circuit at hand with my transformer in the middle. The second is a graph of measurement traces of current and voltage on primary and secondary side and there is a legend in the top right corner so that the traces can be distinguished from each other.Quote from: T3sl4co1l on August 22, 2019, 09:31:09 pmWell, so can I, but in this case I don't have all the numbers and I neither can I obtain all the numbers that would fit nicely in a plug-n-play formula. There is a degree of uncertainly due to unobtainable numbers and information, and I personally don't mind that, it's workable for me.

author: KurtK
author_rank: Contributor
timestamp: August 23, 2019, 09:21:19 am
content: Quote from: MagicSmoker on August 22, 2019, 09:21:14 pmI know you keep saying that, but I still don't understand it. By "instantaneous" I mean "in that given instant" and since the traces in graph are made up of samples at a given instant, I can multiply voltage with current at a given instant and obtain power at a given instant. I do not have to adjust the product of voltage and current for phase displacement because it is already present at that given instant. That is what I think, but I could obviously be wrong.Quote from: MagicSmoker on August 22, 2019, 09:21:14 pmAnd in this case the transformer is driven by a exponential decaying pulse.Quote from: MagicSmoker on August 22, 2019, 09:21:14 pmI sure sense your frustration all the way across the pond. If you look at the graph in my OP you can see the voltages and currents involved. And I couldn't very well calculate average power before I there were some agreement to how it should be done.Also to get to the bottom of something you need to have a dialogue, just like we have now. I can't just present a massive wall of information to being with, that would be highly confusing for the reader as most of it would probably be irrelevant for this topic.Quote from: MagicSmoker on August 22, 2019, 09:21:14 pmIf it was only a question of shame, I would certainly do that, in fact I would outsource everything.

author: KurtK
author_rank: Contributor
timestamp: August 23, 2019, 09:40:10 am
content: Quote from: tautech on August 22, 2019, 10:08:14 pmThank you very much for googling that one for me. But as you can see in thread, MagicSmoker and I agreed to call it "SMPS transformer" because that was what MagicSmoker felt was the most appropriate term and I didn't mind that. But that does not infer that I'm building a SMPS although it may have some elements of a SMPS. As I told MagicSmoker I would have preferred to call it a "step-down transformer", but then probably some helpful soul would google 50HZ sine AC transformer design for me, and I really wanted to avoid that.I get the distinct feeling that this post is starting to go sideways, so maybe I should just bow out now and thank all for their kind help so far, what do you think? Would that be a wise decision of me?

author: Berni
author_rank: Super Contributor
timestamp: August 23, 2019, 11:14:26 am
content: Quote from: KurtK on August 23, 2019, 09:21:19 amThe graph shows no units so this could be kA or nA of current flowing trough it. If we assume they are always base units of A and V then the powers involved are far above 50kW.Additionally if the Y axis is assumed to have only base units then one would assume the X axis also to be in seconds and in that case the pulse duration of the first hump is 15000 seconds long (4 hours) so that would require a transformer designed to operate at 66 microHertz. Such a transformer would likely be the size of a continent at these powers. So this makes one assume the graph is not using base units.So can you explain in what way it was clear from the graph that this is a huge 50kW pulse transformer?I feel like such a thing should be mentioned in the first 5 sentences of the post or even the title. People on forums can only be as helpful to you as much as you are helpful to them in providing the information about your situation.

author: MagicSmoker
author_rank: Super Contributor
timestamp: August 23, 2019, 11:20:42 am
content: Quote from: KurtK on August 23, 2019, 09:21:19 amA transformer primary exhibits both magnetizing and leakage inductances and current lags voltage in an inductor, right?It doesn't matter what the voltage waveform applied to the primary of a transformer looks like, the current through it will be a time (phase angle) delayed version of it*.* - EDIT - and integrated... forgot that part.In a lossless transformer with finite magnetizing inductance and the secondary unloaded, the only current flowing in the primary will be for magnetizing the core and this is a purely inductive current so it will be phase shifted by 90 degrees from the voltage. This is a measurable current and the voltage is of course also measurable, but if you simply multiplied those two parameters together without accounting for the phase shift you get apparent power, VA, not real power, W. To find real power you have to multiple VA by the cosine of the phase angle displacement which in this case is 0.If this "almost ideal" transformer is loaded on the secondary, the primary current will increase - because of the current reflected from the secondary through the turns ratio - and the phase shift will decrease, because the magnetizing current with its 90 degree phase shift will become a progressively smaller fraction of the total primary current. In a real transformer there are both iron and copper losses, of course, and the iron losses will be fixed for a given applied primary volt*seconds while the copper losses will vary with the square of load current. Hence why I said that actually measuring the efficiency of a transformer is tedious and not typically worth the bother.Quote from: KurtK on August 23, 2019, 09:21:19 amDoesn't matter.Quote from: KurtK on August 23, 2019, 09:21:19 amPictures accompanying a post should help illustrate what is written, not substitute for it. Look at all the forums - including this one - where images were hosted on Photobucket or the like and are now no longer available. IMO, it's better to present too much information in the OP than to trickle out details in subsequent replies.Quote from: KurtK on August 23, 2019, 09:21:19 amYou may have misinterpreted me. I could certainly design your transformer but I almost certainly couldn't make it myself - not even a prototype. Thus I would specify the core size and material, number of primary and secondary turns, wire gauges for each, layer stackup and other pertinent details like what UL or EN classification it needs to meet, vacuum impregnation, etc., and then let the transformer manufacturer worry about the rest... like how to wind thick ribbons of copper around the coil former or achieve a 10kV or so functional insulation level.

author: KurtK
author_rank: Contributor
timestamp: August 23, 2019, 12:55:40 pm
content: Quote from: Berni on August 23, 2019, 11:14:26 amThe title of the graph is "voltage and current" usually that means "unless otherwise noted you are dealing with the base units of Volts and Amperes. If you had doubts about that, all you have to do is ask, I'm right here. Yes I could have written that, yes I could have written many things - but I didn't, so I'm sorry.And regarding the X-axis I already made a public apology that I forgot to mention it was nano-seconds, and I will happily apologize again to you; I'm sorry - I forgot, sometimes my short-term memory just doesn't work properly.

author: Berni
author_rank: Super Contributor
timestamp: August 23, 2019, 01:08:30 pm
content: Quote from: KurtK on August 23, 2019, 12:55:40 pmYeah, its best to make it clear you want a massive power transformer to begin with.According to the graph this involves 500kW peak of instantaneous input power into the transformer. The sort of transformers that will handle this sort of power efficiently are likely going to the the size of a minifridge and too heavy for a single person to move. That being said air core is used for some of these specialty high pulse power transformers since it might indeed be difficult to find the massive ferrite core for one of these.

author: KurtK
author_rank: Contributor
timestamp: August 23, 2019, 01:19:35 pm
content: Quote from: MagicSmoker on August 23, 2019, 11:20:42 amI'm wondering, could it be that actually are saying the same thing but are talking past each other because you think I'm talking about only multiplying three integers and expecting to the result to be average power? Because that is not what I mean, I'm talking about using numerical methods, also know as Digital Signal Processing?Quote from: MagicSmoker on August 23, 2019, 11:20:42 amIf I had a stringently defined project with all parameters clearly defined I would have done that, yes. I believe you yanks have a suitable phrase for that "different folks, different strokes" - I can't make everyone satisfied no matter how much I try. But your preferences are duly noted and I will try to accommodate you.Quote from: MagicSmoker on August 23, 2019, 11:20:42 amGreat, if you can design the transformer, I can certainly make it or have it made and then we could close this thread, seems like win/win. We are talking about one transformer only for a prototype device so the only standard it has to meet for now is "no arcing", and if it can sustain a few minutes of operation without core meltdown that would be fine too.

author: KurtK
author_rank: Contributor
timestamp: August 23, 2019, 01:22:25 pm
content: Quote from: Berni on August 23, 2019, 01:08:30 pmAre you sure about that? In the legend in the top right corner it says "current * 100" which usually means "the current you are seeing is scaled by a factor 100", so before you calculate anything with that current you should divide it with 100 first.Edit: You are right, I just calculated - there is a 500kW peak of instantaneous input power.

author: ArthurDent
author_rank: Super Contributor
timestamp: August 23, 2019, 02:48:51 pm
content: So this thread starts with a request for help in designing a “pulse” transformer with little pertinent information given. After multiple requests for more information plus some good suggestions from other posters it turns out that the OP isn’t open to any input and the “pulse” transformer has morphed into a 50KW monster.This appears to be an example of the Dunning-Kruger effect at work and the OP should hire someone who has expertise in the field to evaluate and design whatever the OP thinks they need instead of taking the chance of injury or worse. If this is for a business, you want it designed correctly. If it isn’t for a business but just for the OP, you have to ask what the hell they would need such a transformer for.

author: KurtK
author_rank: Contributor
timestamp: August 23, 2019, 03:18:08 pm
content: Quote from: ArthurDent on August 23, 2019, 02:48:51 pmSo you are spooked by the mentioning of 50KW, and for that I deserve a personal attack, thank you very much. Well I guess I had it coming and can only thank myself, I must have exceed some maximum allowed avereage power for posting on this forum that I wasn't aware of.To the rest of the contributes in this post, I tank you for your input.Cheers!

author: patrick1
author_rank: Contributor
timestamp: August 23, 2019, 03:35:28 pm
content: Funny i just fixed a bunch of analog panel meters. - this would likley be your best way too measure efficiency in my pamphletget yourself a sweep signal generator... and let it run one, till you find your best numbers... even if you decide too take another route, - my thinking is, you can never have too many spare panel meters... and @ $3~ each... dosnt hurt too bad.- this is a video i made yesterday about efficiency in transformers. you may fund it interesting ;=D

author: MagicSmoker
author_rank: Super Contributor
timestamp: August 23, 2019, 05:10:37 pm
content: Quote from: KurtK on August 23, 2019, 01:19:35 pmCan you be any more specific about what you are trying to do either here or in a PM? According to the pictures in the OP and other comments by you, the turns ratio is 5.78:1 and the waveform applied to the primary is an exponentially decaying sinusoid, or "ring down" waveform, with a half-cycle period of ~9us, a peak voltage of ~9kV, and a peak current of ~75A; correct?

author: Berni
author_rank: Super Contributor
timestamp: August 23, 2019, 06:05:01 pm
content: Quote from: KurtK on August 23, 2019, 03:18:08 pmI'm pretty sure there is no such power limit on this forum.Just that saying its a pulse transformer generally means its a transformer designed for wide bandwidth and typically high speed operation. These are typically small low power transformers such as used for signal isolation (Ethernet, SPDIF), driving floating transistors, baluns etc... These transformers are typically the size of a sugar cube and don't handle any sort of high power levels. I know its a bit odd to call these pulse transformers but i suppose pulses are wide bandwidth when you look at there frequency content.So describing it as that gave a large majority of users trying to help the completely wrong idea of what you want, hence leading to all round confusion at conflicting specifications.This is the sort of specialized transformer you would find in some sort of physics experiment rig to produce some weird high energy effect or driving some ridiculously high pulse power radar equipment or something along those lines. Stuff that you really want someone on your team that knows what they are doing. Especially since the construction of such a transformer that performs at its best can also be quite tricky and costly.The number one thing for efficient transformers at these powers is to keep leakage inductance as low as possible and have hugely thick wingdings to provide the low resistance required to keep the pesky P=I^2*R at bay despite the huge currents. The core material is not that hugely important since you can simply use enough turns to keep the BH curve where it is happy. Even laminated steel can operate operate into the KHz range with some careful design. Ferromagnetic cores mainly give you benefit by reducing the number of required turns and thus those resistive losses that are your biggest enemy, core loss should be tiny compared to them.Quote from: patrick1 on August 23, 2019, 03:35:28 pmWhat you describe in the video is one of the big reasons why pretty much all large switch mode power supplies are using topologies that drive current in both directions trough the transformer such as the PushPull or Full Bridge designs (And additional tricks are used to make these even more efficient). Driving the transformer both positive and negative makes for more efficient use of the core because its traversing across all of its available BH curve rather than just the top half of it. The 50/60Hz mains transformers operate in this way. But the reason that small power switching supplies don't use this superior way of driving a transformer is that the electronics are more complex for it, so it becomes more cost effective to reduce component count with the Flyback topology and simply make the transformer a bit larger (But since its a tiny transformer to begin with that's not much of a cost penatly)Oh and as for the solar power thing. The fact that the voltage is rising too high is the fault of an overloaded power distribution network. The voltage drop on the wires to the transformer and perhaps the transformer itself are under rated so they are not transferring the power back into the grid. Because the current is flowing the other way means any voltage drops are actually added on top of the voltage. But typically grid tied solar inverters don't tend scale back production according to voltage, they will output as much power as they can suck from the panels. Its only once the voltage gets too high a safety cutoff will kick in and shut it down to prevent damage. Voltage regulation is the job of the grid, not the inverter. Due to the way AC power grids work if there is more power being generated than is being consumed what will actually go up is the frequency, not the voltage.