Is the DC resistance of a transformer winding DIRECTLY related to the volt drop? - Page 1
Forum section: electronics / projects

author: Deactivated-1
author_rank: Frequent Contributor
timestamp: December 29, 2023, 07:20:45 am
content: Reading various articles on transformer winding, most throw out these "rules of thumb", for example "add 5% voltage to account for losses"But why not just directly calculate the estimated wire length, it's resistance from that, and the voltage drop across that?Or is it not that simple?Also does the wasted power from the core count towards a parasitic voltage drop across the primary DC resistance?

author: moffy
author_rank: Super Contributor
timestamp: December 29, 2023, 09:33:39 am
content: Because you have leakage inductance as well as the resistance, and the leakage inductance drops voltage as well. Core losses look like a resistor in parallel with the magnetising inductance, so they don't drop voltage directly except for the current drawn through the leakage inductance and winding resistance.

author: moffy
author_rank: Super Contributor
timestamp: December 29, 2023, 09:44:04 am
content: Quote from: ELS122 on December 29, 2023, 09:38:35 amLeakage inductance is very transformer/winding dependent. Normally it's a few percent of the magnetisation inductance, but it can vary a lot depending on the overall permeability of the core. In some situations you get a high leakage inductance to help increase the current handling, when you introduce air gaps.

author: Kleinstein
author_rank: Super Contributor
timestamp: December 29, 2023, 09:55:13 am
content: For smaller transformers the main part of the voltage drop is from the wire resistance, both on the primary and secodanry side. There are a few transformers specially build with added series induction, e.g. for microwaves and welders or old style battery chargers. With large transfromers (e.g. 100 kVA+ range, like the distribution transformers) they tend to add leakage inductance to limit the short circuit current. This is not with an air gap, but usually added space between the primary and secondary, that can than also be used for cooling.For the transformers with the secondary winding on top of the primary one can reasonably estimate the leakage inductance. One has the seconday winding number and the distance (including part of the thickness) to make up the effective coil area. From turns and coil area one can calculate an inductance for an air coil. The solution with separate winding chambers it would be harder to calculate, but I would expect a similar to slightly higher inductance.

author: soldar
author_rank: Super Contributor
timestamp: December 29, 2023, 06:02:29 pm
content: "Voltage drop" needs to be better defined in this instance. What type of load? A purely resistive load is not the same as a rectifier bridge followed by a capacitor.

author: Vovk_Z
author_rank: Super Contributor
timestamp: December 29, 2023, 06:06:30 pm
content: Quote from: ELS122 on December 29, 2023, 07:20:45 amFor the 'usual' power transformer, you may pretty accurately use windings DC resistance as the main and the only voltage drop ingredient.Use primary DC resistance divided by n^2 to get a full transformer resistance (where n is a transformation coefficient).

author: johansen
author_rank: Super Contributor
timestamp: December 29, 2023, 06:52:57 pm
content: Quote from: ELS122 on December 29, 2023, 09:38:35 amyou can use rules of thumb, but they are also dependent on the permeability of the iron core. at around 100KVA, the efficiency of a transformer is around 98 to 96% and the inductive impedance is about equal to the resistance, so they add vectorially and the total impedance will be around 4 to 5% on most 100kva sized 60hz transformers of average construction.As the transformers get larger, the leakage inductance increases with the 4th power of the linear dimension allegedly, and so does the power capacity (neglecting the lack of surface area for cooling, so they have to fall away from that limit). as a result a 1 GW transformer may be 99.7% efficient but its impedance is 20%. this is apparently.. typical. part of it is due to the physical distance and size of the 500KV coils and how far spaced out they have to be. part of it may be due to manufacturing dimentional limitations so the cores can be shipped. a 1 GW toroidal transformer would almost certainly be in the less than 10% leakage inductance territory and would be around 99.9% efficient. but it wouldn't be able to be made to handle 500Kilovolts unless you put the primary on one side and the secondary on the other, at which point the leakage inductance would likely be back to the 20% range.Small transformers.. they are just plain terrible. Because the resistance is so high, manufacturers have to run the cores completely saturated.. so that the primary coil's copper losses can be "regulated away" by the saturation of the core.So effectively the impedance of the transformer becomes just the resistance of the secondary coil. this is why a 12v, 1 amp 60hz transformer will have a 25% voltage drop under load instead of 50%, and it will have a no load loss of probably 2 watts instead of just 0.05 watts which is what the bulk iron would lose at reasonable flux densities by weight.Toroidal transformers seem to do a lot better than cheap E core wall wart sized transformers of similar capacity, i'm not sure why, but they also typically have real temperature rises under load that are far less than small E cores. Read typical voltage regulations here: a 1kva toroid only drops 2% but a 30va drops 12 to 14%.https://www.hammfg.com/electronics/transformers/power/1182.pdfI know from experience, small 100 watt toroidal transformers can be run at near double their "nameplate" ratings without burning up. you can't do that with an E core wallwart.Quote from: ELS122 on December 29, 2023, 07:20:45 amyes it does sort of: a typical microwave oven transformer has a 1 ohm primary coil, and the 4 amps it draws drops 4 volts, wasting 16 watts of power. this is 3% of 120vac primary, that will not make it across the core to the secondary windings at all.the iron loss of 100 watts on top of the 16 watts does not directly drop the voltage on the primary coil.You can think of it more like 4 amps of magnetizing current and 1 amp of wasted hysteresis losses add vectorially to the square root of 17, which is like 4.1 amps.so if you put a superconducting coil of the same turns in the same core, you would have a 3% higher flux density in the core and the losses would be even higher. the 16 watts of copper losses would be gone but now the core is going to have 3% higher flux density.

author: johansen
author_rank: Super Contributor
timestamp: December 30, 2023, 01:23:52 am
content: I know for a fact the 4th power effect is real, i just dont understand yet how it affects the leakage inductance.For a given transformer, when you scale it up twice the linear dimension, the voltage per turn increases 4 times, because the cross section of the core increased 4 times, but so does the cross sectional area of the area for the wire. So if you keep the same number of turns per square inch, not only did your voltage per turn rise 4 times, but you also have 4 times as many turns. So 16 times as much voltage, and at least 64 times as much inductance because the inductance follows turns squared times the radius squared. Oh, and now the windings are twice as far apart...so maybe its that simple.But only 8 times the losses, because the volume of the copper only increased 8 times as did the iron. -the problem being the surface area only went up 4 times. So for a given temperature rise, the 16 fold increase in power has to be reduced.So when you look at the datasheets and compare a 98% efficient 1kva toroid that weighs 16 pounds and has a 40C temperature rise.. with a 1.5 pound 50 watt toroid that is 86% efficient and has a 40C temp rise..there's your answerI have a 12kva 120/240: 208 transformer, it weighs about 100 pounds. I could attempt to separate the resistance from the leakage inductance if anyone cares.It may be that if you were to make a 1GW transformer with less than 1 percent of the coils's diameter, distance between any two primary and secondary conductors, you could keep the impedance down to 1%

author: NiHaoMike
author_rank: Super Contributor
timestamp: December 30, 2023, 04:55:42 am
content: If you have an inductance meter, the leakage inductance is easily measured by shorting the primary and measuring the inductance of the secondary. Note that for ratios other than 1:1, the value would be different on the primary and secondary, for mains transformers usually the secondary leakage inductance is used as it allows modeling the transformer as an ideal AC source with series resistance and inductance.

author: johansen
author_rank: Super Contributor
timestamp: December 30, 2023, 05:42:42 am
content: Quote from: NiHaoMike on December 30, 2023, 04:55:42 amYou need to measure it at 60hz and full voltage.Its not as simple as a desktop 1khz meter.

author: TizianoHV
author_rank: Regular Contributor
timestamp: December 30, 2023, 09:43:48 am
content: I tested a small transformer.As someone said, for such small transformers, the windings DC resistance (R1dc ~ Rk) makes most of the series impedance (Zk). But the leakage reactance (Xk) must still be taken into account.What do you mean with regulation? What equivalent circuit are you using?Regarding the MOT's, they have magnetic shunts (not permanent magnets). Manufacturer use them to increase Xk (leakage reactance). In that way they reduce the short circuit current (and maximum power delivery).

author: Zero999
author_rank: Super Contributor
timestamp: December 30, 2023, 12:06:17 pm
content: Quote from: TizianoHV on December 30, 2023, 09:43:48 amRegulation is simply the % change in the secondary voltage, with no load vs full load. Generally the secondary voltage of small mains transformers are specified at full load, so an isolation transformer will have slightly more turns on the secondary and 1:1 and a slightly higher unloaded voltage to compensate for the voltage drop, under load.

author: jonpaul
author_rank: Super Contributor
timestamp: December 30, 2023, 01:28:55 pm
content: Dear ELS122:Reading your posts regarding mains transformers, magnetic lamp ballasts, core and laminations, we are working in magnetics since1970s.0/ power handling of a transformer depends on the core weight, wire gage, cooling, insulation class , hotspot at ful load etc.Our experience is to go by the size and weight.1/ Most commodity lighting ballasts, microwave oven HV, low V trsf use the cheapest grade 0.014" laminations, not grain oriented but plain old silicon steel. The complete specs are available from the steel lamination mfg like Tempel, Hatachi, Magnetic Materials.2/ DC resistance of a winding will depend on wire gauge, length of wdg and temperature.3/ Losses in magnetics are generally "Pcu " copper losses = I sq R resistive, skin effect and prox effect at high freq,PLUS"Pfe" iron core losses = hystersis losses deoends on flux density, frequency, voltage.Pcu varies with square of RMS IPfe varies with applied voltage and frequency, but NOT current.4/ Total losses can be very high in low duty cyc low cost like MOT oven, 20%Losses are generally decreasing with size, thus a 5 w wall trsf can have 10..20% , a 1 kVa PSU perhaps 1..2 % and a utility distribution transformer 100 kVa <0.5%If you like to learn more about transformers, magnetics and transformer design, highly recommend the classic texts like Static Electromagnetic Devices by Hunt and Stein, 1963 (my CCNY professors, still have the book)https://www.ebay.com/itm/314589964115A fine CAD program was written in 1995, Intusoft Magnetics Designer. DOS/early Win but still available and still works.http://www.intusoft.com/mag.htmBon ChanceJonthe internet dinosaur

author: NiHaoMike
author_rank: Super Contributor
timestamp: December 30, 2023, 04:06:53 pm
content: Quote from: jonpaul on December 30, 2023, 01:28:55 pmI read somewhere that iron losses actually decrease a little bit with load for some designs.

author: The Electrician
author_rank: Frequent Contributor
timestamp: December 30, 2023, 05:01:02 pm
content: Quote from: johansen on December 30, 2023, 05:42:42 amIt would seem that applying full voltage to one winding while the other winding is shorted might not be a good idea. Could you explain the details of the procedure?

author: jonpaul
author_rank: Super Contributor
timestamp: December 30, 2023, 05:49:20 pm
content: Llkg:Short secMeasure primary L on bridge, 1V/10mALmag: open secMeasure primary L on bridge, 1V/10mAModel of trsf is ideal ratio with Llkg in series and Lmag in parallel.The power engineers and utilitiesuse an OC/SC test in a different manner:Open sec, apply full rated V at Pri, measureVsec, Vpri IpriShort sec, with ammeter or clamp-on current meter,USE VARIAC, slowly increase V pri 0..low value OBSERVE IsecStop pri V increase when I sec = rated I sec.Read I pri.Jon

author: johansen
author_rank: Super Contributor
timestamp: December 30, 2023, 06:14:52 pm
content: Quote from: NiHaoMike on December 30, 2023, 04:06:53 pmyes because the primary winding voltage drop decreases the excition in the core.Quote from: The Electrician on December 30, 2023, 05:01:02 pmOne method i thought of this morning is to compare the voltage delivered by the mains through a resistor divider with the output of the transformer under full rated load.Or, you could use a smallnon over excitedtransformer, and a variac, to get the voltage just right to exactly match the test transformer, and put it in anti-series with your test transformer, such that you can use an oscope to measure the difference between the two transformers. Then when you apply full rated load to the transformer, the oscope will measure the voltage drop and the phase angle accurately.The mains are going to drop a little with load, and there are the lack of 3rd and 5th harmonics present* due to the predominance of cheap rectifier loads. So these tests can be difficult to make super accurate if you don't have 1% or better precision rms volt meters.*I say it that way because, yes there are 3rd and 5th harmonics present.. not because rectifier loads create them, but rather, consume them in a sense. so the tops of the grid sinewave is typically clipped off with a rising flat top.if you actually add 3rd and 5th harmonics to the grid, the tops of the sine wave would peak, not get flattened.

author: The Electrician
author_rank: Frequent Contributor
timestamp: January 01, 2024, 08:58:41 am
content: Quote from: johansen on December 30, 2023, 06:14:52 pmSo, having got the voltage drop and phase angle, how do you calculate the leakage inductance from that?

author: johansen
author_rank: Super Contributor
timestamp: January 01, 2024, 07:57:33 pm
content: the phase angle and voltage drop is a vector sum of the resistance and inductance. its just simple math.

author: The Electrician
author_rank: Frequent Contributor
timestamp: January 02, 2024, 12:33:34 am
content: Quote from: johansen on January 01, 2024, 07:57:33 pmWhy would the result obtained in this way not be the same as the result measured by shorting the 120 volt winding and measuring the impedance at the 12 volt winding with an LCRZ meter at 1 kHz?

author: johansen
author_rank: Super Contributor
timestamp: January 02, 2024, 05:29:39 am
content: Quote from: The Electrician on January 02, 2024, 12:33:34 amWhat makes you think the inductance will be the same?

author: uer166
author_rank: Super Contributor
timestamp: January 02, 2024, 05:52:06 am
content: Quote from: johansen on January 02, 2024, 05:29:39 amLeakage inductance is a function of the windings geometry only and not a function of core. The core is the only potentially nonlinear element, so the excitation voltage level should not matter in determining the leakage inductance.In fact you can remove the core entirely and still get the same result via shorting the secondary and measuring primary inductance.

author: The Electrician
author_rank: Frequent Contributor
timestamp: January 03, 2024, 12:20:41 am
content: Quote from: johansen on January 02, 2024, 05:29:39 amI didn't offer an opinion about whether it's the same or not.You are making assertions, such as in reply #10: "You need to measure it at 60hz and full voltage.Its not as simple as a desktop 1khz meter."I'm only asking questions.The usual convention in argumentation is that the person making assertions must offer proof if they want their assertions to be believed; the person asking questions doesn't have any obligation to offer proof of anything, or even offer an opinion.

author: johansen
author_rank: Super Contributor
timestamp: January 04, 2024, 06:09:16 am
content: Quote from: uer166 on January 02, 2024, 05:52:06 amQuote from: The Electrician on January 03, 2024, 12:20:41 amI want to see the above poster show his work, why dont you harass him.Dunno what you do for a living but in my life experience, unless you did the actual work yourself, it is often wrong.Proclaiming that the leakage inductance of a 60hz transformer is the same as it is at 1khz and 1% or less of its nominal flux density.or declaring you could get the same results if you removed the iron core entirely..Is like telling an RF engineer that some rf chokes for a wifi router can be measured and characterized with 2 meter ham equipment.

author: TizianoHV
author_rank: Regular Contributor
timestamp: January 04, 2024, 12:04:53 pm
content: "In fact you can remove the core entirely and still get the same result..."That's complete garbage. Leakage inductances represent how well two windings are coupled. If you remove the core there will be almost no coupling between the two windings and the leakage inductance will be much higher.Us electric engineers, to measure leakage inductance (and reactance), we put a short on the low voltage winding and, via a variac, we let flow the nominal current (at nominal frequency) on the high voltage winding. We use a power analyzer to measure voltage, current and power drawn by the HV winding. Then we can use the impedance formulas to get all the equivalent parameters:Zk = V / IPF = P / (V * I)PhaseAngle = cos^-1(PF)Rk = Zk * PFXk = Zk * sin(PhaseAngle)LeakageInductance = Xk / (2 * pi * Frequency)Just like I've shown in my previouspost.

author: uer166
author_rank: Super Contributor
timestamp: January 04, 2024, 11:21:11 pm
content: Quote from: TizianoHV on January 04, 2024, 12:04:53 pmThat is some serious confidence.. You're mixing up terms: there is leakage inductance, and there is coupling coefficient. This isn't just semantics either, it's an important distinction.The leakage inductance as a number (e.g. 10nH, 1uH, etc) does not change when you increase or decrease the amount of mutual inductance. It is purely a windings geometry thing, and the only way to reduce it is to couple the windings more closely via say interleaving, bifilar windings, or for the absolute lowest leakage coax windings. You can also increase leakage when needed (e.g. LLC transformers) by separating the windings into 2 banks.When for example designing a flyback, the leakage inductance energy is lost in snubber, and everything else being equal, a core with higher permeability WILL NOT reduce those losses, only a windings geometry change will.What actually represents how well the pri/secondary is coupled is *coupling coefficient* (e.g. 0.9, 0.999, etc), and that is now a function of both the mutual/magnetizing inductance, and the leakage inductance. You can increase coupling given a fixed leakage inductance by increasing magnetizing inductance, or you can increase it by decreasing leakage inductance as well.

author: TizianoHV
author_rank: Regular Contributor
timestamp: January 05, 2024, 10:28:32 am
content: I'm sorry for being rude but I do not agree with your theory:"the excitation voltage level should not matter in determining the leakage inductance." That's true."In fact you can remove the core entirely and still get the same result via shorting the secondary and measuring primary inductance." But I can't agree with that. Maybe we are talking about two different things. Are you telling that the two transformers in the photo have same leakage inductance since it isn't influenced by the core?

author: uer166
author_rank: Super Contributor
timestamp: January 05, 2024, 10:48:58 am
content: I'll have to think about that one. Here you might be right, a magnetic shunt would be a special case since it's equivalent to adding an entire second inductor in series. This isn't really a "transformer" anymore, but rather a series inductor combined with transformer on one core. In this case excitation level would also change the leakage inductance as well, since the magnetic shunt can saturate.I was talking about normal transformers in general though, and will maintain that the leakage inductance does not change whenever core is removed. That would apply to split bobbin, interleaved, non interleaved etc windings in cases where you're trying to control or minimize leakage. The context was specifically to show that the excitation level doesn't really matter in those cases.

author: TizianoHV
author_rank: Regular Contributor
timestamp: January 05, 2024, 06:06:57 pm
content: Such cores are used in Neon Signs transformers and Microwave Oven Transformers. I used it as an extreme example to show clearly where I'm pointing.Before going in deeper I want to show how I convert magnetic circuit to electric circuit: Fig.1, starting from a draw of the transformer, you visualize the fluxes, reluctances andmmf's.Now, to convert to electric equivalent, you need to swap series/parallel connections (step 3, 4) and rearrange the equivalent electric in a more intuitive disposition(5,6).Since Xfe is much larger than Xa1, Xa2 we can say that the leakage reactance Xk =~ Xa1 + Xa2.In Fig.2 I've removed the core. And yes, Xa1 and Xa2 are still there, but how can you measure them when there's Xa (Xa < Xa1,Xa2) shunting them?Edit. And that's just a theoretical representation, in practice you have to immage leakage fluxes excaping from the core andeverywere

author: najrao
author_rank: Regular Contributor
timestamp: January 18, 2024, 08:16:01 am
content: As an old school 'power' electrical engineer, I am amused by the extreme views expressed in this thread.Be that as it may, this forum caters largely to power electronics users, and they would be concerned with how less-than-ideal a practical transformer would work in their circuits. A most common application would be a 'mains' transformer feeding a capacitor 'input' rectfier. In dozens of other posts in our forum, I have read, with amusement, it expressed that the dc output voltage of such rectifier would be thepeakof the applied ac, less for diode drops. In a practical situation, theaveragedc voltage would be far lower, perhaps closer to the effective ac itself. And it is just not due to the Idc drawn dropping voltage in the transformer resistance! Or even leakage 'inductance'.Yes, the transformer impedance is the culprit, but what multiplies it is not the Idc at all, but the many times largerpeakcap charging current. After the serious drop, the cap will charge to much less than the 'peak' taken for granted. The large fast rising charge current is much more curtailed by the inductance than the resistance of the transformer, because it has much higher frequency components.Even with 'small' transformers where resistance may predominate over reactance, the latter is not negligible at all in cap input rectifier application. Wise designers use practical 'curves' to substitute rules of thumb. Or end up with having to up the transformer specs after test!