Transfomer core loss - Page 1
Forum section: electronics / general

author: innkeeper
author_rank: Unknown
timestamp: October 20, 2017, 12:28:14 am
content: I though i knew how to calculate core loss....but its coming out wrong for me.calculating total loss is pretty easy VA in - VA out = VA Loss ... ok easy enough.power loss primary and secondary should be I2*R = copper lossso total copper loss should be primary and secondary loss added.the total VA loss - copper loss should equal the core loss... but in doing this in a real world scenario, i end up with a negative core loss.i measured the cooper resistance right after measuring the current, so i don't think temperature change is a factor.i figure i must be understanding this wrong or i am making some stupid mistake, but if so i have no clue where.

author: Jay_Diddy_B
author_rank: Super Contributor
timestamp: October 20, 2017, 12:38:27 am
content: Hi,Copper loss = Core loss is an equation used (in the past) to minimize the total losses.It is not a requirement. I am also sure that modern transformers are optimized for minimum cost not minimum loss. Minimum cost is more steel less copper.Regards,Jay_Diddy_B

author: innkeeper
author_rank: Unknown
timestamp: October 20, 2017, 12:47:54 am
content: well now that can't be right that would mean it would be an ideal transformer with no hysteresis loss and eddy current loss.or do you mean that cooper loss is approx the same value as the core loss. that would mean i am supper far off

author: T3sl4co1l
author_rank: Super Contributor
timestamp: October 20, 2017, 12:56:14 am
content: Yes, equal losses.Speaking of cost, a lot of larger transformers opt for aluminum wiring (copper clad, so it looks the same, but doesn't feel the same!). Aluminum is about twice the volume for the same conductivity, needing that much more iron. Somewhere inbetween lies the balance. (For microwave oven transformers, high losses are acceptable, so they might not change anything else. Welder transformers have used aluminum for many years, and get away with a low duty cycle, for the same reason.)FYI, you're not looking at VA -- that includes reactive power, and will unfairly count magnetizing current and leakage flux against you. Real power is the only thing that matters.(If you meant in the sense that P == V*A, that's true, but be careful -- it's the same way that torque and energy (work) are N*m, but one is at right angles, and the other is the same direction. Reactive power measures V*A, but it's actually a unit of potential energy.)Tim

author: Jay_Diddy_B
author_rank: Super Contributor
timestamp: October 20, 2017, 01:00:38 am
content: If you take a given core. As you increase the number of turns the copper loss goes up (longer wire) and the core loss goes down, because the flux density goes down.When the core loss is equal copper loss that is the most efficient configuration for that core.To measure the lossesThe core loss is measured by energizing the core with the secondary winding(s) open and measuring the real power (W) not VA. Since the current is low, the copper losses are very small in this test.To measure the copper loss you short the secondary and apply a low voltage to the primary. You adjust the current until the nominal secondary current is reached.Again you measure real power (W) not VA.In this configuration there is very little flux in the core and the losses are dominated by the copper losses.Remember copper has a relatively high temperature coefficient of Resistance. So you should measure copper loss at the operating temperature or calculate the effect of temperature.Jay_Diddy_B

author: innkeeper
author_rank: Unknown
timestamp: October 20, 2017, 01:02:03 am
content: i know really this is academic.... but..how does one calculate the I2R losses and calculate total non I2R loss that should total up to the total loss.

author: innkeeper
author_rank: Unknown
timestamp: October 20, 2017, 01:12:16 am
content: i should note, my tests were done into a restive load so the PF should be 1 so probably shouldn't have used VA and confused things.I'm not seeing were my calculations or methods went wrong yet.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: October 20, 2017, 01:22:41 am
content: Do you have a feel for how precise your measurements and calculations are? Remember that this is a difference method, so it is very sensitive to small errors. Careful measurements are necessary!Tim

author: innkeeper
author_rank: Unknown
timestamp: October 20, 2017, 01:27:05 am
content: I was using a 120,000 count meter with true rms.I didn't use 4 wire measurement on the resistance, probably should have.

author: Jay_Diddy_B
author_rank: Super Contributor
timestamp: October 20, 2017, 01:42:37 am
content: Hi,What size is you transformer?Efficiency is a function of size. Bigger transformers are more efficient. They can be as high as 98%.Jay_Diddy_B

author: innkeeper
author_rank: Unknown
timestamp: October 20, 2017, 01:48:47 am
content: This is a small 30VA R-core transformer.the overall efficiency isn't that great, mostly due to I2R losses.i do expect the core "iron" losses to be extremely low considering the type of transformer, and the calculations came out only slightly negative. so, to me, it is plausible this is simply an measurement error.hmm... i can see Tim's point on the difference method now that i think on this...but can anyone comment on if my methods / calculation methods are faulty? if not then its got to be measurement error, that is the only logical conclusion.I'm starting to think that my chasing resistance measurements on a hot transformer is not wise. maybe as an experiment i'll do this with everything cold so it is easily repeatable and i am not chasing varying resistances. doing very careful measurements.

author: Zero999
author_rank: Super Contributor
timestamp: October 20, 2017, 07:45:55 am
content: Quote from: innkeeper on October 20, 2017, 01:12:16 amAssuming a PF of 1 should be fine for a power transformer, as long as it's fully loaded, as the leakage inductance will be small. However one can't make that assumption with a transformer which is designed to limit the current, such as a welding or neon sign transformer, which will have a much greater leakage inductance and appear inductive, even when the secondary is driving a short circuit.

author: The Electrician
author_rank: Frequent Contributor
timestamp: October 20, 2017, 08:41:30 am
content: You can't determine core loss from VA measurements. You need to use a true wattmeter, something like this:https://www.ebay.com/itm/Simpson-Single-Phase-300-Watt-Wattmeter-Panel-Meter-Case-Cracked-/302168522960?epid=866061284&hash=item465aa5c0d0:g:nkYAAOSwH09ZFPN2but with a full scale of 10 watts rather than 300 watts.Your transformer probably has a core loss of a watt or so.

author: innkeeper
author_rank: Unknown
timestamp: October 21, 2017, 01:13:38 am
content: Why can't i measure it that way if the PF is 1.0?If the load is purely restive on the secondary, is the power not simply Irms * Vrms * 1.0 on both the primary and secondary?1.0 being the power factor of a purely restive load.please explain why wouldn't the apparent power and the real power be the same, as the load reactance is what is reflected to the primary.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: October 21, 2017, 03:22:15 am
content: The transformer will draw reactive power. In effect, you can measure the parameters of the equivalent circuit:Tim

author: The Electrician
author_rank: Frequent Contributor
timestamp: October 21, 2017, 04:30:40 am
content: Quote from: innkeeper on October 21, 2017, 01:13:38 amThe PF will never be exactly 1.0 because the transformer magnetizing current is always there. You said that you expect this transformer to have small core loss because of the core quality. This means that the difference technique you're trying to use will be even more difficult.The only feasible way to get a good value for core loss is to use the technique described in reply #4. Leave the secondary unconnected and apply power to the primary through a wattmeter.

author: innkeeper
author_rank: Unknown
timestamp: October 21, 2017, 09:07:12 pm
content: So ok i get it, there is going to be some inductive reactance even if the load is purely restive. So were never going to reach exactly 1.0 i defiantly can see that.ok so, help me out here, what would be a viable method to actually measure this using typical lab tools. scope, meters, etc.Again, this is all purely academic exercise. its more to learn how to do this and where i am going wrong.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: October 21, 2017, 09:33:56 pm
content: Measuring power is kind of hard. You really need an instrument for the purpose -- a Kill-A-Watt (or clone thereof) does a surprisingly good job, and is cheap. The problem is multiplying voltage and current together instantaneously, then averaging to get real power. (Averaging or RMS-ing voltage and current separately, then multiplying, gets apparent power.)Otherwise, you need a good current probe (low phase shift), and a scope with multiplication math function.BTW, note that core loss is due to applied voltage, not load current -- you get the most sensitive measurement with zero load, so that the real power is due to only primary DCR (which you can subtract) and core loss.Another way is to null the reactive power, then assume apparent power = real power. This misses harmonic power, but if the transformer is far from saturation, it will be small. (Saturation means a peaky current waveform, at the primary, with no load on the secondary.) You can reach a null by adding capacitance in parallel with the primary, until the input current is at a minimum. At that point, the current will be real, and V*I will be real power.Or you can use a completely different method. Measure the transformer temperature, after a few hours time to let it stabilize. Measure the temp rise, unloaded, at rated primary voltage. Then, apply DC to the primary, such that it reaches the same temperature rise. (This can take a while, because you're adjusting the DC, waiting a few hours to see if you got the right temp rise, and repeating.) DC is real power by definition (there is no reactive power at 0Hz), so by matching the power dissipation, you're using the transformer itself, and a thermometer, as a wattmeter.(In all cases, subtract primary DCR loss. This actually underestimates the copper losses, because there will be some eddy current / skin effect / proximity effect loss or ACR increase, even at 50/60Hz. If it's not a huge transformer, it should still be within 10%, though.)Tim

author: Kleinstein
author_rank: Super Contributor
timestamp: October 21, 2017, 10:15:13 pm
content: The core loss is essentially load independent, so one will usually measure it with no load. Due to the magnetization curve the current waveform is not necessarily sine anymore. The core loss also depends on the voltage, usually with something like U^2.5. So the actual line voltage might matter.Likely the thermal method is the best bet, if you don't have a suitable special instrument.One may not need to get exactly the same temperature with the DC power - one can usually interpolate. This might speed up things quite a lot.Eddy-current loss in the core mainly depends on the voltage and it is core loss, so no problem here. Eddy currents and thus skin effect in the copper are usually rather small for the usually thin wires in the primary of this size transformer. So there is no large error in correcting the DCR loss. Unless running at rather high field, the magnetizing current is also rather small in toroidal transformers in the size. So DCR loss will be rather low at no load.

author: The Electrician
author_rank: Frequent Contributor
timestamp: October 22, 2017, 07:06:45 am
content: The thermal method of comparing temperature rise with rated primary voltage applied and waiting for equilibrium, then heating with DC on the primary isn't very good. The reason is that with rated primary voltage applied, the heat is generated in the core itself. WIth DC applied to the primary, the heat is generated in the primary copper and this heats the core indirectly, through the thermal resistance of the bobbin and various insulating materials. When heating the copper, a lot escapes from the outside of the winding and is not communicated to the core.Since I've been in the business for a while, I have a number of wattmeters of various manufactures, including some designed for low power factor measurements such as this one:https://www.ebay.com/itm/YOKOGAWA-Portable-Single-Phase-Low-Power-Factor-AC-Wattmeter-Type-2041-cos-0-2-/191818503441?epid=0&hash=item2ca9461511:g:UtYAAOSwWTRW2BZtbut this is out of reach of the typical hobbyist.A while ago in another thread the question arose as to what means a hobbyist on a budget could use to measure so-called "vampire" loads. I came up with a very reasonably priced solution described here:https://www.eevblog.com/forum/testgear/yokogawa-cw-10/msg409877/#msg409877This gives the user a true wattmeter for very low power levels at a very reasonable price. This is a great way for a hobbyist to be able to measure core loss in small transformers.Here's another post about the performance of this mod:https://www.eevblog.com/forum/projects/measuring-standby-power/msg413813/#msg413813

author: T3sl4co1l
author_rank: Super Contributor
timestamp: October 22, 2017, 08:47:42 am
content: Quote from: The Electrician on October 22, 2017, 07:06:45 amSo put a sock on it.Keeps the heat in, keeps it more uniform, and raises the temperature so the thermometer is more sensitive.Neat idea with replacing the shunt, just be careful never to plug in a load that draws inrush current! Put a bright sticker on it or something, "low power only".Tim

author: The Electrician
author_rank: Frequent Contributor
timestamp: October 22, 2017, 02:59:55 pm
content: Quote from: T3sl4co1l on October 22, 2017, 08:47:42 amAs was described in one of the threads, a couple of diodes across the .2 ohm shunt solves that problem.