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If a transformer has efficiency of 96%. It means for a 500W capacity and load, the input would be 520W, output 500W so 20W would be emitted as heat. If load is 250W, then input would be 260W and 10 watts would be emitted as heat.

But if there is no load (zero load), how do you compute the wattage that would be lost as heat during the initial stage when magnetic field and flux is being built with no load on the secondary? Is it more than the inefficient percentage lost with load?

I'm computing sealed enclosure heat transfer for a design I'm having and I need to know the above.

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  • \$\begingroup\$ If you know the inductance and the resistance of the transformer (on the primary side) then you can easily calculate that information. \$\endgroup\$ – Harry Svensson Oct 28 '18 at 23:18
  • \$\begingroup\$ The efficiency metric would normally be measured at a particular load (e.g., maximum load). From there you could estimate the equivalent resistance of the transformer. You will also need the inductance to be able to calculate the no-load losses. \$\endgroup\$ – Edgar Brown Oct 28 '18 at 23:21
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Transformers are not rated by "efficiency", they are rated by losses, although the efficiency at maximum load can be defined. So you cannot assume that at 250W of load the transformer will dissipate 10W if at max load (500W) it dissipates 20W, it will dissipate more than 10W. This is because transformers have two types of losses, "load losses", and "no-load losses". Even at zero load a transformer will run warm.

Load losses depend on wire resistance of coils, and are proportional to load.

No-load losses are independent of load, they occur due to continuous magnetizing and de-magnetizing of transformer core, with two basic mechanisms - hysteresis losses, and eddy losses. These losses depend on quality of core construction and materials. To reduce eddy losses, the cores are made of thin sheets, so the eddies can't spread over much of core volume. Hysteresis losses depend of magnetic alloy composition. Some portion of magnetic field escapes the core and induces eddy currents and warm up surrounding conductors (mounting hardware, bolts and brackets), which also adds to no-load losses. Without knowing precise details of core construction and manufacturer's data, you can't "compute" these losses. For example, cheap knock-off transformers for Christmas decorations can stay pretty warm even when the lights are off.

ADDITION: I happen to have an old 200 VA auto-transformer 115:230V, model SU-38, made by TODD SYSTEMS. In idle (no load) mode, with ambient of ~25C and sitting on a pack of papers, its core gets to ~40C, see the thermal image.:

enter image description here

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  • \$\begingroup\$ For this Hammond 172D 500VA 230v-115v step down isolation transformer. hammfg.com/files/parts/pdf/172D.pdf I'm measuring 50 Celsius surface temperature at 30 Celsius ambient temperature. What kind of core or magnetic alloy do Hammond uses? For the most premium small isolated transformer. Can you measure just say 35 Celsius surface temperature at no-load? \$\endgroup\$ – Samzun Oct 29 '18 at 1:28
  • \$\begingroup\$ @Samzun, are you measuring 50C at no load? Or what? \$\endgroup\$ – Ale..chenski Oct 29 '18 at 1:48
  • \$\begingroup\$ yes, I am measuring 45-50Celsius at no load. What is the normal or for efficient brand? \$\endgroup\$ – Samzun Oct 29 '18 at 2:12
  • \$\begingroup\$ In the Hammond, the difference between zero load and half full load is only about 3 C (surface temperature)? How accurate are temperature in estimating core and winding temperature? And at full load, does it become like 10 C difference (based on actual tests of other brands)? From estimate, at half load, do you have maybe 80% of losses already compared to full load since it's not 50%? I need to know this because I'm estimating the heat emitted by the transformer at half load compared to full load. So if it's not half, then maybe 80%, typically? \$\endgroup\$ – Samzun Oct 29 '18 at 6:44
  • \$\begingroup\$ @Samzun, as I said, no-load loss is impossible to calculate, there is no "typical". You do have the transformer at hands, why don't you apply a known load and measure the total power using, say, "KillAwatt" tool, and answer all your detailed questions? \$\endgroup\$ – Ale..chenski Oct 29 '18 at 6:53
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There are two main loss mechanisms in transformers

a) Core losses, which are dependent on the input voltage

b) Copper losses, which are dependent on the load current

Designing a transformer to be low loss is a compromise between these two. In applications where the transformer may spend most of its time at no load, it's worth designing for low core loss, at the expense of high copper loss, by using a low core flux and a lot of turns.

Unless the transformer has a very thorough data sheet giving you losses at several loadings, it's not possible to compute the loss at any loading, it must be measured.

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  • \$\begingroup\$ How about application the opposite or where the transformer may spend most of its time at near full load? Then it means design for bigger core (high core flux) and fewer turns? What kind of core that has zero loss almost like superconducting material? \$\endgroup\$ – Samzun Oct 29 '18 at 6:18
  • \$\begingroup\$ @Samzun, if you want to start designing your own transformer, here is a good place to start: en.wikibooks.org/wiki/Electronics/Transformer_Design \$\endgroup\$ – Ale..chenski Oct 29 '18 at 7:12
  • \$\begingroup\$ Before I design one. I need to know what is the best core alloy material in the world that is almost equal to a superconducting wire. I want to use this core so at no-load, the losses would be minimal.. \$\endgroup\$ – Samzun Oct 29 '18 at 7:55
  • \$\begingroup\$ @Ale..chenski Oh boy, is that wikibooks article a mess. I wouldn't know where to start to clean it up, it needs a total rewrite. I mean, it's not actually wrong, it's that it's just not clear enough to guide a newbie into the design process properly. \$\endgroup\$ – Neil_UK Oct 29 '18 at 9:40
  • \$\begingroup\$ It's not useless link.. there I learnt the best core material in the world was called Vanadium-cobalt-iron core. Where can you buy 500Va transformer that uses it? I want to compare it with the Hammond to see if the temperature would become much lower (maybe just a few degree above room temperature). Of course I need to actually compare actual stuff before I even think of designing my own.. but my focus is really the circuits connected to these transformers. \$\endgroup\$ – Samzun Oct 29 '18 at 13:09
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Transformers are rated in VA rather than W, where V and A are RMS. VA and W are the same only for a resistive load. There is loss in the windings due to the load current, as I squared times R. There is loss in the core due to eddy currents and hysteresis loss. Core loss does not change appreciably with load.There is also some loss in the copper due to the magnetising current. I presume you are talking about a mains frequency transformer. Other factors apply at higher frequencies.

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  • \$\begingroup\$ All "other factors" you've stated also applied for mains transformers of 50/60Hz especially when the power ratings are in KVA ranges. \$\endgroup\$ – soosai steven Oct 29 '18 at 0:59
  • \$\begingroup\$ I was thinking of skin depth and proximity effect. Given that the skin depth in copper is about 9 mm at 50 Hz and the questioner gave a 500 VA transformer as an example, I didn't want to add detail that wouldn't really be applicable to him/her. Perhaps I should have said these factors are not significant at mains frequencies at the power level asked about. \$\endgroup\$ – Steve Hubbard Oct 29 '18 at 5:32

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