Transformer Idle consumption  - Page 1
Forum section: electronics / beginners

author: Jwillis
author_rank: Super Contributor
timestamp: March 26, 2021, 06:12:49 am
content: I've googled this and can't seem to get a definitive answer. Is there a way to determine the No secondary load consumption of the primary winding of a transformer aside from direct measure ? Theoretically it should be zero , but since theirs no such thing as a 100% efficient transformer considering copper losses and other parasitic effects . But even a transformer that is 95% efficient shouldn't consume 5% on a no load condition.It would nice to know if theirs a rule of thumb or a formula to determine idle losses before purchasing.Reason I'm asking is I would like to use a by-stable soft latching switch for the mains on my project instead of a conventional switch. I considered a capacitor voltage drop but I don't like the lack of isolation in case something goes wrong. So I decided to use a small transformer instead . I only need around 50mA on start up and the CD4013 should only draw a couple micro amps on idle .

author: srb1954
author_rank: Super Contributor
timestamp: March 26, 2021, 08:23:23 am
content: Quote from: Jwillis on March 26, 2021, 06:12:49 amAll real mains transformers will draw a current known as the "magnetising current" even with no load on the secondary. The magnetising current sets up a flux in the core which in turn produces a back EMF in the primary to counter the applied mains voltage. The magnetising current is 90 degrees out of phase with the mains voltage so the power factor is zero and no real power is consumed by the magnetising current.This magnetising current is not a produced by lossy elements in the transformer but it is determined by the inductance of the transformer primary. Typically, the magnetising current will be around 10% of the full load current but this varies greatly with transformer design. Very small transformers, such you might be considering here, generally have a higher magnetising current of 20-30% of the full load current.Although the magnetising current itself through the transformer primary inductance produces no energy wastage there will be some power loss due to the resistive losses in the winding and hysteresis losses in the core material. This power loss also varies greatly with the transformer design and is generally worse with smaller transformers. This power loss shows up as a non-zero power factor if measuring the transformer with no load.The magnetising current is sometimes stated in a manufacturer's data sheet but the best way is to measure it directly. Remember to use a true-RMS meter as the current waveform won't be a pure sine wave. In the absence of a suitable meter you could measure the primary inductance and calculate the magnetising current from the impedance at mains frequency.

author: Benta
author_rank: Super Contributor
timestamp: March 26, 2021, 10:55:58 am
content: If you don't want to measure current and phase, the other option is to measure temperature rise. This is per definition pure loss.As a rule of thumb: the smaller the transformer, the higher the relative loss.

author: woodchips
author_rank: Frequent Contributor
timestamp: March 26, 2021, 05:41:19 pm
content: Not really.Off load there are two principle losses, magnetizing current and hysterisis loss.The magnetizing currentis, as as been said, at 90 degrees to the voltage but it is still real current, and will result in a current loss, reactive power. It is quite definitely not loss free.The hysterisis loss is the energy consumed sweeping the magnetic flux around its BH curve. This also is not loss free.A small transformer, 50VA, shouldn't consume more than 10% of its rated load in these losses. If it does then it is a poorly designed and built transformer. A toroidal transformer will be lower, it is more efficient.Just leave a transformer connected but off load, after some time you can feel that it is warm, may be reactive current but still losses.On no load the power factor will be close to zero, reactive losses, and will improve as the load comes on and real power is transmitted through the transformer.Almost any book published prior to 1960 will explain all of this. Anything newer seems to be full of differential equations, which unless you can weild them are not very obvious.

author: Kleinstein
author_rank: Super Contributor
timestamp: March 26, 2021, 06:40:46 pm
content: The no load loss is generally considered to be from the iron core causing lossed from the hysteresis and eddy currents in the core. The magnetization current can cause losses in the winding resistance - this is especially the case with small transtormers like < 3 VA. Usually the lowest loss is with some 3 VA power rating. There are special type with lower loss, which use lower peak magneization and thus way get slightly lower loss at the cost of larger size and higher príce.Very small transformers like 1 VA behave quite a bit different from simple theory: the magenization current is significant and may become comparable to the nominal load current. The voltage drop from the winding resistance also gets significant. So the no load load loss can be quite high - possibly higher than nominal power rating. It also does not help that the small transformers are often build to a price.Especially at the low end it is difficult to calculate calculate the losses without detailed information. So better look up data sheets - they may tell the no load loss.

author: Benta
author_rank: Super Contributor
timestamp: March 26, 2021, 06:43:48 pm
content: OK, two answers explaining magnetizing current, although the OP is completely aware of this already.Why?

author: Jwillis
author_rank: Super Contributor
timestamp: March 26, 2021, 08:22:23 pm
content: I'm probably overthinking this problem but Thank You all . That clears things up much better . I think I'll change my approach and move the switch to the secondary side , if the transformer I'm using won't consume much power in a standby state anyway.The idea is to have the unit monitor and data log over a long period of time and be able to go into a standby condition when a predetermined condition is met. And it may have to do this several times without the need to babysit it.