current capability of mains line tranformer; Loading only one of two windings? - Page 1
Forum section: electronics / beginners

author: julian1
author_rank: Frequent Contributor
timestamp: August 30, 2021, 06:26:09 pm
content: Consider a 100W 50/60Hz mains audio transformer, with two 18VAC * 2.7A secondaries.The transformer is intended for typical AC audio application, and thus a constant max load on both secondaries up to 100W (ignore derating).Now consider, a circuit application where significant current is only drawn from one winding at a time (ie. a AB amplifier sourcing positive or negative DC output).In terms of core-saturation, and heat dissipation from wire resistance - my initial thought is that the 100W transformer should be able to supply 2.7A * 2 = 5.4A from a single 18VAC winding, if the other is unloaded.But that seems wrong given the current spec that applies per winding - and so perhaps the magnetizing/mutual inductance is limited per secondary?

author: Vovk_Z
author_rank: Super Contributor
timestamp: August 30, 2021, 06:41:26 pm
content: I guess you can't get double current from one winding if other is fully unloaded, but yes, it can be loaded a bit higher.

author: bdunham7
author_rank: Super Contributor
timestamp: August 30, 2021, 06:46:13 pm
content: Are you referring to the peak current or RMS current? And is the other winding actually unused or are you just considering it unused because this is a half-wave setup and only one winding has current at a time?If this is a typical bridge rectifier/filter cap setup for a split supply, the actual answer is that you have to derate from the AC rating because of the harmonic power factor. You likely could get no more than 2A, if that. A drawing of your intended configuration would be helpful in case I've misunderstood your question.

author: bob91343
author_rank: Super Contributor
timestamp: August 30, 2021, 06:49:49 pm
content: If the winding resistance were zero you could indeed load full power on just one winding. That not being the case, the heat dissipated is proportional to the square of the current and so you could overload a single winding by 40% to achieve maximum safe current.

author: julian1
author_rank: Frequent Contributor
timestamp: August 30, 2021, 08:03:18 pm
content: QuoteApplication is typical split rail, fully rectified, large reservoir caps. My understanding is PF=>90%, VA~=W. 100W/2 = 50W. 18VAC RMS * 1.414 = 24VDC at 2A. for rated output of single winding.QuoteThanks. I completely under-estimated the extra heat due to a fixed winding resistance and P = I2 × R.

author: bdunham7
author_rank: Super Contributor
timestamp: August 30, 2021, 08:10:42 pm
content: Quote from: julian1 on August 30, 2021, 08:03:18 pmDo you have a circuit diagram just so we're completely clear?I don't see how those numbers can work. If you expect a DC voltage output greater than the AC RMS, your PF is going to be lower. A lot lower. Like 0.4, or 40%. And it certainly can't be anywhere near 1.4xAC RMS unless it is trivially loaded.

author: David Hess
author_rank: Super Contributor
timestamp: August 30, 2021, 08:42:14 pm
content: Double the current from one winding increases copper losses by 4 times. To only double the copper loss, which would be the same loss from using both windings, the current could be increased by 1.414 times which is what I would use for the upper limit in a conservative design. The power dissipation is more concentrated but I would not consider the difference significant.

author: Ian.M
author_rank: Super Contributor
timestamp: August 30, 2021, 09:23:34 pm
content: Even that *MAY* be pushing it a bit hard as the thermal resistance to the core and to ambient is likely to be somewhat higher for a single winding than for both together. You can use the DC resistance of the winding as a proxy for its average internal temperature. The temperature coefficient of resistivity of pure copper is 0.00393 K-1, of annealed commercial copper wire, is about 0.00385 K-1and of hard drawn commercial copper wire can drop as low as 0.00375 K-1, which is about a 5% range, As magnet wire is typically fully annealed and is only stressed moderately during winding, temperature measurement by this method, assuming you use the annealed figure, should give you the average winding temperature to within a few degrees C. You can use this calculator entering the hot and cold winding resistances directly in place of resistivity,(as you don't know the length or cross section area of the wire in the winding),and your room temperature as original temperature and leaving the filed 'Final Temperature' blank, and it will calculate the temperature for you.http://www.endmemo.com/physics/resistt.phpHowever, the average winding temperature is only of academic interest, unless you know the transformer's insulation temperature rating, and also have a hi-res thermal camera to warn you if there are any signs of hot-spotting.To compare secondary load capabilities, you don't even need to do the calculation. If you've got a decent four wire ohmmeter, protected against inductive transients(as the resistance will be too small for two wire measurement with any semblance of accuracy), simply run the transformer with its rated load on both windings, stopping briefly at intervals to measure each secondary's DC hot resistance till it plateaus, then try David's suggestion of x1.4 the winding rating on a single winding, the other of course unloaded. If the hot resistance increases, you are pushing it too hard.

author: julian1
author_rank: Frequent Contributor
timestamp: August 30, 2021, 11:56:58 pm
content: Quote from: bdunham7 on August 30, 2021, 08:10:42 pmYou're right. I have a sheepish understanding of power factor and linear supplies. Reading more, I believe PF theta with a purely resistive load is 1. PF theta with a capacitive load is 0.6 due to phase lag. But I am not sure if a capacitive load is representative of what is seen through a bridge rectifier, and when current is chopped up/distorted by V > than the charge on the storage caps.My rectifier, is typical of split-rail linear supplies. eg.https://www.tubecad.com/2018/02/10/Full-Wave%20Bipolar%20Power%20Supply.pngIs there any rule-of-thumb/ heuristic PF for such a circuit?I believe the VAC RMS of the secondaries stated by manufacturers are the loaded down voltage outputs. At least, I always measure higher when unloaded.Edit,It looks like there's a relationship between output ripple, and PF - whereby with more allowed output ripple (smaller caps), there is less current spiking due to more time to charge, and lower distortion/better PF. I suspect one just needs to size an appropriate transformer and experiment a bit.

author: julian1
author_rank: Frequent Contributor
timestamp: August 31, 2021, 12:03:50 am
content: Quote from: Ian.M on August 30, 2021, 09:23:34 pmThis is excellent advice. I will end up derating everything. But it will be good to test out how much margin there actually is.

author: David Hess
author_rank: Super Contributor
timestamp: August 31, 2021, 01:21:11 am
content: Quote from: Ian.M on August 30, 2021, 09:23:34 pmThat is right, but I usually count on the transformer being overbuilt somewhat to make up for the small difference.Hammond probably has an application note discussing it to go with their application notes about recommended derating of transformer power for different rectifier configurations.QuoteThat is a great idea, and you can run the calculations backwards after making the measurements to determine the thermal characteristics of the transformer.Be prepared to take some time making the measurements though. Big transformers can take hours to reach operating temperature.

author: bdunham7
author_rank: Super Contributor
timestamp: August 31, 2021, 02:06:59 am
content: Quote from: julian1 on August 30, 2021, 11:56:58 pmIt's not a capacitive or inductive load at all, it's something else. Theta only works for pure RLC circuits, for all R PF is one, for all C or all L it is zero and theta is either +90 or -90. Harmonic power factors are different, they don't reflect a simple phase shift.Your circuit is a common center-tap full-wave split-rail setup and current flows in each half of the winding all the time (every half cycle), so you are not using one at at time. And you need to use the 'loaded down' voltage in your calculations because at full output, it will indeed be 'loaded down'. There are precise ways of calculating all this if you have all the transformer parameters, but since I don't here's a brief summary. Your DC output current will be the same as the average output current from the transformer, but the heating of the transformer, and thus its max rating, is based on RMS current. Since shorter periods of conduction with the same average current result in larger RMS current, the available DC output current will be less than the rated AC RMS current. How much depends on the size of the filter caps, what the output voltage is, etc. In short, a very rough guess if you use a typical 2-5mF filter cap is that you might safely get 2A at 18VDC with a PF of 0.5.

author: Vovk_Z
author_rank: Super Contributor
timestamp: August 31, 2021, 02:27:05 am
content: Quote from: bdunham7 on August 31, 2021, 02:06:59 am+1.To TS:Transformer winding rated current is given for sine-wave current. The real secondary current is non-linear. So it is hard to find the accurate allowable DC load current, but we have approximate calculations for transformers loaded with C-filtered rectifiers (I don't have them at my hands).The base approximation is that for 1 A DC load current the transformer secondary must be rated for 1.41 A AC current (for a bridge rectifier), if I'm not wrong.