Transformer voltage for dual power supply (+12V / -12v) - Page 1
Forum section: electronics / beginners

author: belzrebuth
author_rank: Frequent Contributor
timestamp: August 25, 2018, 05:14:10 pm
content: Hi all,I've got a 2 x 12VAC toroid transformer and was wondering if the 12VAC will be enough for a symmetric 12V power supply.I've already got the pcb for the dual power supply and it requires LM317/LM337 regulators so I can't use low dropout regulators (except maybe if the pinout is the same).Can I use this transformer to make a dual 12V power supply?

author: ocset
author_rank: Super Contributor
timestamp: August 25, 2018, 05:32:39 pm
content: Yes 12vac is 17vpk.subtract 2v for the diode bridge.....down to 15v.....subtract a bit for winding voltage drop.....your down to 12V...so you should be ok as long as your winding voltage drop is not too much.

author: belzrebuth
author_rank: Frequent Contributor
timestamp: August 25, 2018, 05:36:56 pm
content: Thanks!I'll try it then.Here's the power supply I'm talking about..http://musicfromouterspace.com/analogsynth_new/POWERSUPPLY2009/POWERSUPPLY2009.phpIt states that'll work with a 2x12 transformer for 12 volt output.I'm a bit skeptical about the 10,000uF capacitors though..I find their value a bit too much for the power supply's amperage rating..I want the lowest ripple possible but still..I thought that 4x4700uF at 25V would do the job nicely..I think there was a formula for calculating the required capacitance for the max current draw..

author: Zero999
author_rank: Super Contributor
timestamp: August 25, 2018, 05:48:58 pm
content: I'd say it's marginal, especially for any sizeable current draw. Use Schottky diodes to reduce the voltage drop and you stand more of a chance, even with the LM317 and LM337.

author: drussell
author_rank: Super Contributor
timestamp: August 25, 2018, 05:51:15 pm
content: Quote from: belzrebuth on August 25, 2018, 05:14:10 pm\$12.0 * {\sqrt{2}} = 16.97\$(As others have stated while I was trying to discover how to post with MathJax...)You'll really need to check the actual voltage. If it is slightly higher than 12 at load, you'll probably be fine. The old 12.6v filament transformers were usually just enough to get actual 12 volt DC regulated at full load with a 7812, etc. non-LDO regulator. YMMV.

author: IanB
author_rank: Super Contributor
timestamp: August 25, 2018, 05:56:36 pm
content: Quote from: belzrebuth on August 25, 2018, 05:14:10 pmIt depends on how much current you want to supply. If you limit the current to, say, 500 mA or less you might be OK.

author: drussell
author_rank: Super Contributor
timestamp: August 25, 2018, 06:09:23 pm
content: Quote from: IanB on August 25, 2018, 05:56:36 pmWell, that depends on how beefy the transformer is, of course.If the OP wants, say, 1 amp and the transformer only puts out 12.0 volts AC at the 1 amp (plus whatever is needed to account for losses) load, then it is going to berighton the edge. If it is a nice beefy 100 VA torroid or something, though, it is bound to have plenty of oomph left with a couple of typical 1A or 1.5A regulators on it.My advice to the OP, measure it under load and see if it will suffice at your intended load. That's the only way to really be certain whether it will work or not. If it is close but not quite and the OP really wants to use that transformer that they already have, if it is an open-centre type, a couple more turns could always be wound around it, perhaps even just the leads themselves if they're long enough... Depends on the number of turns.To the OP: Do you know the rated current output or total VA rating of this transformer you want to use? This would give us a better guess as to how much drop there will be (though torroids tend to be quite good in that regard compared to small EI core general purpose transformers.)

author: Ian.M
author_rank: Super Contributor
timestamp: August 25, 2018, 06:26:02 pm
content: Edit: Drussell also asked the transformer's rating, while I was typing this.@O.P.Whether or not this can possibly succeed is heavily dependent on the exact transformer parameters.How many VA is the transformer, or what is its RMS secondary current rating?(if you haven't got a clue, post its dimensions and weight and we'll try to make an educated guess)Also, we need some idea of its regulation (% voltage drop under load). If you don't have a datasheet for it, to calculate that, we'll need its unloaded secondary voltage, and the voltage with a known load on it. As its got 12V secondaries, car bulbs or 12V halogen light bulbs are convenient as the load. Put one DMM on AC amps in series with the load and the other on AC volts directly across the secondaries. You should parallel both secondaries for this test (be careful of phasing) and load them to between 80% and 100% of the transformer's rating.Another possible issue is mains supply variations. Its not usually a big issue in urban areas in 1st world countries unless you are right next to heavy industry, but if you are in a rural area or somewhere with a weak grid, and you make the measurements at some time of day when the grid is lightly loaded and the mains voltage is above nominal, you could find that you have 10% to 15% less mains voltage when the grid is heavily loaded, with a corresponding reduction in your transformer's secondary voltage, which you just don't have margin for.

author: ocset
author_rank: Super Contributor
timestamp: August 25, 2018, 06:27:13 pm
content: ooops sorry i missed that you are using non LDO regulators......sorry, then 12vac is indeed too marginal

author: belzrebuth
author_rank: Frequent Contributor
timestamp: August 25, 2018, 06:35:45 pm
content: The transformer is rated at 3A and it's a 100VA toroid.I could use low voltage drop rectifier diodes..(I need to find a LDO bridge rectifier package)I do not plan to exceed the 1.5A rating of the LM317.Typical current draw will most likely be 700mA to an amp..

author: IanB
author_rank: Super Contributor
timestamp: August 25, 2018, 06:37:15 pm
content: Quote from: drussell on August 25, 2018, 06:09:23 pmIt kind of does, but since the design is apparently going to use a simple LM317 regulator we already can have a good idea of the operating parameters. People often get very optimistic about how much current is reasonable for an LM317 supply. I would say 500 mA is nicely within a safe and reasonable operating area.QuoteIMHO, a 12 V, 12 VA transformer cannot be made to produce a 12 V, 1 A linear power supply. More generally, you cannot reasonably design for the same DC output voltage as the AC input voltage. The AC input voltage needs to be higher than the DC output voltage to allow proper regulation and ripple rejection.

author: belzrebuth
author_rank: Frequent Contributor
timestamp: August 25, 2018, 06:56:28 pm
content: The transformer is rated for 3.5A and it ouputs 12.9V unloaded.Could I use a simple high wattage resistor as a load load directly to the AC?I could load each 12V winding with say a 5 Ohm 100W resistor to see how much the AC voltage drops under load.Or I could use 2 bridge rectifiers, a capacitor and 2x 12V DC bulbs one for each winding to test the (somewhat loaded) unregulated DC voltage.Then I could subtract 1.5V from that voltage (for the LM317/337) and see what I'm left with..I don't have any big capacitors here and I won't be having any until Monday so I can't directly test the power supply since the PCB is unpopulated atm.

author: Zero999
author_rank: Super Contributor
timestamp: August 25, 2018, 07:32:26 pm
content: Quote from: belzrebuth on August 25, 2018, 06:56:28 pmHave you measured the voltage, with a load attached? That would enable you to work out the impedance of the transformer. Z = (VUNLOADED-VLOADED)/IIt's also acceptable to assume it will output 12V, with a 3.5A load, but that will only be when the mains voltage is at the secondary rating. If it's lower, then the secondary voltage of course will be less.It's generally recommended to allow a voltage overhead of 3V for the LM317 and LM337, but it's possible to get away with less, especially for low output currents.

author: Ian.M
author_rank: Super Contributor
timestamp: August 25, 2018, 08:07:37 pm
content: Don't bother load testing with bridge rectifiers etc. Just connect your 12V *INCANDESCENT* bulbs or power resistors direct to the secondaries. You cant use LED bulbs for this.N.B. metal case power resistors *MUST* be heatsinked otherwise they are only good for typically less than 10% of their full rating.Your 3.5A secondary will only be good for 2.17A max DC current. See Hammond (transformer division)Design Guide for Rectifier Use- Full wave bridge, capacitor input load.As others have said, its unlikely that there will be enough headroom for a clean ripple-free output of more than a few hundred mA. Get us some real measurements for voltage drop on load and we can tell you if its going to be worth building this.

author: mariush
author_rank: Super Contributor
timestamp: August 25, 2018, 08:18:36 pm
content: The peak DC voltage will be approximately Vac x 1.414 minus two times the voltage drop on the diodes that form the bridge rectifier, because at any point in a bridge rectifier, there's two diodes conductingThe voltage drop will vary with the current, the heat of the rectifier... if you oversize the bridge rectifier (for example use a 20-40A bridge rectifier that can be easily found both as new or in some dead atx power supplies) then the voltage drop on these diodes will be around 0.8v... but for simplicity we can just use 1v per diode.... The max AC current will be approximate 0.62 x Iac .You have 100VA toroidal, so that's 50VA per secondary or 50VA/12 = 4.16A ... which means you'll have approximately Idc = 0.62 x 4.16 = 2.58 A so your transformer is sized properly to handle 1.5 A, the maximum for your LM317.So you'll have Vdc peak = 12v AC x 1.414 = ~ 17v - 2 x 1v (drop on diodes) = 15vDepending on the transformer quality, at very low loads, the transformer may actually output more than the nominal 12v AC so it's best to be aware of that and pick suitable components with this in mind. For example, knowing this I would not assume that just because my estimated maximum DC voltage is 15v it is safe to use a capacitor after the bridge rectifier that's rated for maximum 16v, I'd use one rated for at least 25v, ideally 35v.Also, the transformer's output may sag a bit below 12v in certain conditions (like mains voltage dropping a bit, combined with high current output)The linear regulators need to have the input voltage always higher than output voltage by some amount, in order to regulate the output properly. In the case of LM317 regulators, in order for them to regulate well across the whole range of 0.. 1.5A the datasheets say you must have the voltage at around 1.5v above the output voltage.So after the bridge rectifier, you must use a capacitor that will buffer the voltage and make it so the voltage there will be at least 13.5v as much as possible.To estimate capacitance you can use this formula which approximates the size :C = Current / [ 2 x AC frequency x ( Vdc peak - Vdc desired) ]So let's say you want 1.6A of current (a bit more than maximum 1.5A the LM317 is supposed to output), and let's say you want at least 13.5v all the time, and you assume the peak DC voltage will be 15v DC and let's say you're in US where the AC frequency is 60Hz ... put the numbers in formula and ...C = 1.6A / [2 x 60 x (15-13.5) ]= 1.6 / 180 = 0.0088888 Farads ... or around 8888 uF ... so you would go with the next higher standard value that's easy to buy ... 2 x 5600 , 2 x 6800 , 8200uF , 2 x 8200 uF , 10000uF ... etcMore capacitance doesn't hurt, up to a point ... way too much capacitance can "stress" the bridge rectifier or the fuse in front of the transformer ... the capacitors basically act a short circuit "sucking" a lot of energy in them, which makes the bridge rectifier and the fuses "see" more current for a very brief period of time, while the capacitors fill up with energy.

author: drussell
author_rank: Super Contributor
timestamp: August 25, 2018, 08:21:42 pm
content: Quote from: IanB on August 25, 2018, 06:37:15 pmI would agree if it were a 12 VA transformer (well, 24 since we're talking about a 12/24 volt dual secondary unit), but it is a 100 VA transformer, and it is a torroid, which generally have pretty good load regulation compared to a typical EI core unit. Since the OP is unlikely to draw more than about 50 VA absolute maximum with a couple of 1A class regulators, I have a funny feeling that it will beJUSTenough in this particular case.The fact that it is only putting out 12.9 volts AC unloaded instead of 14-15 or so leads me to suspect that it will be pretty stiff (load-regulation wise) since it is rated 3.5A at 12 volts AC per secondary winding.Like others here, I will be interested to see the actual test results.

author: Zero999
author_rank: Super Contributor
timestamp: August 25, 2018, 08:26:34 pm
content: Quote from: Ian.M on August 25, 2018, 08:07:37 pmYes the load must be linear, resistive: no rectifiers, or LEDs.Using a fullwave bridge rectifier results in current surges being draw, rather than a continuous current, which will cause increased heating in the transformer, hence the de-rating, and will distort the voltage waveform, which is why a resistive load must be used, to determine the transformer's impedance.Quote from: drussell on August 25, 2018, 08:21:42 pmPossible, but I wouldn't be too sure. Another issue is enough headroom should be provided to ensure it works, when the mains voltage is on the lower end of the tolerance band, which can ruin your day.

author: David Hess
author_rank: Super Contributor
timestamp: August 25, 2018, 09:24:12 pm
content: 12 volts AC * 1.414 = 17 volts peak17 volts peak - 2 Vdiode(silicon) = 15 voltsThat leaves only 3 volts of headroom for the 317 and 337 regulators which is not enough. So replace the standard diodes with schottky diodes.17 volts peak - 2 Vdiode(schottky) = 16 voltsThat might be acceptable but 1 volt of input ripple at 1.5 amps requires a lot of input capacitance.8200 microfarads*volts/amp * 1.5 amps / 1 volt = 12,300 microfarads15,000 or 22,000 microfarads of input capacitance is a lot for only 1.5 amps but acceptable if a 12 volt AC transformer is all that you have to work with. The capacitance could be halved by using low dropout regulators allowing an increase of the input ripple to 2 volts.Do not forget that the input rectifiers have to handle the surge current to charge the excessively large capacitors. The transformer helps by saturating which limits the surge current but use big rectifiers anyway. And big rectifiers will boost the voltage a little by having less voltage drop.Oh, and if you have two separate AC windings, then you do not need a positive and negative regulator unless you want a bipolar supply with only one control. Two positive regulators can be used and connected only at their outputs; just treat them as two separate floating regulators.

author: Zero999
author_rank: Super Contributor
timestamp: August 25, 2018, 10:44:38 pm
content: Quote from: David Hess on August 25, 2018, 09:24:12 pmIf a bipolar power supply is required, it's better to configure the transformer as centre tapped and use one bridge rectifier, than a bridge rectifier on each secondary, because it will have one less diode drop.

author: David Hess
author_rank: Super Contributor
timestamp: August 25, 2018, 11:34:41 pm
content: Quote from: Hero999 on August 25, 2018, 10:44:38 pmThat is a great point. And still use schottky rectifiers to reduce the voltage drop further.

author: drussell
author_rank: Super Contributor
timestamp: August 26, 2018, 09:06:44 am
content: Quote from: Hero999 on August 25, 2018, 10:44:38 pmIndeed. It would be silly to use two bridge rectifiers in this case and it was my assumption that the OP intended to use just one since a +12/-12 supply was specified. Obviously if you need dual floating supplies (which can be easily done with a transformer with two secondaries rather than one centre-tapped winding, without modification) then you need them separate but in this case, a single bridge is the hot set-up.

author: belzrebuth
author_rank: Frequent Contributor
timestamp: August 26, 2018, 09:54:36 am
content: Some great points here; thanks everybody.Thing is that I've already have the PCB for this power supply so I can't use two bridge rectifiers or 7812/7912.I believe the reason the designer used LM317/337 is to get as close to 12V as possible.The way I see it if I end up using this transformer is that I should up my capacitance to 12,000uF and use schottky diodes for rectifying..Does anybody have a recommendation for a schottky bridge rectifier I can use in place of the KBU401?I often buy from TME.eu but I don't think I've found something there..

author: belzrebuth
author_rank: Frequent Contributor
timestamp: August 26, 2018, 10:34:12 am
content: I'm going with this capacitorhttps://www.tme.eu/gb/details/lgu1v153melb/105c-snap-in-electrolytic-capacitors/nichicon/Price is good considering its size.Most bridge rectifier packages state Vf of about 1V.I've already have a KBU4J and FBIU4D7M1 here..The KBU4J has a Vfmax=1V.

author: Ian.M
author_rank: Super Contributor
timestamp: August 26, 2018, 11:44:31 am
content: Post a photo of the PCB. I'd bet the best option is to use a chassis mount bridge or paired Schottkys in TO-247 or similar packages on silpads and either wire them back to the bridge footprint on the PCB or link out the bridge to turn the PCB's AC in terminals into + and - DC in.N.B. packaged Schottky bridges are rare like hens teeth. Generally when you need that high current, low Vf and fast recovery its a switching supply with a CT secondary so dual common cathode Schottky diode packages are pretty common. High current common anode ones (that would be used for a negative rail) are a lot rarer, and series (half bridge) are rarer still.I couldn't find an appropriate common anode one in stock at Mouser: quite a few show up in the search but when you check the datasheets they are actually common cathode.Mouser have ONE type of series pair in stock:https://www2.mouser.com/ProductDetail/IXYS/DSSS30-01AR(100V 30A) $6.18 each, with a max Vf drop of 0.5V @3A @25 degC. Its Vf drop decreases significantly with increasing temperature, so ideally you'd let it run rather warm, with the heatsink area determined by your max load current.However at that price (and you'd need two) and relatively high Vf, you'd probably be better off with four individual Schottky diodes even though they'd be far more of a PITA to mount.

author: belzrebuth
author_rank: Frequent Contributor
timestamp: August 26, 2018, 12:00:56 pm
content: PCB photos here:http://musicfromouterspace.com/analogsynth_new/POWERSUPPLY2009/POWERSUPPLY2009.phpI could use four individual diodes but yes, it would certainly be uglier and more difficult to mount.

author: Ian.M
author_rank: Super Contributor
timestamp: August 26, 2018, 01:10:46 pm
content: Yes, as I thought, the best option is to move the bridge or discrete diodes off-board. Just link the outer pins of the bridge footprint, each to the nearest middle pin and Sec1 becomes bridge +, and Sec2 becomes bridge -Using individual Schottkys isn't too horrible if they are isolated tab packages - just bolt them down with a dab of heatsink compound, and wire them up to a nearby tagstrip. Its probably worth putting Kapton tape on the chassis under the diode legs so they cant short to the chassis if they ever get bent down.

author: belzrebuth
author_rank: Frequent Contributor
timestamp: August 26, 2018, 01:22:07 pm
content: Could you recommend a particular schottky package that would fit my needs?I've never used those so I'm a bit clueless on what to get.I've only seen them on SMPS power supplies I think..

author: Audioguru
author_rank: Super Contributor
timestamp: August 26, 2018, 02:09:48 pm
content: I haven't looked at the waveform of my mains electricity for many years. I betcha it has its peaks squashed by all the full wave rectifiers and filter capacitors in nearly everything I have turned on most of the time. Then the formula about the peaks are 1.414 times the RMS voltage goes out the window and your LM317 powered from a 12V transformer and rectification will have poor voltage regulation.Does your mains electricity have a voltage drop each time the refrigerator or air conditioner turns on? Mine doesn't because the transformer feeding mine and a few other homes is next door.

author: Audioguru
author_rank: Super Contributor
timestamp: August 26, 2018, 02:13:25 pm
content: Quote from: belzrebuth on August 26, 2018, 10:34:12 amFor each diode. A bridge rectifier hastwo diodes in seriesso the voltage drop is 2V.

author: belzrebuth
author_rank: Frequent Contributor
timestamp: August 26, 2018, 02:20:04 pm
content: I've just discovered that I also have two identical toroids (also 12V).These output 13.8V unloaded.They're 50VA each so I could maybe try those two for a 2x12V substitute.I think I have other transformers laying around too but they're not toroids..I really need to get the psu board populated in order to perform some hands-on tests (I could monitor the voltage while turning the air conditioner on for example).Getting a new transformer is not that expensive but I wouldn't mind trying if what I've already have works..

author: Ian.M
author_rank: Super Contributor
timestamp: August 26, 2018, 02:20:55 pm
content: Quote from: belzrebuth on August 26, 2018, 01:22:07 pmThis looks good:https://www.tme.eu/gb/details/stps1045fp/tht-schottky-diodes/st-microelectronics/45V 10A Schottky in a two pin Iso-TO-220 package.Bend their legs up at right-angles just beyond where they neck down, bolt four down on a smear of heatsink compound using a plain washer + a thin belleville or crinkle washer on top for even clamping pressure (a spring washer is a poor but usable substitute) and wire up as follows:Use solid wire for the links and sleeve the +DC and -DC ones. It should be reasonably neat.

author: David Hess
author_rank: Super Contributor
timestamp: August 26, 2018, 02:23:20 pm
content: Something like a1N5822which is 3 amps would be my first choice. Use aterminal stripto wire the diodes in the air and mount them to the chassis.

author: belzrebuth
author_rank: Frequent Contributor
timestamp: August 26, 2018, 02:23:52 pm
content: Thanks Ian.Much appreciated

author: Ian.M
author_rank: Super Contributor
timestamp: August 26, 2018, 02:30:44 pm
content: Quote from: belzrebuth on August 26, 2018, 02:20:04 pmTime to do some load testing. 12V 20W G4 halogen bulbs will do nicely for test loads assuming you are shooting for 1A DC out from the finished PSU. You can hook them up with croc clip leads and put them on a ceramic tile if you are only doing a short test, but its worth buying some ceramic G4 bulb sockets if you want to do a longer test or for future use.

author: belzrebuth
author_rank: Frequent Contributor
timestamp: August 26, 2018, 02:36:19 pm
content: First thing tomorrow I'm off to buy those bulbs.It's Sunday I won't be finding any shops open today..And I've loaned most of my high wattage resistors to a friend that needed to load-test some PSUs so I only have a 100W 20Ω here

author: Ian.M
author_rank: Super Contributor
timestamp: August 26, 2018, 02:45:18 pm
content: N.B. bare halogen bulbs blow if you touch them before use. They have a quartz glass envelope, and if you leave a finger print on the glass, the oils in your sweat carbonise and cause the glass to overheat + the salts in your sweat fuse with the glass and lower its melting point. Always handle with a tissue and wipe with methylated spirits or IPA after handling.

author: mariush
author_rank: Super Contributor
timestamp: August 26, 2018, 04:19:31 pm
content: Just go to a car or motorbike parts store and you should easily find 10w/20w/40w incandescent light bulbs , and cheap...Otherwise, any regular store should have cheap bulbs ... for example my desk lamp uses G4 bulbs , which are 12v 20w :https://www.amazon.com/CBconcept-10XG412V20W-Halogen-20-Watt-12-Volt/dp/7542904760/ref=sr_1_3?ie=UTF8&qid=1535300104&sr=8-3&keywords=g4+bulbsJust use some paper napkin or something if they're the halogen kind... otherwise the grease can cause problems.Oh... and it's overkill for this project, but when you really must reduce the voltage drop on diodes an alternative would be to get rid of the diodes completely and use a chip like Linear (now Analog?) LT4320:http://www.analog.com/media/en/technical-documentation/data-sheets/4320fb.pdfGet this chip in a DIP package along with 4 through hole mosfets on a prototyping board and you got yourself a bridge rectifier with maybe 0.1-0.2v voltage drop, depending on your mosfet selection... the only downside is the cost... a few dollars for the chip and less than a dollar for the 4 mosfets.Or get this in a surface mount package along with 4 surface mount mosfets and you can basically have a whole bridge rectifier in a TO220 style package, as the losses would be very small at 1.5-2A current output.

author: floobydust
author_rank: Super Contributor
timestamp: August 26, 2018, 05:06:12 pm
content: Measure your main's voltage, all this math can be wrong.I did a linear (audio) power supply like this and it worked great in the lab.At the venue, the line voltage was ~5% lower and the musicians had hum coming out of the power supply, worse when stage lighting was busy. Had to change the transformer from 12.6V to 14V. Schottky rectifiers can give you another volt - all assuming your line voltage is not high. It also changes throughout the day.The PCB you are showing does not have ideal connections to the transformer center-tap, so expect more ripple on the -ve rail.To get less than 5mVpp output ripple, you have pay attention to the PCB trace layout.

author: belzrebuth
author_rank: Frequent Contributor
timestamp: August 27, 2018, 01:01:17 pm
content: Okay I load-tested the transfomers, here are the results..100VA transformer with 20W g4 halogen bulb on one of the secondaries-> 12.6Vac50VA transformer with 14W g4 halogen bulb on one of the secondaries-> 13.2VacThe parts for the PSU pcb should be here within 3 days so I will post results then..I've ordered 4 of the low drop diode packages IanM suggested.I have a KBU4J here I could also test.I'll need to get find heatsinks though..

author: belzrebuth
author_rank: Frequent Contributor
timestamp: August 27, 2018, 01:16:11 pm
content: Should I have measured both the secondaries of the 100VA transformer?I thought about loading each with a 14W bulb.That's 1.2A at 12V.I could also use the 20W bulbs,then it's 1.67A for each winding but then I'm too close to the max current rating..I could do that for a short period though since I will still be within the transformer's tolerance.If I still get more than 12V I think it's going to be ok if I use the schottky diodes..I'm going to probably try to turn a heater or an airconditioner on while still monitoring the voltage..

author: Ian.M
author_rank: Super Contributor
timestamp: August 27, 2018, 01:55:44 pm
content: For transformers with two secondaries load *BOTH* secondaries! If you don't, the losses in the primary are far lower, and its impossible to calculate the true equivalent resistance referred to the secondary.You aren't going to damage the transformers running them at 100% load for 10 minutes in free air. In fact they'd probably be OK with 150% load (50% overload) for that long starting from cold as there is a fair bit of thermal mass so no part of the winding is likely to exceed its maximum insulation temperature. Its therefore good practice to aim for 80%-100% of full load when testing.For the 50VA transformer please use 2x 20W bulbs, one on each secondary, as 20VA@12V is 1.67A RMS which, when you apply a factor of 0.62 for bridge rectification with capacitive input load, corresponds to a DC load current of 1A, which is right where we want to be operating as max load for a LM317/LM337 based PSU.Normally you'd load the 100VA transformer with 2x 20W bulbs PER SECONDARY, but as you specifically are using regulators with a max rating of 1.5A DC, there is no point in going over 2.4A RMS which is 30W @12V.Due to the non-linear resistance vs voltage curve of the bulbs, it would also be helpful if you measured the actual load current of one of the bulbs for each transformer.Also, for our convenience and to guard against mains voltage fluctuations with time of day or week, please repeat the unloaded voltage measurements immediately before and after the loaded measurements and post all the results in the same post.So far it looks like the 100VA transformer is likely to be the best bet.Turning on your heater or air-conditioner etc. is unlikely to have much effect unless your house's wiring is in particularly poor condition. Problems start when a high proportion of the population turn on a high load appliance at the same time. The classic case was the U.K's Coronation Street soap opera on ITV, which regularily aired late afternoon. During its commercial breaks, you could bet that a high proportion of its viewers (over 9 million households in the '60's and '70's) would turn on a 3KW kettle.https://en.wikipedia.org/wiki/TV_pickupDinorweg pumped storage powerstation in the UK was built to time-shift peak loads, but as it has the fastest spin-up time for a major power station in the UK, its currently mostly used to supply the rapid demand surges caused caused by TV pickup.https://en.wikipedia.org/wiki/Dinorwig_Power_Station(Its pretty incredible - 12 seconds to go from near 0 to 1320 MW)

author: belzrebuth
author_rank: Frequent Contributor
timestamp: August 27, 2018, 03:08:43 pm
content: So to summarize:For the 100VA transformer I need to attach a 20W bulb per winding and actually log the current draw as well as the voltage.I'll leave it wired and test at 4-5 different times a day.Correct?I didn't quite understand what you've said about the 50VA transformer though..The 50VA transformer is 1x12V so I only have one secondary.I have two identical units so essentially they could be used as one 2x12 that's why I mentioned it.I may need to focus on the 100VA transformer; having two transformers on such a small case might be inconvenient.For reference this power supply will be mounted on a wooden case used for a modular synthesizer so there's no metal chassis or too much space for huge heatsinks etc..I also wonder what kind of heatsinks I could use for my LM317/LM337:Mounting them off the pcb will be difficult so I thought about getting somethink like this:https://www.tme.eu/gb/details/sk104-38sts/radiators/fischer-elektronik/Or this:https://www.tme.eu/gb/details/hs-132-38/radiators/stonecold/

author: IanB
author_rank: Super Contributor
timestamp: August 27, 2018, 03:19:51 pm
content: Quote from: belzrebuth on August 27, 2018, 03:08:43 pmThis will be awkward since wood is an insulator. If you put a heatsink on a part inside a wooden box it won't be a heatsink any more since the heat will be trapped inside the box and can't escape. You might need to consider drilling vent holes in the box and having a small fan to move air through it and carry the heat away. Heat sinks always need some kind of air flow to work, either natural or forced convection.

author: belzrebuth
author_rank: Frequent Contributor
timestamp: August 27, 2018, 03:27:56 pm
content: Fan is not an option unfortunately.It's awkward enough as it is.This a studio instrument so no fans is absolutely essential..Anyway, I've tested the transformer again with a 20W bulb attached to each of the secondaries.Current draw is 1.75A and the voltage is 12.5V (per secondary).I can try it with 2x 20W in series but wouldn't I exceed the rating?Is the 3.5A rating per winding or for the whole transformer?!?

author: Ian.M
author_rank: Super Contributor
timestamp: August 27, 2018, 04:06:56 pm
content: For the 50VA transformer, 1x 20W bulb would be an appropriate test load for its single winding. I expect the result will be somewhere around 12.9V loaded.100VA/(2*12V)=4.17A so 3.5A per winding would be a pretty conservative rating, (or a 100% rating if its actually an 85VA transformer)2x 20W in PARALLEL per winding (4 bulbs total) would put it right at or possibly slightly over its max rated current. Not a problem for 10 minutes per hour, but not something you'd want to do 24/7. However you only need to test it to about 50VA which you've already done.For 1A out per rail, assuming your 100VA transformer, neglecting unloaded losses in the primary and core, input power is 2*12.9V*1A/.62=41.6W. Output power is 2*12V*1A=24W. The difference of 17.6W must be got rid of.That's enough to make a medium sized closed wooden box pretty toasty after a few hours running, and it gets a lot worse if a significant portion of that 24W is also used inside the same box. I'm with IanB - its either going to need a fan or an externally mounted heatsink. As you are dead set against a fan, the only real option is a rear mounted external heatsink with its fins vertical for better convection.That PCB layout is very stupid from a heatsinking point of view. The regulators and the bridge rectifier should have been mounted along one board edge so they could bolt to a vertical heatsink. The best option, if its got plated through holes would be to mount the regulators under the board with their legs bent up. However that doesn't mesh well with your possible heatsinking options, so you may have to bite the bullet and mount the regulators off-board, with 0.1uF to ground from input and output right at the regulator so it doesn't break into RF oscillation due to wiring inductance.N.B. LM317 tab is Vout and LM337 tab is Vin so they both need isolated mounting kits if you want to use a common heatsink. A minimum thickness mica washer, thermal grease, and an insulating bush on the screw is the best option.

author: belzrebuth
author_rank: Frequent Contributor
timestamp: August 31, 2018, 03:14:06 pm
content: I've built the power supply with a standard bridge rectifier (although I've bought the diodes mentioned by Ian.M).I can draw about 0.6A without any voltage drop but at after 1.0A I get a voltage drop of about 0.2 to 0.5V which may or may not matter for the modules I'm going to power.I better be on the safe side and have 12,000 at all times if possible.I was thinking instead of replacing the diodes and mess with the outboard wiring and heatsinking couldn't I just find a pin to pin replacement LDO regulator for the LM317/LM337?I need low-noise low need for regulation voltage (about half of the that of the LM317 would be nice) and about the same or more max current rating..Is there something that fits these specs?I'm not particularly concerned about the cost since it's only two pieces I'm interested in.

author: Ian.M
author_rank: Super Contributor
timestamp: August 31, 2018, 03:50:41 pm
content: Have you scoped the output rails under load? Your 0.2V to 0.5V voltage drop may actually be a ripple breakthrough of double that! If you don't have a scope, put a 1uF non-electrolytic cap (or nearby value) in series with your DMM on AC V and multiply the result by 2.8 to approximate the peak to peak ripple voltage.Quote from: belzrebuth on August 31, 2018, 03:14:06 pmYou could, but you'll have a hard time finding one that will 'drop in'. Even if the pinout is correct, LDOs typically have far tighter constraints on input and output capacitance and especially the ESR range required, than LM3x7 regulators and if you get it wrong there's a risk of them breaking into oscillation, which I've seen bad enough cases of in the HF band to disturb other electronics over 30m away.Depending on the load you are powering, it may be acceptable to tweak the Ajd pin to Gnd resistor values down a bit to drop the regulated voltages to a symmetrical 11.5V to give the regulators more headroom. That's certainly better than letting them droop on their own as it avoids ripple breakthrough.The diodes I proposed wont get as hot as the existing bridge rectifier. Bolt them down on a sheet of Aluminum that's big enough for a 1" margin round them and they'll do fine.Another option if its an open center toroidial transformer would be to add two overwinds to put in series with the existing secondary windings to boost the voltage. Start by putting 10 turns of hookup wire round its core, through the center. Measure the AC voltage across the hookup wire ends, then calculate how many turns you need to get the voltage headroom you need. Post a photo and we can suggest how to wind it.

author: David Hess
author_rank: Super Contributor
timestamp: August 31, 2018, 06:00:54 pm
content: Quote from: belzrebuth on August 31, 2018, 03:14:06 pmThe LT1086/LM1086/LD1086 is a low dropout improved 1.5 amp replacement for the LM317.The LT1085/LM1085/LD1085 is a low dropout improved 3 amp replacement for the LM317.The LT1033 is a low dropout improved 3 amp replacement for the LM337.

author: floobydust
author_rank: Super Contributor
timestamp: August 31, 2018, 06:12:05 pm
content: I found no benefits to increasing the filter capacitors past 10,000uF because the noise of the LM317/337 dominates. Some manufacturers LM317's have a lot of popcorn noise. LT I found the best and Fairchild the worst for noise performance.There is a performance limit with these convenient 3-pin parts. To do much better, you need more than 3-pins in an IC Vreg package, in order to connect a reference filter capacitor.Adding huge output capacitors 1,000uF-4,700uF to the LM317's is what I see commonly done, as a compromise.

author: David Hess
author_rank: Super Contributor
timestamp: August 31, 2018, 06:19:11 pm
content: Quote from: floobydust on August 31, 2018, 06:12:05 pmBypassing the lower resistor in the divider with a capacitor goes a long way to reducing the output noise but I agree; better performance requires a different part or circuit.

author: hli
author_rank: Frequent Contributor
timestamp: August 31, 2018, 09:47:49 pm
content: Quote from: David Hess on August 31, 2018, 06:00:54 pmOnly partially. I recently stumbled across a crucial difference between the LT1085/86 and the LM317. I have a PSU with a 24V transformer (which is good enough to deliver 24V/2A regulated with about 1V headroom for additional ripple), where I wanted to replace the LM317 with a LT1085 (to actually get the 2 amps).The difference is that the LT108x enforce their SOA much earlier than the LM317. so when you are starting into a bigger load, the differential voltage between input and output can get big enough that the regulator starts enforcing its SOA, and limits the output current to about half an amp (see the 'Short-circuit current' diagram in the datasheet). It took a while until I discovered what the problem was. The datasheet claims that this should not happen, but my experience was otherwise

author: belzrebuth
author_rank: Frequent Contributor
timestamp: August 31, 2018, 09:52:39 pm
content: Since the transformer is able to drive bigger currents and the suggested regulators ( LT1033/LT1085) are low dropout with higher current capability maybe I could replace the LM337/LM317 combo with the LT1033/LT1085 combo.I'm really interested in any circuit suggestions and maybe try things out and see how it works.I see that the LT1085 has very similar circuit configurations (even the 240Ω resistor which I've read somewhere that should be 120).Maybe I should just swap or add only a few components and my PSU will upgrade significantly current-wise..The PCB is large enough for any component additions or modifications..In this page there is the schematic and PCB layout of the PSU.http://musicfromouterspace.com/analogsynth_new/POWERSUPPLY2009/POWERSUPPLY2009.php

author: David Hess
author_rank: Super Contributor
timestamp: September 01, 2018, 12:40:02 pm
content: Quote from: hli on August 31, 2018, 09:47:49 pmHow does the datasheet claim it should not happen while including SOA data in both the specifications and curves?The difference is about 7.5 versus 15 volts but the LM317 has almost twice the junction-to-case thermal resistance making up for this. Under the same conditions, the current limit curves are actually better for the LT1083/4/5. The guaranteed tested curves are lower but I suspect this reflects a more conservative attitude because the LM317 datasheet does not even include them.The above reflects my biggest complaint about 3 terminal and most integrated regulators. Without a separate current limit adjustment, the peak current is much higher than the sustainable continuous current in a variable output application. I would be nice to be able to parallel multiple regulators to lower the junction-to-case thermal resistance without raising the total current limit.Quote from: belzrebuth on August 31, 2018, 09:52:39 pmThe LT1033 and LT1085 datasheets have some application notes worth studying.QuoteThis resistance sets the current through the output divider. 120 ohms is commonly used because 1.25 volts / 120 ohms is 10.5 milliamps which is higher than the minimum required load (10mA for the LT1085 and 5mA for the LM317) under all conditions. If an external load is always present, then this resistance may be higher.

author: hli
author_rank: Frequent Contributor
timestamp: September 01, 2018, 08:06:08 pm
content: Quote from: David Hess on September 01, 2018, 12:40:02 pmTo quote the datasheet:Like any of the IC power regulators, the LT1083 has safe area protection. The safe area protection decreases the current limit as input-to-output voltage increases and keeps the power transistor inside a safe operating region for all values of input-to-output voltage. The LT1083 protection is designed to provide some output current at all values of input-to-output voltage up to the device breakdown.When power is first turned on, as the input voltage rises, the output follows the input, allowing the regulator to start up into very heavy loads.I understand this as "you can start into a heavy load and it should work".Quote from: David Hess on September 01, 2018, 12:40:02 pmFrom what I can see the LM317 actually does not enforce its SOA. It will limit over-current, and otherwise will reduce output current only when it gets hot. This means that you can, short-term, violate its SOA (or stated otherwise: it has no defined SOA)It might be that I dug my own hole, though, since I have an additional capacitor at the ADJ pin - this might slow down the rise of the output voltage just enough to exhibit this problem in the right circumstances. I did not experiment with this too much, since I had a LM350 at hand which behaves like the LM317 in this regard.

author: dzseki
author_rank: Frequent Contributor
timestamp: September 01, 2018, 08:50:12 pm
content: I've had some bad experience with LT1083 as well... I wanted to build a wide range yet simple adjustable supply, the transformer was 24VAC, and at 1.2V output the max current was embarrasingly low, like 100mA or so. The other day I wanted to test the output current limiting, so I shorted the output, the IC instantly blew up, go figure.Then I abbandoned this project and went back to good old uA723

author: David Hess
author_rank: Super Contributor
timestamp: September 01, 2018, 11:57:18 pm
content: Quote from: hli on September 01, 2018, 08:06:08 pmThe exact same thing applies to the LM317 and LM337 and it is discussed in some old application notes. They can start into heavy loads as the input voltage rises but an output short can result in the regulator latching off at high input-to-output voltages.QuoteI know the LM317 enforces its SOA at low temperatures but it may be not as picky as the improved Linear Technology parts which have their current limit trimmed during production.QuoteA partial solution is to use a higher current part but this comes with the disadvantage that I mentioned of raising the current limit which may not be desirable. Another solution is to add an input cascode transistor to absorb excessive input-to-output voltage. And another solution is to use an input transistor to boost the output current of the regulator. Of course adding external power transistors comes with the risk of damaging the power transistor under adverse conditions if care is not taken with the circuit design.