Transformer voltage for dual power supply (+12V / -12v)

[belzrebuth]

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?
 

[ocset]

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.

[belzrebuth]

Thanks!I'll try it then.
Here's the power supply I'm talking about..
http://musicfromouterspace.com/analogsynth_new/POWERSUPPLY2009/POWERSUPPLY2009.php

It 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..

[Zero999]

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.

[drussell]

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?

\$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.

[IanB]

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?

It depends on how much current you want to supply. If you limit the current to, say, 500 mA or less you might be OK.

[drussell]

It depends on how much current you want to supply. If you limit the current to, say, 500 mA or less you might be OK.

Well, 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 be right on 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.)

[Ian.M]

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.   

[ocset]

ooops sorry i  missed that you are using non LDO regulators......sorry, then 12vac is indeed too marginal

[belzrebuth]

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..

[IanB]

Well, that depends on how beefy the transformer is, of course. 

It 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.

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 be right on 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.

IMHO, 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.

[belzrebuth]

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.

[Zero999]

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.

Have you measured the voltage, with a load attached? That would enable you to work out the impedance of the transformer. Z = (V_UNLOADED-V_LOADED)/I

It'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.

[Ian.M]

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.

[mariush]

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 conducting

The 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) = 15v

Depending 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 ... etc

More 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.

[drussell]

IMHO, 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.

I 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 be JUST enough 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. 

[Zero999]

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 a bit over 2A DC.  See Hammond (transformer division) Design Guide for Rectifier Use - Full wave bridge, capacitor input load.

Yes 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.

IMHO, 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.

I 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 be JUST enough 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. 

Possible, 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.

[David Hess]

12 volts AC * 1.414 = 17 volts peak
17 volts peak - 2 Vdiode(silicon) = 15 volts

That 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 volts

That 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 microfarads

15,000 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.

[Zero999]

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.

If 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.

[David Hess]

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.

If 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.

That is a great point.  And still use schottky rectifiers to reduce the voltage drop further.

[drussell]

If 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.

Indeed.  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. 

[belzrebuth]

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..

[belzrebuth]

I'm going with this capacitor
https://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.

[Ian.M]

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.

[belzrebuth]

PCB photos here:

http://musicfromouterspace.com/analogsynth_new/POWERSUPPLY2009/POWERSUPPLY2009.php

I could use four individual diodes but  yes, it would certainly be uglier and more difficult to mount.