Electronics > Beginners
Transformer voltage for dual power supply (+12V / -12v)
IanB:
--- Quote from: drussell on August 25, 2018, 06:09:23 pm ---Well, that depends on how beefy the transformer is, of course. :)
--- End quote ---
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.
--- Quote ---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.
--- End quote ---
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:
--- Quote from: belzrebuth on August 25, 2018, 06:56:28 pm ---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.
--- End quote ---
Have you measured the voltage, with a load attached? That would enable you to work out the impedance of the transformer. Z = (VUNLOADED-VLOADED)/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.
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