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Registered Member #190
Joined: Fri Feb 17 2006, 12:00AM
Location:
Posts: 1567
A transformer has a current rating, like 36vct @ 1.5A. I don't exactly recall, but I am guessing this is the current you can draw without exceeding a certain amount of voltage drop-out. Is it 10%?
Anyway, I am taking the 18v from the center-tap and rectifying it. Without any load is is near 25v. This goes to a 15v voltage regulator so I have plenty of headroom. If I draw more than 1.5A the voltage from the transformer sags, but I still have plenty of voltage for my 15v regulator.
So the question is what does the industry recommend for an adequate current rating for a transformer? Would it be adequate to draw 2 or 2.5A since I have enough voltage to meet the regulator's needs, or does one not do this because of excessive transformer heating, and possibly a shorter working life?
Registered Member #1232
Joined: Wed Jan 16 2008, 10:53PM
Location: Doon tha Toon!
Posts: 881
The transformer rating 36v @ 1.5A means exactly that. It will output 36 volts when you draw the full rated 1.5A from it's secondary. With less than the full rated current from the secondary the voltage will rise. How much it will rise above 36 volts depends on the transformer's "regulation" specification. If it is specified as 10% regulation (a fairly typical value) then the voltage can go as much as 10% higher than the full-load voltage when you remove the load completely.
The "over-voltage when lightly loaded" behaviour is due to leakage inductance (less than ideal magnetic coupling) in the transformer and the IR drop due to the winding resistance.
You typically wouldn't draw more than the rated current from a transformer winding for any amount of time as this is eventually likely to overheat the transformer. This is particularly so if it is already operating in a high ambient temperature and is quite a small transformer to begin with.
However, for a big transformer >1kVA rating (like pole pigs) there's so much iron and copper in there that it won't overheat quickly even with considerable over current. For instance if you draw twice the rated current the I²R heating will quadruple! But the transformer still has a thermal time constant that prevents it from overheating instantly. It may or may not burn out after several hours of overload though, depending on how over-designed it was to start with.
Registered Member #2463
Joined: Wed Nov 11 2009, 03:49AM
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Posts: 1546
Geordieboy wrote : The "over-voltage when lightly loaded" behaviour is due to leakage inductance (less than ideal magnetic coupling) in the transformer and the IR drop due to the winding resistance.
To expand this statement :
The unloaded voltage of a transformer secondary reflects the actual turns ratio of the winding, which is adjusted to provide the correct voltage after rated load is connected. The ratio is the only means of controlling the output voltage once the flux level of the core is set by primary turns. Current can limited by altering flux paths.
Registered Member #162
Joined: Mon Feb 13 2006, 10:25AM
Location: United Kingdom
Posts: 3140
The short answer; If the transformer gets uncomfortably hot OR the voltage drops too much for regulation then you're drawing too much current, otherwise should be ok.
Are you using both sides of the center-tap? i.e. full-wave using two diodes, one from each 'end' This will allow more current than half-wave.
You can use this applet to see variations. (I use it) Check out the benefits of adding an inductor between the rectifier(s) and capacitor
Registered Member #2463
Joined: Wed Nov 11 2009, 03:49AM
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Posts: 1546
Fullwave, 2 diode, CT connection is very wasteful in that the copper loafs for half the time and the winding window, hence the core, has to be bigger. Your output voltage is only one half what the turns ratio could provide. The output current demands bigger copper. than would needed for equivalent wattage with a full wave bridge.
Registered Member #1875
Joined: Sun Dec 21 2008, 06:36PM
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Posts: 635
radiotech wrote ...
Fullwave, 2 diode, CT connection is very wasteful in that the copper loafs for half the time and the winding window, hence the core, has to be bigger. Your output voltage is only one half what the turns ratio could provide. The output current demands bigger copper. than would needed for equivalent wattage with a full wave bridge.
It's better than only using half the windings at all!
Registered Member #2099
Joined: Wed Apr 29 2009, 12:22AM
Location: Los Altos, California
Posts: 1716
ScotchTapeLord wrote ... But drawing half the voltage should mean that you could pull up to twice the rated output current without stressing the core, shouldn't it? In that case it would depend on the secondary coil's wire thickness...
Yes, but only with your qualification "stressing the core". Power loss in the core depends only on voltage, and establishes the transformer V rating. Core loss actually goes down slightly as current increases, because that reduces the internal voltage.
Power loss in the windings (copper loss) depends only on the current, and determines the transformer A rating.
Transformers have no stress (or loss mechanism) that depends on the product of volts and amps, or on the true power delivered to a load. A transformer operating at both its V and A ratings will have a tolerable temperature rise due to core + copper loss. Both loss mechanisms are nonlinear. Reducing current will allow hardly any more voltage. Greatly reducing voltage may allow a bit more A.
You -can- trade off V for A, within a given VA product, when you design the windings for a given core part number. For example, a transformer catalog could have 120V 10A and 240V 5A products, using the same core size but different tradeoff between wire thickness and number of turns.
You can also trade off V for A by using matched dual primary and/or secondary windings, which can be connected in series or in parallel.
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