I understand the basic idea of an autotransformer, using a tapped winding to raise or lower a voltage and no isolation (which isn't necessary for my application anyway).
If I have a 60VA 240V primary mains transformer (the secondary voltages are for this application, as I understand it, irrelevant as long as I don't draw current from them) could I reasonably expect to be able to draw, say, 40 watts from the centre tap to provide a 120V output or does the 'misuse' of the centre tap derate the transformer in some way (if it does, I'd be interested to know the reasons) It's for a simple AC motor in a Christmas decoration. Thanks..
To be conservative you don't want to draw any more than the rated current out of any one terminal, to avoid overheating that set of windings. The 'real' allowable draw is somewhere between that amount and an amount that causes the total winding dissipation to be equal to the transformer used it it's maximum rating -- but without knowing the details of how it's wound and how good it's thermal conductivity is, you can't tell.
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Tim Wescott
Wescott Design Services
http://www.wescottdesign.com
Do you need to implement control loops in software?
"Applied Control Theory for Embedded Systems" was written for you.
See details at http://www.wescottdesign.com/actfes/actfes.html
40 watts sounds fine to me. Try it and check after a while to see if the transformer is getting hot. Too hot to touch, which is ballpark
55C, might be a reasonable limit.
An autotransformer is generally rated at twice the VA as the same device used as a transformer (with no common connection). If the center tap is on the primary and it is designed to work at the same VA rating with 120 VAC input, then that half of the winding should be able to handle the current for 60 VA at 120 VAC, or 0.5 A.
If you have 240 VAC on the primary, and a 60 VA load on the center tap, there will be 0.25 A flowing in the top half of the winding, and 0.25 A flowing out of the common connection back to the source. The center tap will provide a total of 0.5 amps into the load, with 0.25 amps in each winding. The vector sum of all currents at each node will be zero.
The only thing that might limit this capability is the size of the wire to the actual center tap of the transformer, but usually this is provided by the two windings being brought out to a terminal or a flexible stranded wire which is probably more than adequate to handle the current.
If you do an LTSpice simulation (as follows), you will see that with V(IN)=240 VAC and R=240 ohms, and a transformer consisting of two coupled 10 H inductors:
I(IN) = 0.25 A @ 240 VAC = 60 VA I(L1) = I(IN) = 0.25 A I(L2) = 0.25 A (180 deg) I(R1) = 0.50 A @ 120 VAC = 60 VA
I should clarify that you can get 120 VA if you have a 60 VA transformer with two 120 VAC windings. For a center tap you only get the same VA as rated.
Yes, that'll probably work (assuming there IS a center tap). The 60 VA rating, for primary and secondary current flowing in the winding resistance, is lower than what you'd expect from the autotransformer connection. If the primary winding is 4 ohms, 60 VA at 240V creates 0.25W of resistive heating in the primary, and another 0.25W in the secondary.
60VA as a transformer yields half a watt of heat.
60VA as an autotransformer, there are two 2 ohm branches carrying
0.25A each (to make 0.5 A at the 120V tap), which yields a quarter watt of heat.
It would seem that, on heat-generation alone, the autotransformer rating would be higher than the transformer number.
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