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#1 |
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diyAudio Member
Join Date: Jan 2006
Location: Denver
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I am hoping to verify I have this concept down correctly....
If I take a 220V primary, 2 x 35V secondary transformer and connect it to 110V I will have about 2 x 17.5 at the secondaries. Right? Thanks in advance! |
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#2 |
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diyAudio Member
Join Date: Dec 2008
Location: Victoria, BC
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Yup!
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#3 |
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diyAudio Member
Join Date: Sep 2002
Location: Lakewood, Ohio
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However the output power will be cut in half.
That is: If we had a 220VA transformer at 220 Volt and 1 Amp. Then at 110 Volt and 1 Amp we have 110VA.
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Kevin |
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#4 |
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diyAudio Member
Join Date: Nov 2010
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I'd guess you'd be able to draw off more current but regulation wouldn't be as good as a pukka 110 volt one as primary winding will have higher resistance.
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#5 |
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diyAudio Member
Join Date: Jul 2004
Location: Scottish Borders
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no,
cut the supply voltage by half and you cut the VA rating by half. The transformer windings, both primary and secondary are rated for current. This does not change, even though input and output voltages can be changed.
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regards Andrew T. |
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#6 |
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diyAudio Member
Join Date: Nov 2010
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Well, with all due respect to Andrew, he and I don't agree.
The windings will tolerate more current, but, as I said before, regulation won't be as good. I'm NOT saying you will get the same rating, but something in between. I can only suggest you try it and keep an eye on the temperature. |
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#7 |
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diyAudio Member
Join Date: Apr 2010
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The majority of loss in a loaded transformer is due to winding resistance (copper losses) so the maximum power out is more or less proportional to the input voltage, at least up to the point where saturation comes into play. Half volts in equals (same current) half power out.
Regulation will be worse at lower voltages because copper losses are relatvely higher. Barry |
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#8 | |
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diyAudio Member
Join Date: Sep 2006
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Quote:
At its nominal voltage, the transformer takes a total P=Pout+Pcu. The useful output is Pout=P-Pcu. At half that voltage, the current remains the same and P'=P/2. Pcu remains the same, therefore P/2=P'out+Pcu, and P'out=P/2-Pcu. It is easy to see that 2*P'out < Pout: P-2*Pcu < P-Pcu. |
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#9 |
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diyAudio Member
Join Date: Apr 2010
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This is true but given that the copper losses are normally a small proportion of rated power, in practical terms it can be ignored. If you run so close to maximum that the extra loss becomes significant a hot day is going to blow you away!
Barry |
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#10 |
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diyAudio Member
Join Date: Nov 2010
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Looks like some practical testing is called for....
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