Subject: Re: PSU's: Exceeding transformer rated current
Date: Tue, 7 Jan 2003 14:26:27 -0000
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"Richard" wrote in message
> "Richard" wrote in message
> > Don't know whether this forum is appropriate, but here goes.
> > In a half-wave rectifier circuit, a purely resistive load, no filtering, could
> > double the rated secondary current? The transformer will not get any hotter than
> > being loaded at rated current output, but would the secondary windings be prone
> > fail for overcurrent? TIA. Rich.
> Actually, I'm not so sure that you double the current to have transfomer supply
> maximum power. Whatever the current is I'm seeking to figure it out so that the
> transformer supplies it's rated maximum. For a transformer secondary rating of 10v
> (rms) and 1A (rms) the maximum load would be 10 watts, or I think that should be
No, I think you do double current.
Say secondary is 10.5v open circuit, 0.5R winding resistance. When 1A rms (ac) flows
the voltage drop in winding is 0.5v, thus we have 10v to supply. The load will be
10R and 10 watts/VA is dissipated.
If now the current flows for only half the cycle we must half our values. Watts will
be 5W. Wattage is a function of time the current flows; given the supply is 10v rms,
a peak current of 1.414A produces the heating effect of a dc current of 1A. That is
true for every half a cycle. So, if we half the number of cycles the average power
dissipated is half a full rms current.
Now I think this average of a half applies also to the voltage drop across the
winding resistance. A 2A rms ac current flowing thru 0.5R produces a voltage drop of
1v rms. So output of winding is 9.5v rms. If you half the cycles then you have to
average the voltage drop, and the average is a half. So, on average the voltage drop
is 0.5v rms. Because really, 1A rms *ac* is equal 2A rms "dc* in these terms.
As to copper losses, these stay the same.
So, on the face of things, to have the transformer supply it's maximum rated power in
a half wave rectifier circuit as mentioned above you can double the current.
So, I think the only reason you could not do so is something to do with saturating
the core. I don't think the windings would melt for eccess current. Copper losses
stay the same on the face of things.