[ARC5] Receiver AC Power Supplies

[Brian]

Hello Wayne,

I think you're stirring!

Hmmm, it may be a slip in your understanding of the Power Law, not Ohm's 
Law; it is the actual current, not the increase in current that is 
important. Copper losses go up as the square of the current times the 
resistance - and heating goes up with the duty cycle - this last bit you 
seem to have missed. Also, in a full-wave rectifier circuit, only half of 
the secondary winding is under load at any one phase of the mains cycle; so, 
the duty cycle is 50%; actually, it's a bit less than half the DC duty cycle 
because of rectifier Volt drop.

Be very, very, very careful of marketing 'claims'. A transformer rated at 25 
V and 2 A, may not be designed to deliver both at once. The actual 
deliverable VA depends on the design specifications, not the marketing PR. 
You need to pay attention to the core area, the type of magnetic material, 
the core lay-up, the current-carrying capacity demanded of the wire, the 
insulation, the potting, the airflow ....... and the nature of the load. So, 
your naive multiplication of 25 by 2 to arrive at 50 W may not be correct.

You could do us all a favour by removing unwanted tails from your response.

73 de Brian, VK2GCE

On Saturday, December 13, 2014 1:31 PM, HWHall said:

Something that was bugging me today....Ohmic copper loss increases as the 
square of the increase in current (power equals I-squared x R). Windings 
carrying 4amps will dissipate 4 times the heat they did at 2amps. So, 
wouldn't a transformer rated 25vct 2amps (delivering 50watts output) run 
even hotter if you used it to obtain 12.5volts at 4amps (still delivering 
50watts)? Might spell eventual trouble for the insulation around them, if 
not for the wire itself.

Wayne
WB4OGM