> Generally, the power supply formula to find out the final
> DC current the supply
> is capable of is to multiply the AC rms current by 0.62 or
> Iac x 0.62 for a full
> wave bridge. That would give 0.25 x 0.62 = 0.155 A
> available to use. That's
> the power supply formula most all transformer
> manufacturers give.
I think you are using the approximation where a full wave
rectifier center tapped transformer is used in a bridge.
You would
> have then about 2517 Vdc at 155 mA available to use at no
> load.
How do you get 155mA at no load?
With load
> you can figure a drop of about 12% to 13% in voltage, or
> that's the figure I
> always use. 2517 x 0.155 = 390 watts. I think you need a
> little more umph
> than that.
The figure you always use is wrong. It makes no sense at
all.
You have to know the ESR and load capacitance to calculate
voltage drop under load. The values you gave are
meaningless, exact for the 1.414 times RMS for no load
voltage.
The higher the ESR, the less important the capacitance and
the poorer the regulation.
73 Tom
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