Subject: Re: PSU's: Exceeding transformer rated current
Date: Tue, 7 Jan 2003 23:03:55 -0000
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"Richard" wrote in message
> "Richard" wrote in message
> No, I think you do double current.
> Say secondary is 10.5v open circuit, 0.5R winding resistance. When 1A rms (ac)
> 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
> be 5W. Wattage is a function of time the current flows; given the supply is 10v
> a peak current of 1.414A produces the heating effect of a dc current of 1A. That
> 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
> 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
> is 0.5v rms. Because really, 1A rms *ac* is equal 2A rms "dc* in these terms.
Actually the matter is more complicated, a doubling the current is not the only value
that will produce 10VA in a load.
If the diode is perfect, if you only see the primary as a perfect source with
resistance in series, if the load is pure resistance, then it seems the current
flowing in the half wave circuit described to get 10VA dissipated in a load, is
dependent upon the internal secondary resistance.
For an internal resistance of 0.5R with an EMF of 5.25v rms, I've worked out you need
a 1.6R load carrying 2.5A rms
Since voltage is only available for half the cycle, the open circuit voltage is 10.5v
rms /2 = 5.25v rms.
R int = 0.5R
R load = 1.6R
R total = 2.1R
I rms = 2.5A
V drop R int = 1.25v
V load = 4v
Watts = 10VA
My goal was to see what current you would need to draw to get a dissipation of 10VA.
I see that it all depends on internal resistance of the primary. Quite clearly if a
transformer had no inductance, core saturation etc, and it was rated at 10v 1A you
could never get above 5.25v rms on the load given the circuit as described. Depending
on internal resistance you can get more that 10VA dissipated.
But, things are far removed from reality, because we have inductance, core saturation
etc. Next step to add inductance parameter at least and some realistic rectifier
resistance and see what happens.