" KAPP's diagram "
Pencil and Paper , and a good revision of the theory.... thats all you need.
We are Radio Amateurs, aren't we .
Have fun.
Jos on4kj
----- Original Message -----
From: "Ron" <w8ron@stratos.net>
To: "AMPS" <Amps@contesting.com>
Sent: Friday, August 29, 2003 7:32 AM
Subject: Re: [Amps] PS Theory ?
>
>
> Hmmm .
>
> Let's see ......the question was " How can I tell the power capability of
a transformer?".
>
>
>
> I have seen in the literature or handbook where one could resistively load
> the transformer so that the unloaded voltage would drop by 10%. The
loaded voltage and current would be the rated voltage and current for that
paticular transformer.
>
>
> In doubler service , you wouldn't expect the same supply output current
but 1/2 if you could draw smooth current.
> I think that this was what the fellow was asking by using that example and
was a little confused. I don't think that all peak numbers add anything to
the answer to the question.
>
> As far as peak instant currents and service ....well that another issue.
>
>
> Your not going to drive the core into saturation as the primarys are
running
> fixed rated voltage so the only worry is IR heating in the windings.
Drawing
> 4x currents into a capacitive load is acceptable in that the heating is
also
> short and the windings and core will shed the heat during non conduction
times.
>
> If you have such a large filter where you are running conduction times of
> 1 mS out of 8 mS, you might want to consider adding a series inductance
before the diodes or filter to lower harmonic currents and correct the power
factor instead of re designing the transformer you have at hand.
>
> Last rule of thumb.
> If you can fry an egg on it ...it's too hot :-o
>
> ---
> Ron
>
>
>
>
>
>
>
>
>
> rlm wrote:
>
> >
> >
> >>Actually in doubler service I think the transformer peak current
> >>is about four times the DC output current in a capacitor input filter.
> >>
> >>
> >
> >9x to 10x is closer to it since the conduction period is c. 1mS per
> >half-cycle.
> >
> >
> >
> >>A transformer should be designed for doubler service.
> >>
> >>
> >>
> >Amen, Mark
> >
> >
> >>On Tue, 26 Aug 2003 11:38:31 -0400 Ron <w8ron@stratos.net> writes:
> >>
> >>
> >>>When you feed it to a doubler , the current is not .5 A but .25A.
> >>>
> >>>THe remainder may be due to CCS and ICAS duty.
> >>>---
> >>>Ron
> >>>
> >>>
> >>>Mike, K8LH wrote:
> >>>
> >>>
> >>>
> >>>>Please excuse the dumb question... I'm trying to figure out how to
> >>>>determine the power handling capability of transformers (I did take
> >>>>
> >>>>
> >>>a
> >>>
> >>>
> >>>>quick look through the ARRL Handbook last night)...
> >>>>
> >>>>Questions:
> >>>>
> >>>>How can you determine the power handling capability of a
> >>>>
> >>>>
> >>>transformer???
> >>>
> >>>
> >>>>For example, the secondary on the transformer in my Heath SB-201 is
> >>>>rated at 800 VAC at 0.5 amps... I would think this transformer
> >>>>
> >>>>
> >>>would
> >>>
> >>>
> >>>>only be good for about 400 watts of power (800 x 0.5) but it seems
> >>>>
> >>>>
> >>>that
> >>>
> >>>
> >>>>if you feed this secondary into a full-wave bridge rectifier and
> >>>>capacitors you would get about 1,000 volts and 0.5 amps under load
> >>>>
> >>>>
> >>>(500
> >>>
> >>>
> >>>>watts), or feed it into a voltage doubler circuit, like that in the
> >>>>SB-201, and you get about 2,000 vdc at 0.5 amps under load (1,000
> >>>>watts)...
> >>>>
> >>>>I'm confused... What am I missing??? Thanks...
> >>>>_______________________________________________
> >>>>Amps mailing list
> >>>>Amps@contesting.com
> >>>>http://lists.contesting.com/mailman/listinfo/amps
> >>>>
> >>>>
> >>>>
> >>>>
> >>>>
> >>>>
> >>>_______________________________________________
> >>>Amps mailing list
> >>>Amps@contesting.com
> >>>http://lists.contesting.com/mailman/listinfo/amps
> >>>
> >>>
> >>>
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> >>
> >>
> >
> >
> >
> >
>
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