Re: Calculating DC Output Current From Unregulated AC Transformer

emailaddress@xxxxxxxxxxxxx wrote:
On Jul 10, 1:50 am, ehsjr
<e.h.s.j.r.removethespampunctuat...@xxxxxxxxxxxxxxxx> wrote:

There is no "magic formula". You need to learn a whole
lot more about power supplies to calculate with precision
and to understand what needs to be considered and why.


Ok, but this should be resolvable to a reasonable level given the
example context of it being a typical E-core transformer and basic
silicon bridge rectified and capacitive filtered circuit. The output
at that point is the question. Even if some formula has an unknown
variable, or two or one hundred, the start would be to resolve those
which requires a working equation in which to place those variables.



I'll mention some general things.

First, with a bridge rectifier and capacitive filter,
figure the current you can supply to the load at
a little over 1/2 the transformer rating, in your
case, a bit over .9 amps.


This seems to counter the majority of small transformer examples out
there, does it not? Consider products powering just about anything
that uses a wall wart.
Consider a sub-20VA transformer rated in a wall wart for 12VDC, 1A
output. Granted, we could call that a little over 1/2 the rating but
just how much or little is the crucial issue, what variables determine
how much or little and how to express those mathematically.



Next, figure the ripple voltage, Vr. To figure Vr, seehttp://ocw.mit.edu/NR/rdonlyres/Electrical-Engineering-and-Computer-S...



Thank you for the link, though since I am not mass producing equipment
and so not overly concerned about small differences in component cost
or size, I'm essentially going to consider ripple current a constant
resolved later as would apply to any one design, while I am asking
about a general formula for conversion without regard to variables
that would change in different designs beyond the basic assumption of
a capacitor large enough to achieve acceptibly low ripple for example.


Next, find out how much headroom your regulator needs, usually
available from the data***. Then, the maximum voltage at
the output of the regulator will be
(Vsec*1.41) - 2Vf - Vr - Vheadroom = Vmax where Vsec is the voltage
at the secondary. Vf is the diode drop, which you can find
on the curve of Vf vs I on the data***. You should figure Vsec
based on worst case line voltage and worst case Vsec sag under load.

By the way, the 12.3 volt figure is odd - how did you arrive
at it?


This is what the example circuit would need at a bare minimum by
calculating similarly to what you have above plus other drops in the
circuit @ expected current levels, to still retain the necessary
minimum input voltage, plus or minus a margin of error as median
data*** values where used.

However, I am not concerned about the entire circuit, not about drop
over a regulator, I am only concerned about resolving how the AC
transformer rating relates to DC output before anything further in a
circuit beyond a typical bridge rectifier and capacitor sufficient to
smooth to a hypothetical 0V ripple, and I am also accounting for the
(Often negligable) different in forward voltage over the rectifier(s)
at different current levels, but this too can be expressed
mathematically.

Essentially, I don't want information more applicable to one project
than to another. Only what remains true mathetically for all projects
which employ AC spec'd transformers of typical design and through
bridge rectification by the most common silicon diodes. In other
words, as you'd see in the vast majority of electronics already if
they're not using switching PSU.

I don't want to argue - that seems your intent.

I know you want a magic formula - you have made that
clear. There is no single formula for you. Those who
have replied have made that clear.

We've tried to show you some of the factors that are
involved. You insist on ignoring them. So, we can't
help you, no matter how hard we try. Sorry.


Ed

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