klem kedidelhopper wrote:
The second supply will consist of a bridge rectifier off the 24V tap
connected to a suitable filter cap of say 1000UF.

With a full-wave bridge rectifier, if you use an 8300 uF capacitor, the
volts of ripple will equal the amps of load current. (Don Lancaster
taught me this.) The proposed 1000 uF capacitor will have about 8 volts
of ripple at a 1 amp load current. This is still above what the
regulator probably needs as a minimum input, but you might want to
consider a bigger capacitor.

I could add some series resistance either before or after the bridge,
but the voltage drop across this resistance would vary depending on
the total load and I'm not sure how well such a scheme would work.

I have a USB power supply that plugs in to the 12 V in my car that does
this. The 12 V DC goes through a power resistor (around 5 W rating) and
then into the input of a 7805 regulator in a TO-220 case. I'm not sure
if they did this to move some of the heat dissipation from the regulator
to the resistor, or just wanted to provide some current limiting for
some failure modes of the regulator, or what. I've never put a meter on
it but it seems to work OK. (The USB socket provides 5 V DC at 0.5 A,
or 2.5 W.)

If you don't have any better specs on what the parts will do, I'd try it
and see. Install the transformer, bridge rectifier, and capacitor, but
leave out the regulator. Load the 24 VAC output to 4 A. (If you don't
have some power resistors sitting around, car tail light bulbs in series
are good for this and you probably already own some.) Load the output
of the bridge rectifier to 1 A. Plug the transformer in to the lowest
line voltage you expect it to work at (use a Variac if you have one).
Measure the DC at the output of the rectifier - that is the lowest DC
input to the regulator you can expect. If you want, you can then
disconnect the load on the 24 VAC output and on the rectifier, plug into
the highest line voltage you expect (Variac again), and measure the DC
again, to get the highest DC input to the regulator that you can expect.

Then, you need to know the minimum DC input you can have to your
regulator board for it to still regulate. If you already have a
variable DC power supply, this is easy to figure out. If you don't, you
have to guess; if it was a 7812 the standard answer is that you need
about 13.5 to 14 V DC minimum for 12 V DC out. Your board may be
designed for more than this, though.

Once you know the lowest DC voltage you can expect from the rectifier,
and the lowest DC voltage you can supply to the regulator, you can
figure out how many volts maximum you have to drop in the resistor.

Just on paper, if I guesstimate a minimum DC input to the regulator of
14 V, and a 1 A load, I have...

Peak secondary voltage 24 V * 1.414 = 33.9 V
Minus two diode drops 33.9 V - 2 V = 31.9 V
Ripple with 1000 uF @ 1 A 31.9 V - 8 V = 23.9 V
Minus regulator minimum 23.9 V - 14 V = 9.9 V
Resistor needed 9.9 V / 1 A = 9.9 ohms
9.9 V * 1 A = 9.9 watts

I would probably then look for something like an 8 ohm, 15 watt or
better resistor. In perfect conditions, this would provide a little
more DC to the input of the regulator, and in the real world, it would
also account for getting less than 24 V DC from the transformer,
slightly more diode drop, slightly worse ripple, etc.

So I was thinking about employing a voltage divider at the output of
the filter.

The series resistor *is* a voltage divider. The regulator is the
bottom resistor in the divider.

Matt Roberds