I am trying to design a power system for a project with two voltages.
I need 1A@5V for the PIC and some other components, and 3A@24V for a
solenoid. The solenoid will be activated for very short periods of
time (roughly .2 seconds) rather infrequently (let's say an average of
once every minute). My good friend the 7805 just isn't cutting it for
me here.

The power system must (in no particular order):
* provide 1A@5V continuously
* provide 3A@24V for very short periods of  time (solenoid)
* connect to a US 120V wall socket
* be as inexpensive as possible

I have explored these options:

Wall AC->DC adapter 4A@24V with a 7805 for the 1A@5V
   I probably got a few guffaws from this one. Yes, the reg got a bit
warm. Plus, a wall transformer of this caliber is quite expensive.

Wall AC->DC adapter 4A@24V with a switching regulator for 1A@5V
   This could possibly work, but the cost is still rather high for the adapter.

Open Case Dual Supply Switching Regulator
   Anything I found that could handle 3A@24V was $50+.

Lower current 24V rail, but with a lot of caps to store up charge
   This seems a bit dangerous, and by my calculations it would require
quite a bit of capacitance to supply that much current over any
reasonable timeframe.

I don't think a switching regulator will work for the 24V rail because
they generally have minimum current requirements. I don't want to have
to draw 3-4A at all times as the 24V will be needed so infrequently.
Also, I'm not sure switching regulators can provide the responce time
needed for a solenoid.

Is there a better way? Is there a way to get high current wall
adapters cheap? This seems like the kind of problem that should have a
simple solution.

One thing I'm not completely clear on is if I do use a switching
regulator, will the current at the input be the same as the current on
the output (I'm pretty sure the answer is no). If not, how can I
figure out how much current I will need to be able to supply the
regulator at a given output current? Is it as simple as power at the
input = efficiency * power at the output?

Thanks for reading,

Jake
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