Your solution started out as bearable (5V at 100mA) but ended up completely unacceptable at 500 mA. You say that your "wall wart" is rated at 300 mA. When you supply a voltage using a linear regulator the current in is the same as the current out - the regulator drops the difference in voltage. So here if you draw 500 mA at 5V you must supply 500 mA at 12V or 24V. The transformer will be overloaded in either case.
If the ratings are as you say then a potentially acceptable solution is to use a switching regulator (SR) operating from 24V in. \$5V \times 500 mA = 2.5 W\$.
\$24V \times 5 W =~ 210 mA\$. If the SR is 80% efficient (easily achieved) that rises to 260 mA. As that is liable to be an occasional requirement the total current at 24V will probably be acceptable with a 300 mA supply - depending on how many solenoids you wish to maintain on.
If you switch only one solenoid on at once the current drain with N activated is \$20 \times N + 20 mA\$. The surge current is essentially immaterial.
If you wanted more than 3 or 4 solenoids then current drain at 5V may need to be limited.
e.g.
- 10 solenoids at 20 mA = \$200 mA\$
- Balance = \$300mA-200mA = 100 mA\$
- Available current at 5V at 80 % efficient = \$ 100 mA \times \frac{24}{5} \times 0.8 = 384 mA\$, say \$400 mA\$.
Note that when a switching regulator is used, using a higher input voltage will result in less input current drain. Hence it is better here to use the full 24V supply.
Note also that if the transformer is a genuine 24 VAC then the rectified DC will be about \$24 VAC \times 1.414 - 1.5V - \$ "a bit" \$~= 30 VDC \$
Because:
\$VDC_{peak} = VAC_{RMS} \times \sqrt{2} ~= VAC \times 1.414 ~= 34 V\$.
A full bridge rectifier will drop about 1.5V.
34 VDC is peak voltage and available DC will be slightly lower - depends on load. There will be "a bit" of ripple and wiring loss and transformer droop and ...
At 80% efficiency this gives a 24VAC to 5V DC current boost of \$ \frac{30}{5} \times 0.8 = 4.8:1 \$
e.g.
- for 48 mA at 5V you need 10 mA at 30V.
- for 480 mA at 5V you need 100 mA at 30V.
So you about get 10 solenoids plus almost 500 mA at 5V DC :-)
One solution of many:
There are many SR IC's and designs. Here a simple buck regulator will suffice.
You can buy commercial units or "roll your own". There are many modern ICs but if cost is at a premium you could look at ye olde MC34063. About the cheapest switching regulator IC available and able to handle essentially any topology. It would handle this task with no external semiconductors and a minimum of other components.
MC34063. $US0.62 from Digikey in 1's. I pay about 10 cents each in 10,000 qauntity in China (about half Digikey's price).
Figure 8 in the datasheet referenced below happens to be a "perfect match" to your requirement. Here 25 VDC in, 5V at 500 mA out. 83% efficient.
3 x R, 3 x C, diode, inductor. It would work without alteration at 30 VDC in.
Datasheet - http://focus.ti.com/lit/ds/symlink/mc33063a.pdf
Prices - http://search.digikey.com/scripts/DkSearch/dksus.dll?Detail&name=296-17766-5-ND
Figure 8 in the LM34063 datasheet shows ALL component values except for the inductor design (inductance only is given). We can spec the inductor for you from Digikey (see below) or wherever and/or help you design it. Basically it's a 200 uH inducor designed for general power switching use with a saturation current of say 750 mA or more. Things like resonant frequency, resistance etc matter BUT are liable to be fine in any part that meets the basic spec. OR you can wind your own for very little on eg a Micrometals core. Design software on their site.
From Digikey $US0.62/1. In stock. Bourns (ie good).
Price:
http://search.digikey.com/scripts/DkSearch/dksus.dll?Detail&name=SDR1005-221KLCT-ND
Datasheet:
http://www.bourns.com/data/global/pdfs/SDR1005.pdf
Slightly better spec
I'm wondering if I would need any type of isolation ...
That would be an isolating DC-DC converter and would be at least as big as the wall-wart which, presumably, has been sized for the load so you would lose the benefit.
... or noise reduction, ...
Some local capacitance might be a good idea but, in theory, shouldn't be required.
... and would the voltage fluctuate as I plug and unplug devices?
This depends on the transient response of the power supply and should be specified on the power supply datasheet.
One thing that you are assuming is that they all share a common ground connected to the 12 V common. This may not be the case.
simulate this circuit – Schematic created using CircuitLab
Figure 1. Any device with full-wave rectifier reverse polarity correction will have its internal ground raised by 0.7 V.
Introducing a device with input protection as shown in Figure 1 could result in problems. The internal ground will be 0.7 V above your common 0 V line. Connecting, for example, a USB lead to another device may correct the problem but now all the power return will be through the USB lead. If this isn't rated for the current the voltage may rise on that line. You could spend a lot of time debugging this and potentially causing some damage.
Proceed with wisdom and caution.
Best Answer
There are lots of power supplies that deliver +12V, 2A and +5V, 2A; that is a common power requirement for external hard disk drives. I'd use one of those, they're assembled and tested and available whenever an external drive dies...