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
A typical use of the power supply schematic in the O.P. is for powering one (1) analog circuit, which needs both positive and negative power supply rails. In principle, you can power two (2) completely independent circuits from it (+9 to 0 and 0 to -9). This would be a peculiar scheme, though, because the ground of one circuit is at 9V (plus or minus) w.r.t. ground of the other one. Still, if the circuits are in fact independent, such scheme would work. In practice, this scheme is not used in general purpose desktop power supplies.
A classic desktop power supply (like the one in the YouTube video linked in the O.P.) is of a "dual positive type". Each channel has its own independent secondary winding (or even a separate transformer) and its own rectifier. The channels can float with respect to each-other, and that allows to connect them in series.
Best Answer
If you use switching regulators then the simplest solution is a single secondary winding producing 12VAC or more, rectified and filtered to produce >15VDC and powering both regulators in parallel. However designing and building those switching regulators may not be as simple as you would like.
3 terminal linear regulators are easier to implement, but for efficiency you need to avoid large input-output voltage differences. Powering the 5V regulator from 15V would be extremely wasteful, and running it in series with the output of the 12V regulator is not much better (plus it would increase the current load on the 12V regulator).
The solution is to feed the 5V regulator from a lower voltage. Normally this would require an additional transformer winding, rectifier and filter capacitor. However you can get full-wave rectified low and high voltages from a center tapped transformer and a single bridge rectifier, and then only one extra capacitor is required.
The trick to getting good linear regulator efficiency (and a cool running power supply) is to maintain a low voltage differential. The transformer, rectifier, smoothing capacitor and regulator are all 'overrated' to reduce their voltage drop under heavy load. That way the voltage differential can be kept low without risking output glitches due to ripple.
Here is the circuit I use to power external disk drives and small computers that may need up to 1.5A at +5V and +12V. Diodes D1~D4 are actually a 25A bridge rectifier...
NOTE: C1 really should be rated for at least 25V for safety, as 16V is right on the limit.