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 would NOT design anything that would require direct AC power if I were you. Really. You can design your circuit to take 12/24VDC or whatever you need from a wall wart and then use that to charge your D cell.
Seriously. Do not go near the AC outlet. There are lots of design safety issues to consider, the first of which is using an isolating transformer.
Oh one more thing. Do not design anything that takes AC power. Don't.
As I said you can get a wall wart, or use any of the adapters you have sitting around if it supplies the DC voltage/amperage you need (which it should be able to) just use that.
There are lots of NiMH charger circuits out there you can use for reference. Your charging current is a function of your battery capacity, often expressed in terms like 0.1C meaning if your battery is 1Ah you want to charge it at 100mA.
You want a smart charger because NiMH are sensitive to charging current and voltage and have differing requirements depending on how discharged they are. If you ignore those you'll damage your battery.
I am fairly certain you should be able to find a commercial charger that can handle your D-cell. It may not be physically compatible in which case a set of alligator clips/wires can help you connect the two.
Best Answer
First of all the 7812 (positive voltage) and 7912 (negative voltage) regulators need about 2v higher input than the output in order to operate properly (that is at least 14v for the 12v regulator and -14v for the -12v regulator) .
The AC voltage in the input is rectified by two diodes to generate a symmetric unregulated supply (that is a positive and a negative voltage with reference to the ground) which feeds the regulators inputs and you get +12v and -12v regulated rails at the output. If you feed the input with 12v DC there is no way for the circuit to generate the negative rail voltage so this can't work.
Here is a link that explains the operation of the diodes in the circuit you show
http://metroamp.com/wiki/index.php/Half_Wave_Voltage_Doubler
Depending on the internal implementation of the AC-DC adapter it may be possible to use it.
If it is the classic transformer type rather than a switching type (you can easily judge it by size and weight) then you can omit the output circuitry (you'll need to open the adapter case to do that) and connect the AC output of the transformer to the AC input of the O2 power supply circuit.