Electronic – AC to DC rectifier, smoking

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

• Added:

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

• $US0.75/1.

• Surface mount.

• Bourns.

• http://www.bourns.com/data/global/pdfs/sdr1305.pdf

I know you said you don't want to run outlets everywhere, but I think a pile of wall warts is exactly what you want. Jameco has a large selection. 9 V sounds pretty good and that is one of the standard voltages, but you might want to consider 5 V (see below). Get something in the 500 mA to 1 A range.

These supplies are inherently isolated from the line, are usually short-circuit protected (check to make sure, you definitely want that), and draw so little AC power that you can string a bunch of outlet strips together without harm.

In the end, I think this will cost less and provide a better experience for the kids. I remember when I was a kid tinkering with this stuff how frustrating it was to have batteries run down, especially when you're not aware of when you are asking a lot from them. You also feel a lot less guilty abusing a power supply than running down consumable batteries.

You can start a fire with almost anything. If you do just the right thing, even a 9 V 500 mA supply can catch something on fire, but no more so than a 9 V battery and without the chance of the battery itself doing something bad and causing chemical burns.

If you are worried about LEDs getting damaged by them getting hooked up backwards, maybe you should get 5 V or 6 V supplies. Most LEDs can handle 5 V in reverse, and some 6 V. With 5 V supplies, you can eventually run logic cicuits directly without having to use a linear regulator. Lots of stuff will run well from 5 V. If you get 5 V supplies, get at least 1 A capability. That will be useful for running small motors. 5 W total power really isn't all that much.