I am looking for a power supply circuit that fulfills the following specifications:

  • 12VAC or 15VAC wall wart input
  • Symmetric +/-12V output (7812/7912 combo)
  • Should easily be able to provide 500mA on the output

I looked at a few power supply designs and they all have their advantages and disadvantages but for me it's hard to chose one. I though maybe some power supply designers here could give me a few hints.

Thank you very much!

  • \$\begingroup\$ Maybe you could make it narrower by choosing the best design you have found so far and proposing it as a first attempt. \$\endgroup\$
    – clabacchio
    Jul 9 '14 at 13:36

Here is a simple topology. This is opimized for simplicity at the expense of efficiency and effective use of the transformer's capacity:

With just a bare transformer secondary, you only get half-wave rectification to make each of the + and - rails. A center tapped secondary would allow full-wave rectification of both rails.

For a short time once per power line cycle, D2 conducts and charges up C2. Likewise, D1 conducts on the opposite polarity of the AC voltage and charges up C1. C2 and C1 hold up the rails between the bursts of current delivered thru the diodes. The two regulators then produce the nice and clean +12 and -12 volts. C3 and C4 are small ceramic caps that work better at high frequencies than the larger electrolytic C2 and C1. These help keep the regulators stable. C5 and C6 are generally required by such regulators, and again help keep them stable.

The only real brain cycles required here is to calculate what C2 and C1 need to be. At 60 Hz, each will be charged up every 16.7 ms. For sake of analisys, let's simplify this to assume C2 is charged up instantly to the full voltage once every 16.7 ms with nothing being added to in at other times. That's actually a little worse than what really happens, so is a good thing to design to.

You say the maximum output current is 500 mA. A 15 VAC transformer will put out 21.2 V peak. Let's be pessimistic and say the diode drops 1 V at the full current. That leaves 20.2 V that C2 gets charged up to. 78xx regulators need a lot of headroom, let's say 2.5 V (your job to check the datasheet). That means the voltage on C2 can't be allowed to drop below 14.5 V. 20.2 V - 14.5 V = 5.7 V that C2 is allowed to drop in the 16.7 ms between getting recharged. (500 mA)(16.7 ms)/(5.7 V) = 1.5 mF. That should be at least a 30 V cap. The same logic holds for C1.

This should all work, but you also need to look at the power dissipation. This simple circuit won't be very efficient. From above, we can see that the average voltage on C2 at full current is 17.4 V, which means each regulator is dropping 5.4 V. That times the 500 mA current is 2.7 W each. A TO-220 package can handle that, but will require at least a small heat sink.

  • \$\begingroup\$ Amazing Olin! Thank you very much for your answer. This was also the design that I would have preferred. Voltage drop for the 7812 is actually 2V and for the 7912 it is 1.1V (both at 1A), but yeah I really like conservative calculations with power supply designs. I have another question please: You say that "this simple circuit won't be very efficient". What would be your suggestion for a more advanced one? A switching type power supply? Also, how am I too stupid to produce line breaks here? :( \$\endgroup\$
    – user35474
    Jul 9 '14 at 12:54
  • \$\begingroup\$ @user: Two switchers would be more efficient such that you wouldn't need heatsinks. However, they would be significantly more complex, so too much for a quick answer here. You should be able to find off the shelf chips that do most of the work. Just follow the datasheets. As for line breaks, that can't be done in comments. \$\endgroup\$ Jul 9 '14 at 13:01
  • \$\begingroup\$ This is a bit old by now but I have another question: Sometimes this circuit won't startup. I can assume it is probably when the "wrong" halfwave arrives first. Is there anything that can be done to prevent that? \$\endgroup\$
    – user35474
    Jul 23 '15 at 19:34

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