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I am building a circuit that needs to operate on +10V and -10V DC, and of course, a ground connection. It should be good enough to deliver just 1 amp. If I use a 120V to 10V AC transformer along with a bridge rectifier and capacitors, I should expect around +-13V DC. I will then use regulators to get it down to the voltage I need. Here is a quick power supply I threw together.

schematic

simulate this circuit – Schematic created using CircuitLab

Now, while this solution does work, I don't want a bulky transformer either internally nor externally of the device. Is there any easier way to do this that would work with either an AC or DC wall-power barrel connector? Or will I have to stick to this circuit to get the -10V? I'll stick close by to answer any questions you may have to help me come up with an ideal solution. Thanks!

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  • \$\begingroup\$ How much current do you need for both the positive and negative voltages? \$\endgroup\$ Commented Jan 26, 2020 at 22:32
  • \$\begingroup\$ Sorry, I'm still figuring out how this site works. Would have answered much sooner! Apologies aside, I'd say 1 amp would be enough. I will update the schematic. Thanks. \$\endgroup\$
    – zvolk4
    Commented Jan 26, 2020 at 23:15
  • \$\begingroup\$ @zvolk4 Ebay and places like that will have a "AC-DC Dual Output Switching Power Supply" that provides +/-12 V or +/-15 V outputs at an amp or more. You can use a simple regulator from there to get down to +/- 10 V and not a lot of wasted heat. Plus, not a big transformer. Is there a reason you don't want to go that direction? \$\endgroup\$
    – jonk
    Commented Jan 27, 2020 at 5:34
  • \$\begingroup\$ I'll look for those online. The main reason I didn't want to use it was because eventually when my units sell, I want a small unit and the case is already metal so I don't want it to be any heavier. \$\endgroup\$
    – zvolk4
    Commented Jan 27, 2020 at 13:28
  • \$\begingroup\$ What is the device to be powered? \$\endgroup\$
    – bobflux
    Commented Apr 5, 2021 at 8:48

4 Answers 4

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Your question is closely related to already discussed question. Please visit the following link How make a dual +-12V supply from a 24V SMPS

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Maybe buying a 24v AC - DC Power supply, and with a resistor ladder divide in half the voltage at the input, hence creating a virtual ground at 12V. But this 12 V will be the ground supply for your circuit.

enter image description here

For voltage stability you can put a 7812 to regulate +12v with respect to virtual ground, and a 7912 to regulate -12v with respect to virtual ground.

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    \$\begingroup\$ That's not a very good virtual ground. Suppose there is a load of 1 k Ohm between +12 and GND? What is the result? \$\endgroup\$
    – jonk
    Commented Jan 27, 2020 at 1:41
  • \$\begingroup\$ You are right, virtual ground would end at 16V. \$\endgroup\$ Commented Jan 27, 2020 at 3:56
  • \$\begingroup\$ I've changed plans, and now am going with something like the virtual ground schematic. What would be the best way to do this then? \$\endgroup\$
    – zvolk4
    Commented Feb 12, 2020 at 22:44
  • \$\begingroup\$ I'ts been a few years since asked but to complement this old answer I would like to add that buffering the virtual ground output would be the way to go with this design, but you would need an opamp capable of handling the current your circuit needs. \$\endgroup\$ Commented Apr 7, 2023 at 8:44
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The connections on your bridge rectifier are completely backwards. You have AC from the transformer connected to the rectifier's + and - outputs and you have the DC outputs connected to the rectifier's AC inputs.

All the grounds in the circuit should be connected together.

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The old solution was to use an AC output wall wart and two half-wave rectifiers to generate positive and negative voltages in the device. But half-wave rectifiers need 2x the amount of capacitance, and that takes up space.

You can use a switching wall wart that generates +/- 10V but that will need a special connector, and maybe hard to find.

If you want a standard wall wart and barrel connector, then a solution is to use a switching converter to convert the positive voltage into negative. That's a compact solution.

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