# Dual power supply, AC to ±12V DC, based on LM7812 + LM7912

I'd like to build a power supply, based on the LM7812 and LM7912, that outputs both +12 V and -12 V. I found this circuit online:

According to the datasheets, the LM7812 has a typical dropout voltage of 2 V, and the LM7912 is 1.1 V. As I understand it, a 12 V transformer (i.e., an AC-to-AC wall wart) is rated in RMS voltage. So, it should deliver a peak of ~16.9 V. With the 1 V drop across the 1N5402, that leaves only ~15.9 V peak on the regulator inputs. I assume, therefore, the large capacitance on the input side of the regulators is to ensure that the ripple is significantly low enough to avoid dropping below the dropout voltage...

Is this ok? Would I be better off going with a higher voltage transformer (14 V maybe? Something higher?) and lowering the input-side capacitance?

Finally, how can I calculate the current draw on the wall wart? I've had a hard time simulating this circuit... I had to tweak a bunch of .options to get it to simulate in KiCad; LTspice fared better. But both tell me the current from the transformer is zero, after an initial in-rush to charge the capacitors, even when I add 12 Ω load resistors from +1 2V to ground and -12 V to ground (i.e., should be drawing 1 A each).

Note: my intention is to buy a wall wart for the transformer because I don't want to mess with the “mains” side of this circuit. I've never seen a wall wart with a center tap, which is why I'm using a half-wave topology.

Thanks!

Edit: I forgot to mention that this design does call for heatsinks on both regulators. I'd like to design for 1A on each rail. And I'm open to alternative power supply circuits as long as the ripple/noise is low as I intend to use this to power audio equipment.

• How much current do you need? Commented Jun 16 at 13:48
• A 12VAC wall-wart might be specified to deliver 12V RMS at a specified current. Its no-load output voltage will likely be significantly higher than 12V...which should help you out. However, your simulation should include an estimate of source resistance from that transformer. Commented Jun 16 at 13:50
• Do you want a ±12 V PS, or do you want to build something with ancient school voltage regulators? Commented Jun 16 at 13:57
• Hearth: I'd like to design for 1A on both rails - actual current draws will likely be lower. greybeard: I'm open to alternative suggestions for a ±12V supply. Design requirements are 1A on each rail, and minimal ripple/noise, as I intend to use this supply for audio applications. Commented Jun 16 at 14:40
• @bmatcuk I would give some serious thought to using an offline switching converter. You won't need huge heatsinks and can use a smaller (i.e. cheaper) transformer. If you keep the switching frequency out of the audible range (not hard) you won't have to deal with any audible ripple or noise. Commented Jun 17 at 4:01

Even if peak voltage charged to capacitor is around 16V, with the 3x4700uF capavitors and 1A load the capacitor voltage only drops by abouy 1.2 to 1.4V (depending on 50 or 60 Hz mains) so capacitor voltage should be always high enough for 78xx regulators - i.e. more than 2.5V over the regulator.

As the voltage in to regulator is on average about 15V, it means that at 12V and 1A output, you will waste 3W of power in heat. As the TO220 case without heatsink has thermal resistivity of about 50 °C/W, it will heat up by 150 °C from the ambient and because it is more than the 125 °C allowed it will run into thermal shutdown.

So you need a big heatsink.

If you make capacitors smaller and make the peak voltage higher, you need to do the same calculations about what is the average input voltage and how much it heats up and is it within sane limits. Leaving capacitors too high while increasing peak voltage means average voltage is higher and so is average power dissipation.

• Ah, yes, I forgot to mention: this design does call for heatsinks on both regulators. Commented Jun 16 at 14:35
• Thinking more on this, no matter how I change the transformer or capacitance, I'd still need a heatsink, right? Because I need the average voltage on the regular input to be at least 2V (datasheet says dropout voltage can be as much as 2.5V for the LM7812) over the 12V out. At 1A, that 2.5V becomes 2.5W of power dissipation which pushes the LM7812 to the 125℃ limit. Commented Jun 16 at 14:49
• @bmatcuk If you at least use a more modern linear regulator, you can get dropouts less than 1 V. But really, if you don't want to use huge heatsinks, you need a switching converter. Commented Jun 17 at 4:03

(For 2×12 W, the transformer better be 50 W or more - such used to be commonplace with "halogen lighting", if not as a wall wart. (and often with just 11.4 V nominal output - you'd want low(er) drop voltage regulators, and maybe Schottky diodes.))

Few audio circuits profit from a regulated voltage, some do profit from a well filtered supply.

What really rubs me the wrong way about the design presented is half-wave rectification, large input capacitors, significant&short current "spikes".

Options include:
• Full-wave rectification (or other means (SMPS…) to get 24 V$$\_{DC}\$$) and creating a virtual ground (centre point of an active 1:1 voltage divider)
• Using a centre-tapped transformer and a full wave-rectifier for a symmetrical supply

For microphone&pickup amplifiers, batteries are still one option to consider.