Electronic – Stepping down 12VDC linear power supply to 9VDC

power supplyvoltage-regulator

I have a +/-12VDC power supply that I have built up for testing simple audio circuits. It works just fine for me now but I would like to add a permanent +9VDC leg to it. I have never seen a scenario where one linear regulator feeds another which is why I haven't simply connected a 7809 to the 7812's output. The 9VDC leg I need would also be for audio circuits (but they won't need the bipolar supply just the +9 and GND). Current draw of the 9V leg won't exceed 100mA. The +/-12V supply never needs to put out anymore than 100mA as well. I am trying to determine a smart way to drop the 12V to 9V while keeping the supply clean and quiet (audio circuits) and not generating a lot of heat (anymore heat than a normal linear regulator generates).

schematic

simulate this circuit – Schematic created using CircuitLab

Best Answer

The first approach is to connect the 7809 directly to the positive output of the full wave bridge. The peak voltage there should be about 17 V. That means the 7809 would drop 8 V, which will dissipate 800 mW at 100 mA out. A TO-220 standing up in free air at room temperature should be able to handle that. Check the datasheet.

One nice thing about the 78xx regulators is that overheating won't damage them, they'll just shut down, cool off for a little while, then come back on again. You can therefore just try it and see what happens. Although I think it's unlikely it will get so hot as to shut down at only 800 mW, you can add a small heat sink if you run into problems. There are heat sinks specifically made for clamping or bolting to TO-220 packages.

The other possibility is to add the 7809 after the 7812. This will spread the dissipation across both devices. However, it decreases the current available from the +12 supply relative to the -12 supply, and could now cause the 7812 to overheat. With 100 mA at 12 V, and another 100 mA for the 9 V regulator, the 7812 will now dissipate up to (17 V - 12 V)(200 mA) = 1 W. Since that gives you a worse worst case than the other way, and makes the 12 V supplies assymetric, I'd put the 7809 directly on the unregulated 17 V line.

Either way the same overall power will be dissipated as heat at the same current.

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