Stepping down 12VDC linear power supply to 9VDC

Question

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).

^{simulate this circuit – Schematic created using CircuitLab}

Answer 1

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.

Answer 2

You can either connect the 7809 to the 7812 output or the 7812 input. There are advantages and disadvantages with each approach.

Connecting it to the input:

• ripple will be higher since it sees the power supply ripple across the 1000uF caps

• the power dissipation in the 7809 will be the unregulated input voltage minus the 9V output voltage times the output current, plus about 5mA times the unregulated input voltage. The 7812 will be unaffected.

Connecting it to the output:

• ripple will be lower since it sees a regulated input

• the power dissipation in the 7809 will be 3 volts times the output current plus about 60mW (5mA times 12V)

• the power dissipation in the 7812 will be higher by the unregulated input voltage minus 12V, multiplied by the 9V output current plus about 5mA. Thus the total dissipation in the 7812 will be (Vunreg - 12V) * (Iout12 +Iout9 + 5mA) + (Vunreg * 5mA)

To put numbers on this, if you're drawing 100mA from the 12V supply and 50mA from the 9V supply, and the unregulated voltage is 17V then you'd have (if I did the arithmetic right, E&OE):

Case 1:

7812: 0.585W

7809: 0.485W

Case 2:

7812: 0.860W

7809: 0.210W

In the first case you can get away without heatsinks on either regulator in most situations, in the second case the 7812 should probably have a small heatsink (especially when you consider the possibility of high line voltage).

In general, if you're connecting a regulator to another regulator you also have to worry about the dropout voltage. In this case, there is 3V difference available which is fine, even for a 7809, but if you tried to run a 7805 from the output of a 7806 there would not be enough voltage for the 7805 to operate correctly.