Voltage reduction

I believe a simple 9V DC regulator will achieve what you need. You will be able to feed the 18V into this and it will produce the 9V output to run the boost pedal.

L7809 9V/1.5A regulator

The device above would be once such option, the data sheet for it will give you guidance on how to use it. You will need to ensure the correct size heatsink is used dependent upon how much current the boost pedal requires.

The other option would be to use three batteries if you have the room in the case for them. Two for the electric mistress and then the third for the boost pedal.

Dependent upon the current drawn, if you are using two batteries for the electric mistress, a third option would be to just tap off 9V from one battery to feed the boost pedal, but obviously, it will reduce the life of that battery in comparison to the other one.

Kind regards

Andy,

I would not know without looking at the schematic, but sometimes it is possible to run circuits from higher voltages (i.e. it might be possible to run both circuits from the same source).  You would have to determine if all of the circuits on the low voltage board could be driven from the higher voltage (i.e. transistors, etc.). Also you would need to look at any biasing circuits to see if they would function at the higher voltage.

Good luck,

Gene

Andy,

I found the schematic that you had posted awhile ago (TL061 op-amp), and it looks like the circuit will operate at 18V.  The only thing that you might need to check are the voltage ratings on you capacitors.

Gene

Hi Andy,

I see two possibilities. #1 you said that the unit runs on (2) 9 volt batteries in series to make 18 volts. If you tap into the connection between the two batteries you will have the required 9 volts. One battery will be running both devices so it would likely fail sooner. #2. If you are going to use a voltage regulator I would recommend that you use the DC to DC converter type as these are more efficient than the linear 7809. Check out the following link

https://www.newark.com/recom-power/r-78e9-0-0-5/dc-dc-converter-9v-0-5a/dp/84Y3692?st=9v%20dc%20power%20supply

With a battery powered device it is a good idea to conserve as much energy as possible. If you used the 18 volt source and had one 12 volt converter and one 9 volt converter both of your batteries would see the same load.

John

The simple, small voltage regulator that Donald recommended is probably all you need.  Another suggestion is something like this:

https://www.vellemanusa.com/products/view/?id=528425

I like it because it has some protection built in, and I can dial the voltage with a screwdriver to exactly what I want.

Using a dc-to-dc converter should be the preferred option as it is more efficient. Dropping 18 V to 9 V using a linear regulator means loosing half the power to the heat sink.

Maybe both circuits can run off the 13 V from the electric mistress circuit. If you post the circuits, I'm sure a simple solution can be found.

It seems that the electrical mistress works with a single voltage of about 13V. ? -13V a little outside the norms, near 12V or 15V ,,, hm ....
Well, then you can do a split voltage - two 9V batteries in series, the middle contact supplies the booster.
How much voltage does the booster withstand, or what are the tolerances of the supply voltage? Up to 13V or 14V would be the optimum.
Important - how much current do these modules need during peak loads?
Well, the lower battery is slightly more loaded than the upper one.
That's when the batteries are used as splitter voltage.

Or supply both modules of the 18V, each with their own voltage regulators (DC-DC PS with two outputs).
But then you have for the 9V a little more heat in the controller because of power loss.
For this, the batteries are not charged differently, which is better for the batteries.
Two regulators in serial is the same as two batteries in serial circuit. Here the 13V regulator will be more loaded, similar as the lower battery.

In both options you have almost the same mechanical effort, either more batteries, or a regulator.

But with a one-voltage battery (15,,18,,,3000000) you are a bit more flexible with the following circuits.

With a dual-voltage power supply you have twice as much effort to solve.

Edit:
LDO regulators would be a good alternative to the switching regulators.
So if you choose these with very small drop-voltage, then they are really alternative and also very small in space.

Best Regards

Gerald

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Have a look at the efficiency of this cheap switching circuit at 10mA. It probably isn't much better than using a linear regulator.

This is audio equipment. Batteries are nice and quiet. Why spoil things by putting a switcher in the same box?