Dropping voltage from 18v to 9v with a switch?

[cthilley1]

I am planning to build a two-in-one overdrive that runs on an 18v power supply. One circuit is designed to run at 18v, and the other *can* run either 9v or 18v.

The circuits I'm using are the Mach 1 (9v or 18v) and the Super '64 (typically a 9v in and a charge pump boosting it to 18v internally - I would rather try out building one without a charge pump and use true 18V in, since I've read charge pumps can often cause extra noise and my JHS Superbolt is somewhat noisy.)

I would like to put a switch on the pedal that allows me to choose either 9V or 18v for the Mach 1. I imagine it would just involve a resistor or resistors to impede the voltage, but I don't know what values or how to arrange them or anything like that. How would I do this, and are there any things to watch out for when running two pedals internally at different voltages from the same supply?

 

[Brett]

The easiest solution is to use a board designed specifically for the task: https://www.pedalpcb.com/product/3pdt-chargepump/

You could wire the switch rather than board-mounting if you're set on having the switch externally accessible.

 

[Brett]

I realize that you're going to be turned off by the solution I've provided because it uses a charge pump, but it's going to be a little more elegant than tacking a voltage divider and capacitor network on a toggle switch.

If you really want to go the voltage divider route, you'll need to determine how much current the Mach 1 draws and choose appropriate resistor values for your voltage divider and add bulk/filtering capacitors. You may also experience some nasty behavior (pops) if you activate your switch with power applied.

 

[jwin615]

This doesn't answer your question, but...
IMO, there's no reason to want to run the mach 1 at 9v and little benefit to run it at 18v, so do what's easier.
The first gain stage is close to unity gain. The second is a little over 2x gain.
That means you won't be clipping the opamp. Usually, when you see a high-end opamp used, like a 2134, the circuit is designed to not clip it. It's there to be a quiet source of gain only.
18v will give you more headroom to ensure you don't clip the rails but You would have to send a real hot signal in to do that even at 9v.
You'll likely end up with a switch that sonically does nothing but make a loud pop.
Apologies if any of this comes off as gruff, in a hurry fixing dinner...

 

[JTEX]

Easiest way to drop 18V to 9V is to use a 9V Zener diode in series with the 18V supply. Mind the polarity. The Zener's negative goes towards +18V. You could use a switch across the Zener to bypass it for 18V (and a big pop).

 

[cthilley1]

Thank you guys for your responses, this has been super educational for me! The Zener diode is interesting, I'll do some more research on that.
For now I've decided to just run both circuits at 18v. I knew the OPA2134 was a "high quality" / unlikely-to-clip opamp, but the info about the amount of gain in the Mach One circuit stages / reassurance that the 18v/9v wont make a huge difference is helpful.

I do need to read up more on how charge pumps work and the different ways they can be used. In the meantime I appreciate the help!