Can I do this with 9V, LT1045 and TLE2426 and get +-12V?

[Boner]

Assuming the rest of the circuit met all circuitcurrent specifications, can you get +- 12V from a 9V input at Vcc and Vss with respect to Vref?

Vcc is referenced to ground, and Vref is referenced to -8.6..... that rubs me the wrong way. If thats a problem does that mean I can split 16.2 at best and not 24.8?

[antonis]

I'm a bit confused about what you really want, conserning Vref..
In case you wish to use it as a 1/2 Supply virtual ground you have to use Vcc & Vss of equal value and reverse polarity..
As it is, Vref should measure about 3.8V (by heart..) 

[Boner]

Right, so Vref would be 3.8 with respect to -8.6 and would be a virtual ground for the rest of the circuit.

So if you treated 3.8 as "ground" (Vref), would Vcc and Vss be +-12.4 with respect to Vref or would this not work?

[PRR]

> if you treated 3.8 as "ground"

Standalone, yes. If the "9V" is shared with other pedals (as is now common), it won't be happy.

What does +/-12V do that you really CAN'T do with +/-9V?

[Rob Strand]

Two diodes drop off 9V will give you 1.3V drop and a 9V-1.3V = 7.7V  Drive that into the rail splitter then 7.7V/2 = 3.85V.    The regulation from the diodes and 9V is probably going to be stiffer and cleaner than the rails out of the charge pump.

Doesn't fix the thing PRR mentioned.

Are you trying to get +/- 12V referenced to the 0V of the 9V rail?   Maybe do it with charge pumps and no rail splitter.

[Rob Strand]

I like PRR's +/- 9V if you can get away with it.


Some LED driver circuits have a x1.5 mode.   Here's an interesting idea doing a similar thing with diodes,

https://www.electronicdesign.com/power-management/article/21808621/simple-charge-pump-boosts-input-voltage-by-50

I have no idea how well it works.

[Boner]

all that makes total sense, thank you so much! I knew something didn't smell right.

[Boner]

> if you treated 3.8 as "ground"


What does +/-12V do that you really CAN'T do with +/-9V?

would you go with Vcc being 9 and Vss as -8.6 and Vref at 0? Something like this:



Output jack would connect Vss with ground turning the circuit on/off

[antonis]

[PRR]

> would you go with Vcc being 9 and Vss as -8.6

Yes, 8.6 is 9.0 for most practical purposes.

If you really hunt you can find "5V to +/-12V" modules which accept up to 9V (but maybe not the 11V of a raw wart). They tend to cost $20 in singles from "reputable vendors".
https://www.hicomponent.com/1w-isolated-dc-dc-converter-dual-output-1.5kv-isolation.html
"..input range can be 4.5-9VDC, 9-18VDC, 18-36VDC, 36-72VDC, and output power is 1Watt. ..output regulated ±5V, ±12V, ±24V" (select when ordering) (I have no idea who  HiComponent Shenzhen Brightwin Technology Co is)

[Boner]

> would you go with Vcc being 9 and Vss as -8.6

Yes....

> if you treated 3.8 as "ground"


What does +/-12V do that you really CAN'T do with +/-9V?

would you go with Vcc being 9 and Vss as -8.6 and Vref at 0? Something like this:



Output jack would connect Vss with ground turning the circuit on/off

I misspoke.... Vcc would be 9, Vref=Vss=0 and Vee or Vdd or any other naming scheme would be -9 correct?

That would allow me to have 18V of rail to rail power for any and all booster stages correct?