When's it okay to omit a charge pump?

[Joben Magooch]

Long story short, I'm running out of 9V outputs on my power supply but it does have an 18V output that I'm not presently using. I have a handful of pedals with a charge pump so I'd assume that connecting them to the 18V out of the supply wouldn't be good, but at the same time some of them step up a 9V input to 18V internally - so I'm wondering if there's any cases where I could pull the charge pump and connect directly to an 18V supply.

I know I've seen it mentioned for a few different pedals (that you can pull the pump and run at 9V, at least) but didn't know if there was any hard and fast rule for determining if/when it's doable.

Presently I'm looking specifically at the Son of Ben, Informant, and/or Cleaver. If any of them can have the charge pump left out and ran at straight 18V, I'd be golden...

 

[fig]

Son of Ben has no charge pump.
I believe the Informant and Cleaver would run without it (though I don't know how they would perform).

 

[Joben Magooch]

Son of Ben has no charge pump.
I believe the Informant and Cleaver would run without it (though I don't know how they would perform).

Oh yep, you're right. Must've got it mixed up with another though not sure which. Maybe I was just thinking of running it at 18v...I don't remember

So I guess my question then would more or less be...Would there be any appreciable difference (for either the informant/cleaver) running at 18v - no charge pump - vs running at 9v, stepped up internally to 18v via charge pump?

 

[Dan M]

I did this with the Powersound Overdrive. I tweaked the filtering as discussed here:

Thread 'Am I correct on: How to remove a charge pump?'
https://forum.pedalpcb.com/threads/am-i-correct-on-how-to-remove-a-charge-pump.8833/

The only issue I can think of is if the pedal has some parts of the circuit operating at 9V, and some parts operating at 18V. If you remove the pump and plug in to 18V, the whole thing is now running 18V.

I’m not sure if any Pedalpcb circuits do this, but I would at least check each schematic first.

 

[Dan M]

Hmm. I guess in all cases, the LED resistor is assuming 9V. So you may want to increase the resistor to drop the LED brightness. If I’m doing the math correctly, it would need to be a little more than double to keep the same brightness.

 

[Feral Feline]

[Would there be any appreciable difference (for either the informant/cleaver) running at 18v - no charge pump - vs running at 9v, stepped up internally to 18v via charge pump?]


Maybe in terms of mA power draw, but how they'll run/sound should be the same — 18v is 18v... or is it, fig?

 

[fig]

[Would there be any appreciable difference (for either the informant/cleaver) running at 18v - no charge pump - vs running at 9v, stepped up internally to 18v via charge pump?]


Maybe in terms of mA power draw, but how they'll run/sound should be the same — 18v is 18v... or is it, fig?

Indeed.

 

[Bricksnbeatles]

Oh yep, you're right. Must've got it mixed up with another though not sure which. Maybe I was just thinking of running it at 18v...I don't remember

So I guess my question then would more or less be...Would there be any appreciable difference (for either the informant/cleaver) running at 18v - no charge pump - vs running at 9v, stepped up internally to 18v via charge pump?

the difference between 9 and 18v for the son of ben is incredibly subtle to me.

 

[giovanni]

Another good hint is if the original can run at 9V, 18V or both (that is, unless the original itself is using a charge pump).

 

[Joben Magooch]

the difference between 9 and 18v for the son of ben is incredibly subtle to me.

Yeah, I'm wondering if that's not the simplest solution. Just run it at 18V, keep the others "stock" at 9v w/ charge pump and not mess with pulling anything.

 

[Locrian99]

I did this with the Powersound Overdrive. I tweaked the filtering as discussed here:

Thread 'Am I correct on: How to remove a charge pump?'
https://forum.pedalpcb.com/threads/am-i-correct-on-how-to-remove-a-charge-pump.8833/

The only issue I can think of is if the pedal has some parts of the circuit operating at 9V, and some parts operating at 18V. If you remove the pump and plug in to 18V, the whole thing is now running 18V.

I’m not sure if any Pedalpcb circuits do this, but I would at least check each schematic first.

You’d also need to be mindful of cap value. The original bom might not denote a cap for instance needing to be 25v. Because even with the charge pump it may have only been seeing 9v. I’ve thought about this some myself as my 9V section on my power supply is full and my 18v plug is just sitting there. Thought my mystery meat when I get back to finishing that up will run at 18v.

 

[Joben Magooch]

So, just for sake of learning...I'm curious.

I'm a total noob when it comes to reading schematics. More or less I understand what the various components are, but only have the very most tenuous grasp of how things interact with each other and how to follow connections.

So I'm looking at the schematic for the Informant.
If I'm looking at it right, it looks like it's broken out into two sections, which I would guess is an "audio" path and a "power" path? In this instance, the top would be the audio (I can see where it's passing through the TL072s, A100k potentiometers, etc) and the bottom would be the power (LT1054 charge pump, LED, etc). But what I guess I'm trying to figure out is like...how can you tell where the two sections interact/meet?

As some have mentioned above you may have instances where parts of the circuit are operating at 9V and parts (post-charge pump) seeing 18v. In this instance, it looks like the LED is expecting 9V (as it comes before the charge pump in the circuit, from what I can tell) but beyond that it looks like the rest is post-charge pump. But I guess I'm just trying to figure out how to read it and tell where the power "meets" the audio portion of the circuit. I see that post-charge pump it looks like it shows a connection to both IC1 and IC2, so maybe that is the spot? Or maybe I just have a fundamental misunderstanding of how it all interacts from the get-go....

 

[giovanni]

In that schematic, you can see where the ICs are connected to the power in the power section (after C104). The ICs are the only things that need power. Then you can see the voltage divider with a node marked as Vref. That same voltage is referenced in the signal path (it’s half the total power voltage), meaning there is a connection there. Now if you wanted to remove the charge pump, you would remove all components between C100 and C104. If you do that, the ICs will be powered directly by the 9V and the voltage divider will now be 4.5V rather than ~9V. You want to keep D100 (protection diode) and all components from C101 on (RF filters plus the voltage divider).

 

[Joben Magooch]

In that schematic, you can see where the ICs are connected to the power in the power section (after C104). The ICs are the only things that need power. Then you can see the voltage divider with a node marked as Vref. That same voltage is referenced in the signal path (it’s half the total power voltage), meaning there is a connection there. Now if you wanted to remove the charge pump, you would remove all components between C100 and C104. If you do that, the ICs will be powered directly by the 9V and the voltage divider will now be 4.5V rather than ~9V. You want to keep D100 (protection diode) and all components from C101 on (RF filters plus the voltage divider).

That's super interesting to me. Thanks so much for the explainer.
Probably making myself sound a bit silly. It sounds a little obvious now as I understand some components can function just fine passively, but I guess having never really given it any serious thought I kind of just assumed everything was receiving power at some point.
Makes total sense, just never really stopped to think about it!

So again back to the Informant. In that particular circuit then, I'm guessing C100 is for some sort of filtering? Directly after that I'm seeing SW2. I am correct in assuming this is Footswitch Lug 2 and that it's showing when it's closed/activated it's then sending power thru the charge pump IC? My best guess, anyways...

And so then to bypass the charge pump you would just pull IC100 (charge pump), D101, C103, and D102? As mentioned also it looks like the LED and resistor are coming pre-charge pump so they would be "expecting" 9V input - so I'd wanna more or less double the value of the LED resistor if i'm sending it 18V instead?

Would anything around the charge pump IC need jumpered for it to work okay while removed, or is the power circuit still "intact" even with the aforementioned components pulled from it?

As above I will probably just power another one of my pedals from the 18v tap on my PSU so not sure that i'll end up messing with this one or not, but I'm at least curious to learn!

 

[swyse]

Some circuits like the greengage for example will have the op amp referenced to ground instead of vref because its making -9v with the charge pump IC. If you remove the -9v and replace it with ground you will probably have most op amps stop working as you'll be at the voltage rails. On the greengage you would have to change where the resistors R3. R4. R8. R12 go and add a vref for them to go to which would be a messy operation.

 

[giovanni]

That's super interesting to me. Thanks so much for the explainer.
Probably making myself sound a bit silly. It sounds a little obvious now as I understand some components can function just fine passively, but I guess having never really given it any serious thought I kind of just assumed everything was receiving power at some point.
Makes total sense, just never really stopped to think about it!

So again back to the Informant. In that particular circuit then, I'm guessing C100 is for some sort of filtering? Directly after that I'm seeing SW2. I am correct in assuming this is Footswitch Lug 2 and that it's showing when it's closed/activated it's then sending power thru the charge pump IC? My best guess, anyways...

And so then to bypass the charge pump you would just pull IC100 (charge pump), D101, C103, and D102? As mentioned also it looks like the LED and resistor are coming pre-charge pump so they would be "expecting" 9V input - so I'd wanna more or less double the value of the LED resistor if i'm sending it 18V instead?

Would anything around the charge pump IC need jumpered for it to work okay while removed, or is the power circuit still "intact" even with the aforementioned components pulled from it?

As above I will probably just power another one of my pedals from the 18v tap on my PSU so not sure that i'll end up messing with this one or not, but I'm at least curious to learn!

C104 can also be removed iirc. There are like three other capacitors for filtering (including C100) that may or may not be necessary. It probably won’t hurt to keep them. You do need to jumper D101 and D102. I think you just need one jumper to connect the anode of the first to the cathode of the second.