Charge pump limitations

[Outlaws]

I just built a standalone charge pump for my -9v pedals. I used this Variant of the ICL7660.  IC is still in shipping so it's only a socket for now.
I see the data shows 45mA. Is that the limitation irrespective of the voltage? Or is the mA spec lower at 9v than 12v?  I don't know how ICs work so maybe that's a dumb question.
I was hoping to be able to run two -9 Fuzz pedals off this, but I guess that might not be doable.

[GibsonGM]

I'm not sure how (or even if) "Things change) with a charge pump if you go from 12 to 9V....but why should you need more than 45mA for two fuzzes??

The big draw on them is the LED, and with the new high-efficiency ultra bright ones you can shave their draw down quite a bit.   I think 45mA should be PLENTY....each probably only draws a couple mA! 

[Transmogrifox]

The limitation is a combination of internal FET resistance and heat.  Most likely you will start to lose voltage regulation before you burn out the chip.

These work by switching the capacitor onto the Vin and charging it up.  Then the switches reconfigure the capacitor connections to the -9V so it discharges into the -9V, keeping the -9V output cap topped off.

The internal FET switches look like resistors to the output.  As you draw more current, you get more voltage drop across the resistors.  This makes the -9V sag toward -8V, -7V, and the difference is burned up inside the IC.  At some point it will smoke.

You can put several in parallel if you need more than 45 mA.

[Outlaws]

I'm not sure how (or even if) "Things change) with a charge pump if you go from 12 to 9V....but why should you need more than 45mA for two fuzzes??

The big draw on them is the LED, and with the new high-efficiency ultra bright ones you can shave their draw down quite a bit.   I think 45mA should be PLENTY....each probably only draws a couple mA! 

I don't have LEDs on them, but I looked up Tone Bender and someone said the MKIII was like 30mA.  I haven't tested my personal MKII or MKIII.

[Outlaws]

The limitation is a combination of internal FET resistance and heat.  Most likely you will start to lose voltage regulation before you burn out the chip.

These work by switching the capacitor onto the Vin and charging it up.  Then the switches reconfigure the capacitor connections to the -9V so it discharges into the -9V, keeping the -9V output cap topped off.

The internal FET switches look like resistors to the output.  As you draw more current, you get more voltage drop across the resistors.  This makes the -9V sag toward -8V, -7V, and the difference is burned up inside the IC.  At some point it will smoke.

You can put several in parallel if you need more than 45 mA.

Ok, so all the +9 and grounds can be two long rails, and then separate the middle sections and then I could reconnect the -9v outs to a shared rail for -9v with all the amperages totaled up and available?  I like that idea.

[duck_arse]

I've got a Mk III-alike on the BB right now, and from the voltage readings, it doesn't even draw 800uA. who is this "someone" that said "like 30mA"?

[Outlaws]

I've got a Mk III-alike on the BB right now, and from the voltage readings, it doesn't even draw 800uA. who is this "someone" that said "like 30mA"?

Was a quick google search lol. That's excellent info.

[GibsonGM]

Probably 29 mA on the LED, 1mA on the circuit, LOL.

These things usually don't draw squat, Outlaws     45mA is a BEAST amount of current available for distortions etc. 

[bluebunny]

...and there's no reason to be powering an LED from the charge pump anyway.

[GibsonGM]

...and there's no reason to be powering an LED from the charge pump anyway.

Just curious - if you were running a fuzz on -9V, how would you choose to run the LED?  OP says he wants the pump to get -9V for those...

[bluebunny]

Not sure what question you're asking, Mike.  Forgive me if I answer the wrong one!    The charge pump exists only to give you -9V (for the fuzz) from the otherwise +9V power supply.  It doesn't have to power the LED.  Use the beefier +9V supply to power the LED.  The fuzz and the LED are entirely separate circuits.  (Am I missing something?)

[GibsonGM]

Mmm, I think I see what's up - if you put the charge pump IN the pedal, you have both polarities available  

I was thinking more like this would be found in a pedalboard, and you'd have to run a line for your -9V to a more distant location.  Then, adding the + would be more cumbersome.  But apparently that's not the deal here.    The desire to 'run two pedals off this' threw me....somehow he needs to run a wire to the next -9V fuzz, making it more complicated (but not very), if they're not in the same enclosure.   
Just thinking of it from a different perspective!   

[Outlaws]

I don't have a LED in the pedals, but the charge pump isn't going in either pedal. I am putting it in a 1590LB with two 3.5 jacks for my -9 outs.

[Transmogrifox]

Ok, so all the +9 and grounds can be two long rails, and then separate the middle sections and then I could reconnect the -9v outs to a shared rail for -9v with all the amperages totaled up and available?  I like that idea.

Yes, nice and straight-forward.

As others have raised you probably won't need more than 30 mA.  The nice thing is if you lay out your circuit with space for more, then you can build exactly what you need when you need it.  If doing perf board, you just leave a nice large section blank.

Really fancy is to add a current sensing circuit to the -9V rail wired to an LED that comes on when your circuit takes too much juice, then you have a quick visual telling you when you need to add more.  Because you're doing it with an external box it might not be a bad idea to put some kind of a current limiter on the box to protect against shorts -- or if you accidentally connect into a positive ground circuit that has the diode protection across the rail it might damage the charge pump.  A socketed charge pump is a good start since it's easy to replace if it dies.

Although I don't expect you to need 30 mA unless you have a huge pedalboard full of positive ground pedals.

[GibsonGM]

I don't have a LED in the pedals, but the charge pump isn't going in either pedal. I am putting it in a 1590LB with two 3.5 jacks for my -9 outs.

That's exactly what I thought you were doing, and why you'd have to run the LED on the -9V, IF you were including one.  You still have mA to spare      But for those putting the circuit IN the box, they'd of course have both polarity voltages present to play with.

As said - man, those fuzzes are VERY low draw - you have tons of clearance, no worries.  Only if you really start doing a lot with the -9 would you need to start adding it all up to see what you're drawing...