Charge pump with ground isolation

Started by jul059, June 08, 2014, 01:23:40 PM

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jul059

Hello everyone,

I am trying to come with a design for a charge pump that could output +15v and -15v rails from a single 9v source. Basically, I'm planning on implementing this circuit (http://www.geofex.com/circuits/+9_to_33.htm) with an additionnal voltage doubler on the negative rail. I will then regulate both rails to +15v and -15v with LM317L regulators in a way similar to this : http://web.archive.org/web/20060704142724/http://myweb.tiscali.co.uk/nuukspot/decdun/gc/snub.reg.psu.png.

First, do you see any obvious problem with this?

Then I read about the issue of high current switching causing problems on the signal ground (http://www.geofex.com/circuits/+9_to_-9.htm). Unfortunately, I don't really understand how to implement it. Won't the grounds meet somewhere on the circuit anyway?

R.G.

Quote from: jul059 on June 08, 2014, 01:23:40 PM
I am trying to come with a design for a charge pump that could output +15v and -15v rails from a single 9v source. Basically, I'm planning on implementing this circuit (http://www.geofex.com/circuits/+9_to_33.htm) with an additionnal voltage doubler on the negative rail. I will then regulate both rails to +15v and -15v with LM317L regulators in a way similar to this : http://web.archive.org/web/20060704142724/http://myweb.tiscali.co.uk/nuukspot/decdun/gc/snub.reg.psu.png.

First, do you see any obvious problem with this?
It's awfully complicated for what you're trying to do. You're working hard at combining blocks of circuits to get to some end result, but my feeling is that you'd be better not combining blocks and instead redesigning from scratch. A small ferrite transformer running from a TI UCC38xxx power controller would give the direct result with better conversion efficiency and simpler debugging.

Then I read about the issue of high current switching causing problems on the signal ground (http://www.geofex.com/circuits/+9_to_-9.htm). Unfortunately, I don't really understand how to implement it. Won't the grounds meet somewhere on the circuit anyway?
Yes, they will. But you're on the point of understanding something about grounding.

The big issue with grounding is that wires are not "short circuits" in that they are not zero-ohm connections. They're resistors, albeit low-value resistors. If you have current pulses of several amperes through 20 milliohms, you get voltages across those conductors of several times 20mV. That's easy to get up into the 100mV level of a guitar signal, and suddenly you can hear those current-caused voltages. There are two ways to keep from having ground-current-induced noise. One is to fill the universe with copper in an attempt to keep the resistance low(er). The other is to be devious about what currents have to run through which wires. If a conductor - even a 1K resistor - has zero current flowing through it, the voltage across it is zero. So for your sensitive, low-voltage stuff you arrange for ground wires to carry little or no current, and you can set the signal reference voltage to really be where you want it.

If you separate the high-current "sewer" grounds that take used electricity back to the power supply for recycling, this doesn't have to affect the signal ground voltages if it travels on its own "ground" wires. Any side-effect current that the wires make can't affect signal ground.

If you're careful about what current flows on what wire, you can then have connected high current and signal grounds and not have noise. But this requires that you *know* what currents flow where to make intelligent choices.

In designing power supplies, it's easy to get big currents flowing for short times in odd places. That's why I put in the comment about "high pulse currents".  Charge pump circuits have high pulse currents too; they just happen when you connect a capacitor across a voltage source and let the irresistable force (the voltage source) meet the immovable object (the capacitor) and let them workint out.

Power circuits that do ground isolation have their uses. I have designed isolated 9V to [whatever] mini-supplies before, and they're tricky, but kinda fun too. And they can give you both isolation and much freer and simpler power conversion than charge pumps. Knowing when to change over requires knowing how much current and voltage you're delivering.
R.G.

In response to the questions in the forum - PCB Layout for Musical Effects is available from The Book Patch. Search "PCB Layout" and it ought to appear.

merlinb

Like RG said, if you're thinking of using some sort of switching converter (charge pump) and a linear regulator, you're doing it wrong. You should be making the converter work for you by doing the regulation too.  Maybe look into an integrated DC-DC converter. These take an input voltage and provide one or more isolated output voltages, usually regulated enough that you won't need to regulate it further:
e.g. http://www.farnell.com/datasheets/1520590.pdf

jul059

Quote from: merlinb on June 08, 2014, 02:44:03 PM
Like RG said, if you're thinking of using some sort of switching converter (charge pump) and a linear regulator, you're doing it wrong. You should be making the converter work for you by doing the regulation too.  Maybe look into an integrated DC-DC converter. These take an input voltage and provide one or more isolated output voltages, usually regulated enough that you won't need to regulate it further:
e.g. http://www.farnell.com/datasheets/1520590.pdf

My reasoning was that a LM317L does not take much space and is rather cheap. This circuit will be at the beginning of a pedalboard, so I want to introduce as little noise as possible. I also absolutely want to use 9v DC as supply, so I guess a transformer like RG suggested is unusable.

But then I might be wrong to use +18v and -18v. perhaps I could simply use +30v regulated by LM317L? This will be running a buffer that I want to have as much headroom as possible. Is there any advantage to use a dual rail supply vs. single rail for an opamp buffer? I'm thinking there might be some noise advantage (on the ground), but as you can see, I don't know much about electronics...

TheWinterSnow

If you are using an Opamp Buffer and you just need enough headroom to drive a guitar signal, you don't need anymore than a single pump charge setup in a +9v and -9v rail.  Opamps are great in the sense that you aren't going to hear an increase in distortion because you used a lower supply voltage that you would with transistors or vacuum tubes simply because, opamp buffers are feedback amplifiers and are extremely linear up until the point of clipping. 

Considering even the hottest of the hot pickups on the market peak no greater than 11Vpp to 12Vpp, a total supply voltage of 18v (+9v/-9v dual rail) is overkill.  Dual rail 12v, 15v, 18v and 24v are more common for much higher, line level applications where you want preferably 20dB of headroom due to the fact you aren't just using unity gain op amps.

PRR

Any pickup that delivers 12V p-p needs to be turned-down.

No guitar amp input will take more than a few Volts cleanly.

Adjust your audio levels into the 0.1V-1.0V zone. Don't adjust power rails to absurd voltages that later stages can't swallow.
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commathe

I think he meant 1.1Vpp to 1.2Vpp. 11Vpp would be too big for any stompbox to handle!

TheWinterSnow

#7
There are a few high gain pickups that put out slightly higher levels than Active pickups, typically active pickups are limited to about 8.5V p-p when running on 9V supplies.  I can say that I haven't played a guitar that hasn't put out less than 7v p-p and that would be my Fender Strat single coil, stock.  My bass had a class-G style 20dB boost amplifier but I removed that because it fell apart -literally- and I was tired of bumping it and it would get way up high, no amp can handle that without landing you in nasty clip city.  I have confirmed those levels with both O-scope and line level meter and DAW clips with a 1M load resistor on the guitar (typical impedance of the amps I use).  Might have to design a peak detecting circuit to confirm, would make for a good myth-busting day.

Can't say I have personally had issues with pickups that hot, then again I rarely even play clean stuff, and if I do, the bottleneck in the system is the active pickup clipping from not having enough supply voltage.  Been needing to give the guitars the ol' 18v mod.

I am talking about absolute peak voltage too, as in the peak transient when you strum as hard as you reasonably can without the guitar sounding like crap.

Quote from: commathe on June 08, 2014, 11:51:23 PM
I think he meant 1.1Vpp to 1.2Vpp. 11Vpp would be too big for any stompbox to handle!

You're right, 11v p-p would be too high for the average stomp, and can be a problem if you have a pickup that hot, I have heard from guys with some high gain pickups that were distorting clean effects like delays and choruses.  Me having most of my guitars with actives running on 9v I am on the boarder of having issues.  Then again, I play hardrock/metal and high gain amps most of the time, so its never been an issue.

commathe

I've never used active pickups, but none of my guitars can even spit out 2Vpp no matter how hard I hit them. 

TheWinterSnow

Quote from: commathe on June 09, 2014, 12:10:14 AM
I've never used active pickups, but none of my guitars can even spit out 2Vpp no matter how hard I hit them. 

What models are you running?  I just went back to my strat to double check and I was getting just shy of 2.5V p-p, my other guitar with actives, round about the 3.18Vrms (~9v p-p) they are rated at.

That's the big reason that I do all my designs with a pump charge bipolar design of +/- 9v, you don't want something that is supposed to be clean distorting because the pickups were too hot.  Hot pickups pushing medium to high gain guitars further into distortion is desired, but clipping an active DI, EQ, or any other effect/device that is supposed to be clean is never desirable.  It also makes good to keep transistor circuits as hi-fi as possible if you are going for that kind of sound.

jul059

Quote from: TheWinterSnow on June 08, 2014, 11:16:53 PM
If you are using an Opamp Buffer and you just need enough headroom to drive a guitar signal, you don't need anymore than a single pump charge setup in a +9v and -9v rail.  Opamps are great in the sense that you aren't going to hear an increase in distortion because you used a lower supply voltage that you would with transistors or vacuum tubes simply because, opamp buffers are feedback amplifiers and are extremely linear up until the point of clipping. 

Considering even the hottest of the hot pickups on the market peak no greater than 11Vpp to 12Vpp, a total supply voltage of 18v (+9v/-9v dual rail) is overkill.  Dual rail 12v, 15v, 18v and 24v are more common for much higher, line level applications where you want preferably 20dB of headroom due to the fact you aren't just using unity gain op amps.

Is there an advantage in using a +9v/-9v dual rail vs. +18v single rail?

TheWinterSnow

Quote from: jul059 on June 09, 2014, 01:27:43 AM
Is there an advantage in using a +9v/-9v dual rail vs. +18v single rail?

The only thing I have been able to gather is that if you have an amp with multiple opamp stages, a bipolar supply reduces part count.  With one or two opamp stages, it really doesn't matter.  You have to bias each stage with a single supply, and so for the best quality sound you have to pay attention to the coupling capacitors, there are a few traps that can fall into place with both ceramic and electrolytic caps and the right value has to be chosen to get the best frequency response.  That is not much of a problem with a DC coupled bi-polar supply as if you get an op amp with low enough input offset voltage you don't need to worry about biasing the opamps and if the offset is horrid enough where you need to use some coupling caps anyway, a 0.1uF 25v film cap will do the job without as much fuss as having a larger DC bias.

samhay

Quote from: jul059 on June 08, 2014, 01:23:40 PM
I am trying to come with a design for a charge pump that could output +15v and -15v rails from a single 9v source. Basically, I'm planning on implementing this circuit (http://www.geofex.com/circuits/+9_to_33.htm) with an additionnal voltage doubler on the negative rail.

It sounds like you are probably not going to go down this road, but I have used a single 7660S to generate a +/-17V(ish) supply before. The voltages will sag if you load it down much, but it can work for a few op-amps.
Take a look at the 'bipolar supply doubler' example on pp11 of the 1054 data sheet:
http://cds.linear.com/docs/en/datasheet/1054lfg.pdf
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