An Idea for a Universal Input Buffer/Booster (onboard)

[mzy12]

I don't want the permanent voltage drop from a series diode

I mean, technically, there is a reverse current leakage from that diode to ground, reducing your voltage with it

But it realistically isn't that much haha

 

[JTEX]

The series diode makes more sense with an AC adapter. I think of battery power by default, which is why I avoid the voltage drop. It wastes some battery life.

 

[PedalBuilder]

The series diode makes more sense with an AC adapter. I think of battery power by default, which is why I avoid the voltage drop. It wastes some battery life.

Have you looked at using a P-channel mosfet for reverse polarity protection instead? It has virtually no voltage drop. The downside is that it uses three components instead of one, but it avoids the issues that you get when a parallel diode starts conducting.

 

[JTEX]

I don't like complexity. I use a beefy enough diode so it can easily withstand being across a 9V battery for the short time it takes someone to figure out that the clip goes in the other way. As for AC adapters, the diode can handle 2A indefinitely, and the AC adapter should protect itself against short circuits or overloads unless it's a total hack job. I protect my circuit against the outside world, you protect yours

 

[lowpitch]

Excuse my ignorance, but on your website it says the middle position sets the buffer to unity gain and flat frequency response regardless of dip switches 4-8, but dip switch 8 puts C8 in parallel with R8 in the feedback path. Is it of no consequence to the sound because of the large value or why can it be disregarded? Still learning

 

[JTEX]

Excuse my ignorance, but on your website it says the middle position sets the buffer to unity gain and flat frequency response regardless of dip switches 4-8, but dip switch 8 puts C8 in parallel with R8 in the feedback path. Is it of no consequence to the sound because of the large value or why can it be disregarded? Still learning

Great catch! I didn't think anyone would ever notice.
In the middle position R8 might as well be a short circuit and you'd still have unity gain, so it doesn't really matter if a cap also ends up across. The reason the cap is left connected in that particular situation is the limitations of the On-On-On toggle switch used. It's not a true DP3T, but more like a DPDT with a special middle setting where one half goes up and the other goes down. I had to get creative and keep some unnecessary but inconsequential connections in the middle, due to a lack of switching options. Ideally R8 (and the cap) should be shorted out for unity gain, but there were no contacts left on the 3-way switch for doing that. Dead short or 2k + maybe the cap, it ultimately does not really matter* unless you have OCD. I do, and it does bug me a little when I think about it. Thanks for reminding me

*except for a very minor noise penalty from the resistor noise

 

[lowpitch]

No problem Just so I understand you correctly, if I wanted to try out this circuit for unity gain only I'd just jumper pins 2 and 6 without any other components in between?

 

[JTEX]

Yes. To make a unity gain noninverting op-amp buffer you mostly just need a short between the output and the inverting input.

 

[Tantalus]

Jerry, thanks as always for the great work and for sharing it! Coupla questions:

You suggest leaving the passive tone control in place before the buffer. Why would one do that? It would still load the pickup, wouldn't it? And with an unstable (variable) load at that. I would use the tone control after the buffer, so the pickup would live in a pristine, stable, load-free environment, and probably a shelving EQ instead of just a 1930s era variable bleed.

Also, what about a 2nd buffer after the tone/volume controls? Even lower - and again, stable output impedance. Cable capacitance totally tamed!

 

[JTEX]

@Tantalus: I didn't mean to suggest leaving the tone control in the usual place, I merely wanted to clarify that you can use a typical one in front of the buffer if you want to. You could use a no-load tone pot (open circuit at full CW) if you don't want it to load the pickups at all when fully open.

I don't see a need for a second buffer after the volume pot. I use a 10k pot. The worst case output impedance is with the pot at 50%, where it's 5k down to ground and 5k looking up into the low impedance opamp output. The cable capacitance sees the two 5k's in parallel if I'm not mistaken, so that's an RC filter formed by a 2.5k resistor and, let's say, a 2000pF capacitance for a really long cable (20 meters). Works out to -3dB cutoff at 30kHz. I wouldn't worry. Not even if you use a much more common 25k volume pot.

 

[Tantalus]

Understood - pre-buffer tone controls could make sense for those who want to minimize wiring changes, but I think they'd be missing an opportunity to really free up the pickup (now on sale: "Free up the Pickup" T-shirts!)

You're right on the effect of the 10k pot on cabling, of course (I knew we could count on you to do the math ), but logically I would think the cable sees impedance from the direct signal path only; the 5k to ground is not supplying signal so should be irrelevant to coupling impedance. I'm just a self-educated amateur, though, so I could easily be wrong...

 

[JTEX]

The 5k to ground from the pot set at 50% is just as much a part of the circuit's output impedance as the other 5k from the op amp's output to the pot's wiper. I'm not very good at theory, but in a nutshell, any resistance between the output and ground, or between the output and any other low impedance node (such as the op amp's output pin, or even the positive supply rail) is part of the total output impedance, and all such resistances appear in parallel as far as the cable sees "looking back into the output of the circuit". I assure you that the circuit's output impedance with a 10k output pot at 50% is 2.5k, and this is the worst case scenario. Any pot position other than 50% will yield an even lower output impedance, all the way down to almost zero at min and max. Even with a 25k volume pot, the worst case output impedance would still be a fairly low 6.25k. If you find this hard to accept intuitively, you can try it in a circuit simulator. I just did, because I wasn't 100% sure I didn't talk out of my behind