Varitone: Pull DOWN resistors=misnomer?

[rjb]

Not trying to start an argument- just trying to avoid re-re-re-inventing the wheel....

Everywhere I see information on the Varitone circuit (or it's cousin, the multi-capacitor low-pass filter), I read advice to wire 1M pulldown resistors in parallel with the caps to keep the unselected caps DISCHARGED so they don't pop when selected.

But every Gibson schematic I've seen shows the resistors connected to "hot" (signal)- not to ground. I would think Gibson would have connected them to ground if they were meant to be connected to ground.

Here's what I guess is going on:
For whatever reason, the signal has some DC offset from ground; that is, the signal "wiggles" around some voltage other than 0V. The pull UP resistors keep the unselected capacitors CHARGED to the offset voltage- because if they DISCHARGED (due to natural leakage), selecting a cap would cause a pop when sudden inrush current charged that cap to the offset voltage.

My questions:
1) Does using pull DOWN resistors really stop "cap pop"?
2) If so, how?

It would be nice to avoid the extra (albeit tiny) signal load of all those parallel resistors- but I don't see how pull DOWN resistors can do any good.

Thanks,
-rb

[Chuck H]

First, the Gibson scheme probably does pop. But that would have been tolerated at that time.

Unless I'm wrong it's because the caps will HOLD a charge that they pop. Since the pickups have a 0V reference you need the cap to have the same 0V reference to avoid the offset. You don't need to use 1M resistors. Three in parallel would be a 333k load in parallel with, say, a 250k pot (and most amps typical 1M input resistor, but we'll ignore that for now). Effectively changing your guitars load from 250k to as low as 143k!!! (at some frequency determined by the cap values, less profound at other frequencies) No good. Try using 10M resistors instead. Same 0V reference but no appreciable added load. Or better yet, if your building this on a rotory switch just use a "make before break" or "shorting" switch. That should reduce the pop to where it can be ignored and perhaps this is how Gibson did it.

Chuck

[David Schwab]

The Gibson schematic shows 10M resistors. The 10M resistors bleed off the caps so they wont pop.

I've made varitones without the resistors, and you do get some popping.

[Chuck H]

+++

I didn't expand the schematic

That said, I guess great minds think alike

Chuck

[rjb]

First, the Gibson scheme probably does pop.

I dunno. I've never played a guitar with a "real" Varitone. But others on this forum surely have. Hey folks, does the Gibson scheme pop?

But that would have been tolerated at that time.

Again, I dunno. Gibson went through the trouble of using a humbucking inductor. But Varitone clones made with unshielded audio transformers are tolerated today.

You don't need to use 1M resistors.

Oops. I meant 10M (same as in the schematic).

[rjb]

I may not have stated my question clearly. Why would the Gibson engineer(s) bother adding bleed resistors to the circuit, then connect them to the "wrong" place? In every (puportedly) Gibson schematic I've seen, the bleed resistors are tied to signal. Here's a schematic that shows this more clearly (please ignore the obviously mis-drawn volume pot wiring in the lower left corner). If the idea is to discharge the caps, why connect the bleed resistors to signal instead of to ground?

But it occurs to me I may be asking the wrong question altogether. I just want to build & install a multi-cap rotary switch in a PBass, am a little fuzzy on why Gibson configured the bleed resistors as they did, and hoped to minimize "futzing around time". Maybe instead of worrying about electronic theory or trying to read the original designer's mind, I should ask something like this:

Can anyone out there tell me "I built a Varitone with the bleed resistors tied to ground, have been using it for x years, and it works great, without a pop, no matter what I plug it into (amp, direct box to PA mixer, etc.)"?

[Chuck H]

Same difference as tying them to 0V. The new schem makes it clear. This is a similar arrangement to how caps are switched in and out in most amplifier situations. The cap is tied to it's proposed termination point via a large resistor. This keeps the cap at the correct voltage potential for where it will eventually land, therefore no pop. The big ass resistor keeps enough of the caps pass effect out of the circuit while still keeping it charged (or discharged) as needed. This arrangement also prevents additional loading to the circuit. Do it this way if you can instead of loading to 0V.

Chuck

[rjb]

I suck at written communication. I didn't really ask the question I meant to, and I think that (for the most part) we agree. Let me back up and try again.

Say you want to build an inductor-less "MultiTone" (my term- I think you know what I mean) with 10M bleed resistors to prevent pops. Where do you install the resistors?

Some folks advise to install "pull down" resistors parallel to the caps, from each switch position to ground. I look at the Gibson schematics and see bleed resistors between each switch position and Switch Common, and Switch Common connected to "hot". I would call those "pull up" resistors.

In another thread on this forum, someone states that, even in bypass position, Varitones "suck tone". It is speculated that the only "tone suck path" is via the parallel bleed resistors.

This made me wonder why Gibson installs the 10Ms where they do. Installing them parallel to the caps would be simpler to assemble, and the purported "suck path" would be eliminated. So why install them anywhere else? I think Gibson had good reason for their configuration- which you've mostly explained.

Same difference as tying them to 0V.

I think there is a subtle difference.

With "pull down" resistors, an unselected cap discharges to 0V across the cap.

With "pull up" resistors, the DC voltage across an unselected cap stabilizes to the DC component of the "hot" signal (which, given ideal conditions, equals 0V). A tiny AC wiggle (the current, limited by the big-ass resistor, times the cap's reactance) rides over the DC level.

The cap is tied to it's proposed termination point via a large resistor. This keeps the cap at the correct voltage potential for where it will eventually land, therefore no pop. The big ass resistor keeps enough of the caps pass effect out of the circuit while still keeping it charged (or discharged) as needed.

Gotcha, I think. The cap is charged to the same DC level as the signal at the termination point- but I don't think the cap tracks the instantaneous signal voltage.

Here's what I'm wondering: Defining the pickup to be referenced to 0V, is the DC component of our "hot" signal always exactly 0V? If no DC offset ever exists, I see no advantage to "pull up" resistors over "pull down" resistors. That's why I speculated the hot signal may sometimes have a DC offset. (What if, say, the capacitance of a dirty selector switch in the 2nd schematic becomes high enough to act as a DC blocker?)

This arrangement also prevents additional loading to the circuit. Do it this way if you can instead of loading to 0V.

I'm not sure what you mean by this. "Pullup" resistors (which, I think, is what you are advising) load the circuit, but not by much. "Pulldown" resistors wouldn't load the circuit- but I'm not convinced they would work under all conditions.

Thanks for your insight, Chuck. I think I've now over-thought this enough to proceed with going with my original plan (Gibson Varitone-syle "pullup" resistors)!

-rb

[Chuck H]

OK, I'm a broad strokes kind of designer (helps me to avoid math). Any offset doing the resistors either way can be neglected because it's mice nuts. I'm not really inclined to do the math it takes to figure the exact size of mice nuts. That said... If the goal is to build sans inductor then you really must redesign AND use pull down resistors as you are intending to load at specific frequencies via a capacitor (probably through a resistor to limit the effect). As far as that goes I think you should just use an inductor. He he... Read on...

I know (because I've looked) that it's nearly impossible to find an appropriate inductor for the job. I'm just guessing that this is one reason you want to build sans inductor. Well don't fret (no pun intended) Dan Torres did the work for us. He sells a "varitone" circuit for guitars and basses that uses a small audio transformer, readily available at Mouser, as the inductor. pn: 42TL-021-RC

http://www.mouser.com/catalog/specsheets/XC-600127.pdf

Just snip off the leads for the secondary and CT and use the remaining leads as your inductor.

Mouser currently shows that this part is on order (???) Hope that doesn't mean it's discontinued.

Cheers

Chuck

[David Schwab]

Say you want to build an inductor-less "MultiTone" (my term- I think you know what I mean) with 10M bleed resistors to prevent pops. Where do you install the resistors?

With those circuits, one side of the cap is already connected to ground. I'm not sure you need the bleed resistor.

In another thread on this forum, someone states that, even in bypass position, Varitones "suck tone". It is speculated that the only "tone suck path" is via the parallel bleed resistors.

That's because of the 100K resistor in series with the signal. I don't use those when I make Varitones.

[David Schwab]

I know (because I've looked) that it's nearly impossible to find an appropriate inductor for the job. I'm just guessing that this is one reason you want to build sans inductor. Well don't fret (no pun intended) Dan Torres did the work for us. He sells a "varitone" circuit for guitars and basses that uses a small audio transformer, readily available at Mouser, as the inductor. pn: 42TL-021-RC

http://www.mouser.com/catalog/specsheets/XC-600127.pdf

Just snip off the leads for the secondary and CT and use the remaining leads as your inductor.

Mouser currently shows that this part is on order (???) Hope that doesn't mean it's discontinued.

Craig Anderton used the 42TM-019 in his Varitone circuit in Electronics Projects for Musicians.

You can also get them here:

Http://www.smallbearelec.com/Categor...art=9&total=17

[rjb]

OK, I'm a broad strokes kind of designer (helps me to avoid math).

I only resort to the math (and text & reference books) when something doesn't seem to make sense to me. Like "Why'd they do it that way when it would be easier to do it this way? Maybe they had a reason- let's get out the books and plug in some numbers..."

If the goal is to build sans inductor then you really must redesign AND use pull down resistors as you are intending to load at specific frequencies via a capacitor

I know the cutoff frequency is determined mostly by the pickup's inductance and the tone cap (plus the cable's capacitance, at higher Fc's). Since I don't know the pup's inductance, I'm just going to get a bunch of caps and tune by ear.

(probably through a resistor to limit the effect).

I'm just going to run the caps through a pot (replacing the inductor in the Variac schematic)- which will usually be set at 0. In other words, it'll be like a normal tone control with a selection of caps. The pickguard already has a hole where the previous owner had a toggle switch for 2 caps- so I'm just replacing it with a rotary and a bunch of caps.

(As far as that goes I think you should just use an inductor.

Nope, don't want to. I want the frequency boost just below cutoff. I stumbled upon this clip and said "I gotta make me one of those".
YouTube - 12-way, variable cutoff, resonant peak, passive low-pass tone control (prototype).

Besides (if I understand the texts- which is a big "if"), the Varitone notch frequencies would be determined more by the pickup's inductance than by the inductor- so it's likely I'd probably have to redesign anyway.

I know (because I've looked) that it's nearly impossible to find an appropriate inductor for the job.....transformer, readily available at Mouser, as the inductor. pn: 42TL-021-RC

That's another thing- I don't understand what constitutes an "appropriate" inductor. I look at the specs for that thing and see input impedance of 4Kohm +/-10% at 1KHz. Do the math thing... Z = 2PI x f x L, solve for L... looks like 0.64 Henry. Isn't that kinda low for this application? I must be doing something wrong, but the designs I see floating around the web don't seem to "work on paper" - I get bandwidths over 2KHz. I don't have any real test equipment, and don't wanna mess with it (at least not right now, and not for an old PBass).

Thanks again,
-rb

[rjb]

You can also get them here:

Http://www.smallbearelec.com/Categor...art=9&total=17

It's always good to have an alternate source. It seems all the "sanctioned by someone or another for Varitone use" transformers, plus the most popular rotary switches, are continually on backorder at Mouser.

[rjb]

With those circuits, one side of the cap is already connected to ground. I'm not sure you need the bleed resistor.

Yea, but the cap is just sitting there with nothing to discharge it other than its own leakage current.

That's because of the 100K resistor in series with the signal. I don't use those when I make Varitones.

Un-unh. The bypass switch position shorts the 100K.

However... if you leave out the 100K resistor, the depth of the notch theoretically is reduced by about a factor of ten, BUT, using the "Anderton" transformer (10K input @ 1KHz) and estimates of 6H and 12Kohm for the DiMarzio PBass pups,....... scribble, scribble, scratch,... that looks like a bandwidth of about 250Hz. Damn.

Thanks a lot, David. Now I *have* to buy an audio transformer and a DPDT pot!

[Chuck H]

I can't read many electronic formulas. Never learned the language complete. But IIRC when I was a boy I did buy a kit from Dan Torres and it had the 019 transformer, not the 021. I got the transformer # online in a post I dug up. I figured that was the inductor Dan must have switched to for some reason. Could have simply been incorrect info. On David's good advice and my recollection I'd go with the 019, also available at Mouser on the same spec page I linked.