Basic Varitone on my Belt

[Rossington88]

Hey, All  !!

I'm new to the board. As a first post, I figured I'd ask a few questions about my first DIY pedal project. Basically, I'm going to build a varitone circuit, mount it inside a small pedal box, attach a clip and wear it (Instead of, y'know, putting holes into my 335!!). I've included schematics to be as helpful as possible.

1.) First and foremost, will my schematic even work?

2.) Is the Varitone wired in-PARALLEL with the hot wire (As in my crude pedal schematic) or is it basically a bleed circuit (As in the original Gibson schematic)?

3.) Being a Gibson-guy, I know almost everything about Gibson's. In the 1950's, Gibson made their own 1.8 Henry inductors for the Varitone (I'm not sure what they did but those things were about the size of a 1/2-pack of smokes!!). In the 1970's, they actually coiled wire around a nut and bolt ( !!). My question here is, WHAT should I be using as an inductor? And WHERE could I find it? I've only found a Bill Lawrence Q-Filter to be similar but that only has two leads and could not be grounded if the circuit must be tied back into the hot wire...

4.) Finally, will either of these values NOT work with my application? Caps rated @ 250V. Resistors 10M @ 1/4W 5%.

PEDAL SCHEMATIC


MY VARITONE CIRCUIT + GUITAR SCHEMATIC


ORIGINAL GIBSON STEREO VARITONE SCHEMATIC

[Digital Larry]

I'll admit that I don't quite know what is meant by the red and green circled parts of the schematic.

You don't seem to have a selector switch to choose which series cap is going to be connected between the inductor and your guitar's hot lead.

The 10M resistors on each cap are to keep DC voltage from building up on them to prevent a huge POP when you switch from one cap to the next.  As drawn, with the 10M resistors in series, no it won't work. [unless that green thing coming over is part of a rotary switch, which is not clear from your drawing] 1/4W or 1/8W 10Meg resistors should be fine.  250V caps should be fine, unless you touch a high tension line while you are playing.

The combination of L and C forms a resonant circuit.  At the resonant frequency the impedance will be very low, shunting those signals to ground.  The 100K resistor before the LC will make that notch deeper.  Taking that resistor out will make the notch shallower.

I did some SPICE analysis on the Varitone circuit, which I posted over here.

[Rossington88]

Hey, Larry!! Thanks for the response!!

Ah!! You are right, I failed to draw a 6-way rotary in there. It helps if I have a way to switch between the caps, doesn't it  ?

I understand what you are saying about the 10M resistors but I'm not sure what you are saying to do to fix the issue. How should they be wired in with the capacitors?

The red and green circled areas were for my own reference when this idea first came about. I don't like to just go into something without any kind of understanding of what I am doing so I wanted to figure out what each of those components did. The 100K resistor was a treble-bleed added by Gibson to darken the bright-sounding ES-335.. The green phantom line is just the hot wire if the Varitone were not in the circuit. The red circled inductor, admittedly, I'm still a bit sketchy on. I kinda understand the function of a choke but I'm not at all sure why it is present here (Newbie to electronics, remember?).

Also, should this thing tie back into the hot line and still be grounded or just be grounded?

Thank you!!

Post Script: That is some interesting data you have graphed, there. The pot does seem like a pretty interesting way to dial in the notch.

[Digital Larry]

Sorry I wrote this out of order since I couldn't see your post while I was typing.

If the rotary switch contacts are those things between the 10M resistors and the caps, then you're OK.

A resistor cannot be a treble-bleed as it is not frequency dependent in the audio range.  My assertion is that the series 100k is to make the notch deeper and wider, but I wasn't there when it was designed so that is just a guess.  Certainly it would sound different without it (less effect).  It also reduces the loading of the tone controls when the switch is not in position 1.

When the switch is in position 1, the varitone is effectively switched out.  The 100k resistor is shorted out by the switch and there's no connection to a cap.  Look at the Gibson schematic very carefully.  You have missed a few things.  

The lead of the inductor (choke) that is shown connected to ground should go to ground.  The pole of the rotary switch should go to your hot lead.

The thing that is going to be a little different is the pickup tone controls.  On the Gibson, the tone controls are on the other side of the varitone from the pickup and will have less effect when the varitone is on (because of the 100k resistor in series).  On your schematic, the tone controls are between the pickup and the varitone and their effect will not be lessened when you engage the varitone.  You can compensate for this by ear of course by just adjusting the tone controls.

Take the simulation graphs with a grain of salt.  I did not model the pickups accurately at all.  Still, it gives you an "idea" of how it works, if not exactly.

[Rossington88]

Hey, Larry!! Thank you for another response.

I think I am understanding you now. I've apparently been looking at the schematic backwards . The flow of the signal should go: Lead -> Rotary Pole -> Capacitor -> Terminal -> 10M Resistor -> Inductor -> Ground. The 100K should be after the branch of the varitone.

That makes a bit more sense. I was looking at the schematic thinking that the signal would have to be pulled through the resistors, then caps towards the pole and then into the inductor. I would say I HAVE made a bit of an error. Thank you for pointing that out to me!!

That about solves everything!! I have been looking up how to calculate and make a proper inductor so that is my last piece of the puzzle.

Thank you for your help !!

[Digital Larry]

Ah well just to be sure put up your revised drawing and I'll have a look see.   

[Rossington88]

Hey, Larry!! Thanks for all your support!!

I've included a drawing of the final schematic. It's formatted differently to show the flow of the signal in a way that makes sense to my brain. Forgive me if it seems a bit crude...




A final question. Should I be lining this box with some type of shielding? I'll be using 20 AWG vintage-style cloth wiring with a wax coating.

[Digital Larry]

#1 The 100k resistor goes between pin 1 of the rotary switch and the common (pole) of the switch.  Your diagram shows a wire across the 100k and this will not work.

#2 compared to your first drawing, you put the 10M resistors between the inductor and the pots.  This won't work either.  The 10M resistors are to keep any voltage from building up on the caps when they aren't connected.  The best way to think of this is that the 10M resistors go across each one of the caps (parallel) rather than in-line (series).

The original varitone schematic is not very well drawn and so it's quite easy to get confused by it.  Nevertheless I would print out a copy of the Gibson schematic and focus on the switch connections and the little resistor/capacitor assembly.  Use a colored highlighter and mark each point (component ends/connections) on the original schematic and find a corresponding point on your schematic.  

Another build strategy that might help you is to do it in stages.  I.e. make sure that the bypassed (position 1) works fine.  For that you will not need to attach the inductor or caps.  Then add the inductor and cap for position 2.  Make sure that works.  etc. etc.  Put on the 10M resistors last as all they are there to do is to help reduce pops when switching.  Those resistors should go between switch contacts 2-6 and ground or the hot lead (switch pole), it shouldn't matter.  Note that this is a little different than the way the Gibson schematic is laid out, so to put yourself at ease, follow that exactly.  The end result will be the same.

I think if you proceed this way then you will be able to find problems as they occur rather than building the whole thing and then wondering what the heck went wrong.

I don't see any need for extra shielding although it can't hurt.  There's no gain in this circuit so the main thing that could cause problems would be a missing or flaky ground connection (or any other connection).  Ah, let me update this remark.  The coil will be sensitive to picking up hum from external magnetic fields just like a single coil pickup would.  To kill this you need shielding of a magnetic variety, like a steel case.  I've also seen descriptions of humbucking varitone coils.  Now what you might want to try along those lines is getting some old humbucker pickups (say for free or cheap from your local guitar shop) and pry the magnets off.  I've never tried it, so take this advice with a grain of salt.


Good luck!

DL

[PRR]

> The original varitone schematic is not very well drawn and so it's quite easy to get confused by it.

+42!

Caps can be 3v rating. (In 1959, values this size usually didn't come any smaller than 250V; even today you may not find smaller than 50V?) 20% tolerance is not wrong; use 10% if the price is all the same.

All 10Meg resistors could be 0.001W, +/-500% (it's really non-critical). The 100K should be at least 20% spec. Yes, 1/4W 5% make more sense than microscopic loose-spec custom parts.

That coil.... yes, it is unlikely to be a stock part. The value does not have to be 1.800H; 2H isn't much difference. However all the suitable coils you will find are both large, and specified for DC current filtering. Iron-core coil inductance can vary a lot from nominal (DC filtering) current to this zero-DC audio filtering application.

Oddly, a pickup's worth of very fine wire on a bolt may be the easiest path. If you got a signal generator and AC volt-meter, you can calibrate it yourself.

Anyhow....

The basic idea: a coil and a cap together (a "tank") will resonate at one frequency.

At that frequency:

* Parallel resonant tank tends to infinite impedance
* Series resonant tank tends to zero impedance

Here we have series, so impedance tends to zero at resonance, higher everywhere else.

We rig the tank to-ground.

The "zero" may be several thousand ohms. That may not be low enough to short-out guitar signal. To give it a fighting chance, we put 100k before the tank to ground.

We get a notch. Larry's plots look about right to me. In real life, the coil will have a few hundred or thousand ohms series resistance (spoiling the bottom of the notch, musically insignificant), and output loading may narrow the notch a bit. Gibson didn't run it through a computer... I suspect it started as a mis-wire with a treble cap and an extra pickup, and they vamped on that.



For different notch frequencies we use different coil-cap values. Because caps are cheap and coils are dear, we switch several caps to the one coil. As Larry's plots show, this does not give "the same" notch over a wide range of coils. The high-end response tends to be the same, while the lows vary a lot. This is musically acceptable, and the alternative (multiple, tapped, or adjustable coils) is commercially ugly.

For flat (non-notched) response, we cut out the tank, and for good measure short-out the resistor. Rotary switch position "1" does that.

When caps are switched-OUT, they may gather a charge, which will POP when you switch them in, especially on hi-gain amplifier inputs. Throwing some large-value resistors in gives them a path to bleed-off. The value should be much more than the ~~100K impedances in the signal circuit, much less than the cap leakage and whatever source they may charge from. Here, unlike plate-circuit caps, there really isn't anywhere they should get a charge. It may work fine without the 10Meg resistors. OTOH, it's cheap insurance against going pop-pop all night if you get a gig next to a neon sign or Tesla coil. Or wear silk and spandex in dry air.

[Digital Larry]

Over here there is a discussion about using the Fasel wah wah inductor in a Varitone circuit.

http://guitarnuts2.proboards.com/thread/1872

If you put two of them in series the inductance doubles.  You might have to adjust the values of your capacitors to accommodate that.

The product of L (inductance) and C (capacitance) should stay the same to keep the resonant frequency the same.  Suppose the target inductor is 1.8H and the capacitor is 0.01 uF.  If the inductor you can get is 1.0H (e.g. 2 x 500 mH in series) then the cap should be 0.018 uF.  It's not a standard value but 0.015 uF is close enough.

And if you wired two inductors out of phase but positioned pointing the same direction, then you might get some level of humbucky-ness.

[Rossington88]

It's funny you would mention the Fasel. I was looking into going that route. I would need three in series for the total of 1.5H but it would be nice to know that I don't have to TRY to build an inductor. As far as the values go, I am trying to stay as close to the original specs as possible. Also, I'm not just using regular parts, I'm probably going to spend double what I should on this to make it the absolute best it can be (CTS rotary, Switchcraft jacks, Mullard Tropical Fish caps, Allen Bradley carbon composite resistors, waxed vintage-style cloth wire-). I have found the original values in these period-correct components so I think I will just stick with that. Thanks for the input, though.

[Ellerbrock Electronics]

> The original varitone schematic is not very well drawn and so it's quite easy to get confused by it.

+42!

Caps can be 3v rating. (In 1959, values this size usually didn't come any smaller than 250V; even today you may not find smaller than 50V?) 20% tolerance is not wrong; use 10% if the price is all the same.

All 10Meg resistors could be 0.001W, +/-500% (it's really non-critical). The 100K should be at least 20% spec. Yes, 1/4W 5% make more sense than microscopic loose-spec custom parts.

That coil.... yes, it is unlikely to be a stock part. The value does not have to be 1.800H; 2H isn't much difference. However all the suitable coils you will find are both large, and specified for DC current filtering. Iron-core coil inductance can vary a lot from nominal (DC filtering) current to this zero-DC audio filtering application.

Oddly, a pickup's worth of very fine wire on a bolt may be the easiest path. If you got a signal generator and AC volt-meter, you can calibrate it yourself.

Anyhow....

The basic idea: a coil and a cap together (a "tank") will resonate at one frequency.

At that frequency:

* Parallel resonant tank tends to infinite impedance
* Series resonant tank tends to zero impedance

Here we have series, so impedance tends to zero at resonance, higher everywhere else.

We rig the tank to-ground.

The "zero" may be several thousand ohms. That may not be low enough to short-out guitar signal. To give it a fighting chance, we put 100k before the tank to ground.

We get a notch. Larry's plots look about right to me. In real life, the coil will have a few hundred or thousand ohms series resistance (spoiling the bottom of the notch, musically insignificant), and output loading may narrow the notch a bit. Gibson didn't run it through a computer... I suspect it started as a mis-wire with a treble cap and an extra pickup, and they vamped on that.



For different notch frequencies we use different coil-cap values. Because caps are cheap and coils are dear, we switch several caps to the one coil. As Larry's plots show, this does not give "the same" notch over a wide range of coils. The high-end response tends to be the same, while the lows vary a lot. This is musically acceptable, and the alternative (multiple, tapped, or adjustable coils) is commercially ugly.

For flat (non-notched) response, we cut out the tank, and for good measure short-out the resistor. Rotary switch position "1" does that.

When caps are switched-OUT, they may gather a charge, which will POP when you switch them in, especially on hi-gain amplifier inputs. Throwing some large-value resistors in gives them a path to bleed-off. The value should be much more than the ~~100K impedances in the signal circuit, much less than the cap leakage and whatever source they may charge from. Here, unlike plate-circuit caps, there really isn't anywhere they should get a charge. It may work fine without the 10Meg resistors. OTOH, it's cheap insurance against going pop-pop all night if you get a gig next to a neon sign or Tesla coil. Or wear silk and spandex in dry air.

Hello. I don't understand what you mean by this:

"The "zero" may be several thousand ohms. That may not be low enough to short-out guitar signal. To give it a fighting chance, we put 100k before the tank to ground."

The reason I ask is because, for the life of me, I don't uinderstand the purpose of the 100K series resistor in the Varitone circuit. Thanks!

[Ben N]

You know Craig Anderton's inductor solution - one side of a cheap, common, *small* 4TM019 mini audio transformer. https://www.diystompboxes.com/smfforum/index.php?topic=72431.0

[antonis]

I presume @Ellerbrock Electronics is asking about the purpose of 100k resistor which simply forms a part of resonant LPF..

P.S.
Welcome, also..

[PRR]

Welcome!

Please study "voltage dividers". Everything is a voltage divider.