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So it was like a speed bump or a curb and this is where you fell?


If you eliminate the tone pot, you do have some interesting options. You could use two volume pots alone, or you could use two push pull pots for the volumes.

Hmm, there be switches attached. What could we use them for? I know, to switch in a tone setting. If you almost never use the tone, have one setting do nothing (or select a 240K resistor in series with the existing tone cap) and the other a resistor value equal to the tone pot setting when you actually aren't almost never using it.

In other words, you could have four tone settings (two switches - four combinations) a la' the Gretsch tone switches.


Now, here's a possible fly for your ointment; do these pickups have a single conductor plus shield wiring? If so, there may be some issues with series and reverse phase wiring (one pickup's shield and metal cover will not be grounded, but connected to the output (reverse series), or the other pickup's output (series)).


Either way, touching the pickup's metal cover will generate some interesting "buzz-tones".

There are two kinds of blend pots; the first are the pan pots sold as blend pots, and the second are the true blend pots used in active bass guitars. Blend and Pan Pots

In either case, you would be able to blend between both pickups to varying degrees, but have no overall control of the volume.

I might suggest that you use a dual concentric shaft pot such as the EP-4586-000 for two volumes in one. This way you can still adjust the relative level and the overall volume.

Using individual volume pots with the T-Riffic will require some redesign.

I recently did a design that does what the T-Riffic does but with two volumes and two tones. It also had two push pull switches per pickup for coil selection. While considerably more complex than what you need, I'll link to it since, if the push pull coil selection stuff is removed, it shows how I use individual volume pots in a series/parallel switching structure.

P-Rail Switching Scheme

If you don't "see" how to reduce this for what you want, let me know and I'll modify the drawing since it's much easier to take stuff out than to add it in.

Also note that I use a different combination sequence than John A. I feel that tonally, the N*-B belongs next to the bridge selection and the N*B belongs next to the neck selection.

Anyway, the ordering is arbitrary.

For mice I use the plastic resettable traps. I've found that I need to anchor them with nylon cord (mice will eat organic string) as the victim tends to wander off whilst "engaged".

They're very easy to deal with; I just open the trap over a waste bag and then place it "back in service".

For larger vermin I use the "Havaheart" spring cage traps.

For ground hogs, a rag soaked in ammonia stuffed into the burrow works well. They move on.

Not having a garden works wonders.......


I have made repellers using a tweeter and an audio oscillator (not formal test equipment) at about 22 to 40 Khz. If you have dogs, they will notice these. Of course, I first made these for dogs.....A neighbor had dogs that were put out at 7 am and barked forever. I set my oscillator to go on at 3 am.....

I used an LM386 audio amplifier IC and a Mr. Coffee coffee pot timer (these were a few bucks at every surplus place 10 years ago).

www.national.com/ds/LM/LM386.pdf

You have it in mind, but need to see it.


a blend pot .......

which are not.



Pan away.



I could, but, these mods abound a'web. Visit the Stew Mac and Seymour Duncan sites.

The Strat already has a master volume. Remove the (now) master tone circuit from the 5-way switch and attach it to the hot end terminal of the volume pot. A pan pot to blend in (well, in this case it IS blending in) is just a series connection from the pickup to be blended in to the hot terminal on the volume pot (or 5-way switch).

The standard pan pots sold as blend pots will solve your handed problem (you won't have to rewire) as well as do the blending function.


You'll have to fix the knob knumbers yourself.

Uh oh, what does this mean?

I'll take a look at the Fender drawings.

;D ;D ;D ;D ;D

And I have. You've been bushwhacked by the "handedness" of things guitar. The "Left Hand" wiring arrangement ( ) that you mentioned is too true to be good!

Some might call it marketing.

While the wiring to the pots is indeed reversed (mirror-image) between the two wiring diagrams, the pots ARE NOT. Both 013 MexiStrats that you reference use the 0013446000 250K audio taper pot. In both right-handed AND left-handed guitars, these are right-handed pots.

Furthermore, the 5-way switch terminals also are mirror imaged from the right-hand to the left-hand guitar. Both use the same 0053291000 switch.



When one connects audio taper pots backwards it makes linear taper pots look real special. Your pots are wired backwards. Your switch is likely wired ok.


Some fart-for-brains in their drafting department just mirror imaged the drawing for the left-hand guitar. This would work fine if the pots were left-handed (reverse audio taper). (It wouldn't hurt if there were reverse-numbered knobs too.)

(I remember when d'afting departments actually checked drawings prior to publishing them.)


You will have to follow the right-handed wiring for the pots. BTW the left-hand guitar shows the switch correctly, the right-hand guitar drawing is incorrect in the terminal positioning. If one followed the terminal sequence wiring order for the right-hand drawing of the switch, it would still work.


If you'd had the sensibility to use linear pots to begin with, this wicket wouldn't be sticky. ;D ;D


OK, I said that I had a way around right-handed pots, pan you see it.

If you want to know more, let me know.

Some questions come to mind.


First, a simple but not always obvious one. Are you left-handed and turning the pots backwards as compared to a right-handed operator? If you ARE left-handed and want controls that turn backwards (CCW is "10"), I do have a simple electronic solution for this, but you're on your own concerning the "knob knumbering".

Why do you believe the 250K pots are linear?



Yes, it will affect the overall brightness at the full-off rotation of the tone control as well as the loading effect of the volume control. A neat practice is to always use a 500K pot for tone. It will be at 250K at around "7-8" and be at "12" (brighter) when full on.

The taper (linear or audio) is the rate of change of the pot when rotated. This affects the rotational response of the control.


I always test every component that I use. They're easier to just not use than to unsolder and replace.

Test the pots. Measure the end terminal to end terminal resistance. Write it down and on the pot side with a black Sharpie marker.

Set the pot to exactly half its rotation. Measure the resistance from the wiper terminal to EACH end terminal.

If it is Linear, you will read about half the end to end resistance on each side of the element. Linear pots are often marked with a "B" for taper, but this varies.

If it is Audio, there will be a substantial difference (around 5 to 1 or more) in resistance on each side of the element. Audio pots are often marked with an "A" for taper.

Write an "A" or "L" on the side of the pot.


A linear taper volume pot turned from "0" to "2" has the same resistance ratio as an audio taper pot turned from "0" to "6". If the volume pot, when turned clockwise from "0", turns on abruptly, I suspect a linear taper.

A similar effect will occur for a linear tone control. The tone changes will occur abruptly and then not change much at all.


There are those a'planet that claim that linear controls are better for volume and/or tone, but I suspect that this is related to the blatant overuse of a Marshall full-stack over extended periods of time.

My 85 year old Father-in-Law calls this effect "Huh, eh?"

Knob knob?

Please keep in mind that the order that you want things in may adversely affect the ease of how you may actually have things realized.

path

Let's start with the same pole assignment on the 4P5T lever switch. We have four poles; we'll assign one to Bhot (Bnot is locally grounded), one to Mhot, one to Mnot, and one to Nnot (Nhot is connected to the output).

With this structure, we can trivially do any parallel combination since we determine where each of these four leads go (in a simple one dimensional parallel structure the choices are no connect, and either output or signal common.


Let me make it clear that I am not calling anything "ground" since no lead coming from a pickup is EVER a ground except for a shield connection which needs to be locally grounded.

These are AC generators with two leads, one is only ever ground if we externally connect it to ground. In any series structure, all pickups CANNOT have a "ground" lead since all not-HOT leads are not connected to ground.


In a series structure (two dimensional), these leads go to other places in addition the those mentioned for parallel. Realizing any collection of combinations on a 4P5T switch thus assigned is trivial. In fact, every possible combination (for three single coils, 17 excluding OFF and phasing) is possible except M*(B+N) which requires 5 poles since Bnot or Nhot must leave its permanent connection to realize this case.


What is not trivial is the reassignment of the terminals in each position to alternative connections. We need to have a "x"PDT switch for reassignment. I would focus on this reassignment while leaving the 4P5T pole assignments in place.

NORMAL: B, B+M, M, M+N, N

SUPER: B*M, B*(M+N), B*N, B+M+N, B+N

Let's look at reassignments;

1. B to B*M; this requires the selection reassignment of Bhot, Mhot and Mnot. M's are inactive in normal and we could leave Mhot connected to output in both cases so this requires two poles.

2. B+M to B*(M+N); this requires the selection reassignment of Bhot, Mnot, and Nnot. This requires three poles.

3. M to B*N; we could leave Mnot connected to ground in both cases. This requires the selection reassignment of Bhot, Mhot, and Nnot. This requires three poles.

4. M+N to B+M+N; this requires one pole.

5. N to B+N; this requires one pole.

Let's add them up; 2+3+3+1+1. Hmmm, that would be 10.

Now,

positions 1 & 2 both change Bhot from going to out to going to Mnot. Therefore, one pole can be used to realize both of these variations. -1

positions 4 & 5 both add in the bridge in super. Therefore, one pole can be used to realize both of these variations (the reassignment of Bhot). -1

That would be 8. Therefore, this scenario IS realizable with a 4P5T and an 8PDT.

There may well be other optimizations as well that further reduce the pole count. If two more are found, a 6PDT is needed (which is commercially available in a rotary switch).


I'll take a quick stab at it.



Oh woe and misery, I also can only find use for a 3PDT switch, but I believe that I realize your desired combinations in the desired order.

Now, it's late and I've had a full day so I may still be lost in musing on interbrane quantum phasing, but I'll revisit this on the morrow since it makes way too much sense now.

This is the path I would (and do) follow for most all designs.

/path


Now, you may have done exactly this, but I don't "see" it.


Done is.




No. The original Mike Richardson design is the second one of his that I posted. It did B+N and B*N in the middle position. This is the one that I drew for my design.

The first Mike Richardson design that I posted (the modded mod) is the second one that he did adding M and B*(N+M) in place of B+N and B*N.

..or a power chord.

tedfixx

I presume that you meant;

NORMAL: B, B+M, M, M+N, N

SUPER: B*M, B*(M+N), B*N, B+M+N, M*N


Your desired combinations are nearly identical to a Mike Richardson design of yesteryear.

It does the following:

NORMAL: B, B+M, M, M+N, N

SUPER: B+M+N, B*M, B*(M+N), M*N, B*M*N

It might be fairly simple to modify the B*M*N into a B*N, you'll have to look at it.

I don't want to disappoint you, but it only requires a 4P5T super switch and a 2PDT.




This was his original version,


and my implementation of it,
Mike Richardson wiring with phase

in here.
The Padouk Caster

One of the things that can be done to determine two of the leads on a transistor is to measure the diode drop using a multi-meter that has a diode detector (this is usually a position on the selector switch with a diode symbol).

Most silicon diodes will have a drop of about 0.65 VDC and most germanium diodes will have a drop of about 0.35 VDC. The operative word is "about".

You need to know the voltage polarity coming out of your meter when using this function. Use another meter set to 20 VDC (if your meter has a 9 VDC battery) and connect it to your meter while it is set for diode detect.

Measure the polarity of the emitted voltage. Keep track of which test lead is positive.

Once you know this, you can detect diodes such as the base to emitter junction inherent in every bipolar transistor by measuring from lead to lead to lead both ways (this is 6 tests).

Now, you know for relative certain that the third lead (the one that doesn't belong to a diode) is the collector. You also know that the other two comprise the base to emitter junction.

If it's an NPN, the base will be the lead connected to the positive voltage lead from your meter and the emitter will be the lead connected to the negative voltage lead from your meter.

If it's an PNP, the base will be the lead connected to the negative voltage lead from your meter and the emitter will be the lead connected to the positive voltage lead from your meter.


This doesn't tell you everything, however;

if you know the base material polarity (NPN/PNP) it tells you the pinout, and

if you know the pinout (CBE) it tells you the base material polarity.

Search me.

(Well, go search the web.)

That's because, in my experience, I haven't found a supplier that actually KNEW the schematic representation of their switches.....(or what the words meant).