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Post by ashcatlt on Apr 10, 2013 9:42:51 GMT -5
Oh it'll probably be nasal. You know, I think my Rick will do something like that, though the cap values are a bit different. I'll have to look into it...
Couple things on John's graph:
1) He's got both pickups exactly the same. Like, to a point of precision that you would only get out of a factory on accident. They are "wound" exactly the same and are both putting out the same amount of energy.
2) The signal being "picked up" is exactly the same for both pickups, and perfectly in phase between the two. The closest you'll get in the real world is maybe two coils of a SC-sized humbucker.
3) All this to say that the graph above show the frequency response of the filter created by the pickup network. It answers the question of what happens electronically, but doesn't necessarily tell you what you'll actually get out of it. If this is between a neck and a Bridfe pickup, they will each be contributing a significantly different signal in terms of harmonic content, and likely at different overall levels. Those signals get mixed, with all of the positive and negative reinforcement from the phase differences at different harmonics, and the that signal is affected by the filter curve as shown.
4) Since both signals are the same, they add to somewhere close to +6db where neither is affected by the filter. On the red line it looks like the bridge pickup is contributing nothing up to about 1KHz. Not only are the lower frequencies noticeably attenuated, but they're all coming from the neck pickup. Or at least that's my read on it.
Don't know if any of this helps...
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Post by ashcatlt on Apr 4, 2013 16:01:16 GMT -5
But it's just easier to speak of these things using words like "ground", so long as we all understand that these are not absolutes in a guitar circuit (in a DC circuit, like the innards of a stompbox, the terms are absolute, and "hot" and "ground" cannot be interchanged). This is getting a bit beyond what the OP is looking for, but since you went there... I'm not completely comfortable with what you've said here. The difference between DC and AC is that with DC one wire is always more positive than the other. Either (or neither) of these may be used as ground, depending on the needs of of the circuit. "Ground" is really a term we use for a reference voltage, or a common node in a circuit. Doesn't really matter whether that circuit is AC or DC. We normally think of "hot" as being more positive than "ground", but this is not always the case. There are a number of pedal designs where everything is referenced to the "top" of the battery - the +9VDC node of the circuit - and are called "positive ground" circuits. Some are bipolar - for example +/- 9VDC supply with ground halfway between (at 0V). In fact, most active circuits need to be bipolar in order to work on the entire AC waveform so even when using a "single supply" 0V - 9VDC the audio signal will be biased (referenced) to some voltage halfway between the the two extremes of the battery. This is often called Vref - meaning "reference voltage" - but is nearly as often called "audio ground".
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Post by ashcatlt on Apr 3, 2013 9:51:49 GMT -5
What newey said, but... ... move that ground wire to "hot" to silence to (sic) opposite coil. I know what you meant, and you know what you meant: Replace the ground wire on the switch with the pickup's "hot" wire going to the 5-way.
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Post by ashcatlt on Apr 3, 2013 9:42:35 GMT -5
Take another look at that bridge V there. It should be wired like the neck's. Right now it's just a static resistor to ground.
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Post by ashcatlt on Apr 1, 2013 10:57:35 GMT -5
Yeah, if there's a locator tab in your way just break it off. The picture above is a bit too blurry to tell exactly what's going on.
The "ground" wires can be collected anywhere you find convenient. I hate soldering to the back of a pot, so I usually gather these at some solder lug somewhere. The case of the pots and switches should be connected to that mess also for best practice shielding purposes, but it's not essential to the function of the circuit. It might be a slight bit noisier without, but it'll work fine.
But you asked a specific question... I've never actually used a P/P pot like that, but I have heard from others that you need to be careful about soldering to the case around the switch as too much heat for too long can mess up the insides. This is true of many components, but this kind of switch seems to be especially delicate.
Some of these have an extra solder lug hanging off the side which can make things a bit earlier. If you don't have one of these, my suggestion would be to either find a a ring terminal which fits over the pot shaft/threads, or just solder a wire to the washer on the pot. When you get everything together this will conduct to the pot/switch shaft and accomplish your shield ground.
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Post by ashcatlt on Mar 31, 2013 11:51:45 GMT -5
The red and green wires in that diagram come from the pickup. It's like newey said above - the pickup mode switch (s/p/split) has four wires in and two wires out. In this case the two outputs would be jumpers connected directly to two of the input wires. So Dimarzio took the liberty of using those same wires as the two inputs to the phase switch and jumpering from there. It works fine, but I can see where its confusing.
So, red and green come from the pickup and don't need to go anywhere else. The "to hot" and "to ground" are the two outputs and go to the V pot. Interestingly, these two wires are interchangeable since the phase switch basically flips them around anyway!
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Post by ashcatlt on Mar 30, 2013 11:24:00 GMT -5
I don't have a multimeter so I can't really check it but is there another way to be dead sure? Battery and LED or small DC bulb? You should just get yourself a meter. The cheap one from your local hardware or auto parts store will be good enough for most guitar projects. But there's no such thing as "upside down" on this type of switch.
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Post by ashcatlt on Mar 27, 2013 10:45:26 GMT -5
Absolutely go grab JohnH's GuitarFreak spreadsheet from our Reference Section. You can change the pickup capacitance value to simulate adding a parallel cap, and just change the V pot value to show what happens when you add the parallel resistor.
I believe that what you'll find is that the resistor damps down the resonant peak at the high end without changing the cutoff point, while the capacitor moves the peak without much changing its amplitude.
But that won't really tell you how it sounds. You can test these components outside of the guitar. As long as you leave the V pot on 10, the order of the components don't matter. Use some alligator clips and try it out!
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Post by ashcatlt on Mar 26, 2013 18:34:27 GMT -5
The orchestral version was much more recognizable, and more pleasant to my ears.
Not sure why there's a Cyrillic watermark, but it says "BNT Archive".
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Post by ashcatlt on Mar 26, 2013 18:23:27 GMT -5
I believe that any output that is cut with a resistor will disproportionately be from the high end of the spectrum... Yes, way disproportionately. The resistor will have to start to get close to the value of the pickup's internal resistance before you start to notice much broadband volume reduction. By that point you'll be getting little more than luke warm mud out of the thing. Of course, since treble content often fools us into thinking a sound is closer and/or louder, there may be a loss of "percieved" volume, but technically...
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Post by ashcatlt on Mar 25, 2013 22:50:18 GMT -5
tl;dr (err...w?)
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Post by ashcatlt on Mar 22, 2013 9:13:50 GMT -5
For some reason I can't see the pictures right now, but here's a couple things:
1) Choose the split coils for hum-cancelling and then install the pickups so that those coils are physically where you want them. If the original diagram actually gives you the RWRP coils, but those foils are not closest to the neck, then rotate the pickups.
2) Putting a resistor parallel to the volume pot actual increases the load on the pickup. When we say load, we generally mean current draw. Less resistance = more current demand for a given voltage = greater load. It will reduce treble as you expect.
3) Splitting the coils of a humbucker in series is completely different from splitting them when it's otherwise in parallel. Any S/P switch for the bridge pickup will have to "re-wire" the 5-way also, in order to avoid dead spots. Seems to me like somebody around here figured out how to do it with a DPDT in the not too distant past, but I cant remember who or where.
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Post by ashcatlt on Mar 20, 2013 10:10:26 GMT -5
If what we think is happening is actually happening, then you should also lose the neck pickup when in normal (in phase) series mode. Didn't you say that volume is reduced in that case? It should also be a bit noisier with bridge only if you flip the phase switch.
You could diagnose with a meter across the cable, but I'm pretty confident that this is the issue. It's most likely not in the switching itself, but in the fact that the cover is sharing a wire with the coil. The pickup cover needs to be separated from the coil wire. This will probably mean some rather delicate surgery. First you need to open the cover and remove that connects to it, then add a separate wire.
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Post by ashcatlt on Mar 19, 2013 14:26:27 GMT -5
Do the covers have a separate wire of their own, or are they sharing the "bottom" coil wire?
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Post by ashcatlt on Mar 17, 2013 3:13:04 GMT -5
At the end of a Big Black tribute for the Tides of March.
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Post by ashcatlt on Mar 14, 2013 15:22:34 GMT -5
Nosir. I made a sort of A/B booster thing a while back, but not exactly what youre looking for.
Edit to add - I don't think the led should need to share any connections with the audio part of the circuit. Specifically I don't think you need to ground the bottom of the battery or the led. In fact, in my mind it seems like isolating the led citcuit would have less chance of causing audible pops from sudden surges through the LED. I could well have it backwards, though.
If, after adding the LEDs, you end up with switching pops which were not previously there, you may need to add a capacitor to slow things down a little. In that case component order will matter. You'd probably want to go: battery + > resistor > switch > LED > battery - , and then put the cap from where the resistor meets the switch to the battery - so that the cap is NOT switched when the LED is. It might behoove you to wire it this way to begin just in case you end up needing to add the caps.
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Post by ashcatlt on Mar 14, 2013 0:55:20 GMT -5
As long as the led, resistor, and switch are in series between the top and bottom of the battery (with the led "pointing" the right way) it doesn't make any difference what order they come.
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Post by ashcatlt on Mar 13, 2013 10:15:52 GMT -5
I don't own any horses. Or any women for that matter. This weekend begins the Ides of March Baccanalia here in Duluth. We don't wait for Lent to pass...
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Post by ashcatlt on Mar 10, 2013 22:17:33 GMT -5
Yep! Fine job going from my intentionally vague description to a perfectly functional diagram.
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Post by ashcatlt on Mar 10, 2013 12:58:50 GMT -5
Wow, I said I'd be back "tomorrow night" three days ago! Sorry about that. You figured it out anyway. That won't give you the "No Load" option that you wanted. Luckily you've got plenty of poles to work with! Look at Sw4a+b, and Sw5a+b from this scheme that's in my Rickebacker: They are Neck V and Bridge V, respectively, and both do "No Load". Edit - Those are wired "backwards" for some reason that I'm sure made sense at the time. You could (probably should) swap the two commons to get normal V control action with shorting of the jack when off.
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Post by ashcatlt on Mar 9, 2013 14:02:34 GMT -5
Don't think it's a ground loop. The jumper between the two jacks should be about exactly the same as with nothing plugged in at all. It sounds to me like some kind of problem with the sleeve of the jack. Either it's somehow not contacting the sleeve of the plug when it's inserted, or its solder connection has come loose inside there or something.
Got a meter? Stick a cable in the input and another in the return jack and measure resistance between the two sleeves. It should be very close to 0, but I'm betting it'll be closer to infinite.
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Post by ashcatlt on Mar 8, 2013 13:11:27 GMT -5
Pretty sure you could just wire the six lugs of the blend pot exactly like the six lugs of the DPDT switch that's typically used. It'll probably act like a funky volume control, with severe attenuation (and I'd guess a good deal of noise) in the middle of the rotation.
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Post by ashcatlt on Mar 7, 2013 4:34:05 GMT -5
Welcome aboard! Interestingly, my first post on this forum was also about this type of "stepped attenuator". I don't have a link right now...
The way you've drawn it looks like it'll work, but it's far more complicated than necessary. You can accomplish the same thing with one pole of the 5way (two I guess if you want the "no load" position) and a bunch less resistors. Take it as a hint, see if you can work it out. If you don't get it before tomorrow night, and nobody beats me to it, I know I can dig up something.
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Post by ashcatlt on Mar 6, 2013 21:54:09 GMT -5
Can you explain how the 56K stops a resistance from going above 1M? Is that because its in parallel with whatever load there is from the in side and the wiper which always makes it overall less than 1MEG? But then there is 1MEG to go in series afterwards at the volume pot? Or am I still a way from getting my guitar electronics diploma? Well, the V pot is kind of supposed to do that. Consider what would happen without that resistor, though. Leave the V pot all the way up, and turn the T pot halfway down. Now, independent of the high cut action, we've got a voltage divider composed of 500K on top and 1M on bottom. That's a noticeable broadband volume drop! The 56K changes the taper of the pot (what sg said) and keeps the interaction with the V pot a little more predictable.
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Post by ashcatlt on Mar 6, 2013 9:11:13 GMT -5
Yep, looks like the Jag strangle cap strangles everything if it strangles anything. The old Rickenbackers supposedly had a cap permanently in series with the bridge pickup.
That potentiometer style of T pot wiring is pretty strange. Don't really see that about anyhere. I'd think that 56K resistor helps keep there from being upwards of 1M resistance between the pickups and the jack!
Neither the Rickenbackers nor the G&L with bass cuts do this, so I don't think it has anything to do with that...
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Post by ashcatlt on Mar 6, 2013 7:10:25 GMT -5
Yeah. I still get confused as to which lug is 10 and which is 0. I guess I had it backwards for the V control? That's probably partly because your original diagram had the strangle caps on the other wires...but I guess only two of us know that...
Frankly, though, that just makes it worse. At least one lug of each V pot and one wire from each pickup needs to reach the jack sleeve via straight wire!
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Post by ashcatlt on Mar 5, 2013 19:55:17 GMT -5
Did I mention that I'm kind of a flake? I sure am sorry that I missed the opportunity to hang with ya'll, but... well, we all have excuses. This whole week is pretty well screwed for me, but if you haven't passed on too far west by Monday I may be able to catch up with you for a while.
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Post by ashcatlt on Mar 5, 2013 19:47:25 GMT -5
This is the reason we try not to edit things in place, especially not after they've been commented upon. Ruins the continuity of the thread and leaves people wondering what the heck I was talking about.
Are the pots marked T meant as typical high-cut tone controls?
The problem I see is that you've got the T pot, its cap, the presumed bass-cut cap (caps 1 and 2) and the pickup all in series with the V pot parallel to the pickup and its wiper going to the switch. The pickup only finds "ground" through all that crap. With V all the way up and T all the way down you have both caps in series acting as bass cut. With V all the way up and T all the way up you've got the two caps plus the very large resistance of the T pot between the pickup and ground. With V all the way down and T all the way up you've got the pickup hanging "upside down" from hot. With both pots all the way down you'll have silence in those high frequencies passed by both caps, but low frequency noise (like our old friend Buzz) going to the output.
That is all kinds of yuck!
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Post by ashcatlt on Mar 5, 2013 7:10:08 GMT -5
Well there's a number of issues there. The T controls are wired all crazy and will work as bass cut and probably volume to some extent, but not typical hi-cut tone control. For very similar reasons the V controls will not behave in a satisfactory manner.
I could have sworn that we were wondering whether there would be a difference between a cap per pickup and having one master bass cut. A couple of different perspectives have been given. I am prepared to be proven completely wrong, but I definitely think that one of us needs to Spice this up and post the results.
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Post by ashcatlt on Mar 4, 2013 22:16:27 GMT -5
I thought I was pretty careful to specify that I was trying to (over?) simplify the question by specifying that I was talking about active mixing and active filters to avoid insertion losses. Maybe I went a bit too far there?
In order to understand why two pickups mixed passively don't add up to double the voltage of either one, I think it's pretty easy to just look at it like a resistive voltage divider.
Consider the case where we have two identical pickups connected in parallel. We're not adding the low-cut filter yet, just flipping a Tele switch to the middle position.
Analyze the contribution of the neck pickup to the overall output. It generates some voltage Vn. This can be thought of as going into a voltage divider with the impedance of the neck pickup as the "top" resistor, and that of the bridge parallel to any controls and the amp/pedal input as the "bottom". Assuming that the controls and amp impedance are chosen to be acceptably high, they will only reduce the impedance of that "bottom" part by a very small amount, so that our voltage divider is essential made from two equal impedances. This will divide Vn down to half.
Do the same for the bridge pickup and you get the same answer. It's contribution is one half of Vb.
If we assume that for some given frequency Vn = Vb, then Vout= 1/2 Vn + 1/2 Vb = 1/2 Vn + 1/2 Vn = 1/2 (2Vn) = Vn.
Yes, there is an inductive (and also a capacitve) component to each coil's impedance which makes the Z change for different frequencies, but you'll recall that we specified that these two coils were identical, so that the impedance at any given frequency will still be equal, and we still have a 1:2 voltage divider, so inductance and capacitance doesn't really impact this answer in this one specific (and not particularly realistic) instance.
The phase difference between harmonics caused by different positions along the string does impact our perception of how the output changes, but it doesn't actually change the action of the filter, and it changes depending on where on the string you're fretting, and is difficult to predict.
TBH, I think that the actual low-cut action will be approximately the same either way.
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