.01 capacitor

[cassius987]

I was thinking of replacing the .0047 capacitor in my push-pull 4003 with a .01 to see what that sounded like... There are .01 and .1 capacitors in the 4002 schematics. But are all caps created equal? Is there a certain quality or rating I would need to verify before purchase?

They are cheap enough, so I'm not too worried about what'll happen if I don't like the sound.

[basmansam]

I would make sure the pots are correct for what you are doing.

[cjj]

Theoretically, capacitors are capacitors and all you have to worry about it the capacitance value and the voltage rating.

Then there's reality. For the most part, just about any non-polarized capacitor will work (you do not want an electrolytic cap, they are mostly for power supply filtering and not for signals). The main thing you get into is the linearity of the capacitance value over frequency. Ceramic caps are not the best in this area (I'm not sure you would really notice the difference), mylar, polyester, and mica types are better and what is usually used.

As for voltage rating, it shouldn't really matter since there aren't any high voltages present, at least not with modern amps. Voltage rating dictates the physical size of the cap, and since there's plenty of room, you don't have to get a really small one.

[cassius987]

I would make sure the pots are correct for what you are doing.

Can you expand on this?

[cassius987]

I would make sure the pots are correct for what you are doing.

I have been looking at the 4002 schematics since you made this comment and the only place of concern I can note is the tone pots. To be honest, they look like they are wired very differently from the 4003 schematics. I don't understand the point of having a 10M resistor AND a .01 cap on the way to the selector switch; but obviously the 500K pots make sense for humbuckers. Now, I'm not sure if the 250K pots for volume are as orthodox (probably make the bass brighter?), but still, fairly germane.

What the heck are those two non-variable resistors (10M)? Not potentiometers obviously. I guess it's just a part you put in the signal path between wires?

Also, why are both of the 0.1 mfd caps placed between the tone pots and the GROUND? (Which I'm assuming is to the butt-side of the pot itself.)

Of course another difference is that the pickups go straight to the tone pots, not the selector switch like in the 4003, but that can be done any number of ways I would think...

Finally, is there a difference between a capacitor labeled with two parallel lines and one labeled with a line and an open semi-ellipse?

Thanks all for any clues, this is a neat education for me.

[cjj]

Finally, is there a difference between a capacitor labeled with two parallel lines and one labeled with a line and an open semi-ellipse?

Not really. In the old days, the semi-ellipse was used for polarized capacitors and denoted the negative end (there may be other uses as well), but these days, the two tend to get used pretty much interchangeably. It's still more common to see polarized (electrolytic) caps with the curve, but it's fine to draw them with parallel lines and a '+' symbol on the positive side. In most cases where a polarized cap is shown, regardless of the symbol used (curved vs. straight), a '+' is placed on the positive side.

[cjj]

What the heck are those two non-variable resistors (10M)? Not potentiometers obviously. I guess it's just a part you put in the signal path between wires?

Also, why are both of the 0.1 mfd caps placed between the tone pots and the GROUND? (Which I'm assuming is to the butt-side of the pot itself.)

Of course another difference is that the pickups go straight to the tone pots, not the selector switch like in the 4003, but that can be done any number of ways I would think...

Thanks all for any clues, this is a neat education for me.

OK, I took a look at the 4002 schematic. As for the tone pot & the 0.1uF cap, the "butt end" of the pot is not grounded (it is on other Rick designs). A pot is kind of just like what's shown in the diagram, a resistor with 2 ends and a variable connection into the middle. In this case, one end is connected to the cap and the signal comes into the center. On a 4003, the signal comes into one end and the other end as well as the center is connected to ground.

Ignoring everything else going on, looking just at the pot & the cap, it's really equivalent, it doesn't matter which side of the pot the cap is on. That sounds a bit cryptic and I could prove it with math, but that would bore you to death. It's easier to explain how the tone control works. Essentially, the capacitor shorts the higher frequency parts of the signal to ground, so if the pot is set so that the signal has to go all the way through the resistor, less high frequencies get shorted to ground, if set towards the cap end, more gets shorted out. So, whether it goes through the resistor and then the cap or through the cap first, the effect is the same. It's really a lot more complicated than that, what you're doing is adjusting the resonant frequency of a filter circuit, which changes what passes on and what doesn't, but that's harder to think about.

As for the 10M resistors, those are 10 Meg Ohm, or 10 million ohm. Here again, you are looking at a filter. I'm not sure what they're trying to do here, it's been too many years since I've done a lot of filter calculations. I'll have to think about that one for a bit.

[johnhall]

The cap and resistor are just a basic RC circuit for tone control.

The pickups were NOT humbucking, at least not in the dual coil sense.

[cassius987]

]In this case, one end is connected to the cap and the signal comes into the center. On a 4003, the signal comes into one end and the other end as well as the center is connected to ground.

So one of the far lugs of the pot goes to the 0.1 mf cap which then goes to ground? I guess I just didn't expect that to really be what was going on so it was hard for me to understand what the schematic was indicating... but now that you say this I guess it's indicating exactly what it says.

Far lug: 0.1 cap-->earth
Middle lug: pickup
Far lug: 0.01 cap & 10M resistor-->pickup selector

Correct??

I was reading a schematics essay for beginners and it said that all of the ground points are contiguous with one another. Is this so for a Ric bass? If you ground a pickup ground wire to the pickup surround, does that electrical ground still act in concert with other ground points like to the bridge?

Thanks CJ and John Hall. Much respect and gratefulness.

[whojamfan]

Cassius987,
Contact iiipopes, I'm sure he will fix you up with any info you need and maybe even make you a diagram. I know you guys are friends, and you know he knows his stuff, and is willing to share it.
Good luck!

[cassius987]

Absolutely-- iiipopes is a genius! But I want to do some of the work on my own and bring it to him semi-completed rather than just throwing ideas at his feet that may be totally insane! He's got a life and I hate to waste his time.

I have a Cort Jazz Bass copy from the 70s that was given to me by a girl in high school... said it was her dad's... I always had the dream of fixing it up for her because it might make a good way for her to remember him. However, she's long since moved from the area, from what I understand. So now I am restoring it slowly but surely to educate myself. One thing I noticed last night when I was cleaning up the wiring: how the heck do you remove potentiometers from the control plate (or pickguard)? I was not able to salvage ANY of the pots, with their lugs all bent up and covered in excess solder and rust. The jack, which had a big ball of solder over the lugs, was able to be salvaged by just melting the solder off. But the pots... not so lucky. I did get a 0.05 cap out of the salvage though.

Another question: is there a priority to any of the lugs of a pot? I.e., middle lug=primary, other two=secondary, or a 1,2,3, system?

Third, what resistance does a "wide open" pot transmit--max or min? Minimum, I guess? And is minimum=0 or is it a set minimum resistance for the particular pot?

Finally, how do you determine volume or tone functionality for a pot in a circuit? Every theory I come up with gets smashed... for instance in Rickys they tend to be the last pots on the way to the jacks. In a Jazz Bass, however, a tone pot with cap exists between the volume pots and the jack. So how do you assign functionality?

Thanks again. One idea I'm now toying with is putting a 5th knob on the way to the jacks that is a very high-resistance (5-15 Mohm range) to shape the tone, but I don't actually know if this would work--thus my attempt at self-education, and my great luck at having a Frankenbass.

[cjj]

how the heck do you remove potentiometers from the control plate (or pickguard)? I was not able to salvage ANY of the pots, with their lugs all bent up and covered in excess solder and rust. The jack, which had a big ball of solder over the lugs, was able to be salvaged by just melting the solder off. But the pots... not so lucky. I did get a 0.05 cap out of the salvage though.

Well, salvaging electronic components is somewhat of an art. It takes some skill, a lot of care and some good equipment. It can be done with a cheapie soldering iron, but a good quality one, with proper temperature control will help you keep from overheating parts. There's another neat device, a solder sucker, which is a spring loaded piston device that, once the solder is melted, you use to pull the solder off. There's also a product called solder wick which absorbs the solder. If things are too corroded up, you may not have much luck.

Edit: I just realized you asked about removing pots from the pickguard. You remove the knob and then remove the nut from the shaft.

Another question: is there a priority to any of the lugs of a pot? I.e., middle lug=primary, other two=secondary, or a 1,2,3, system?

In general, no, there's no priority. Pots are designed a lot like the schematic symbol, the two end ones are connected to each end of the resistive element, the center one is the wiper that moves along the resistive element to give a variable resistance. In reality, there are some types of pots that are not equivalent from one end to the other, logarithmic or audio taper pots vary the rate of change of resistance to better match the way the ear responds to changes in sound level.

This page is a decent intro to pots:
http://en.wikipedia.org/wiki/Potentiometer

Third, what resistance does a "wide open" pot transmit--max or min? Minimum, I guess? And is minimum=0 or is it a set minimum resistance for the particular pot?

I'm not sure what you mean by "wide open". The nominal value for the pot, 250k, 500k, etc. is the value of the resistive element. If you measure from one end to the other, you get this value and it doesn't change when you turn the shaft. The center wiper moves from one end to the other and, theoretically, if put at one end or the other, gives you zero resistance between the center lug and the lug at the end the wiper is closest to and full resistance to the other end. Many pots don't quite go all the way to the end, so you get some small resistance even at the end.

Finally, how do you determine volume or tone functionality for a pot in a circuit? Every theory I come up with gets smashed... for instance in Rickys they tend to be the last pots on the way to the jacks. In a Jazz Bass, however, a tone pot with cap exists between the volume pots and the jack. So how do you assign functionality?

Ahh, well, here you ask a question that's not always easy to answer. The correct answer is to take a lot of electronics theory courses so you can understand circuit design. In general, it doesn't matter a whole lot the order in which you place the various circuit elements in the design. What you've got in most guitar circuits is a tone control and a volume control. These translate to, generally, a variable tuned filter (tone control, often a "low pass" filter) and a level attenuator (volume). The filter takes out certain frequencies from the signal, which ones and how much by its design and how you have the pot adjusted, the volume control adjusts how much signal gets through to the output. It doesn't really matter whether you filter first and adjust the level second, or adjust the level then filter, it's all how you decide to design the circuit.

In general, it the pot has a cap associated with it, it's probably a filter and is the tone control. The volume pot just takes the signal and, by using the pot, adjusts how much signal can pass through. That's probably hard to understand without more circuit knowledge, I know, but I can't think of a better way to put it just now. Just to confuse things, there may be other caps and resistors in the circuit (as in the 4002) which act as other filter type elements to change the tone too.

Thanks again. One idea I'm now toying with is putting a 5th knob on the way to the jacks that is a very high-resistance (5-15 Mohm range) to shape the tone, but I don't actually know if this would work--thus my attempt at self-education, and my great luck at having a Frankenbass.

Hmm, here again, it all depends on what you want to do and how you do the design. You're talking about another filter here, probably. I'll throw a little bit of theory at you. A "low pass" filter, also called a treble cut filter removes high frequencies and passes low ones. Where it starts cutting the high frequencies is a spot called the "cutoff frequency". This frequency is given by the equation:

Fc = 1/(2*PI*R*C)

Where PI is 3.14159265..., R is resistance in ohms, and C is capacitance in Farads (uF is microfarads, or 1/1,000,000 of a farad, also known as 1e-6 or "one times 10 to the -6th power" farads), Fc is cutoff frequency in Hz (cycles per second).

So, you can see, a larger value of R gives you a lower cutoff frequency, or less treble. A higher value of capacitance does the same thing.

So, for example, using your 0.05uF cap you salvaged and a 5Mohm, you can make a filter with a cutoff freq. of

Fc = 1/(2*PI*0.05e-6*5e6) = 636.6kHz

If you used a 10M resistor, you'd get 318.3kHz. These aren't even close to audible frequencies (20Hz to 20kHz), so you'd need a bigger cap.

Well, this is getting a bit long, and I'm not sure this forum is really the place to teach electronics theory.

The circuits in a guitar are not really too complicated, but how and why they are designed the way they are can be quite involved and takes some knowledge to understand. For me, it's been a lot of years (20+) since I've designed any sort of filters, so I'm a bit rusty. I design digital IC's these days which are a whole 'nother story.

Hopefully, I've managed to explain a few things to help you out. Please, keep asking questions, it's really the only way to learn! I'll give it my best shot at explaining as I'm sure others will too.

[cassius987]

Edit: I just realized you asked about removing pots from the pickguard. You remove the knob and then remove the nut from the shaft.

Right, that's the normal way, and I've actually done it before for students when I was teaching beginning guitar theory out of my house between working jobs in my actual field (chemistry). I guess I've got some funky pots or something--I'll upload pics later. There is no clear way to remove the knob. I may have to just pry them off, but I don't want to break them. They're kinda' cool.

I'm not sure what you mean by "wide open". The nominal value for the pot, 250k, 500k, etc. is the value of the resistive element. If you measure from one end to the other, you get this value and it doesn't change when you turn the shaft. The center wiper moves from one end to the other and, theoretically, if put at one end or the other, gives you zero resistance between the center lug and the lug at the end the wiper is closest to and full resistance to the other end. Many pots don't quite go all the way to the end, so you get some small resistance even at the end.

In spite of not getting what I meant, you answered my question exactly! Thanks. Basically I was wondering if a pot that was at point "0" on a 0-10 scale meant that it was theoretically a 0 resistance resistor... I figured, like you said, there'd be some lingering resistance, so it wouldn't be equivalent to a wiring where no resistor occurred. By wide open, I meant the pot on "10" or full volume.

Hmm, here again, it all depends on what you want to do and how you do the design. You're talking about another filter here, probably. I'll throw a little bit of theory at you. A "low pass" filter, also called a treble cut filter removes high frequencies and passes low ones. Where it starts cutting the high frequencies is a spot called the "cutoff frequency". This frequency is given by the equation:

Fc = 1/(2*PI*R*C)

Where PI is 3.14159265..., R is resistance in ohms, and C is capacitance in Farads (uF is microfarads, or 1/1,000,000 of a farad, also known as 1e-6 or "one times 10 to the -6th power" farads), Fc is cutoff frequency in Hz (cycles per second).

So, you can see, a larger value of R gives you a lower cutoff frequency, or less treble. A higher value of capacitance does the same thing.

So, for example, using your 0.05uF cap you salvaged and a 5Mohm, you can make a filter with a cutoff freq. of

Fc = 1/(2*PI*0.05e-6*5e6) = 636.6kHz

If you used a 10M resistor, you'd get 318.3kHz. These aren't even close to audible frequencies (20Hz to 20kHz), so you'd need a bigger cap.

Hm, it sounds like I have a lot of digging left to do! To be honest I was trying to emulate aspects of the 4002 circuitry while introducing an element of variability with the pot. But that's what led me into the M-ohm region. Now I see that the circuit may not have been designed to filter what I thought it was filtering. I appreciate the equation, I will put it to good use. Thanks SO much for that--I definitely need some raw data and this'll help me keep track of what's really going on, rather than inaccurately speculating about the effect of this or that cap with this or that pot...

Hopefully, I've managed to explain a few things to help you out. Please, keep asking questions, it's really the only way to learn! I'll give it my best shot at explaining as I'm sure others will too.

As long as I'm not annoying the forumites, then sure, I'll keep asking questions! You have been infinitely helpful already, CJ, so thank you so very much; even if this info was all I could squeeze out I'd be happy!

[whojamfan]

You're not annoying at all

[cjj]

Edit: I just realized you asked about removing pots from the pickguard. You remove the knob and then remove the nut from the shaft.

Right, that's the normal way, and I've actually done it before for students when I was teaching beginning guitar theory out of my house between working jobs in my actual field (chemistry). I guess I've got some funky pots or something--I'll upload pics later. There is no clear way to remove the knob. I may have to just pry them off, but I don't want to break them. They're kinda' cool.

OK, pics would be helpful. In general, there are 2 main types of knobs for pots, those that attach with a small set screw and those that are friction fit. I've seen some that the screw hole is plugged up with dirt/whatever and is hard to see. The friction fit ones you just have to pull the knob off. Sometimes it takes quite a lot of force, they can get kind of corroded on. Pots themselves usually have a shaft with either a flat spot (for the screw to hit) or have grooves all around the shaft. The "groovy" type, if metal, usually has a cut down the middle to let the sides flex a bit, if plastic, they usually don;t have the split. Shafts with a flat side can also be used with friction fit knobs, these usually have a small piece of spring metal that lined up with the flat and holds the knob on.

Sometimes, it helps if you can get a flat piece of sheet metal as wide or wider than the knob and cut a slot up the center. Then you can slip the metal under the knob and use it to help pry the knob off. Sometimes though, you end up breaking something...