How to not hear Mylie Cyrus when I am playing guitar?

[dietermoreno]

How to not hear Mylie Cyrus when I am playing guitar?

You know what, now I actually want to not use a guitar for radio receiving and use it to play guitar and although comical to hear Mylie Cyrus after I have played an anti-corporate song, I would like to not hear Mylie Cyrus anymore.

Would some ferrite chokes on my guitar cable help do you think?

This is in the middle of the night, I think that might help the stations come in louder. That also means that I should probably stop playing now and go to bed before the neighbors call the cops. I don't want to play with headphones because they suck and don't sound like the guitar speaker cabinet. Its unfortunate that I started playing at 11:00 at night because earlier I was busy desodering and resodering the ground connection on my guitar output jack.

Just as bad as hearing Mylie Cyrus, when I stand with my guitar perpendicular to the south west I hear a Mexican polka station. When I stand with my guitar perpendicular to the north east thank God I can change the station and I hear Papa Roach on the active rock station 95 WIIL Rock. So now I'm at a total of 6 stations depending on the direction I face, and the directionality appears to not affect WBBM. The navigational beacons OBK and ORD appear to be highly directional, thus why airplanes can use them to find O'hare. I guess that higher frequencies are more directional, so since aviation is higher frequency than FM it is even more directional than FM.

but I would rather listen to 95 WIIL Rock on a real radio rather than on my guitar, so I want to make it a guitar only, and not a radio.

My Grundig muli-band radio receiver is the best quality radio that I have that sounds even better than my car radio. It has a telescopic antenna. It sounds louder if the antenna is made longer. So longer antenna = louder? -- wrong (for FM). Longer antenna = picks up more crap (for FM). I can use a hanger split in half attached with masking tape to the telescopic antenna top faced perpendicular in the direction of the favorite FM station (95.1). If that directionality allows the station to be heard on my guitar, then certaintly it ought to be a massive improvement in quality for the Grundig radio for that station, maybe enough quality to record music from the radio that is even better quality than YouTube (whos gonna stop me). So at least I learned something useful.

Edit: now I tried the coat hanger trick for the radio (I found out that I don't even have to cut through metal all I have to do is untwist the metal and strip it of paint at where it will contact the telescopic antenna) and I found that its accuracy meter reads 30% with trimmer capacitor at minimum clockwise and dipole facing perpendicular to northwest, its accuracy meter reads 40% with trimmer capacitor at maximum clockwise and dipole facing perpendicular to northwest, its accuracy meter reads 55% with trimmer capacitor at minimum clockwise and dipole facing perpendicular to northeast, and its accuracy meter reads 70% with trimmer capacitor at maximum clockwise and dipole facing perpendicular to northeast.

 

[BobK]

I feel your pain.

Bob

 

[duke37]

Try a 1nF (1000pF) capacitor across the amp input to short out the spurious RF signal. Keep the lead lengths down.

 

[dietermoreno]

Try a 1nF (1000pF) capacitor across the amp input to short out the spurious RF signal. Keep the lead lengths down.

Okay, so to achieve this without opening up the amp, would it work to buy 2 guitar output jacks, mount them onto a small card board box about 8 inches x 4 inches wrapped in aluminum foil, and solder in the capacitor?

I am now qualified to solder guitar output jacks (I did it last night to fix my loose ground connection).

I attached a picture so you can see what I have in mind.

I suppose this could also work as the RF removing capacitor in my penny radio.

In fact, I think its the same thing so my drawing is based on a crystal radio schematic.

Maybe if I get this design to work, I can add in a transistor, a battery, 2 switches (one switch disconnects the battery to not waste battery power when not in use and the other switch connects the high pass filter), and a high pass filter to build my own guitar solo pedal (treble boost) and it also works as a pre-amplifier for the penny radio.

I'm sure my dad would gladly give me $10 for these components rather than giving me $150 to buy a so called "supressor" or "noise gate" from Guitar Center or Sam Ash.

I didn't label an in and an out on the picture because I don't think it matters what direction you connect it if it is a passive one capacitor device.

 

[duke37]

You have a very generous father.

The box does not need to be bigger than 1 x 1 x 1 inch, you can mount the two sockets on one side so that the connctions are short.

 

[dietermoreno]

The box does not need to be bigger than 1 x 1 x 1 inch, you can mount the two sockets on one side so that the connctions are short.

Okay, so same drawing but with both sockets mounted on the same side to keep the wires as short as possible.

I'm going to RadioShack today to buy the capacitor on my way to drop my friends off. I'll ask my dad if he can give me $5 more to buy the output jacks from Amazon. I get free shipping on Amazon because I have Amazon prime shipping since I have a .edu email account for my college and I buy all of my college text books from Amazon.

 

[dietermoreno]

Opps, I think I should have told the guy at RadioShack 1nF instead of 1000pF, because he looked at me kind of funny when I asked for a 1000pF capacitor and he said they didn't have one, and the closest they had was 100pF so I bought the 100pF one (actually its sold in a package of 2) for $2.

Will the 100pF capacitor work?

Or will the 100pF capacitor just create a tuned circuit with the guitar pickups making the radio noises even louder?

I didn't take it out of the package yet and I still have the receipt.




Edit:

Well here in this crystal radio schematic, it shows the tuning capacitor with a value of 365pF and the RF removing capacitor at 1000pF, so I should return the 100pF capacitor and get my money back and buy a 1nF capacitor instead. All the 100pF capacitor would do is further the radio receiving effect. If I use the 100pF capacitor, I would be building a radio, and that's the opposite of what I'm trying to accomplish.

I could also try telling the guy 0.001uF if he can't find 1nF.

Yep, here it is online and in store: <<http://www.radioshack.com/product/index.jsp?productId=2062362>>

I wish that guy at RadioShack could have told me that the 0.001uF that they have is the same thing as the 1000pF I was looking for.


When I completed physics I thought I would never again in my life have to do unit prefix conversions. I guess I was wrong. Funny thing is that that's the same manager who said that I'm not what they're looking for when I applied there and he doesn't even know unit prefix conversions.

 

[davenn]

1000pF = 1nF

1000pF doesnt not equal 100pF, you would have to put 10 x 100pF caps in parallel to get 1000pF

 

[duke37]

If you cannot get 1nF, you would be better to go up to say 10nF.

I looked up Radio Shack USA and they have 1000pf 50V ceramic capacitors for $0.79 on line only. They do not recognise nF.

In the UK such capacitors cost about 10p i.e. 10 to £1.

The circuit you show has the 1000pF capacitor to do the identical job to what you want to do which is to remove the RF and leave the audio.

 

[(*steve*)]

When I completed physics I thought I would never again in my life have to do unit prefix conversions. I guess I was wrong.

Very wrong. It should be second nature. You'll be doing it all the time with low value capacitors.

Until fairly recently, it was relatively rare to see nF used. 0.001uF was common, as was 1000pF. And that was about where people changed from uF (or earlier mfd) to pF. It is still very rare to see mF, possibly because there is some risk of confusion with nF or mfd.

These days you'll see plenty of values that are equally reasonably depicted as pF or nF, or nF and uF. e.g. 820pF = 0.82nF, or 68nF = 0.068uF.

The same thing happens with low inductances, but you never(?) see pH -- which is a ridiculously small inductance -- and is also easily confused with the other pH (measure of H+ ion concentration). With inductances, you will commonly see nH, uH, mH, and even H

Funny thing is that that's the same manager who said that I'm not what they're looking for when I applied there and he doesn't even know unit prefix conversions.

Not unusual at all. They have abysmally low technical standards.

 

[KrisBlueNZ]

I'd suggest using the 100 pF capacitors. Electric guitars have relatively high output impedances. With 100 pF, assuming a source impedance of 100 kilohms the cutoff frequency is only 16 kHz. In fact you might even have problems with resonance with the pickup inductance!

Also, you can add it across the connections in the plug at the amplifier end of your jack-to-jack cable. That's not quite as good as putting it inside the amp because it doesn't deal with noise picked up at the amplifier's input socket.

Make sure your plugs and sockets are clean. Use isopropyl alcohol on a cotton swab or tissue paper to clean the metal surfaces. On every plug and socket in the signal path.

 

[dietermoreno]

I'd suggest using the 100 pF capacitors. Electric guitars have relatively high output impedances. With 100 pF, assuming a source impedance of 100 kilohms the cutoff frequency is only 16 kHz.

No, the DC resistance of passive humbucker pickups is around 5k ohms, according to EMG.

You said impedence, but I'm pretty sure its the DC resistance value that matters in the RC filter calculation.

The results of an online RC circuit calculator using a DC resistance of 5k ohms and a capacitance of 1000pF gives a cut off frequency of 31.83099 khz.

In fact you might even have problems with resonance with the pickup inductance!

I'm assuming that the pickup inductance is what is allowing it to receive the local 50kw MW station. The pickups have nothing to do with receiving the local FM stations, because when I turn my guitar volume to minimum it just makes the FM stations come in clearer with less mains hum.

Also, you can add it across the connections in the plug at the amplifier end of your jack-to-jack cable. That's not quite as good as putting it inside the amp because it doesn't deal with noise picked up at the amplifier's input socket.

If I did that, wouldn't that give a different cut off frequency if I put another 1000pF capacitor in series?

1/C serial = 1/C1 + 1/C2

1/1000pF + 1/1000pF = 2/1000pF = 1/500pF

C serial = 500pF

500pF is enough to stop RF (at least today, not so much in the 1920s when that is the frequency range that the Radio Central long wave trans-Atlantic telegraph used), but why do I need 2 capacitors if one works just fine? The bag is sold as 2 ceramic disc capacitors, but I don't know why I would need to use more than one.

Cut off frequency of 500pF with 5k ohm pickups is 63.66 khz.

Actually, Radio Central in the 1920s used 18khz it says here, so the 1000pF capacitor wouldn't have been enough back then to block all RF. Fortunately today audio frequencies are no longer used for radio carrier wave frequencies.

Make sure your plugs and sockets are clean. Use isopropyl alcohol on a cotton swab or tissue paper to clean the metal surfaces. On every plug and socket in the signal path.

Okay, I'll try that. Makes sense to remove any rectifying dirt particles.

 

[(*steve*)]

You said impedence, but I'm pretty sure its the DC resistance value that matters in the RC filter calculation.

*sigh*

 

[KrisBlueNZ]

*sigh*

LOL!

No, the DC resistance of passive humbucker pickups is around 5k ohms, according to EMG.

A pickup is inductive and will form a resonant circuit with the lumped capacitance of the cable. The overall response will have a peak and a rolloff. The volume potentiometer in a typical electric guitar is 100k~500k and its position will have a significant effect on the Q of the tuned circuit (the Q will be highest at maximum volume setting) as well as the low-pass rolloff (which will be least noticeable at maximum volume setting). This is one reason why active electronics inside the guitar will give better performance.

If I did that, wouldn't that give a different cut off frequency if I put another 1000pF capacitor in series?

Adding any capacitor across the cable will affect the resonant and cutoff frequencies. I didn't suggest adding a capacitor in series.

The bag is sold as 2 ceramic disc capacitors, but I don't know why I would need to use more than one.

+sarcasm Well, if they sell them in a bag of two, then that obviously means that you HAVE to use both of them together, right? /sarcasm LOL

Actually, Radio Central in the 1920s used 18khz it says here, so the 1000pF capacitor wouldn't have been enough back then to block all RF. Fortunately today audio frequencies are no longer used for radio carrier wave frequencies.

Fascinating. Utterly fascinating. :-/

 

[dietermoreno]

Oh.... so the DC resistance of the pickups is insignificant compared to the DC resistance of the volume pot on the guitar...

So changing the volume setting on the guitar volume pot effects the cut off frequency.

So if the pot is set at 500k DC resistance and a 1000pF capacitor is used, then the cut off frequency is 381 hz. Yikes! Not quite what I had in mind!

So if the pot is set at 500k DC resistance and a 100pF capacitor is used, then the cut off frequency is 3.18khz.

So if the pot is set at 100k DC resistance and a 1000pF capacitor is used, then the cut off frequency is 1.59khz.

So if the pot is set at 100k DC resistance and a 100pF capacitor is used, then the cut off frequency is 15.92khz.


I guess the crystal radio uses a 1000pF capacitor because the radio signal received has no resistance other than the resistance of the air.

So Yahoo Answers says that the resistance of dry air with non ionizing voltages less than 30kV per Cm is about 10 billion ohms per meter.

So to transmit a radio signal 1km through dry air with non ionizing voltages less than 30kV per Cm, that would result in a 10 trillion ohm perceived DC resistance load. So for a crystal receiver to receive a signal with a 10 trillion ohm perceived DC resistance load, if a 1000pF capacitor is used, that results in a cut off frequency of 0.0000159155 hz.

Hhm...so I don't think a radio signal has a perceived DC resistance load of 10 trillion ohms...

Aha! That's why a radio signal to travel 1km must use thousands of volts greater than 30kV! Or you don't even need a voltage greater than 30kV if you use a spark gap transmitter. My electric guitar connected to the speaker output jack of my guitar amp made a good spark gap transmitter and it was only 40V!

So using the RC circuit calculator to solve for the perceived load of a crystal radio receiving a radio signal:

Cut of frequency: 20khz

Capacitance: 1000pF

Then, DC resistance = 7.958k ohm

So the perceived load of a crystal radio receiving a radio signal is around 8k DC resistance.




Now I'm wondering what to say to the manager if I go back to the same RadioShack and he asks me why I'm back for the 3rd time trying to buy the capacitor that I just returned.

This is like my dog chasing its tail.

Edit:

LOL!

A pickup is inductive and will form a resonant circuit with the lumped capacitance of the cable. The overall response will have a peak and a rolloff. The volume potentiometer in a typical electric guitar is 100k~500k and its position will have a significant effect on the Q of the tuned circuit (the Q will be highest at maximum volume setting) as well as the low-pass rolloff (which will be least noticeable at maximum volume setting).

Oh...............an RC circuit that is a tuned circuit means that the R is impedence, not resistance, by the definition of impedence being the resistance impeding alternating current at a given frequency.


So if you can create a tuned circuit with impedence and capacitance, why not save some money why does crystal radio have to use the dumb inductor? Why not substitute the inductor for a resistor that has the same impedence as the inductor?



Sorry if I'm asking some really dumb questions that are a little out there. I remember when I was in my physics class, I was the only person asking questions and everyone stared at me and thought I was weird for asking so many questions and the professor told me to Google my questions because he's only paid to teach me what's in the text book, and yes we skipped the chapter on this thing that Steve and Chris blue call "impedence". That chapter looked hard with these wierd drawings called "phasor diagrams", so we skipped it because he said he's not required to cover that chapter and he said he's only paid to teach the chapters he's required to cover.

I think I did ask the professor why a radio has to use an inductor instead of a resistor of the same impedence, and he just stared at me and said "because that's how it works, you can ask more questions on Google".

Well here I am, I Googled to find this forum.

I know how to use Google.




Edit 2: What am I even doing. I don't even know. I'm chasing my tail even more. I think maybe I'm having a bad day and should not edit this post or post a reply until tommorow.

 

[KrisBlueNZ]

Oh.... so the DC resistance of the pickups is insignificant compared to the DC resistance of the volume pot on the guitar...

Yes. But at audio frequencies, the important characteristic of the pickup is not its DC resistance; it is its impedance. This is also measured in ohms at a particular frequency, and is higher than its DC resistance. Both the pickup's impedance, and the resistance added by the volume potentiometer, will interact with the total load capacitance (that is, the cable capacitance, plus the amplifier input capacitance, plus any capacitance you add), to form a low-pass filter (i.e. they will attenuate the signal more at high frequencies).

Also, the inductance of the pickup and the total load capacitance will form a resonant circuit (parallel LC circuit) which will cause a peak in the response at a certain frequency. When the volume potentiometer is at maximum, this peak will be most pronounced (the "Q" of the tuned circuit will be highest) because there is minimum resistance between the pickup and the capacitance. The frequency of this tuned circuit will not be affected much by the volume potentiometer's position, but the Q will be.

So changing the volume setting on the guitar volume pot effects the cut off frequency.

Yes.

So if the pot is set at 500k DC resistance and a 1000pF capacitor is used, then the cut off frequency is 381 hz. Yikes! Not quite what I had in mind!

Not exactly. The impedance at the output of the electric guitar is not easy to define. When the volume potentiometer is at maximum, the pickup is connected directly to the output socket, and the potentiometer is just a 500k resistance in parallel with the pickup, which has little effect. In this situation, the output impedance is roughly equal to the impedance of the pickup. (If you have access to information on your pickup, you should be able to find out its impedance at various frequencies.) As you turn the volume control down, more resistance is added between the pickup and the output socket. This will cause the cutoff frequency (due to the R-C circuit) to drop and your high frequencies will be more attenuated. But it also reduces the peak at the L-C resonant frequency because it reduces the Q of the tuned circuit. So there are two effects happening at the same time.

So if the pot is set at 500k DC resistance and a 100pF capacitor is used, then the cut off frequency is 3.18khz.

Not exactly, because when a 500k volume potentiometer is set in between minimum and maximum, it contains two resistances. So the output resistance will be lower than 500k.

So if the pot is set at 100k DC resistance and a 1000pF capacitor is used, then the cut off frequency is 1.59khz.

Again not exactly. The worst case will be when the impedance of the pickup plus the resistance of the potentiometer from the top pin to the wiper balances the resistance from the wiper to ground, which will be somewhere around 7 out of 10 (depending on the impedance of the pickup).

Thanks for the glimpse into your thought processes regarding crystal radios... but let's leave that for another thread, OK?

Now I'm wondering what to say to the manager if I go back to the same RadioShack and he asks me why I'm back for the 3rd time trying to buy the capacitor that I just returned.

A couple of capacitors costs very little. I can't imagine why you returned them. If you're not sure what to tell him, you could always just make up another unconvincing lie...

Oh...............an RC circuit that is a tuned circuit means that the R is impedence, not resistance, by the definition of impedence being the resistance impeding alternating current at a given frequency.

No. An RC circuit doesn't resonate. An LC circuit does. Google LC resonant circuit. You should try to find out the inductance of your pickup, and its impedance at frequencies of interest.

Sorry if I'm asking some really dumb questions that are a little out there.

Do you like gardening? Gardening is great! You could become a gardener, couldn't you? Yeah, gardening. You were born to do it!

 

[BobK]

Dieter,

You are a unique individual. I can see that you really want to learn. But instead of doing the work, you jump to conclusions that are incorrect at every opportunity. Find a discussion of the RLC circuit and really try to understand what is going on. Ask questions if you do not understand something.

Keep being curious, but try to learn in small steps instead of giant leaps, which keep ending up with you falling in the chasm.

BTW: I disagree with Kris's suggestion that gardening might be a good career choice, I would go for avant garde performance artist if I were you!

Bob

 

[dietermoreno]

Yes. But at audio frequencies, the important characteristic of the pickup is not its DC resistance; it is its impedance. This is also measured in ohms at a particular frequency, and is higher than its DC resistance.

Okay, got it. Impedence load is more important than DC resistance load at audio frequencies since impedence is frequency dependent (I think that's what those phasor diagrams in my physics text book were trying to say -- impedence is frequency dependent).

Impedence is also measured in ohms and the total resistance load is the sum of the impedence load and the DC resistance load.

Both the pickup's impedance, and the resistance added by the volume potentiometer, will interact with the total load capacitance (that is, the cable capacitance, plus the amplifier input capacitance, plus any capacitance you add), to form a low-pass filter (i.e. they will attenuate the signal more at high frequencies).

I don't know the capacitance of the amplifier input, but I do know that the standard capacitance of a guitar cable is around 30pF per foot, according to this link.

So if I use a 10 foot long cable, that will result in a capacitance of 300pF.
Not including impedence of pickups and only including impedence of volume pot at maximum impedence (500k), that would result in a cut off frequency of around 1khz. The cable forms its own low pass filter when the volume pot is at maximum impedence. Not including impedence of pickups and only including impedence of volume pot at lowest impedence where the pot wiper is still connected to the circuit (100k), that would result in a cut off frequency of around 5.3khz.

So if I use a 20 foot long cable, that will result in a capacitance of 600pF. Not including impedence of pickups and only including impedence of volume pot at maximum impedence (500k), that would result in a cut off frequency of around 531 khz. The cable forms its own low pass filter when the volume pot is at maximum impedence. Not including impedence of pickups and only including impedence of volume pot at lowest impedence where the pot wiper is still connected to the circuit (100k), that would result in a cut off frequency of around 2.7 khz.

So in order to minimize capacitance that affects tone, the cable should be as short as possible. In addition, in order to minimize impedence that affects tone, the volume pot should be at full volume (wiper disconnected from the circuit so only impedence is the pickup impedence). I thought it was just my imagination when I thought that my guitar sounded bad at low volume on the guitar volume pot. Now I know why my guitar sounds bad at low volume on the guitar volume pot!

So, the high end response is decreased when the guitar volume pot is at maximum impedence when the cable length is increased increasing the capacitance of the cable.

So those cable manuafactures selling "ultra low capacitance" cables for $50 do have a valid claim saying that their lower capacitance cables have a greater high end response.

Also, the inductance of the pickup and the total load capacitance will form a resonant circuit (parallel LC circuit) which will cause a peak in the response at a certain frequency. When the volume potentiometer is at maximum, this peak will be most pronounced (the "Q" of the tuned circuit will be highest) because there is minimum resistance between the pickup and the capacitance. The frequency of this tuned circuit will not be affected much by the volume potentiometer's position, but the Q will be.

RimStar explained that LC resonant circuits have only a peak in response relative to the human ear, and in actuallity how it actually works at the electron level is that when you connect the antenna (or in the case of this thread, the signal generated by the guitar pickups) to a position of less coil turns on the inductor between the antenna and ground, all frequencies at higher coils turns are shorted to ground, and the resonant circuit actually pickups up several radio stations at the same time (not applicable to this thread) because a radio station is not what is selected, it is radio stations that are not selected below a cut off frequency at which all radio stations above the cut off frequency are shorted to ground.

Also, I have read on the internet in other places that the crystal radio can be improved in selectivity to one frequency by having a second LC resonant circuit. So I'm guessing that this second LC resonant circuit improves selectivity to one frequency by being connected to ground the opposite that the first LC resonant circuit is connected and changing the coils turns of both at the same time zooms in on a frequency and the band width value of the selected frequency can be increased or decreased.

So is "Q" the same thing as the "bandwidth" I'm describing?

Not exactly. The impedance at the output of the electric guitar is not easy to define. When the volume potentiometer is at maximum, the pickup is connected directly to the output socket, and the potentiometer is just a 500k resistance in parallel with the pickup, which has little effect. In this situation, the output impedance is roughly equal to the impedance of the pickup. (If you have access to information on your pickup, you should be able to find out its impedance at various frequencies.) As you turn the volume control down, more resistance is added between the pickup and the output socket. This will cause the cutoff frequency (due to the R-C circuit) to drop and your high frequencies will be more attenuated. But it also reduces the peak at the L-C resonant frequency because it reduces the Q of the tuned circuit. So there are two effects happening at the same time.

Got it.

Not exactly, because when a 500k volume potentiometer is set in between minimum and maximum, it contains two resistances. So the output resistance will be lower than 500k.
Again not exactly. The worst case will be when the impedance of the pickup plus the resistance of the potentiometer from the top pin to the wiper balances the resistance from the wiper to ground, which will be somewhere around 7 out of 10 (depending on the impedance of the pickup).

I still don't understand this point. Are you saying that on the guitar volume pot between 7 out of 10 (depending on the impedence of the pickup), there is a selection that will short all impedence including the pickup impedence to ground leaving zero impedence?

Thanks for the glimpse into your thought processes regarding crystal radios... but let's leave that for another thread, OK?

Okay. I'm already confused enough, no need to make myself even more confused.

A couple of capacitors costs very little. I can't imagine why you returned them. If you're not sure what to tell him, you could always just make up another unconvincing lie...

Okay.

No. An RC circuit doesn't resonate. An LC circuit does. Google LC resonant circuit. You should try to find out the inductance of your pickup, and its impedance at frequencies of interest.

Okay. Well I'll go with the RimStar animations of an LC resonant circuit for now. I already understand how it works in the garden hose and bucket analogy I made on another thread.

Okay, I'm going to ask on UG forums for help finding out the inductance and impedence (at audio frequencies of my guitar) of my pickups, since a quick Google search didn't give me what I was looking for.

Do you like gardening? Gardening is great! You could become a gardener, couldn't you? Yeah, gardening. You were born to do it!

Why?




Now that I think of it, I suppose that my guitar penny radio and guitar razor blade radio probably didn't work as well as I wanted them to because I used four 20 foot long low end guitar cables laid on the guitar strings as dipole aerial antennae, so the capacitance is way too high. Static is heard instead of mains hum, but the static is closer to audio frequency than radio frequency apparently considering the huge capacitance values of the cables and taking into account the 500k volume pot.

The cut off frequency with a 500k volume pot and four 20 foot long low end guitar cables (2400 pF) is 132 hz! Lol

 

[KrisBlueNZ]

Okay, got it. Impedence load is more important than DC resistance load at audio frequencies since impedence is frequency dependent (I think that's what those phasor diagrams in my physics text book were trying to say -- impedence is frequency dependent).

Not exactly. The impedance of the pickup is more important than its DC resistance because the impedance is higher than the DC resistance. In fact the impedance _includes_ the DC resistance, so there's no need to use the DC resistance in any calculations. And the phasor diagrams relate to phase relationships, which do depend on frequency.

Impedence is also measured in ohms and the total resistance load is the sum of the impedence load and the DC resistance load.

Close. You can't just add them, because impedance includes the inductive or capacitive component (inductive, in the case of guitar pickups) and the DC resistance is just resistive. This is what those diagrams are for. But you're on the right track.

I don't know the capacitance of the amplifier input, but I do know that the standard capacitance of a guitar cable is around 30pF per foot, according to this link.
So if I use a 10 foot long cable, that will result in a capacitance of 300pF.

Right. That's quite a lot of capacitance, isn't it.

Not including impedence of pickups and only including impedence of volume pot at maximum impedence (500k), that would result in a cut off frequency of around 1khz. The cable forms its own low pass filter when the volume pot is at maximum impedence. Not including impedence of pickups and only including impedence of volume pot at lowest impedence where the pot wiper is still connected to the circuit (100k), that would result in a cut off frequency of around 5.3khz.

Your explanation is right. I haven't checked your calculations.

So if I use a 20 foot long cable, that will result in a capacitance of 600pF. Not including impedence of pickups and only including impedence of volume pot at maximum impedence (500k), that would result in a cut off frequency of around 531 khz. The cable forms its own low pass filter when the volume pot is at maximum impedence. Not including impedence of pickups and only including impedence of volume pot at lowest impedence where the pot wiper is still connected to the circuit (100k), that would result in a cut off frequency of around 2.7 khz.

OK. Bear in mind that the cutoff frequency for an RC filter is not the frequency beyond which you don't hear anything. It's the frequency at which the signal is attenuated by 3 dB (decibels). You will still hear frequencies beyond the cutoff frequency; they will be attenuated. Search for single pole low-pass filter on Google and/or Wikipedia.

So in order to minimize capacitance that affects tone, the cable should be as short as possible. In addition, in order to minimize impedence that affects tone, the volume pot should be at full volume (wiper disconnected from the circuit so only impedence is the pickup impedence). I thought it was just my imagination when I thought that my guitar sounded bad at low volume on the guitar volume pot. Now I know why my guitar sounds bad at low volume on the guitar volume pot!

That's all true, but you should also investigate the effect of the LC resonant circuit that is formed by the inductance of the pickup and the capacitance of the cable. This is a parallel "tuned circuit" that adds a peak in the response. This effect is strongest when the volume control is at maximum, because at that setting, the pickup inductance and the cable capacitance are directly connected together. Google some of those keywords for more information. And try to find out the inductance of your pickup so you can calculate the resonant frequencies with various amounts of capacitive loading.

So, the high end response is decreased when the guitar volume pot is at maximum impedence when the cable length is increased increasing the capacitance of the cable.

Ding! Right.

So those cable manuafactures selling "ultra low capacitance" cables for $50 do have a valid claim saying that their lower capacitance cables have a greater high end response.

Right again.

RimStar explained that LC resonant circuits have only a peak in response relative to the human ear, and in actuallity how it actually works at the electron level is that when you connect the antenna (or in the case of this thread, the signal generated by the guitar pickups) to a position of less coil turns on the inductor between the antenna and ground, all frequencies at higher coils turns are shorted to ground, and the resonant circuit actually pickups up several radio stations at the same time (not applicable to this thread) because a radio station is not what is selected, it is radio stations that are not selected below a cut off frequency at which all radio stations above the cut off frequency are shorted to ground.

In this case, the resonant frequency will probably be in the audio frequency range. Although LC resonance may be explained in relation to tuning a radio receiver, you should forget about the radio analogy. The effect that's important here is at audio frequencies.

Also, I have read on the internet in other places that the crystal radio can be improved in selectivity to one frequency by having a second LC resonant circuit. So I'm guessing that this second LC resonant circuit improves selectivity to one frequency by being connected to ground the opposite that the first LC resonant circuit is connected and changing the coils turns of both at the same time zooms in on a frequency and the band width value of the selected frequency can be increased or decreased.

Sort of. But let's keep the crystal set stuff for another thread, OK?

So is "Q" the same thing as the "bandwidth" I'm describing?

Q and bandwidth are related, yes. A higher Q means a narrower bandwidth, and a taller peak in the response. Google Q and bandwidth.

Got it.

Yes, you're getting it!

Are you saying that on the guitar volume pot between 7 out of 10 (depending on the impedence of the pickup), there is a selection that will short all impedence including the pickup impedence to ground leaving zero impedence?

No. Around position 7 on the volume control is the point where the output impedance of the guitar will be at _maximum_.

A volume control at position 7 is a voltage divider consisting of two resistances. There is the resistance from the wiper to the top end (the end where the pickup is connected), and there's the resistance from the wiper to the bottom end (ground). For a logarithmic pot, which is the type of pot used for volume controls, around position 7 is where these two resistances are roughly the same; around 250k each, for a 500k pot. Assuming zero pickup impedance, these two 250k resistances are effectively in parallel, so the output impedance will be around 125k. That will be the maximum series resistance introduced by the volume pot, so it will be the setting where capacitive loading (cable, amplifier etc) has the most effect and where the cutoff frequency will be lowest.

Okay. I'm already confused enough, no need to make myself even more confused.

Actually, you ARE getting it.

Okay, I'm going to ask on UG forums for help finding out the inductance and impedence (at audio frequencies of my guitar) of my pickups, since a quick Google search didn't give me what I was looking for.

Good!

Now that I think of it, I suppose that my guitar penny radio ...

Please, let's forget the radio stuff for now!

 

[dietermoreno]

Not exactly. The impedance of the pickup is more important than its DC resistance because the impedance is higher than the DC resistance. In fact the impedance _includes_ the DC resistance, so there's no need to use the DC resistance in any calculations. And the phasor diagrams relate to phase relationships, which do depend on frequency.

Oh.... impedence INCLUDES DC RESISTANCE. Wikipedia article on impedence says that Z serial = Z1 + Z2 + ... Zn. Wikipedia article on impedence says that Z parallel = 1/Z1 + 1/Z2 +...1/Zn. Wikipedia article on impedence says that a resistor has no frequency dependent impedence by itself and Z resistor = DC resistance resistor. Wikipedia article on impedence says that impedence of capacitor is 1/jwC. I don't know what the j is. Wikipedia article on impedence says that impedence of an inductor is jwL. I still don't know what the j is. Interesting, the impedence of a capacitor is the reciprocal of the impedence of a capacitor. I'm assuming that w is the period in seconds of the sinusoidal AC wave.

So those phasor diagrams are designed to allow engineers (and physics students) to use algebra instead of have to solve differential equations, according to Wikipedia.

I have no clue why guitar pickup manufactures specify the DC resistance of the pickup, but don't specify the inductance and impedence of the pickup. Maybee its more people like that RadioShack manager who don't have very high technical standards but are very good salesmen so they don't need to have high technical standards because they can fool people.

Close. You can't just add them, because impedance includes the inductive or capacitive component (inductive, in the case of guitar pickups) and the DC resistance is just resistive. This is what those diagrams are for. But you're on the right track.

Right, I can't add the DC resistance of the pickup and the DC resistance of the volume pot to the total impedence, because the total impedence includes DC resistance.

but why is the impedence of the volume pot 500k if its DC resistance is only 7k? I thought Wikipedia article on impedence said that Z resistor = DC resistance resistor?

Right. That's quite a lot of capacitance, isn't it.

Yes it is. So this makes me think that 300pF 10 foot long cable (30pF per foot) and 500k volume pot should already be a low pass filter that attenuates signals at a greater frequency than audio frequencies, and adding further capacitance would just give less of a treble response.

So I guess that crystal receiver schematics should not be applied to guitar electronics.

So the only thing my 1000pF capacitor is good for is building a crystal radio.

So I could wrap my cables in aluminum foil to block out the RF, since we have established that adding more capacitance to a circuit with high capacitance won't do anything other than kill high end response; but then that would just increase the capacitance and kill the high end response, since aluminum foil wrapped around an insulated conductor is I have built a capacitor.

So then my "project" is a waste of time, and there's nothing I can do to remove radio interference if I don't want my treble response to be affected other than (in order of lowest cost):
(1) don't record at night if don't want to record WBBM.
(2) instead of building a low pass filter to attenuate RF, build a noise gate that only allows the strong signals generated by the guitar strings vibrating in the magnetic field of the pickups to pass and attenuates weak noises such as RF noises.
(3) install electric outlets into my basement and record in the basement (probably not a good idea for me to mess around with mains since I'm new to this).
(4) build a one transistor short range super regenerative FM guitar transmitter and receiver tuned to a frequency not used for broadcasting and add in the damn low pass filter to attenuate RF in the radio receiver.
(5) buy a noise gate.
(6) buy a wireless guitar system to eliminate the capacitance of the cable altogether and make sure that the wireless receiver has a low pass filter to attenuate RF and use a 6 inch guitar cable to connect the wireless receiver to the amp.
(7) buy a better amp that was designed to remove RF.
(8) buy out Star 105.5 and WBBM and turn off their transmitters whenever I want to record.

Your explanation is right. I haven't checked your calculations.

I used this online calculator.

OK. Bear in mind that the cutoff frequency for an RC filter is not the frequency beyond which you don't hear anything. It's the frequency at which the signal is attenuated by 3 dB (decibels). You will still hear frequencies beyond the cutoff frequency; they will be attenuated. Search for single pole low-pass filter on Google and/or Wikipedia.

Okay, I searched for that search term and I found that what your saying agrees with Wikipedia. The cutoff frequency does not literally stop all frequencies above or below it, it attenuates the frequencies at 3db per decade (I think decade means 1khz). The graph formed is a line going down like this:

I was curious, and I experimented at home with different cable lengths, and I found that when I removed the guitar from the circuit and replaced it with a dipole antenna made from two guitar strings insulated from each other mounted and one string is connected to ground, a loop of guitar string 4 inches in diameter (inductance of 0.02899 uH connected to the "hot" string of the dipole antenna), a 6 inch long cable for 15pF capacitance (the resonant frequency of the LC circuit is 241.35 MHZ), and a green oxidized penny (a discussion for another thread) so that it has a constant audio signal that is the full hearing range (FM covers the full hearing range), I found that when I plotted a graph of the spectrum in Audacity using longer cables to connect to the guitar amp resulted in the start of the slope line of frequency attenuation moving to a lower audio frequency. And I heard Led Zeppelin on my guitar amp, I could change the station from the pop variety station Star 105.5 to the classic rock station by facing the dipole aerial perpendicular to the direction of the station (WDRV 97.1 and WLUP 97.9 which both have transmitters in the Chicago loop so I faced the antenna perpendicular to the south east from Island Lake), yay no more Mylie Cyrus problem solved and it turns out you don't even need a guitar to make a radio (well problem not solved if I want to play guitar and not build an FM radio) (but I think this is a discussion for another thread). I think I heard air traffic control too that was a mostly one way conversation from probably the transmitter onboard the aircraft is too weak for me to receive but the transmitter at the airport is powerful enough for me to receieve (''...[static]...[indisriminate man 1 talking]...over...[static]....copy that over [man 2]...[static]......[indiscriminate man1 talking]"....UA niner seven [man 2], cleared to land on runway 9 left...over...[static]...). by the way, I live 35 miles away from O'hare. I can only imagine the fun if I lived in one of those houses in DesPlains with the runway starting in the house's backyard.

That's all true, but you should also investigate the effect of the LC resonant circuit that is formed by the inductance of the pickup and the capacitance of the cable. This is a parallel "tuned circuit" that adds a peak in the response. This effect is strongest when the volume control is at maximum, because at that setting, the pickup inductance and the cable capacitance are directly connected together. Google some of those keywords for more information. And try to find out the inductance of your pickup so you can calculate the resonant frequencies with various amounts of capacitive loading.

Okay. Also I need to find the inductance of my pickups. I think my initial search query (what is inductance of switch pickups) was too specific, and when I changed my search query to "what is inductance of guitar pickups" I found that the inductance of guitar pickups ranges from 1 Henry for single coil pick ups to 4 Henry for high output humbucker pickups. More inductance increases the output of the pickup.

Ding! Right.

Good.

Right again.

I like being right. I'm used to everything I post on here is complete nonsense.

In this case, the resonant frequency will probably be in the audio frequency range. Although LC resonance may be explained in relation to tuning a radio receiver, you should forget about the radio analogy. The effect that's important here is at audio frequencies.

Your right! A 1 farad pickup ensures that the LC resonant circuit is in the audio frequency range. We wouldn't want a guitar to have the right number of coil turns to receive broadcast radio frequencies, would we. Also, the right number of coil turns to receive broadcast radio frequencies on the size of a guitar pickup with a ferrite core (maybe 120 turns) would not give a high enough output to drive the pre amp tubes or pre amp transistors in the guitar amp. A single coil pickup uses thousands of coil turns with a ferrite core, and it is still considered to be a relatively low output pickup. I really have to crank the volume on my 15 watt Marshall MG 15 to hear my single coil pickup guitar at all (unfortunately, increasing the volume also increases the volume of the infernal buzz of mains hum, which is why humbucker pickups were invented), compared to I can have the volume on the amp turned down much more to achieve the same volume on my humbucker pickup guitar.

Sort of. But let's keep the crystal set stuff for another thread, OK?

Sound like a good idea. Unfortunately, I think you should have said that at the top of this page because its too late I already typed responses involving crystal radio theory in this reply.

Q and bandwidth are related, yes. A higher Q means a narrower bandwidth, and a taller peak in the response. Google Q and bandwidth.

Okay, I Googled Q and bandwidth and I found that Google says that Q is a unit less concept that is the ratio of (2*pi*f*L ) / R .

The reason for the impedence load increasing at higher frequencies is the skin effect I also found out.

Yes, you're getting it!

I guess...

No. Around position 7 on the volume control is the point where the output impedance of the guitar will be at _maximum_.

A volume control at position 7 is a voltage divider consisting of two resistances. There is the resistance from the wiper to the top end (the end where the pickup is connected), and there's the resistance from the wiper to the bottom end (ground). For a logarithmic pot, which is the type of pot used for volume controls, around position 7 is where these two resistances are roughly the same; around 250k each, for a 500k pot. Assuming zero pickup impedance, these two 250k resistances are effectively in parallel, so the output impedance will be around 125k. That will be the maximum series resistance introduced by the volume pot, so it will be the setting where capacitive loading (cable, amplifier etc) has the most effect and where the cutoff frequency will be lowest.

Oh...so at that position 7, the wiper is connecting the two resistances (the resistance of the pickup and the resistance of the pot) in parallel instead of in series, and in parallel will actually lower the resistance since its the reciprocal.

I'd always wondered why around position 7 on my volume pot it seems to make my guitar quieter than the position counter clock wise to it. Now I know why!

Actually, you ARE getting it.

Good.

Good!

I'll tell you what the inductance of my pickups are when I find out.

Please, let's forget the radio stuff for now!

Okay, I'll try.