stein Posted August 11, 2011 Share Posted August 11, 2011 Yeah, I hear ya brother. I've spent many years teaching IT to people without any idea under the same premise, I'm the biggest advocate of "blaggers guide". As I've said before, if Michael Schumacher couldn't change a spark plug, it wouldn't stop him from being a phenomenal driver - you don't need to now how it works, just that it does work (if you do THIS, THIS and THIS... in that order... don't touch that though) Andy - all modesty aside, I could read stuff like that all evening. I've already downloaded your flv files and PDF'd the text for future. I'm a shameless student, I LOVE hearing from people who know their onions, I don't care if I sound stupid, just really happy to be learning more. Thanks again fellas, hope I can repay the help sometime. Glad to have a happy discussion. I may know what I am talking about SOMETIMES, but I also can sound like I know more than I do. I LOVE electricity, it is fascinating, especially in an audio circuit. Link to comment Share on other sites More sharing options...
Andy R Posted August 12, 2011 Share Posted August 12, 2011 So, that is why a 500k pot is louder than a 250k pot. The 500k has more resistance to a signal, and it would make sense to think the more resistance you have, the less signal you get. But, we wire them NOT to give resistance to the hot signal, it RESIST the hot signal to being a dead short. Hmmmm Me thinks not... There will be a perception of 250K sounding muddier on a humbucker or the perception of a 500K pot sounding very thin on a single coil but I don't think there would be an actual volume difference... Link to comment Share on other sites More sharing options...
stein Posted August 12, 2011 Share Posted August 12, 2011 Hmmmm Me thinks not... There will be a perception of 250K sounding muddier on a humbucker or the perception of a 500K pot sounding very thin on a single coil but I don't think there would be an actual volume difference... I think it can be difficult to be able to tell volume differences changing pots because we have the amp volume which actually has more effect. I mean, unless we are listening for it, we don't notice the difference. Like the difference between a strat and tele. I have tried it and a 500k pot in a strat is louder than a 250k, but, to bright to be worthwhile I think. Consider a linear 500k pot, half volume is 250k. The darker tone we hear when we turn down is the same as what you describe about most thinking a 250k pot sounding too dark, and the 500k in a strat is the opposite. This could get really deep. My brain is smoking as I ponder the reasons why the tone changes from bright to dark as the volume is changed. Link to comment Share on other sites More sharing options...
Guest farnsbarns Posted August 12, 2011 Share Posted August 12, 2011 My brain is smoking as I ponder the reasons why the tone changes from bright to dark as the volume is changed. Think about how the volume pot is connected to the tone put and the effect on frequency response that the more resistive material in the volume pot would have. Some LPs have 400k volume pots. Link to comment Share on other sites More sharing options...
Andy R Posted August 12, 2011 Share Posted August 12, 2011 I think it can be difficult to be able to tell volume differences changing pots because we have the amp volume which actually has more effect. I mean, unless we are listening for it, we don't notice the difference. Like the difference between a strat and tele. I have tried it and a 500k pot in a strat is louder than a 250k, but, to bright to be worthwhile I think. Consider a linear 500k pot, half volume is 250k. The darker tone we hear when we turn down is the same as what you describe about most thinking a 250k pot sounding too dark, and the 500k in a strat is the opposite. This could get really deep. My brain is smoking as I ponder the reasons why the tone changes from bright to dark as the volume is changed. It is a correlated effect between the pickup impedance and volume pot impedance. I see why you think that 250K would be half the volume compared to a 500K and might be some what of a factor on a linear taper pot. Not necessarily on an audio taper pot. I believe it would depend on the complete circuit. The AC produced from the pickup is going to be the same regardless of the Pot and being AC it should still follow the same principle that the signal is going to travel the path of least relative "impedance" I believe would be the accurate term. In other words it wouldn't care if the pot has the potential of 500K of resistance or 250K of resistance. As long as there is an easier path to take it is going to try to take it. Also think about where the wiper is in relation to the center pin out and the pickup in lug at 250K pot at full volume, it is pretty much a pin to pin connection. Then consider the wiper is in relation to the input lug center pin and the center out ( on a linear pot) it would have a longer distance to travel over the resistive element before the signal went out the center pin. Then figure in human hearing and how we perceive volume changes as well as how we perceive loss of high frequencies as the volume is reduced. Also consider that the resistance of the Volume pot is going to have an effect of the pickups impedance and thus changing the frequency response of the pickup. The change in impedance is going to focus the pickup more or less in a certain frequency range. So I maintain that there might be a perceived volume change but it is due to the frequencies that are being passed through not that it is actually allowing the pickup output signal to become louder.... Just my opinion.... Andy Link to comment Share on other sites More sharing options...
stein Posted August 12, 2011 Share Posted August 12, 2011 These are good points FARNS and ANDY, but just for clarification: I am not implying that "the volume is half" on a 500k pot (linear) is half in the center position, I was using it as an example that it is definitely less. I also mean to imply that what we hear when a 500k pot is in a position to provide 250k of resistance, it is VERY similar to what we hear from a 250k on full. DIGRESSIN BREAK: Andy is absolutely correct that the properties of the pup have a lot to do with the effect of the pot, that is why there seems to be a "magic" formula in the value of the pot chosen for each type of pup. I am using the volume difference to explain HOW the pots are wired. They are wired to provide a short at one end of the travel, and a RESISTANCE to short at the other end. The greater the resistance to short, the more current goes through. Link to comment Share on other sites More sharing options...
SHO Posted August 12, 2011 Share Posted August 12, 2011 I'm sleepy, so I might be all confused and wrong! Apologies if so... Why does treble go away as the volume is turned down? Because the pickup, the volume pot, the cable and the amp input stage is all one system that affects each other. Easily put, a guitar pick up can be modelled as this: An AC voltage source to an inductor and a resistance is series, and parallel to both is a capacitor. The AC source is the whole moving string in magnetic field deal, the inductor and the resistance is the coil. Ok. Then comes all the junk after the pickup, the load. This is stuff like the volume pot, cable to amp capacitance, amp input. To make it easy, all this is is just a capacitor and a resistor in parallel. Connect them together and you have a 2nd order low-pass filter. The filter will have a cut-off frequency and a resonance peak frequency (where the inductance and the capacitance resonate). Lowering the load resistance (meaning the parallel resistance the pickup "sees") either by turning the volume pot down or changing it to another value move the filters characteristics downwards in frequency making for a darker sound. I suppose the size of the resonance frequency peak is affected by the load resistance too. The load capacitance is pretty much all cable capacitance, so this is why different guitar cables sound different too! I suck at explaining these things, especially with out pictures (you don't want to see my MS paint skills anyway), but I think it works something like that anyway. Correct me if I'm wrong! I'm sure it's explained pretty well somewhere on-line as this is basic electronics. Look up RC and LC circuits in google and it should all be there somewhere. Link to comment Share on other sites More sharing options...
stein Posted August 12, 2011 Share Posted August 12, 2011 SHO: Thanks for that. I like to try and "hear" electrical characteristics. Or learn about one, and understand what it "sounds" like. Can you explain what a "2nd order low pass filter" is? Link to comment Share on other sites More sharing options...
SHO Posted August 12, 2011 Share Posted August 12, 2011 SHO: Thanks for that. I like to try and "hear" electrical characteristics. Or learn about one, and understand what it "sounds" like. Can you explain what a "2nd order low pass filter" is? A low pass filter is just what it sounds like. Ideally, I lets frequencies below a certain point pass unaffected, and for frequencies above that point it will block. In reality it's not that perfect though. Instead of blocking completely, it will just dampen the signal. Actually, the cut-off frequency of the filter is defined to when the signal has been dampened by -3dB. For a first order filter, the signal will be dampened by -6dB per octave (meaning every time the frequency doubles). For a second order filter, the signal will be dampened by -12dB per octave, so you could say it is more effective. In my example above we have an inductance in series, and that dampens high frequencies. So that's the first low pass filter. The capacitance in parallel shorts high frequencies to ground, so along with the resistances we have a second low pass filter. Each of these alone is a first order filter. Together, they become a second order filter! Since in this case, we have both an inductance and a capacitance, there will be a frequency where we get electrical resonance. At this point, the circuit will actually work as a voltage amplifier, even though it is a passive circuit. But the power doesn't not go up though, as the current will drop. Anyway, this means the filter will have a peak, usually just before the cut-off frequency, so we get a signal boost there. So, in lack of a picture, describing the way the 2nd order low-pass filter looks" when going up in frequency would be: "output = input, output = input, ..., signal increase, resonance (max signal level), signal turning back down, signal level going below input level, signal at -3dB of input level (filter cut-off frequency), signal at -12dB for 2 x cut-off freq, signal at -24dB for 4 x cut-off freq, and so on..." I hope that helps somewhat. Else I could look into doing some sort of pictures sometime. =) Link to comment Share on other sites More sharing options...
stein Posted August 12, 2011 Share Posted August 12, 2011 A low pass filter is just what it sounds like. Ideally, I lets frequencies below a certain point pass unaffected, and for frequencies above that point it will block. In reality it's not that perfect though. Instead of blocking completely, it will just dampen the signal. Actually, the cut-off frequency of the filter is defined to when the signal has been dampened by -3dB. For a first order filter, the signal will be dampened by -6dB per octave (meaning every time the frequency doubles). For a second order filter, the signal will be dampened by -12dB per octave, so you could say it is more effective. In my example above we have an inductance in series, and that dampens high frequencies. So that's the first low pass filter. The capacitance in parallel shorts high frequencies to ground, so along with the resistances we have a second low pass filter. Each of these alone is a first order filter. Together, they become a second order filter! Since in this case, we have both an inductance and a capacitance, there will be a frequency where we get electrical resonance. At this point, the circuit will actually work as a voltage amplifier, even though it is a passive circuit. But the power doesn't not go up though, as the current will drop. Anyway, this means the filter will have a peak, usually just before the cut-off frequency, so we get a signal boost there. So, in lack of a picture, describing the way the 2nd order low-pass filter looks" when going up in frequency would be: "output = input, output = input, ..., signal increase, resonance (max signal level), signal turning back down, signal level going below input level, signal at -3dB of input level (filter cut-off frequency), signal at -12dB for 2 x cut-off freq, signal at -24dB for 4 x cut-off freq, and so on..." I hope that helps somewhat. Else I could look into doing some sort of pictures sometime. =) That explains a lot...I understand a lot of that from your explanation, but I also need to absorb it to truly understand. I'm chomping on that. Link to comment Share on other sites More sharing options...
Andy R Posted August 12, 2011 Share Posted August 12, 2011 These are good points FARNS and ANDY, but just for clarification: I also mean to imply that what we hear when a 500k pot is in a position to provide 250k of resistance, it is VERY similar to what we hear from a 250k on full. This is the point where my logic disagrees with yours. Not that you are wrong.... I just think you are! So we must settle this.... We must know! Lets keep it simple and remove the tone cap, tone pot , switches etc... from the equation. So lets think in terms of humbucker pick-up to volume pot direct then volume to output and ponder from there. My theory is that a 250K pot set wide open and a 500K pot at 250K will sound completely different tonally and in volume. I believe this because that the 250K pot will be a near direct connection between the wiper and the input lug from the pickup. While the signal from the pickup will have had to have already passed through 250K of resistance on a 500K pot before it reaches the contact of the wiper to go to the output. There may be some experimenting going on tonight. Andy Link to comment Share on other sites More sharing options...
stein Posted August 12, 2011 Share Posted August 12, 2011 This is the point where my logic disagrees with yours. Not that you are wrong.... I just think you are! So we must settle this.... We must know! Lets keep it simple and remove the tone cap, tone pot , switches etc... from the equation. So lets think in terms of humbucker pick-up to volume pot direct then volume to output and ponder from there. My theory is that a 250K pot set wide open and a 500K pot at 250K will sound completely different tonally and in volume. I believe this because that the 250K pot will be a near direct connection between the wiper and the input lug from the pickup. While the signal from the pickup will have had to have already passed through 250K of resistance on a 500K pot before it reaches the contact of the wiper to go to the output. There may be some experimenting going on tonight. Andy THIS sounds like a worthy experiment. I expect them to be similer, but I too expect some differences. I might imagine that the equal resistace of a 500k pot set at 250k vs a 250k pot set to 250k would produce an equal effect as a result of the (obviously) but I do wonder about what qualities BESIDES resistance might come into play. We might both be in for some surprises. Now, personally, I HAVE done experiments changing values of pots, (as well as brands) enough to the point of being able to predict the effect of say, changing a 300k to a 500k. I am usually right. (when I am wrong, of corse, that is when I seem to learn the most). A note: One thing I have come to discover is that it is common for a pot to NOT measure the value it should. From my experience, a 500k cts pot will usually actually be between 425k and 510k, and average close to 500k but a little less. A 300k pot though, USUALLY measures well above 300k. I remember from my stash that I actually had some 500k pots and some 300k pots that were the same in value measured for resistance. I never thought to compare what a 300k that measures high and a 500k that measures low (both having the same reading) might sound like compared. Regardless, I appreciate YOU doing the experiments and sharing. I don't mean to be lazy, but I don't have the set-up or resources at the time. Link to comment Share on other sites More sharing options...
Recommended Posts
Archived
This topic is now archived and is closed to further replies.