ComradeFletch Posted September 16, 2003 Report Posted September 16, 2003 can anyone explain what effects the value of a pot has on the sound and fuction it provides? i keep seeing these like '1meg' ones and thinking 'hmmm, what can i add onto the guitar now?' Quote
canuckguitarist Posted September 16, 2003 Report Posted September 16, 2003 500k pots are for volume knobs on humbuckers 250k for volume on single coils and 250k's for tone knobs too i think (not totally sure about the last one) Quote
ComradeFletch Posted September 16, 2003 Author Report Posted September 16, 2003 cheers man, but those i have sorted. its all these wierd and wonderful values tht i'm wondering about, like the 1 meg pots and stuff. Quote
canuckguitarist Posted September 16, 2003 Report Posted September 16, 2003 not too sure about those... Quote
ComradeFletch Posted September 16, 2003 Author Report Posted September 16, 2003 just to prove that its not some product that i've dreamed or sumthin one wierd pot Quote
canuckguitarist Posted September 16, 2003 Report Posted September 16, 2003 from stewmac on 1meg pots... #3481 1Meg-ohm Get the most out of your guitar with a 1Meg pot. All pots bleed a certain amount of the signal to ground, this "attenuation" can be used to warm the sound of your guitar's pickups, but you can also run your guitar "wide open" with these 1-Meg pots! The 3/8 x 32 threaded portion is 3/8" tall. 3/8"-diameter mounting hole required. Split shaft with fine knurling. Quote
ComradeFletch Posted September 16, 2003 Author Report Posted September 16, 2003 cheers dude, i shud really start going on stewmac more Quote
Saber Posted September 16, 2003 Report Posted September 16, 2003 Increasing the pot value to 1M should give you a bit more treble. Quote
krazyderek Posted September 16, 2003 Report Posted September 16, 2003 cheers dude, i shud really start going on stewmac more LOL god no.... only if you absolutely have to.. imo.. the higher the value on the pot the brighter/crisper the pickup will sound Quote
westhemann Posted September 16, 2003 Report Posted September 16, 2003 stewmac is the best place imo to gather detailed info on how to install and use everything they sell.i love them even though they are more expensive than some places because most of what they sell is of such high quality. they also have always shipped the next day and my stuff is always here with no mistakes and carefully packed within a couple of days.and if i order more than once in the same day they combine the orders to save me shipping. some stuff you can get cheaper elsewhere though. Quote
daveq Posted September 17, 2003 Report Posted September 17, 2003 For a volume pot, when it is set to be a short circuit (0 ohms) is this the full volume position or is the 500K or 1M the full volume position? I've never actually paid any attention to this before - I guess I've been getting lucky with my wiring so far. If the full volumne position is the 0 ohm position, then you would need to turn your guitar down a little to hear the difference between a 500K and 1M so maybe it's the other way around? Quote
Saber Posted September 17, 2003 Report Posted September 17, 2003 For a volume pot, when it is set to be a short circuit (0 ohms) is this the full volume position or is the 500K or 1M the full volume position? I've never actually paid any attention to this before - I guess I've been getting lucky with my wiring so far. If the full volumne position is the 0 ohm position, then you would need to turn your guitar down a little to hear the difference between a 500K and 1M so maybe it's the other way around? That's not exactly how it works. A volume pot is connected as what we call a "voltage divider" because it "divides" the pickup signal voltage. What does this mean? Basically, one end of a 1M pot is connected to common ground with the common lead of the pickup. The other end of the 1M pot is connected to the other end (or hot lead) of the pickup so the full pickup signal is applied across the full 1M resistance at all times. When you place the 1M pot's wiper at 25%, for example, the pot becomes 2 resistors connected in series: one is .25M (250k) between ground and the wiper, and the other is .75M (750k) between the wiper and the pickup's hot lead. Since your wiper is at 250k from ground (or 25% of 1M), you're "tapping" 25% of the signal that's across the total pot. If you place the wiper at 50%, you have two 500k resistors and you're tapping into 50% of the signal. So the output signal is not really a function of how much actual resistance you place in series with it but rather a fraction of the total signal determined by the fraction of the pot that you're dividing it's resistance into with its wiper. Theoretically, only the signal level should change as you vary the volume, and not the frequency response. But in practice, because the amp's input impedance loads the signal (that's a whole other story), you lose some highs as you reduce the volume. Why does a 1M volume pot cut less treble than a 500k pot? The pickup coil is an inductor and when placed in series with a resistor (volume pot) they form a low-pass filter when taking the signal across the resistor. Doubling the resistance will theoretically double its cutoff frequency which means more treble. Again the amp's input impedance limits the benefits of this "theoretical" advantage but it does have an effect. I don't know if I made it clear enough but I'll try to include more details and visual aids in a tutorial that I just started and hope to finish one day. Quote
westhemann Posted September 17, 2003 Report Posted September 17, 2003 For a volume pot, when it is set to be a short circuit (0 ohms) is this the full volume position or is the 500K or 1M the full volume position? I've never actually paid any attention to this before - I guess I've been getting lucky with my wiring so far. If the full volumne position is the 0 ohm position, then you would need to turn your guitar down a little to hear the difference between a 500K and 1M so maybe it's the other way around? That's not exactly how it works. A volume pot is connected as what we call a "voltage divider" because it "divides" the pickup signal voltage. What does this mean? Basically, one end of a 1M pot is connected to common ground with the common lead of the pickup. The other end of the 1M pot is connected to the other end (or hot lead) of the pickup so the full pickup signal is applied across the full 1M resistance at all times. When you place the 1M pot's wiper at 25%, for example, the pot becomes 2 resistors connected in series: one is .25M (250k) between ground and the wiper, and the other is .75M (750k) between the wiper and the pickup's hot lead. Since your wiper is at 250k from ground (or 25% of 1M), you're "tapping" 25% of the signal that's across the total pot. If you place the wiper at 50%, you have two 500k resistors and you're tapping into 50% of the signal. So the output signal is not really a function of how much actual resistance you place in series with it but rather a fraction of the total signal determined by the fraction of the pot that you're dividing it's resistance into with its wiper. Theoretically, only the signal level should change as you vary the volume, and not the frequency response. But in practice, because the amp's input impedance loads the signal (that's a whole other story), you lose some highs as you reduce the volume. Why does a 1M volume pot cut less treble than a 500k pot? The pickup coil is an inductor and when placed in series with a resistor (volume pot) they form a low-pass filter when taking the signal across the resistor. Doubling the resistance will theoretically double its cutoff frequency which means more treble. Again the amp's input impedance limits the benefits of this "theoretical" advantage but it does have an effect. I don't know if I made it clear enough but I'll try to include more details and visual aids in a tutorial that I just started and hope to finish one day. i think i'm in love! Quote
Saber Posted September 18, 2003 Report Posted September 18, 2003 I wish I had the same effect on women. Quote
ComradeFletch Posted September 18, 2003 Author Report Posted September 18, 2003 so, muchos treble with a 1meg on the volume..... i have seen a 5 meg i think, but aint tht sure. Quote
Jehle Posted September 18, 2003 Report Posted September 18, 2003 so, muchos treble with a 1meg on the volume..... i have seen a 5 meg i think, but aint tht sure. Well, there's a point of diminishing returns after a while. Going through more resistance to get the output level kills the current level a little. V = I * R Quote
daveq Posted September 18, 2003 Report Posted September 18, 2003 Oops, I guess I should have told you that I'm an Electrical Engineer - I could have saved you a lot of typing. I just never bothered looking at a wiring diagram for a guitar before. I guess it's kind of like the mechanic who drives a p.o.s. Thanks anyway. Quote
Saber Posted September 18, 2003 Report Posted September 18, 2003 so, muchos treble with a 1meg on the volume..... i have seen a 5 meg i think, but aint tht sure. Well, there's a point of diminishing returns after a while. Going through more resistance to get the output level kills the current level a little. V = I * R Current (I) does not really concern us in small signal circuitry as it does in power electronics like in driving speaker loads. We are mainly concerned with how the voltage signal is affected. But it is true that going above 1M won't give much of an advantage because the signal then has to go drive the amp input impedance which itself can be around 1M or even lower in some cases. With your guitar volume at 100%, the amp's input impedance effectively becomes connected in parallel with your volume pot. Since 2 resistors in parallel always result in a total resistance that's less than the lowest one, if your amp's input impedance is 1M, the pickup would be driving an impedance that's less than 1M whether your volume pot is 1M or 100M! And in addition, the linearity of the volume pot will be affected much more if it's much higher than the amp's input impedance. And a higher resistance is more susceptible to picking up noise. So it's a matter of balancing the benefits with the disadvantages. If anyones interested, the equation for 2 parallel resistors is: 1/Rparallel = 1/R1 + 1/R2 or: Rparallel = 1/( 1/R1 + 1/R2 ) Replace R1 by the volume resistance and R2 by the amp's input impedance and that's what you're pickup will "see" when your guitar volume is at 100%. Quote
Saber Posted September 18, 2003 Report Posted September 18, 2003 Oops, I guess I should have told you that I'm an Electrical Engineer - I could have saved you a lot of typing. I just never bothered looking at a wiring diagram for a guitar before. I guess it's kind of like the mechanic who drives a p.o.s. Thanks anyway. Well, hopefully others will benefit from this information, anyway. Quote
Jehle Posted September 18, 2003 Report Posted September 18, 2003 Hey, I'm a physicist and I'm impressed. We normally just blow things up and then deduce what the best thing to use is. Quote
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