250K or 500K vol pot?

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ma	250K or 500K vol pot?		Lead	[-]
Posts: 34 Tue, Apr 21, 2009 09:43:04	Tags : None Dear people at the forum. I suppose should be able to work this out by myself as I have a basic knowledge about how to "leak"away various frequencies to ground to fine tune my pickups but: I have now for the first time in my life got a guitar with both singlecoil and humbucker (a TL290 in bridge position and a minihumbucker made by J. Lundgren in the neck position). So I should typically have a vol pot with 500k for the humbucker but a 250K sounds good with the TL290. Am I correct in assuming that I should go for the 500K pot and simply attach a cap to the TL290 to take some top end of if it gets to much? There are a lot of telecasters on the market that has this type of pickup configuration so I suppose the answer is simple but as I have never had this before it would be great to hear if anyone here has any suggestions Thanks
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PeteMK1		#1	[url]	[-]
Posts: 2180 Tue, Apr 21, 2009 11:22:30 Contributing Palladium Member Resident Electronics Guru	Have you already tried the whole setup with just a 500k pot? Chances are that it works pretty well without any additional components. You can say you to me.
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ma		#2	[url]	[-]
Posts: 35 Wed, Apr 22, 2009 08:29:40	No I have not tried that yet, thanks, I'll do that first and see how it goes. But I also saw a simple but great sollution that I did not think of at first: To simply wire in a resistor to ground (in parallell with the pot) from the lug of the pickup selector where the bridge pickup hot wire is attached. And depending on the resistor value I could get a total load of 250K, or any other value if desired, with the bridge pickup without affecting the neck pickup. Surely this is a much better way to go than adding caps to reduce treble as this will actually have exactly the same effect as having a 250K pot? By the way I dont have a tonecontrol that adds load to the pickups. I use a Stellartone Tonestyler and it is, simply put, a rotary switch with caps but no resistor, i.e. every cap is on full so to speak.
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PeteMK1		#3	[url]	[-]
Posts: 2182 Wed, Apr 22, 2009 09:19:09 Contributing Palladium Member Resident Electronics Guru	Yeah, the resistor solution is a good one, but keep in mind that the reduced load will also be present in the middle position of the pickup selector. That might not be that much of a problem though, since the parallel combination has more high end anyways. Actually my assumption is that the 290 will work well with both setups. If you have a capacitor switch instead of a tone pot, IMO it's the best solution to use a single 250k volume pot without additional resistors. The recommendations from the pickup manufacturers usually assume one volume and one tone pot. The smaller value volume pot has the advantage of cutting less highs when turned down. FWIW, I use 250k volume pots and capacitor switches in all my guitars with a variety of single coil and humbucker pickups, and this setup works extremely well. You can say you to me.
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ma		#4	[url]	[-]
Posts: 36 Wed, Apr 22, 2009 11:22:07	thanks, i'll try both 250K and 500K then and see how it goes. It would be convenient if it worked well with the 250K I've already got in my guitar. But I have also seen some people using pots with inbetween values of about 325K so I just have to have try all options and choose. Thanks again
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ma		#5	[url]	[-]
Posts: 37 Thu, Apr 23, 2009 09:44:18	Just realised that I will probably use a Q-filter with this guitar. So then mayby I should have a 500K pot for the volume and 500K for the Q-filter pot? Any thoughts regarding this? By the way, I read somewhere that one could use matched pots that are measured to be exactly the same value. How could that improve the tone I wonder? I thought that the combination of pots (resitors) will simply affect the load on the pickups and the load is the result of your pots resistance. So how could it be negative from a tone perspective to have, say one 490K pot and one 520K pot? The matching of pots should not mean anyting tonewise unless of course one in extremely picky and think that 500K is the only thing that sounds good? I mean it should be possible to arrive at the same load as using two 500K pots by using two pots where one is lower than 500K and the other is higher that 500K should it not?
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PeteMK1		#6	[url]	[-]
Posts: 2185 Thu, Apr 23, 2009 15:10:59 Contributing Palladium Member Resident Electronics Guru	You're right, matching pots is totally useless in this application. It probably increases the depth of the pockets of the one who offers this service. You can say you to me.
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Bigue		#7	[url]	[-]
Posts: 105 Thu, Apr 23, 2009 19:48:04	PeteMK1 wrote: It probably increases the depth of the pockets of the one who offers this service.
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