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Posted

My next bass build will have passive wiring with single coil Nordstrands. My question is - whilst I can easily find a wiring diagram for vol, vol & tone, I cannot determine whether I need lin or log pots? Can anyone help me on this?
Thanks!

Posted (edited)

Hi Chop

have a look at guitarnuts.com. They give a lot of detail about strat wiring and (if I remember correctly) also go into the log and linear topics as well.

Also, maybe contact Nordstrands ?

hope that helps :)

T

Edited by essexbasscat
Posted

[quote name='Ian Savage' post='821410' date='Apr 28 2010, 01:41 AM']Generally speaking, log for volume, lin for tone. Avoids the 'all-at-one-end' effect that you sometimes get with ill-though-out control setups.[/quote]


Thanks Ian. Thats interesting because I asked the same question on Talkass (Paah!) and I got the opposite - Lin for volumes and log for tone. I guess I can try both options and se which works best.

Posted (edited)

considering the rule of having to 10x the power to double the volume, power to volume is an exponential relationship, and therefore to stop your pot turning output also being exponential, you need the pot to be logarithmic to transform pot angle Vs output into a relationship that's closer to linear.

hooray for A-level maths, lets just hope i've remembered it correctly.

Edit: have thought of it more properly, you start of with something like 10^p=k.V, where p is power and V is volume, with k being some constant, and using log pots changes that to p=log(kV), so your response graph goes from [url="http://hotmath.com/images/gt/lessons/genericalg1/exponential_graph.gif"]exponentional[/url] to[url="http://people.richland.edu/james/lecture/m116/logs/logarithmic.gif"] logarithmic[/url], which is just a bit nicer.

Edited by Zach
Posted

[quote name='Zach' post='822735' date='Apr 29 2010, 09:47 AM']considering the rule of having to 10x the power to double the volume, power to volume is an exponential relationship, and therefore to stop your pot turning output also being exponential, you need the pot to be logarithmic to transform pot angle Vs output into a relationship that's closer to linear.

hooray for A-level maths, lets just hope i've remembered it correctly.

Edit: have thought of it more properly, you start of with something like 10^p=k.V, where p is power and V is volume, with k being some constant, and using log pots changes that to p=log(kV), so your response graph goes from [url="http://hotmath.com/images/gt/lessons/genericalg1/exponential_graph.gif"]exponentional[/url] to[url="http://people.richland.edu/james/lecture/m116/logs/logarithmic.gif"] logarithmic[/url], which is just a bit nicer.[/quote]


Great - it makes perfect sense now! I will try log volumes and lin tone and see how it goes.
Thanks for helping out.

Posted

no problem. It seems wikipedia agrees with me, so that's reassuring:
'The 'log pot' is used as the volume control in audio amplifiers, where it is also called an "audio taper pot", because the amplitude response of the human ear is also logarithmic. It ensures that, on a volume control marked 0 to 10, for example, a setting of 5 sounds half as loud as a setting of 10'

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