Capacitors for WARPIG Bridge.

[WITH FULL DISTORTION]

Hails Guys.
How are yah?
i have a doubt
im feeling that the warpig in my fender is lacking HIGHS and its sustainless on the same higher frequencies.
what should be the problem??

it has a one volume pot Only. 500k. and no switch or toggle thing. and no other pickup with it.
i just soldered it to the Pot, and then to the jack. does it need, as required in a 1 volume 1 Tone potted guitar, to put a capacitor on the Volume pot? If so, which value and wich model of capacitor?
Hails and thanks in Advance.
JP

[FELINEGUITARS]

No capacitor required........unless you have a tone pot

[gwEm]

if you want to open up the highs, try a 1M pot

[WITH FULL DISTORTION]

Ahhhhhh Excellent Then.
Any tips on how to Set the 2 bars of Pole pieces?to acchieve more highs ?
JP

[WITH FULL DISTORTION]

What if i want to put the stock Hexa poles in? mine came with normal pole pieces.
do i need to return it to tim or can i do it for myself?
JP

[gwEm]

Ahhhhhh Excellent Then.
Any tips on how to Set the 2 bars of Pole pieces?to acchieve more highs ?
JP

hails JP!

to do that:

lower the polepieces nearest the neck
raise the polepieces nearest the bridge

i forgot the warpig has 12 screw poles :)

[Yamhammer]

A 1M pot should fix the problem right away.

[f.sardis]

i was always under the impression that a 1M pot only matters if there is a capacitor involved. that will change the tone but simply changing the resistance in the circuit without the capacitor will just decrease the volume and should not change the tone at all.

someone care to explain?

[gwEm]

i was always under the impression that a 1M pot only matters if there is a capacitor involved. that will change the tone but simply changing the resistance in the circuit without the capacitor will just decrease the volume and should not change the tone at all.

someone care to explain?

that would be true if the coils of a pickup were an ideal electrical source. don't forget the guitar pickups are inductive, capacitive and have an resistance which is relatively high.

[f.sardis]

if V=IR and we assume in our case that for the same picking force the V is constant then raising the R will result in lower I. That means weaker current hitting the amp. What is the result of weaker current through the amp?

[gwEm]

if V=IR and we assume in our case that for the same picking force the V is constant then raising the R will result in lower I. That means weaker current hitting the amp. What is the result of weaker current through the amp?

sorry, thats an extremely gross simplification, and incorrect.

but even entertaining your suggestion - the effect is nothing. guitar amplifiers are voltage rather than current driven.

the guitar pickup is a highly non-ideal device, and has an impedance which is loaded by following components. if you increase the load on the pickup, with a lower load impedance (pedals, volume pot, tone pot, amp input) - then the voltage produced at the output of the pickup is lower.

but even this is too big a simplification - pickups are inductive and capacitive - the affect of loading will be different at different frequencies. its extremely difficult to model these. but essentially its the higher frequencies which are attenuated more.

[f.sardis]

Cool. learned something new today  :D

[gwEm]

Cool. learned something new today  :D

:D

[phlip]

if you want to open up the highs, try a 1M pot

And if you want a swift quick and dirty test, just wire the pickup direct to the output jack. If it opens up the way you like then a top quality 1m pot should achieve pretty much (With a tiny bit less top) the same effect.

[Roobubba]

Gwem speaks the truth, although there shoud have been an 'e' instead of 'a' (effect being used as a noun). :)