PRS SE Paul's Guitar - why the output resistor?

[Troisnoir]

I noticed my SE Paul's Guitar has an 820k resistor across the output.

I've not seen this on any guitars over the last 50 years, including my other three PRS's.

Any idea what the function is?

I'm assuming it either maintains the brightness of the pickups, or perhaps tones them down, since this guitar is a simulation of the American HCI pickups, using standard size humbuckers.

Inquiring minds, etc. ....

 

[xjbebop]

PRS patented tone enhancer... ???

 

[Troisnoir]

I'm thinking it would alter the final output impedance of the guitar, or keep whatever amp or pedal you're plugging into from having such a large effect on the tone.

It might be a good idea...but I'm wondering why no one else (including PRS to my knowledge) has done it before?

I changed out the pickups on my SE Pauls Guitar (the originals sounded quite good - but didn't work for my uses). I didn't keep the resistor in.
I think I'll try popping the resistor in the circuit with alligator clips, and see how it effects:
- tone
- vol and tone pot operation
- volume of pickups when split to single coil

They may be onto something...

 

[shimmilou]

Across the output? Maybe you mean across the volume pot?

If you put an 820k resistor in parallel with a 500k pot, you get 300k total (310k)

 

[elvis]

With a cap in series it could be a treble bleed

 

[Troisnoir]

If I recall correctly, the resistor was connected from the volume pot center lug, which is the guitar output, to ground.
It may be there as part of the treble bleed circuit, though. I'll have to reinstall it and see if the volume works nicer.

There are three commonly used treble bleed versions, and this isn't one of them. But if it works, it works.

I'll let you all know...

 

[elvis]

That is super weird.

 

[Troisnoir]

Well, is my face red

I went looking for the resistor to reinstall it, and it was still on the original pot (I replaced it with a Mojotone low friction pot).

And behold - it was soldered to the top lug, not the middle - my bad...

So it's either a cheap way of simulating a 300k pot, or changing the taper. I'll see later this morning...

However, since I've replaced the HCI pickups with a Duncan JB and Jazz neck, I may no longer need or want the resistor. We'll see how it sounds.

As always, the management regrets any inconvenience...

 

[Troisnoir]

OKAY......

Here's what I found:

1. There may be a difference in the volume taper, if so, it's subtle.
2. I noticed a definite reduction in harshness when splitting the pickups with the resistor in the circuit. It's probably a cheaper way of making a 300k pot than actually using one, reducing the production costs.

In my case, I find the resistor mod too dark for my pickups, especially the JB. But I do like the single coil sounds a little better with the resistor in.

So I may rewire my switches to engage the resistor only when splitting. In the case of the original pickups, which were VERY bright, the resistor worked fine in both modes.

That Paul sure is a clever guy!

 

[shimmilou]

...So it's either a cheap way of simulating a 300k pot, or changing the taper...

Seems to be the case.

 

[docteurseb]

Putting a resistor that way changes both the tone's effective value and taper.

The effect on taper is more severe the lower the effective pot value is from the original one.
I was curious about it on the MEV a few months ago since they do this there as well to let you 'switch' between a 500 and 250k volume pot.

Bold line would be a 500k linear pot with 500k resistor in parallel, while the other one is a linear 250k pot.

 

[Troisnoir]

Thanks!

BTW: Excuse my ignorance, but what is the MEV?

 

[shimmilou]

Modern Eagle 5

Example:
https://www.sweetwater.com/store/de...020-modern-eagle-v-electric-guitar-river-blue

 

[Troisnoir]

Thanks!