Suhr HSS strat 500K / 250K schematic question

[fuzzy645]

OK, I know this is a stomp box forum, but I thought I would post this question about an interesting approach (supposedly) that has been used by John Suhr in some of his Strats that mix humbuckers with single coils.  I saw this diagram posted on another forum.  The objective is that the humbucker sees 500K pots and the single coils see 250K.  

I get the simple concept that if you put a 470K resistor in parallel with a 500K pot the resulting resistance is approx 250K.  Simple math.  However, what I am questioning is if this is really happenening in the schematic, or not.

To keep everyone on the "same page" for initial discussion, lets make 4 assumptions:

1.  Lets start by just looking at one of the positions of the 5 way switch, namely position 5 (to start).  
2.  Also, for purposes of discussion, lets assume each single coil pickup measures 6K, and the humbucker measures 12K (a pair of 6K pickups in series).
3.  Lets assume the volume pot is on "max volume" , in other words full 500K volume pot resistance in the connection going from jack to the wiper of the volume pot  to ground, and pretty much zero resistance in the connection going from jack to wiper to the pickup (via the switch).
4. Ignore the treble bleed mod (R1 and C1) for purposes of discussion, make believe they are not there (that can be another post).

When I look at position 5 only, this is what I see starting from the jack.

The jack is connected to the wiper of the volume pot, therefore I see 2 paths in parallel.  Lets call the first one PATH-A which is jack to wiper to ground, and since my assumption is that the volume is on "max" there would be full 500K resistance here.  This is in parallel with the what I will call PATH-B which is jack to wiper to the neck pickup (6K).  At this point since the volume pot is on 10 there is pretty much zero resistance between the jack and the pickup, therefore we can think of this is a straight shot over to the 6K pickup, however that pickup just happens to be in parallel with this 470K fixed resistor (the reason for my post).

So you know where I'm going, is that from a resistance point of view, the 6K pickup will dominate here and the formula for parallel resistance will result in approx. 6K.

Therefore, do you think the use of this 470K resistor will really make any bit of difference in this case?

Thoughts?

[alexradium]

500k pots can be a bit bright on single coils,that's why Suhr puts that 470k in parallel.
Using the super switch he assigns that and the appropriate tone pot to singles and the 500k vol and tone to humbucker.
Speaking of tone difference, I tried all these things and I came to the conclusion that this is system dependable, maybe in Suhr guitars you can hear it because they're a tad brighter than others.
Plus,if you use a lot of combined single positions with 250k vol and tone it can muffle the sound.
Again it all depends on pickups,pots and wood choices.

[PRR]

> assume each single coil pickup measures 6K

Well, but that's true only for DC and low audio frequency.

There is also inductance and capacitance. A tuned circuit. It resonates. Near resonance the impedance is higher, possibly much higher. AM radio tuners bump-up 100X. The bump in a pickup is hard to nail-down, but many pups clearly sound different with 100K than with 1Meg loading. Some ballpark simulated dart-throwing suggests an impedance peak of 100K-300K in the 1KHz-4KHz range.

If the pup winder says to load with 250K or 500K, that's what your side of the job is (to start).

> assume the volume pot is on "max volume"

Then you must include the amplifier (or stompbox) input. Some cruder inputs are 100K or less. Fender is usually 1Meg. Some Ampegs have gone higher. The 1Meg is somewhat significant. 100K makes this 250K/500K adjustment pretty moot (when "max volume").

> the 6K pickup will dominate here and the formula for parallel resistance will result in approx. 6K.

Yes, but.....  we often want a mis-match. My power line from the street is 0.4 ohm impedance (10A load causes 4V sag). My load is 120V 10A or 12 ohms. So why don't I "match" a 0.4 ohm load to my 0.4 ohm line? Because I'd only get half-voltage. My lights would be dim. A matched guitar (if it were possible) would be weak. We want to not-load the source to get high and reasonably constant or predictable output.

[Suhr]

[quote
So you know where I'm going, is that from a resistance point of view, the 6K pickup will dominate here and the formula for parallel resistance will result in approx. 6K.

Therefore, do you think the use of this 470K resistor will really make any bit of difference in this case?

Thoughts?


[/quote]


Old post but to clarify you are making a huge mistake here.
You are assuming the DC resistance is the impedance of the pickup. It is not.
The value of the pot makes a tremendous difference.
You also need to consider the distributed capacitance in the cable and at which end of the cable the load is at.

[aron]

John, thank you for your guitars! Welcome to the forum!

[Paul Marossy]

You are assuming the DC resistance is the impedance of the pickup. It is not.

Yep, that's right. There are several things going on in passive electric guitar pickups. There's a DC resistance, a resonant frequency, some amount of impedance, etc. And like Suhr has said, impedance and DC resistance are NOT the same thing.