Questions re: pot resistance and direct through wiring

[Ro_S]

I have a few questions, please.

So I understand that the resistance value of pots has an affect on the brightness/darkness of the pickups in a guitar; and that higher pot resistance values will brighten the base tone of a pickup and vice versa.

Q1.  
With a 'direct through' mod (aka 'passing lane' mod?), whereby the pots/controls are being bypassed and the pickups thus going directly to the output, why does this make the pickups sound brighter as opposed to darker?   Shouldn't the lower resistance - due to the resistance of the values of the post  being removed from the circuit -  make the pickups sound darker?  I don't understand.

Q2.
How much of an affect/difference, in terms of both tone and volume, does the direct through mod have compared to both/all of the control pots not being bypassed and them being set on full (wide open)?

Q3.
Is the direct through mod regarded as a useful one?

Q4. 
On the earlier wiring schemes of the Fender Telecaster (and the Broadcaster), in the middle selector switch position, the neck pickup alone was active and it had no tone control or tone capacitor engaged.  Similarly,  on the vintage wiring of a Fender Esquire, in the rear selector switch position, the single bridge pickup had no tone control or tone capacitor engaged.

With just the tone control/pot being bypassed/inactive(?): 
a) that seems like a partial direct through wiring, correct? 
b) that would make the pickup sound brighter, correct? 
c) how much of an affect/difference, in terms of both tone and volume, does this have compared with tone pot alone not being bypassed and it being set on full(wide open)?


thanks!

[thetragichero]

"WHAT IS A NO-LOAD POT?
Before we get into No-Load Pots, let’s talk about “Load”: The definition of Load in terms of electricity is anything in a given circuit that “consumes” energy as opposed to sourcing (providing) energy. Take your Tone pot for instance: even if your Tone pot is on “10” (Clockwise) and the pot is not engaged, it is still “sucking up” some of the electricity. The Sweeper (Middle Lug) is still technically on the Resistance Strip, which draws power from the Volume Pot.

On a No-Load pot, there is a break on the Resistance Strip where the wiper is taken completely out of the circuit. It’s like “no man’s land” for the wiper, so much so that the Volume Pot doesn’t “see” the No-Load Tone Pot at all – almost like it’s invisible.

So what does this do to your guitar’s tone? Well, you’ll never know until you hear it for yourself, but, it will make your pickups sound a little more “full-throttle”. They might sound a little bigger, fuller, with added bass and treble. This is all personal taste, and we can take them or leave them, depending on the guitar"
- www.fralinpickups.com/2017/03/03/volume-tone-pots-101/

www.seymourduncan.com/blog/the-tone-garage/250k-pots-versus-500k-pots-going-deeper-into-the-subject

that's a little bit of science, now for empirical experience:
my 5-way tele with 250k vol and 500k tone has a bridge straight to jack option. noticeably louder than any of the other settings (including neck and bridge in series) and brighter (can be icepick-y depending on amp settings). i like it. quite useful if I'm playing with the volume and tone turned down a bit - can be like stepping on a dirt box depending on amp setting

[newey]

I'll second what TragicHero said. The main thing you get with bypassing the pots is not more brightness, but more output. It will be a bit brighter, but mostly it's louder.

I wired my Hofner travel guitar with a bypass setting (The wiring is essentially like the original Esquire wiring you mentioned, with one switch position having a capacitor for a bass-y tone). Here's the thread:

Hofner Travel Guitar mod thread

Unfortunately, the sound clips I posted therein (now 7 years ago) have gotten lost in the ether, as they resided on a cloud account that I no longer have access to. But I like the bypass setting;it's useful for going quickly from a rhythm setting to an instant lead sound with the flick of a switch.

That's why it's often called a "solo switch". We've also had it called a "blower switch" (I guess, from the blower on a dragster). I hadn't heard "passing lane" before.

With a 'direct through' mod (aka 'passing lane' mod?), whereby the pots/controls are being bypassed and the pickups thus going directly to the output, why does this make the pickups sound brighter as opposed to darker? Shouldn't the lower resistance - due to the resistance of the values of the post being removed from the circuit - make the pickups sound darker? I don't understand

.

That's a good question, and one that I am poorly qualified to answer- One of our learned electrical gurus will come along with a more-technically-correct explanation.
But (I believe) that you've basically got it backwards- when you remove the pots, you increase the resistance. This is because you pickups and the pots are all wired in parallel, and it's the total resistance of the entire circuit that matters, not any one component's resistance.

Parallel resistances are calculated as a sum of the inverses of the various values- which means that the total resistance of the circuit decreases as you add more components in parallel.

Let's take an example. Say you have a Strat SC pickup of a typical 5KΩ resistance value. You have 2 500KΩ pots. We'll drop the "K" from the calculations to make the maths easier. Call the 3 as 5, 500, and 500.

The total resistance ("R_t") is given by the formula 1/R_t= 1/R₁ + 1/R₂ + 1/R₃. Plugging in our values, we get 1/R_t = 1/5 + 1/500 + 1/500. Converting to decimals gives us 0.2 + .002 + .002 = 0.204Ω, but this is equal to 1/R_t. So, solving for R_t, we get R_t= 1/0.204, so Rt = 4.901Ω (We can of course multiply by 1000 to add the "K" back in).

But, recall that our theoretical Strat pickup was 5KΩ. If we take the V and T pots out of the circuit, the total resistance is just the resistance of the pickup, i.e., 5K, whereas with the pots it was 4.901K.

In a parallel circuit, the total resistance must always be less than the resistance of any one of the components.

(At least, I think that's the explanation.)

[sumgai]

newey has it absolutely correct. If you didn't know that he never went to electronics school, you'd think he was a genius!

I think this is gonna have to earn my first ever LIKE.





sumgai

[Ro_S]

Thankyou

[newey]

Thanks, sg. But when I joined this outfit about 11 years ago I could never have known, much less posted, about that. It is only because of other members who were willing to share their knowledge (and sumgai, you were certainly one, along with many others) that I learned a little of this stuff.

I'm just passing the baton, really. . .

[sumgai]

^^^^ Better for the environment than passing gas.....

[b4nj0]

The parallel value is always lower than the lowest resistor value too, but I guess that's implicit in the reciprocal calculation.

e&oe...