No Load Pot for Volume?

[loudriver23]

I've done a few searches and am pretty sold on the No Load for the tone controls. But I didn't see anything about anyone using one for the Volume control.
I'm talking Strat here.
Does anyone use a No Load for the Volume, or is this mod best for tone controls only.
Thanks,
Louis

 

[Lewguitar]

Re: No Load Pot for Volume?

A no load pot would break the connection and give you dead silence when turned to "10". Pretty useless.

 

[loudriver23]

Re: No Load Pot for Volume?

WOW...
My lack of electronics is astounding!!
Thanks for the clarification Lew.

 

[MikeRocker]

Re: No Load Pot for Volume?

I just learned this the hard way with the new Tele I built - a No-Load wired in as a volume pot makes a hell of a good on/off switch! :laugh2:

 

[MattPete]

Re: No Load Pot for Volume?

A no load pot would break the connection and give you dead silence when turned to "10". Pretty useless.

Huh? I have a no-load volume on my strat, and when I turn it from 9 to 10, I get a huge jump in output.

 

[Artie]

Re: No Load Pot for Volume?

Huh? I have a no-load volume on my strat, and when I turn it from 9 to 10, I get a huge jump in output.

You must have something wired wrong. A "no-load" volume control can't work.

 

[MattPete]

Re: No Load Pot for Volume?

Hmmmm....are you sure?

So, a tone pot works by resisting the highs from going to ground (the cap acting as a high-pass filter). So, if the tone is on 9 (let's say 240k), there is 240k of resistance. If it is clicked to 10, there in infinite resistance and none of the highs bleed to ground.

A volume pot works the same way (I think). It resists the entire signal from bleeding to ground. So, when it is on 9, there is 240k worth of resistance, and only a little bit goes to ground. In contrast, when the volume is on 0, there is no resistance, and all of the signal goes to ground. If it is turned to 10, there is infinite resistance, and none of the signal goes to ground -- it's as if you've bypassed the volume pot altogether.

 

[SpiderVenom]

Re: No Load Pot for Volume?

Hmmmm....are you sure?

So, a tone pot works by resisting the highs from going to ground (the cap acting as a high-pass filter). So, if the tone is on 9 (let's say 240k), there is 240k of resistance. If it is clicked to 10, there in infinite resistance and none of the highs bleed to ground.

Generally, a tone pot is wired up by running an additional hot wire to it, so that it can bleed some highs off to ground via the capacitor. With a no load pot, clicking to 10 disconnects the wiper, and hence no signal is bled off. However, because the signal doesn't pass through the tone pot, all you've done is taken the pot out of the chain without holding it up. A rough explanation, but I think that's it. Sort of :laugh2:

Futhermore, wouldn't infinite resistence, theoretically, mean no signal makes it through?

 

[MikeRocker]

Re: No Load Pot for Volume?

Well, when I had mine in it worked as an on/off switch. Any setting below the "10" detent resulted in total silence. When I swapped the No-Load for a regular 250K, it worked perfectly, as a volume control is supposed to.

 

[Artie]

Re: No Load Pot for Volume?

The thing is, a volume pot is wired as a voltage divider, (3 terminals). A tone pot is wired as a rheostat, (2 terminals). So the no-load works for the tone control. If the volume pot has the cut at the "top", you'll have broken the connection to the signal the moment you turn off of the "10" mark. If the broken connection is at the bottom, you'll simply have the the resistance of the pot in series with the signal, acting as a current limiter. (Not a voltage divider.) Since a pickup generates so little current to begin with, even a 500k resistor will barely have any audible affect on the signal. Maybe a little, but not much.

 

[MattPete]

Re: No Load Pot for Volume?

Generally, a tone pot is wired up by running an additional hot wire to it, so that it can bleed some highs off to ground via the capacitor. With a no load pot, clicking to 10 disconnects the wiper, and hence no signal is bled off. However, because the signal doesn't pass through the tone pot, all you've done is taken the pot out of the chain without holding it up. A rough explanation, but I think that's it. Sort of :laugh2:

Futhermore, wouldn't infinite resistence, theoretically, mean no signal makes it through?

In this case (at least with my [perhaps naiive] logic), is that infinite resistance means that no signal makes it through to ground, which means that 100% of the signal makes it out of the guitar.

 

[MattPete]

Re: No Load Pot for Volume?

I believe this is the final wiring diagram that I used for my strat. The only difference, that I am aware of, is that I have a small resistor in series with the capacitor on the volume pot.

 

[Artie]

Re: No Load Pot for Volume?

I believe this is the final wiring diagram that I used for my strat. The only difference, that I am aware of, is that I have a small resistor in series with the capacitor on the volume pot.

Exactly. You have three connections on the volume pot. Hot input on the right, output in the middle, and ground on the left. Thats how a volume control works. Are you saying that you used a no-load pot for that volume pot?

 

[MattPete]

Re: No Load Pot for Volume?

Are you saying that you used a no-load pot for that volume pot?

Yup.

Second one from the bottom:

http://www.guitarelectronics.com/category/20electronicparts.2controlpots

 

[Artie]

Re: No Load Pot for Volume?

I just doodled this up on paper. The resistor you have in series with the cap is what's saving you. You have created a very weird volume pot. Here's what's happening: (Lets assume you used a 220k resistor for your treble bleed. Heretoafter refered to as the TBR.)

On 10, you truly have a no-load pot. There's no volume pot at all, and even that cap and resistor are out of the circuit. I imagine that this is a bit bright.
When you roll the volume down to, say, 8, you have the TBR on the top half and the value of the volume pot on the bottom half. Your volume should drop to approximately half. As you continue to turn thr pot down, you get less resistance to ground, but the top half remains at the TBR value. In other words, your total effective volume pot value decreases, with a subsequent slight change in the tone. The same change that you'ld get from using different pot values.

Does that make any sense?

 

[MattPete]

Re: No Load Pot for Volume?

Does that make any sense?

Er, no. To be honest, I don't understand the concept load, and I have a naiive hydraulic model of electricity.

Here's my [naiive?] theory of how it works. First, let's pretend it's a plain-ol 250k volume pot:

1) When the pot is on 10, most of the signal to ground lug is blocked, and hence most of the signal passes out to the middle lug. The TBR essentially has no effect.

2) When the pot is on 2, most of the signal to the middle lug is blocked, and hence goes to ground. The TBR allows some of the highs to bypass the tone pot and go to the middle lug. The resistor in series with the cap attentuates the amount of highs that are allowed to flow through the TBR and bypass the volume pot.


So, now let's pretend it's a no-load 250k volume pot:

3) When the pot is on 10, all of the signal is blocked from going to the ground lug (infinite resistance), and all of the signal passes to the middle lug. The TBR has no effect at all.



So, why does the TBR save me?

 

[Artie]

Re: No Load Pot for Volume?

Give me just a sec . . . this calls for a diagram.

 

[Artie]

Re: No Load Pot for Volume?

Ok, I tried to keep this simple.

Fig 1. Normal volume control at halfway.
Fig 2. Electrical equivalent of Fig 1.

Fig 3. No-load volume control at halfway.
Fig 4. Electrical equivalent of Fig 3.

Fig 5. No-load volume pot at halfway with treble-bleed resistor.
Fig 6. Electrical equivalent of Fig 5.

Here's what you notice:

In Fig 3 and 4, the hot input, (from the pup), is completely disconnected. Ergo, a no-load volume pot simply won't work at all once you roll back from "10".

Note that Fig 6 is exactly the same as Fig 2. This is why you're "saved".
But, there's a catch. In Fig 2, as you roll the volume down, the upper resistor increases in direct proportion as the lower resistor decreases. However, in Fig 6, the upper resistor always equals the value of your treble-bleed resistor while the lower resistor decreases. If we assume the TBR and the pot to be 250k, then at "10" the pup "sees" a 500k load. At "0", it sees a 250k load. And of course, it varies as the volume pot is placed at any midway point.

Thats the difference between a normal volume pot and yours. You are effectively defeating the purpose of the treble-bleed mod, and you have that volume drop as you roll off of "10".

Does that help?

Artie

btw - I left out the treble-bleed cap for clarity. Its not really relevant to this problem.

 

[Chaos]

Re: No Load Pot for Volume?

Er, no. To be honest, I don't understand the concept load, and I have a naiive hydraulic model of electricity.

Here's my [naiive?] theory of how it works. First, let's pretend it's a plain-ol 250k volume pot:

1) When the pot is on 10, most of the signal to ground lug is blocked, and hence most of the signal passes out to the middle lug. The TBR essentially has no effect.

2) When the pot is on 2, most of the signal to the middle lug is blocked, and hence goes to ground. The TBR allows some of the highs to bypass the tone pot and go to the middle lug. The resistor in series with the cap attentuates the amount of highs that are allowed to flow through the TBR and bypass the volume pot.


So, now let's pretend it's a no-load 250k volume pot:

3) When the pot is on 10, all of the signal is blocked from going to the ground lug (infinite resistance), and all of the signal passes to the middle lug. The TBR has no effect at all.



So, why does the TBR save me?

Using your ‘hydraulic model’, you’ve drawn some incorrect conclusions.

Think of the pot as a ‘bleeder valve’ on a pipe.
100% of the flow goes through the pipe when the bleeder is fully shut (infinite resistance)
The resistance of the pipe doesn’t matter. If the flow isn’t bled off, all of the flow is going though the pipe (just like hydraulics)

The more you bleed off (lower resistance) less is available in the pipe.

A no-load pot doesn’t remove the bleeder from the circuit, it removes the pipe from the circuit.

 

[MattPete]

Re: No Load Pot for Volume?

I'm still trying to digest all of this, but in the meantime here are a few questions:

(1) Are you saying that my no-load vol pot is electrically different from a regular 250k vol pot when both are on 9?

(2) Does Figure 3 imply that all of my output is passing through the TBR and none through the pot? If that is the case, that would suggest that all I'm getting in treble-bleed?

(3) Is a 500k vol pot set at 250k electrically different from a 250k pot set at 250k?