# Simulate Pot Load Using DPDT?

Discussion in 'Accessories / Connections' started by Guitarfreak, Apr 22, 2010.

1. ### GuitarfreakWell-Known Member

A friend of mine wants to build a guitar with one pickup and no knobs at all and I told him that pickups don't produce current without a load present. This got me thinking though, he could simulate a resistive load using a DPDT switch to activate the load full on/full off couldn't he?

Using this as a diagram:

He could, say wire the pickup hot to the B and pickup cold to the A, then run the hot output from D and cold output from C. And using a 500k resistor, he would put it.... umm, between B and A? And putting the switch to the other side would activate the terminals with nothing attached, effectively turning the guitar off.

OR, he could do something cool like jumper B to F and A to E and put a 250K between E and F to simulate a different pot load with the switch in the opposite position. Or simulate a 500k pot turned halfway down by putting a 250K between F and E AND another 250K between F and D. This is a real brain teaser, but I'm convinced that it could be done. Right?

2. ### jg49Well-Known Member

Not sure I am folllowing your logic here but a single pickup no knobs wired directly to a jack will produce sound if plugged into an amp. Its just that all your tone, volume, etc will be controllable only at the amp.

3. ### BoswellModeratorDistinguished Member

You are talking about a passive wound pickup on an electric guitar, not piezo or other types on an acoustic guitar, right?

The pickup acts as a voltage source, so it's correct that no current flows without a load. However, the amplifier input circuit constitutes a load, and so current flows into that. No parallel load resistor is needed, unless you want to experiment with difference in tone by loading the pickup differently.

As for wiring the switch, I would not put any switch contacts in the ground (cold) lead. Leave the pickup cold permanently connected to the output cold, but take it to pin E. If you now take the hot and output both to pin C, then connect your experimental resistor between A and E (cold), you have normal unloaded output with the switch in the centre, loaded output when pushed towards the A direction, and muted (short circuit) in the other direction.

4. ### GuitarfreakWell-Known Member

I stand corrected. See, I just thought that by separating the inductor and the load like that would result in a lot of treble bleed because of the capacitance of the cable. Boswell, your example of wiring is simple, yet confusing. You'd have both the hot and hot output connected to lug D correct? You said to wire the pickup cold and cold output to lug C, but then you said to move pickup the cold to E, why is that? Also, how would the circuit shut off in the third position with the resistor jumpering A and E? In my mind that would make an active connection.

5. ### JeemyWell-Known Member

Um why does he want to do this out of interest? I presume he believes he'll get better tone with less in the signal chain?? But then your idea of a switch negates that anyway. IMHO any benefit to tone would be immediately eradicated by the loss of control that a volume switch gives, not just in simple and essential player control of volume which is instrumental to using a nice amp properly, but in terms of actually being able to silence the guitar in live use, which can be very unprofessional if its left to hiss, squeak and feedback between songs. If this is intended to be done with a pedal again I'd say this negates the tone benefit of removing the vol control in the first place. But then I don't know why he wants to do this.....

6. ### BoswellModeratorDistinguished Member

Not quite. The basic connection is for the pickup wires always to go to the output, but the switch can either put a resistor across them or a short for muting. So the hot (pickup and out) goes to C, the cold (pickup and out) to E and the resistor between A and E. I'm assuming your switch is a centre-off type.

7. ### Scott GriffinGuest

Dollars to donuts this is the correct answer. Growing number of guitarists out there that want to get as much pure wood tone as possible.

8. ### djmukilteoWell-Known Member

guitarfreak:
Check out some of the info on this guitar wiring site and it may help you in your quest!

9. ### GuitarfreakWell-Known Member

Thanks for the link, it's good for a basic understanding, but I have the feeling that a few of the major points were dumbed down for transmission to the masses. Still, a DPDT is a little more complex than a resistor or variable resistor circuit.

10. ### djmukilteoWell-Known Member

guitarfreak:
I was trying to show you what a guitar pickup circuit consists of....it's not complex,, it's quite simple...it's just a pickup....
Every guitar pickup circuit consists of those components....a "switch" regardless if it's a SPST, SPDT, DPST, DPDT or 8P8T does nothing more in any circuit than "switch" one thing to another...a switch is not complex at all..in fact it's not a component or active electronic device at all...it's a switch...it's not a "load"....it has zero resistance (hopefully)...and FWIW resistors, capacitors and potentiometers are also not complex components...

11. ### GuitarfreakWell-Known Member

Oh dj, I know what you meant by it. I think (to my knowledge, whatever that's worth) that I have a fairly good understanding of the basics, and I didn't mean to sound like a dick about it.

Boswell, I know you told me how to wire it, but frankly I don't yet understand why this setup wouldn't work. Examine this and let me know what's wrong with it so that I can learn from it. What I am trying to do is have first position simulate a 500k pot full up, pos two be off, and position three simulate a 500k pot turned down halfway. AFAIK, with no firsthand, this should work. And I even took your advice to keep the pickup cold with the output cold, but I kept the pickup hot to the sides because a. I'm thickheaded, and b. I think it makes more sense to me as far as what I plan to accomplish with the setup. Anyway, what does and doesn't work about it? I am thinking that moving the 2 and 6 connections both to 4 that it might be better?

12. ### BoswellModeratorDistinguished Member

Well, several things here. Firstly, this implements a different specification to the one we were talking about (or, at least, I understood you to be talking about). Secondly, the output is always in series with a 250K resistor, and so the amplitude of the output and the tone will be critically dependent on the input impedance of the following amplifier and the capacitance of the lead. Thirdly, the DPDT on-off-on switches I am familiar with all have the centre position as no connection to anything, so there will be no muting action by shorting the output to ground.

There is no possible wiring of an on-off-on switch to achieve all these specs, but you can get near them. I can't edit your admirable diagram, so I will have to describe the changes: Input to pin 1, output from 3 and cold to 4 as before. 250K from 1 to 5, 250K from 3 to 6 and 250K from 6 to 4. This gives full signal in position one, half signal in position three and no signal in position two. There is 500K from input to ground at all switch positions and also 500K from output to ground in position two. The output impedance of the switch arrangement is whatever value the pickup has in position one and around 125K in position three, minimizing the sensitivity to amplifier input and lead capacitance.

13. ### MrEaseActive Member

This circuit will not work as you seem to think.

With the switch in the upper position (1 connected to 3 and 2 to 4) you will add a 500k load in parallel with the amp input. This will look like a 500k pot turned up full.

In the mid position (no connections between switch contacts) you have 250k (3 to 5) in series with the amplifier load. This will slightly reduce volume depending on the amp input impedance. It most certainly will not mute.

In the lower position (5 connected to 3 and 6 to 4) you will have a 250k load in parallel with the amp input. This is not the same as a 500k pot turned down half way.

What you need is 2 * 250k resistors (the 500k is not needed). All cold connections will be commoned using pin 4. Referring just to the "hot" ends, connect the pick up, as now, to pin 1. Connect one 250k between pins 1 and 5 and the other between pins 5 and 4. Connect the output to pin 3. This will work as you anticipate but also be aware that the lower position will not be identical to a 500k pot at halfway, as the 500k pots normally used are log taper and not linear. To simulate the "halfway" position better then use, say, 390k between 1 and 5 and a 120k between 5 and 4. Although this is akin to a 510k pot, the difference will have no discernable effect.

A further note is that the centre off position does not actually short circuit the output as a pot would, it just leaves it open so you may get some breakthrough in the off position. In this case it may be worth adding a 1M between pins 3 and 4 to reduce any breakthrough. There is no easy way round this with this type of switch as the centre position has no contacts made. Using a Telecaster type three position switch would allow you to include a shorted position and also have the "halfway" point in the mid position.

EDIT: Boswell posted in while I was writing! His solution is better for the "off" position although you may still get some breakthrough. If you wanted to adjust the "halfway volume" on his circuit, make resistors from 1-5 and 6-4 390k and the resistor from 3-6 should be 120k.

14. ### GuitarfreakWell-Known Member

OH, I think I see what I did now. What I was trying to do with those 250k's was split the signal path between parallel and series for a total load of 500k in the third position simulating pot half down. Apparently, that jumper wire makes it so that in positions 1 and 3 all 6 lugs are ON, negating the idea completely. Right?

15. ### djmukilteoWell-Known Member

Guitarfreak:
I didn't think you were a dick at all....I was just trying to help you understand how a pickup circuit works and how a switch works....sorry if any of that discussion came out in a negative way!
Listen to Boswell....I think he's got your idea in hand....

16. ### MrEaseActive Member

Still not quite right I think. It occurs to me that you may have got the switch functions mixed up. There are two totally seperate switches here (Double pole) with a centre off position. In the top position the connections are 1 to 3 and 2 to 4 with no connection between 3 and 4. Pins 5 and 6 are isolated. In the centre position ALL the contacts are isolated. In the lower position the connections are 3 to 5 and 4 to 6 again with no connection between 3 and 4. In this case pins 1 and 2 are isolated. So with your jumper (using your diagram) then in switch positions 1 and 3 then pins 1, 3 and 5 will be shorted and not all 6.

With a 500k pot there is always 500k loading the pick-up and the wiper merely picks up the required output voltage so there is no need to try and create series/parallel function with the switch. Of course with a pot, when you turn the output right down, the output is shorted. With the switch you have, you cannot reproduce this as both Boswell and I have pointed out.

17. ### djmukilteoWell-Known Member

I don't think I want to even try and work that diagram out....GF
The best way to do things like this is to draw a schematic with the switch and the resistors, ins and outs etc. and then draw a diagram like that to assemble it...
You end up with a better understanding if you do the schematic first (easier to trace out the paths) and then the illustrated diagram of how to solder things together after the schematic has been flushed out...

18. ### GuitarfreakWell-Known Member

Yeah, I think like MrEase said I was getting the connections wrong. I'm going to map out what Boswell said and see if I can come to an understanding from that.

19. ### djmukilteoWell-Known Member

DPDT
Double pole, double throw.
Equivalent to two SPDT switches controlled by a single mechanism:
"Center Off" are not that common, but I don't know if that's what you have there...

20. ### MrEaseActive Member

Just a small point if you are trying to understand Boswell's explanation. His statement

is not quite right. In fact the 250k is only in series in the centre position as in the up and down positions, the switch shorts out the 250k. Hey we all make mistakes, even mod's!