Wiring Up a 1590B |

Wiring up the off-board components, particularly a 3PDT switch, can be confusing. There are many good explanations of how to do it (see for example geofex.com, tonepad.com, and generalguitargadgets.com) and this one adds to the pile by breaking down the logic of one of the more elegant layouts with a series of detailed figures. I walk through the case where the jacks are not insulated from a metal enclosure, there is an LED to show when the circuit is on, and when the circuit is off there is simple by-passing with the circuit input grounded.
It seems simplest to start with the wiring for so-called true by-pass, a straight connection from the input jack to the output jack. The input jack is on the left and the output jack is on the right. They will be reversed to the usual placement when the stompbox is closed up and turned over.

The input jack pictured here is a stereo Switchcraft jack called the 12B. The output jack is a mono Switchcraft called the 11. You can read a post about these jacks in 1/4″ Phone Jacks and Plugs. The switch is a 3PDT Taiwan Blue. The picture shows
- a white wire from the tip lug of the stereo jack (on the right, for input) to the switch;
- in one switch position, this will connect to the white wire that goes across the bottom of the face of the switch;
- which will then connect to the white wire that runs from the switch to the tip lug of the mono jack (on the left, for output).
So we have a simple connection from one tip lug to the other: true by-pass.
In the other switch position, we want to connect the input lug to the input of the circuit and the output lug to the output of the circuit. These are added in the following image:

The blue wire should be connected to the input pad of the circuit board. The yellow wire should be connected to the output pad of the circuit board, or to the middle lug of a level pot if one completes the circuit. I am leaving the board out of these pictures for simplicity. Any unconnected wires in this tutorial are supposed to connect somewhere on a circuit board.
As it stands, we do not need the 3PDT switch because we are only using 2 poles, one for input switching and one for output switching. The third (middle) pole can do the switching for an LED that lights up when the circuit is engaged (or not by-passed). For that we will also need a power supply, which we will get from a DC voltage supply.

This image shows the LED wiring from another vantage point, with the wires for the guitar signal removed for clarity. The LED switching opens and closes the ground connection for the LED circuit. The ground for the circuit is the sleeve lug of the output jack.
- The DC jack has a direct wire to that lug: the green wire that runs all the way across the middle of the picture.
- The LED also connects to that grounded lug through a resistor (2.2K for limiting current) and two green wires, one that goes from the LED to the switch and another that goes from the switch to the lug. Note that these two wires are connected when the switch is in the “not by-passed” position.
Also note that in this particular setup, the sleeve lugs of both jacks are connected through the aluminum enclosure that holds them. The entire stompbox is grounded through the output cable. So the input sleeve lug is grounded by its connection to the output sleeve lug through the enclosure.
In setups with insulated jacks, one must make these connections with wiring. You should still ground the enclosure in those cases because this improves the ability of the enclosure to shield the circuit from outside radio frequency (RF) signals.
Here is a close-up view of the DC jack connections:

The top lug is the positive power supply connection and the angled lower lug is the ground connection. We will use the third middle lug later when we hook up a 9V battery as an alternative power supply.
The red wire is the positive power supply. This colour is consistent with the leads found on most 9V battery snaps: red is positive and black is negative (or ground). In these pictures, I am using green for ground because it shows up better.
I prefer to use the DC jacks that are fastened with a nut on the outside of the enclosure and that is what is pictured in these figures. DC jacks also come configured with the nut on the inside of the enclosure. I find this inconvenient because it requires me to install the jack in the enclosure before I solder the wires to it. As a result, if I want to remove the circuit from the enclosure then I must unsolder these wires. The input and output jacks, the 3PDT switch, and any pots all have their nuts on the outside. If the DC jack does also, then one can remove the nuts and the whole circuit lifts out of the enclosure completely connected.
Here is a close-up view of the output sleeve lug for ground connections:

At this point, two wires are supposed to soldered to this lug. One wire is coming from the DC jack (not shown) above. The other wire is “flying in” from the switch.
Here is an image of the switch wiring with all of the wires in place: input, output, and ground connections:

Note that there is an additional green wire. This is the short wire that connects the first (input) pole of the switch to the middle (ground) pole of the switch.
- When the switch is in the by-pass position, this short green wire connects the input of the PCB to ground.
- When the switch is in the engaged position, this short green wire does not connect to anything.
This additional wire feeds the stompbox circuit a quiet input signal when the circuit is by-passed. That is the trickiest part of the switch wiring, making a nice use of that otherwise unused lug on the switch.
Now we are ready for the circuit board. Besides the input and output connections, the board needs the positive and ground connections which come from the same places as for the LED circuit: the positive lug of the DC jack and the sleeve lug of the mono output jack.

Wires for those connections appear in the figure above. Generally, it is good to run your positive supply lines next to ground lines. This is true of PCB traces as well. So I am showing the positive board supply wire running next to the ground wire for the DC jack.
I am not quite finished. I still need to add the wiring for the battery. It will take a while to make a figure for that, but in the mean time it is easy to describe. The red battery snap wire connnects to the remaining free lug on the DC jack. The black battery snap wire connects to the ring lug of the (stereo) input jack. That’s the obvious one facing up to the right of the tip connector. This battery snap hookup accomplishes two things:
- The battery negative terminal will be connected to ground only when there is a mono plug inserted into the input jack. In that event, the ring connector is in contact with the sleeve of the mono plug and a ground connection is made through the ground lug of the input jack.
- The battery positive terminal will be disconnected to the LED and the board when there is DC plug inserted into the DC jack.
Because of these properties, the battery will supply power when there is no alternative DC power supply and there is an input for the stompbox. Otherwise, the battery is preserved.
dfx said:
cool guide and visual instructions, bar none….. what program did you use for the 3d pics?
Posted 29.05.2008 at 9:29 pm
gaussmarkov said:
dfx,
oops! i forgot to say. those are all made with POV-Ray, a ray tracer that has been around since forever. there’s a nice windows wrapper for it now.
thanks for the note!
paul
Posted 29.05.2008 at 9:39 pm
gaussmarkov: diy fx » 1/4″ Phone Jacks and Plugs said:
[…] off this post about 1/4″ phone jacks commonly used in stompboxes. This tutorial supplements a recent post on wiring up a stompbox with off-board […]
Posted 04.06.2008 at 9:14 am
Keith VanDen Heuvel said:
I really like your site and the information contained within it. I am also impressed that you used POV-RAY to create all of your images. It must have taken some time to come up with the scripts to create the various components that you have used. Is there a POV-RAY library of componenets like jacks and switches that you were able to utilize? If so, I would be interested in seeing it. In the meantime I will be doing a bit of searching in an attempt to locate any. Keep the great work coming!
Posted 10.06.2008 at 9:33 am
gaussmarkov said:
hi Keith,
thank you for your compliment. the resistor and the LED came out of the eagle3d library. see www.matwei.de. the rest i made myself.
eventually, i will make these available. i have to clean up the code a bit and i want to co-ordinate them with eagle3d.
all the best, paul
Posted 10.06.2008 at 9:48 am
Manny M said:
I have been cutting off the ring, tip and sleeve tabs off the 1/4 plugs. I then solder the corrsesponding wires to the tip snap, the sleeve snap and just use the ground tabs for ground. Is this good practice or not. I seem to get mixed up with the ring tab and solder the neg battery wire to the inout sleeve snap.
The 3-D work you shoed here is nothing short of outstanding work. I tried to do tis with Autodesk Inventor but takes a long time to do.
Great work Guasmarkov.
Manny
Posted 17.06.2008 at 8:10 pm
gaussmarkov said:
hey Manny,
thank you kindly.
i see nothing wrong with cutting off the tabs and soldering to the snaps. in really tight situations, the tabs get in the way and one can cut them off or bend them to suit.
solder on, gm
Posted 18.06.2008 at 6:45 am
Mick said:
Hey GM,
I LOVE this site!
I was looking over the diagram above and noticed that the LED resistor was on the grounded side of the LED, not the power side.
I’m not sure that it will accomplish it’s current limiting purpose there, but I am also not 100% sure either. I just noticed it and thought I’d point it out just in case.
Many Thanks for an awesome site!
Mick
Posted 11.07.2008 at 7:45 pm
gaussmarkov said:
hi Mick,
thanks for the props, man! glad you like it.
as far as the current limiting resistor goes, it does not matter which side of the LED you are on. you can think of a diode as a resistor, just not a fixed one. resistors and capacitors in series give the same current and voltage across the series no matter what the order of the components is. so either way, LED first or resistor first, you get the same current and the current limiting resistor does its job.
thanks for asking! gm
Posted 12.07.2008 at 4:18 am
stef said:
NICE NICE NICE! Very clear and sweet renders! thanks a lot man, love your blog.
Posted 12.07.2008 at 3:15 pm
Mick said:
Hey GM,
I never thought of a diode in those terms… but it makes sense…
Thanks for the reply and again, Many Thanks for a wonderful site! Keep up the good work!
Mick
Posted 19.07.2008 at 7:17 pm
gaussmarkov said:
i should probably add that my explanation only works for DC analysis. the variability in the resistance of the diode means that order does matter with AC.
Posted 20.07.2008 at 4:11 am
Criss said:
By far the best ive ever seen . I realy think this site is easy one of the best .
Thanx for all your efforts
Posted 22.08.2008 at 10:42 am
Dan said:
I’m glad someone else asked about the current limiting resistor on the LED - I had always wondered about that myself. Since we’re asking questions - I notice in this tutorial - that you don’t use the ring or sleeve connection on the input jack. Have they been left the diagram for clarity or do you not use them ? Thanks Again !
Dan
Posted 23.09.2008 at 7:19 pm