Laying out DPDT as SP3T in Eagle

[iamjackslackof]

Hi all,

I'm trying to layout a circuit in Eagle where I have a DPDPT ON/ON/ON switch that switches an input between 3 outputs, as shown here. I've tested this with an actual DPDT switch and it works great, but I'm having trouble mapping the physical representation of the switch to the schematic for use in Eagle.

The DPDT schematic looks like this:


And the physical layout looks like this:


I have added the numbers to the diagram from the link above. Basically I'm trying to figure out which lugs in Eagle map to which numbered physical lugs. My guess is SW1_A is 1, 2, and 3, or the left side of the switch, and SW1_B is 4, 5, 6, or the right side, but I have seen different numbering schemes online. I could just get the board made and find out, but I'd rather save a step and seek the wisdom of this forum instead.

Thanks!

 

[BryGuy]

Basically like you said. Any DPDT is just 2 switches in a single package.

 

[Robert]

Is this what you're looking for?

DPDT ON/ON/ON as 3-way Switch - PedalPCB Wiki

wiki.pedalpcb.com

As far as determining which schematic pin maps to which PCB pad, that's going to depend on your specific component library.
I don't use Eagle, but mine look like this (in relation to the schematic at the link above)

In the case of an ON/ON/ON there are two different types that have to be taken into consideration, and you also have to be aware of which side of the PCB your switch is going to be installed.

 

[iamjackslackof]

Thanks @Robert , that's exactly what I'm trying to do, and I didn't know it was in the wiki! I'll have to start checking it more often.

Do you happen to know how to differentiate between the 2 types of ON/ON/ON DPDT switches? I also hadn't thought about the side of the board thing, thanks for the heads up.

 

[iamjackslackof]

Another question occurred to me as well. I can use this to switch one input to one of three different paths, but each path needs to connect to the same final output. It's basically input -> buffer A/no buffer/buffer B -> output. Is there a problem with just joining the output of each path as the final output? I'm wondering if the output of one path could "go backwards" back into one of the other paths.

Thanks!