can someone tell me how switched jacks work

Started by southtown, January 18, 2007, 07:18:50 AM

Previous topic - Next topic

southtown

like this one here, http://www.banzaieffects.com/1-4-cliff-jack-mono-black-nut-lugs-pr-19690.html i want to make a effect loop in my pedal which when nothing is plugged in it is ignored but if a pedal is lugged in then it will go through the pedal

Thomas P.

Those switches are close. They open when you plug something in simply by lifting the contacts.
The pins on the right have contact to the plug those on the left don't (regarding the photo)!
god said...
∇ ⋅ D = ρ
∇ x E = - ∂B/∂t
∇ ⋅ B = 0
∇ x H = ∂D/∂t + j
...and then there was light

southtown

#2
so let me get this straight, im going to try some ascii here 
-o-------o-----
|
-o-------o-----

if we label the contacts A B C D working from the top left clockwise.

When no jack in inserted there is a connection between A and D, B and C are just empty with nothing on them,
When a jack is inserted The connection between A and D is broken, now its a regular mono switch using just B and C?

or is it D and C are always connected

Thomas P.

If we look from above and the jack is inserted from the right.

--A----B---
|                        <=jack   
|
--D----C---

No! When no jack is inserted then A is connected to D and B to C!
When a jack is plugged in then both connections are open (which means there is no connection between either one of the pins!).
god said...
∇ ⋅ D = ρ
∇ x E = - ∂B/∂t
∇ ⋅ B = 0
∇ x H = ∂D/∂t + j
...and then there was light

southtown

ok and then does it work like AD is tip and BC is sleeve?

idlechatterbox



The Wiki on this site has a fairly good discussion of jacks (and switches). The beginner project also has some really good visuals that help one understand not just how jacks work (that's the easy part in some ways), but how they work in the circuit. Some of the questions that you ask are answered in that thread. It's very informative, and not just for switches.

I guess there's no one way to learn that's appropriate for everyone. But for my money, trying to learn how jacks work through the use of text or ASCII symbols is not the most effective. Again, just my 2 cents, but a better approach would be to read up in the wiki or other websites (there was even a thread a couple days ago about switching jacks). Then, once the reading's done, study some schematics, and grab some jacks (I included some in a family photo above) and a cable. Insert the cable end, and actually look at what happens. Trace the contacts and that gives a really good sense of how they function. Sometimes the light bulb goes on when you actually see how the sleeve of the jack makes contact, or doesn't make contact.

I would offer the same advice for switches. Looking at them in schematics, and reading about them, is only the first step. If you really want to understand them, grab one, get some alligator clips and/or a meter, and start messing around.

While jacks and switches can be sort of confusing at first, look on the bright side: you can stare at a capacitor all day if you want, and that won't tell you squat about how IT works.  :icon_razz:

Good luck!  ;D

southtown

i understand how basic jacks and i know how switches work, this one just confuses me, is it it just an enclosed version of this?


Thomas P.

Quote from: southtown on January 18, 2007, 08:21:47 AM
ok and then does it work like AD is tip and BC is sleeve?

No! Only A is the tip and only B is the sleve. And yes, it is basically an eclosed version of this open jack but it has a switch for the sleve, too.
god said...
∇ ⋅ D = ρ
∇ x E = - ∂B/∂t
∇ ⋅ B = 0
∇ x H = ∂D/∂t + j
...and then there was light

southtown


Thomas P.

god said...
∇ ⋅ D = ρ
∇ x E = - ∂B/∂t
∇ ⋅ B = 0
∇ x H = ∂D/∂t + j
...and then there was light

southtown


Thomas P.

an error occured so i deleted the double post...
god said...
∇ ⋅ D = ρ
∇ x E = - ∂B/∂t
∇ ⋅ B = 0
∇ x H = ∂D/∂t + j
...and then there was light