How do you wire both these things into one circuit so when there is an adapter plugged in the battery does not get used? I only use DC power myself so I never needed to learn it but a friend wants a pedal with both power options available. I'm using this DC jack;


That jack has three connectors, the top is 9v, the left is 0v but the middle is unused for just DC power so I assume that's what is used for this? I also want the power to be switched on when a lead is plugged in the input so please add that information in your explanation.

Dylan

Yes, you hook your battery's + terminal to that one in the middle. You then hook your battery's negative terminal to the ring terminal of a stereo jack to turn off battery when unplugged.

RDV

Ok, battery +V to middle terminal, then wire the +V connector of the DC jack to the +V of the board as normal? Then battery black goes to the ring of the input jack, which will implement power switching. After that I just ground the sleeves of the jacks together as normal?

Quote from: DylanOk, battery +V to middle terminal, then wire the +V connector of the DC jack to the +V of the board as normal? Then battery black goes to the ring of the input jack, which will implement power switching. After that I just ground the sleeves of the jacks together as normal?

YES.


Regards

RDV

Dylan,

For a really good visual of the connections check this out:

http://www.generalguitargadgets.com/v2/diagrams/ge_boost_lo_amz_mosfet.gif

It should clear up any questions that you might have when you warm up the soldering iron.

Derek

Thanks everyone. I think this is one of the basics I should've learnt when I started on electronics a couple of years ago. I ran before I could walk really.  :oops:

The thing to remember about adaptor jacks is that there needs to be a way for there to be *some* default power source at all times.  "Closed-circuit" jacks provide just such a means.  

In a closed-circuit jack there are two contacts that are always in touch with each other unless one of them is moved out of the way.  If you look at a 1/4" or 1/8" closed-circuit phone jack, you will see that there there is a little bent spring-metal contact for the tip of the plug. (you can see one in the picture Rick Vance posted but this is actually the wrong jack)  That bent part nestles nicely in the little furrow between the tip of the plug and the shaft.  The reason it fits so nice and snug is that it is always leaning "forward", and you have to essentially push it out of the way to fully insert the plug.

In a closed-circuit version of this jack, you will also see a kind of little tongue-contact just in front of the contact for the tip of the plug.  When the plug is not inserted, the tip contact leans forward, as per usual, and makes contact with that little tongue in front of it.  When you insert the plug, though, the tip of the plug pushes the tip contact on the jack back just far enough to break the contact with the "little tongue".

Okay, let's attach power.  Wire the V+ from the battery to the solder lug going to the "little tongue", and wire the V+ lead from the board to the solder lug for the top contact on the jack.  With no plug inserted, the battery is connected to the board through the two jack contacts making contact.  Insert the plug, and the contact is broken.  The jack now takes its V+ feed from whatever is now touching the tip contact.  The battery, meanwhile is lifted out of circuit.

Bear in mind that the jack Rick showed is STILL important for enabling/disabling the battery when you plug your *guitar cable* into the pedal, but it has nothing to do with switching over from battery to adaptor.

Okay, back to your scenario....

The adaptor jack you show is for a barrel-style plug.  Since closed-circuit type jacks require that a physical connection/contact be broken by something being pushed out of the way by whatever comes into the plug, the question is "What does the pushing?"  In the case of phone plugs (which E-H and MXR used for a while, but which Boss/Roland rarely, if ever, used), it is the plug's tip which does the pushing.  In the case of barrel-style plugs, there IS no tip, so that cannot be used to do the deed.  Just as important,, in the case of phone plugs it HAS to be the tip since the rest of the plug and jack usually short out against the chassis and ground.  In the case of barrel-style plugs, the jack itself is frequently isolated from chassis ground, and the outside shaft/shank HAS to do the pushing to break the default (closed-circuit) contact.  For this reason, the overwhelming majority of pedals that use a battery OR adaptor with barrel-style plug will use the outside=V+/inside=V- convention.  The outside will HAVE to push the contact which carries the V+ to the board out of the way of its default connection with the battery.

So, how do you know what's what?  Simple.  Get your 3-contact barrel jack, set your DMM on audio beep for continuity testing, and without anything plugged into the jack, touch the probes to the jack lugs until you find two that beep (i.e., indicate continuity).

Now that you know which two are candidates, which is the one that gets "bumped out of the way"?  For this an unused barrel plug is useful.  Insert the plug into the jack, and look for continuity between the plug's solder lug for the outside, and one of the two designated contacts on the jack.  Which ever one continues to make contact with the plug during insertion is functionally equivalent to the tip contact of a phone jack, and the other solder lug is equivalent to the "little tongue" that gets pushed out of the way.

Quote from: D WagnerDylan,

For a really good visual of the connections check this out:

http://www.generalguitargadgets.com/v2/diagrams/ge_boost_lo_amz_mosfet.gif

It should clear up any questions that you might have when you warm up the soldering iron.

Derek

Erm...

About the above schematic... Wouldn't it be better if the -9V from the
battery would go to -9V of the power jack and than the power jack would
go to the input jack sleeve. The way it's wired now no guitar is necessary
to turn the effect on (led flashes anyways) while powering via an adapter.

Just my thoughts, please correct me if I'm wrong...
LM