Hello,

On many circuit designs the INPUT has three legs:

But the input socket I've got has only two legs, one of which I connect to ground. In the image above, the other leg would go down the path with the capacitor.
Where and how do I connect the battery...?

 

What is your input signal?

As drawn, the capacitor provides AC coupling to whatever in the plug is connected to that ring. And with the assumptions in the next paragraph, it blocks any DC.

That design MAY be meant to act as a DC switch in addition to the AC input mentioned above.. In other words, it can be used to turn the device on or off. For that to work with the jack configuration you show, the barrel and second ring from the tip need to be connected in the plug and the device and plug need a common ground. That seems unusual to me. When I have used a plug as a switch, I have done it differently.

On the other hand, the ring connected to the negative battery terminal may simply be providing access to that terminal to whatever is connected to the plug.

As for how to connect the battery, you need to show us the rest of the relevant circuit or decide whether you want the plug to act as a switch.

John

 

Hmm here is the image:

 

Where and how do I connect the battery...?

It depends on what the circuit is.

 

@Moderators
@ScottWang
@DickCappels

Why was my response to the other (earlier) version of this thread deleted? I then reported the later-posted duplicate thread, which is the thread you retained. Please find my post and restore it to this thread.

John

 

As I pointed out earlier, that configuration suggests that the barrel and second ring perform the function of an on/off switch. The first ring and capacitor block DC from either an input or output. If that function is not needed, then you connect the single ring or tip as if it were the second ring.

John

 

As I pointed out earlier, that configuration suggests that the barrel and second ring perform the function of an on/off switch. The first ring and capacitor block DC from either an input or output. If that function is not needed, then you connect the single ring or tip as if it were the second ring.

John

Hi John,

Hmm I don't really understand the tech linguo (what are the rings?), as I am a total beginner here... Could you re-explain this to me, like to a 5-year-old...?

Here is an example circuit:

I'm trying to build something like this on a breadboard. I've got two leads in the input socket: one goes to the ground, which is one of the side columns on the breadboard, and the other one goes in any one of the 5 pin rows, where the capacitor is also connected. Where does the '-' terminal of the battery go...?

 

@Moderators
@ScottWang
@DickCappels

Why was my response to the other (earlier) version of this thread deleted? I then reported the later-posted duplicate thread, which is the thread you retained. Please find my post and restore it to this thread.

John

Hi John:
Sorry about that, I already restored it to this thread on #2.
When I tried to delete the double post, I have had check there was no any other post yet, and I locked the thread fist and then I did the deleted action, I think you just post before I locked the thread, you just chosen at that moment to sent, so I even didn't know that if you didn't said it, it just like our mods tried to delete the same thread at the same time and one of mod will be slower and appeared error.

 

Hi John,

Hmm I don't really understand the tech linguo (what are the rings?), as I am a total beginner here... Could you re-explain this to me, like to a 5-year-old...?

Here is an example circuit:

Where does the '-' terminal of the battery go...?

It usually connects to the sleeve. Don't worry about the rings if yours doesn't have one

 

The jack connects power to the circuit when a plug is inserted.

Bob

 

I think the confusion comes from the drawing in post #7
The V shape farthest from the entrance goes to the input capacitor because it connects to the tip of the male thingy. The male thingy is only the TS type because the sleeve is used for both audio ground and DC ground.