3.5mm Push-to-make

[alband]

Hi again,

I am designing a headphone amplifier to fit inside my guitar and I would like not to add any more external controls. This is so I can plug my headphones (or speakers designed for a 3.5mm jack) straight into the guitar.

I'd like to be able to power it on and off but just plugging in the headphone jack. There are plenty of 3.5mm jacks out there with switching mechanisms, but they are all push to break, and usually involve the actual left/right channels in the switch circuitry (e.g.).

So, this would need a circuit running constantly to check whether this contact has been broken, which then turns on the whole system. My question is;
Has anyone ever coma across a push-to-make switch in 3.5mm jack (so I can just run the power supply through it), or, can anyone think of a more elegant or reliable way of detecting the presence of a plug, than a normal, stereo plug with a tactile switch super glued onto the end (which would eventually fall off).

As always (with me at least ) there are size and budget constraints, but I shouldn't think they'd be too much of an issue here as, these type of parts are very small anyway.

Thanks, in advance.
David.

 

[boriz]

Idea...

Forget the switch. Use the resistance of the headphone coil instead. With the jack unplugged, you have open circuit between ground and (say) the left channel. With the jack inserted, the resistance falls to whatever the headphone coil’s DC resistance is (prolly <32 ohms).

Arrange for the base current to a transistor to be supplied via the headphone coil. So that normally it is off (no power used), but when the headphone is connected the transistor turns on. This transistor could then be made to operate a relay or whatever.

 

[alband]

Would that damage the headphones so would they need a resistor in series to limit the current or should they be able to take that?

Expanding on that idea though;
Headphone is plugged in, this triggers a transistor like you said which wakes up a PICAXE. The PICAXE then switches an FET on that also provides it with power, and switches of the current going through left headphone allowing them to be used.

Could the PICAXE drive itself using an FET this way, and therefore, turn itself off?

Also, how would the PICAXE know when the jack has been unplugged and when to turn off?

One other thing was that I was planning on having the PICAXE monitor the batteries voltage. I was planning on connected a 3.3V regulator to the supply voltage and feed this into an adc pin. Then, the PICAXE could compare this to VDD and thus be able to check the batteries' voltage. Would this work?

Thanks,
David.

 

[hippy]

Most stereo 1/4" jacks and some 3.5mm jacks switch all three contacts, tip, sleeve and ring. One of the ring contacts will supply 0V to the headphones the other contact will short to that 0V when nothing plugged in, become disconnected / floating when plugged in.

That switched ring contact can be used as a simple switch for a PICAXE or power management circuit with a pull-up to +V. The pull-up value can be quite high so minimal current will flow into the headphones as inserted, and won't affect the headphones or audio at all when inserted.

You just have to find the right jack socket to use.

 

[alband]

Wouldn't that slowly drain the battery through the pull-up resistor while the headphones aren't connected?

 

[hippy]

Possibly but 100K would only be 45uAh with a 4V5 supply, and R could probably be higher.

Totally isolating anything from a supply or minimising current drain is relatively difficult. I'm sure the switched ring connector could be used in conjunction with a FET or something and have minimal drain but not my field of expertise. A PICAXE could wake regularly, supply the voltage to the pull-up and check the return and go back into sleep mode.

There's the wider picture; is it at all necessary ? If batteries are rechargeable they'll drain over time, if not, they'll run out with normal use anyway. Is having it all 'ready to go' when you need it even achievable without having to first recharge or put batteries in ?

System and product design is much more than solving one aspect of the problem; usage and practicality play a big part. Many people have probably designed absolutely phenomenal solutions to problems or needs which have been deemed less useful in practice. If modifying a guitar to hold amplifier and batteries then is it really that much more inconvenient or unacceptable to add a discreet on-off switch ?

One always has to ask themselves if a project isn't simply creating an awful lot of work, cost and complexity to solve a problem which isn't really a problem in the first place, actually displacing probems from one place to another. Most people I know would buy a Marshall MS-2 or MS-4 for around £20, and if requiring headphones would use an external haedphone amp.

 

[boriz]

Other idea...

Hot-glue a microswitch to the back of the socket so that it is depressed by either part of the jack plug when inserted, or by the opening of the sockets normally closed switch.

Other other idea...

Use a stereo socket, and a mono audio signal, giving you a spare channel. Just make this channel a short circuit and power the amp through that.

Other other other idea ...

Have the amp circuit powered by batteries in the headset

 

[alband]

Sorry for the delay...

I found another idea which is just the ticket (sorry, but thanks ). I've used (another) spare iPod headphone jack which happens to have a hole on the opposite side of the connector to the main GND pin. I've just added a contact to that hole using an SMD SIM card slot from an old phone (oh the massacre!). I will now have all the GND's in the circuits connect to the original GND pin on the headphone jack, but have the negative of the battery connected to the new GND pin on the jack (if that makes sense).

The only problem now is that I've run out of 3.5mm headphone jacks ; I think I'm going to have to assemble an order on teck-supplies and get one of theirs with some other stuff (sarcastic sigh).

btw, should that method of using a 3.3V regulator and an ADC input pin to measure how much battery is left, work?

Thanks for the suggestions,
David.