Adding 3.5mm jack to bluetooth earphones

[mgpaulie]

HI have a set of blue tooth custom fit ear plugs, ought at the NEC bike show pre covid, used a few times but now won't turn on. I chatted with the guys I bought them from at lasts years NEC bike show about returning them for repair but sounds like I could end up paying close to the cost of a new set. The ear buds fit well and rather than bin them thought about making them wired only ear phones.

See photo - the bit on the left is the on/off volume & battery, the bit on the right is the bluetooth module. Thinking about having them on a 3.5mm jack plug or similar. nsuggestions on the best way to do this?

 

[cookie]

Good luck with that.

 

[Wapping]

Buy a cheap pair of jackplugged earpnes. Cut off the earpieces and attach your moulded earpieces. I hope you are good at soldering very thin wires.

You may even be able to find an inline volume controller or, easier, control the volume from the music playing device itself.

PS The whole thing sounds like a load of faffing about.

 

[Santa-2512]

Do able - but with sacrafices as Wapping says

you may loose the inline volume & deffo the bluetoth module

just cut them close the volume & bluetooth parts


strip the wires back, and solder them in place of a cheap wired pair


you may find there hard to solder, as they will be thin coated wires so you need to remove the glaze to solder

 

[EKIMYRF]

you may find there hard to solder, as they will be thin coated wires so you need to remove the glaze to solder

I have tried doing this, the heat required to melt the solder will melt the thin wires
Be prepared to bin and buy new ones

 

[Wapping]

Or buy a cheap pair of Bluetooth earphones. Cut off the earpieces and solder your moulded earpieces on….. but those wires are very thin, as I say.

Edit

Beaten to it….

 

[Wapping]

Be prepared to bin and buy new ones

Start by buying a suitable bin

 

[Wapping]

Ideas on how to strip the insulation of a very thin wire and how to solder it. The bloke extends the thin wire but the principle will be the same, I guess: