Is it feasible to jerryrig a charger [Jaybird Freedom wireless headphones]to avoid buying proprietary hardware?

[Skystrike7]

I have a pair of wireless Jaybird freedom headphones, but don't have the proprietary charging clip for them. Seeing as the connectors are easily accessible, I figured I'd ask around and see if I could perhaps cut an old USB/ microUSB wire and solder together a working solution? Do device chargers like this really have any complexity to them, or are they just simple connectors? I ran a probe over them, and deduced that one of the "pins" is a hot and all the others are ground, since the one pin is the only one that involves a voltage across terminals, and is a positive value only when my red lead is on it, and not vice-versa. Resistance across all terminals with respect to each other is a mixed bag that I found no useful pattern for and don't want to waste my time with it unless it's to prove that what I ask is not easily done.

 

For reference, here is a picture that shows the connecting zone (5 "pin" terminals). The left is the headphone piece itself that contains the battery, and to the right is a proprietary clip that you plug a microUSB into, then connect it to your headphones via the pins shown. 

 

The charging clip, while also serving as a sort of external battery, is about as expensive as just buying a new pair of headphones, so you can see why I'd rather not buy it straight up.

 

 

[WereCatf]

2 minutes ago, Skystrike7 said:

Do device chargers like this really have any complexity to them, or are they just simple connectors?

There is quite literally no way of knowing with proprietary chargers, unless you're willing to reverse-engineer the whole thing or just take a risk at frying it. It could have some sort of smarts in it, or might not.

[Skystrike7]

Just now, WereCatf said:

There is quite literally no way of knowing with proprietary chargers, unless you're willing to reverse-engineer the whole thing or just take a risk at frying it. It could have some sort of smarts in it, or might not.

Thanks for the reply. I kinda figured that it depends on the exact item, but I'm hopeful that perhaps someone might know enough to say "well duh, the sort of batteries in those headphones need extra circuitry to receive the input 5v of a USB port", or "nah bro the only reason it's like that is to save real estate on the headphones themselves".

[WereCatf]

7 minutes ago, Skystrike7 said:

Thanks for the reply. I kinda figured that it depends on the exact item, but I'm hopeful that perhaps someone might know enough to say "well duh, the sort of batteries in those headphones need extra circuitry to receive the input 5v of a USB port", or "nah bro the only reason it's like that is to save real estate on the headphones themselves".

Well, it's most likely a li-ion battery in there and you'd set it on fire, if you charged it with 5V and that's why there's a charger-circuit that controls the charging-voltage, dropping it from 5V down to a voltage depending on what the current phase of charging is going on. Your problem is, is the charging-circuit in the charger-unit or is it in the headphones? Putting everything or most of it in the charger-unit would save space in the headphone-unit, but then you'd need to implement a similar circuitry yourself to charge your headphones.

 

The quick answer: do not go plugging 5V into those pins, unless you know there's a circuitry inside there that can handle 5V input.

[WereCatf]

The hackish way of charging the headphones without official charger-unit could be e.g. opening the headphone-unit up, checking if it is, indeed, a li-ion battery in there. If so, buy a TP4056-module on eBay and make sure there's a DW01A protection-circuit on-board, and hack together a connector of your liking from the module to the battery. Also, check that the resistor on the TP4056 is set so that you're not charging at more than 1C-current, where C = the battery's capacity in amp-hours.

 

The TP4056 is a lithium-battery charger-chip and DW01A is a protection-circuit to protect against over-voltage, over-discharge etc.

[Skystrike7]

2 minutes ago, WereCatf said:

Well, it's most likely a li-ion battery in there and you'd set it on fire, if you charged it with 5V and that's why there's a charger-circuit that controls the charging-voltage, dropping it from 5V down to a voltage depending on what the current phase of charging is going on. Your problem is, is the charging-circuit in the charger-unit or is it in the headphones? Putting everything or most of it in the charger-unit would save space in the headphone-unit, but then you'd need to implement a similar circuitry yourself to charge your headphones.

Hmm that sounds about right, space and weight are emphasized pretty heavily in this sort of device, since any less weight would make them less prone to falling out of your ears. Probably also why they know they can charge so much for the clip, I can't even find a Chinese knockoff for the 2-year old product. It might be a fun little project to enlist my electrical engineering student friend of mine, to make a little adapter with I suppose a stepdown transformer or some other kind of circuit elements to go from 5 volts to the (quick internet search) 4ish volts a LI-ion battery should be charging at. After hearing this,  I won't be experimenting with plugging straight from a stripped USB to the charging terminals

[Skystrike7]

2 minutes ago, WereCatf said:

The hackish way of charging the headphones without official charger-unit could be e.g. opening the headphone-unit up, checking if it is, indeed, a li-ion battery in there. If so, buy a TP4056-module on eBay and make sure there's a DW01A protection-circuit on-board, and hack together a connector of your liking from the module to the battery. Also, check that the resistor on the TP4056 is set so that you're not charging at more than 1C-current, where C = the battery's capacity in amp-hours.

 

The TP4056 is a lithium-battery charger-chip and DW01A is a protection-circuit to protect against over-voltage, over-discharge etc.

I'll look into that, thanks. 

[Hackentosher]

10 minutes ago, Skystrike7 said:

I'll look into that, thanks. 

Be careful, most TP4056 modules come with a 1.2K program resistor, which means it'll try to charge whatever battery is connected to it at 1 amp. 1 amp is almost certainly way too much for whatever battery is in there, so you'll need to remove the program resistor and solder on a new one. Google the datasheet for the 4056 IC, it has the equation for calculating which resistor you need for the desired charge current. 

 

I have a feeling the headphones have their own charge circuitry on board and the clip thing just provides power to the charger. If I were you, I'd just buy the damn clip for $25 rather than risking my shiny wireless headphones.

[Skystrike7]

11 hours ago, Hackentosher said:

Be careful, most TP4056 modules come with a 1.2K program resistor, which means it'll try to charge whatever battery is connected to it at 1 amp. 1 amp is almost certainly way too much for whatever battery is in there, so you'll need to remove the program resistor and solder on a new one. Google the datasheet for the 4056 IC, it has the equation for calculating which resistor you need for the desired charge current. 

 

I have a feeling the headphones have their own charge circuitry on board and the clip thing just provides power to the charger. If I were you, I'd just buy the damn clip for $25 rather than risking my shiny wireless headphones.

Thank you. The jaybird website says maximum charge rate is 1 A, so I'll stick a slightly bigger resistor in there to tone it down a little. Also, I got them for free without the charger included, so I don't want to spend very much to be able to use them. If I blow these, then the charging clip costs the same as buying an entire new headset, so I don't have much to lose here.

[Hackentosher]

Just now, Skystrike7 said:

Thank you. The jaybird website says maximum charge rate is 1 A, so I'll stick a slightly bigger resistor in there to tone it down a little. Also, I got them for free without the charger included, so I don't want to spend very much to be able to use them. If I blow these, then the charging clip costs the same as buying an entire new headset, so I don't have much to lose here.

I'm guessing that's a quoted 1a from the USB. In reality, it probably wont draw that much to charge what I'm assuming is a <200mAh battery.

[Skystrike7]

2 minutes ago, Hackentosher said:

I'm guessing that's a quoted 1a from the USB. In reality, it probably wont draw that much to charge what I'm assuming is a <200mAh battery.

Oh you're right, it's probably cause the cable is only rated to handle a current of 1 A. What do you suggest might be a suitable current for charging such a battery? How big of a resistor should I solder in?

[Hackentosher]

1 minute ago, Skystrike7 said:

Oh you're right, it's probably cause the cable is only rated to handle a current of 1 A. What do you suggest might be a suitable current for charging such a battery? How big of a resistor should I solder in?

Best practice for charging batteries is 1C, or 1 amp for every 1 amp hour of capacity.