I think I fried my charger, possibly batteries

[ArgusMcJohnsten]

So, first I admit I'm an idiot. I somehow left 2 Samsung 25R batteries on my unplugged xtar vp2 for about 2 weeks. Remembered about them today, tried plugging it in.
Left bay reads 3.6 v with a green led for both batteries I've tried. Right one reads .8v and flashes red. Think it has to recover both or did I fry something?

Thank you for reading

 

[Baditude]

From what you say, it seems the right bay is probably toast.

Not sure about the left side. 3.6 volts on both batteries is within normal operating range of the batteries. However, they should be reading 4.2 volts at full charge. It's difficult to know if its just the charger not reading voltage correctly, not charging the batteries fully, or both.

I'd probably invest in another charger, and maybe new batteries. The old batteries might still be ok, but you won't be able to tell until you get a new charger.

I had a great Pila charger for a couple of years, when one of the bays stopped working and just flashing red LEDs on that side. The other side worked fine. But I had multiple batteries to charge at that point and went ahead and got a 4 bay Nitecore Intellicharger.

 

[Ryedan]

So, first I admit I'm an idiot. I somehow left 2 Samsung 25R batteries on my unplugged xtar vp2 for about 2 weeks. Remembered about them today, tried plugging it in.
Left bay reads 3.6 v with a green led for both batteries I've tried. Right one reads .8v and flashes red. Think it has to recover both or did I fry something?

Thank you for reading

IMO, your charger is good, but that 0.8V battery is toast.

When you leave a battery in an unplugged charger there is generally some small drain on it. Here's a review of your charger from lygte-info.dk . Scroll down to the 'Measurements' section and you'll find this: "Without power it will discharge the battery with about 12mA on usb channel and 5mA on the other channel".

You've got 2500 mAh (milliamp hours) in each of those batteries when they're fully charged. If you discharge one at 12mA for one hour you'll take 12mAh from the battery.

14 days = 336 hours. 336 hrs * 12mA = 4032 mAh. That is I think the battery that was in the usb channel slot and is now reading 0.8V.

336 hrs * 5mA = 1680 mAh. That is the 3.6V battery.

The charger will not charge the 0.8V batt because: "Between 0.15 volt and 2.0 volt the battery symbol will flash and voltage will be updated, but it need 1 volt to start this mode (Led is red)". IMO a Li-ion battery taken down to 0.8V is unsafe to recharge or to use again because of chemical changes that can cause internal shorts at any time without warning causing the battery to vent.

 

[Monotremata]

Doesnt the VP2 have the ability to recover a battery thats been discharged too low? Or is 0.8v just too far gone? My MC2 is older than the VP2 model, and claims it has "0v Activation" for dead batteries, although Im not about to run one of my Samsungs down to 0v to test it hehe..

 

[Baditude]

Doesnt the VP2 have the ability to recover a battery thats been discharged too low? Or is 0.8v just too far gone? My MC2 is older than the VP2 model, and claims it has "0v Activation" for dead batteries, although Im not about to run one of my Samsungs down to 0v to test it hehe..

According to Battery University, any IMR battery less than 2.5 volts is probably beyond recovery for any charger.

 

[Monotremata]

According to Battery University, any IMR battery less than 2.5 volts is probably beyond recovery for any charger.

Good to know. Im glad Ive never had to use that feature haha.

 

[Kemosabe]

OP, perhaps check the cells' voltage with a multi-meter to verify the charger is reading the voltage accurately. that way you can determine if the charger is bad, if the cells are, or both.

regarding recovering overly discharged cells.. in my earlier days, i suffered a mishap and accidentally discharged a VTC3 down to 1.58v. i used a VP1 to "recover" it, and it appeared ok on paper, but during vaping, it was poor. i didnt have proper equipment to test it, but it must have it lost mAh and the internal resistance must have been increased or worse.