Opus BT-C3100 2.2 Problem

[Jesse James]

Hello all. I am not sure if anyone is familiar with this lithium Ion charger. I have had discharging issues that turned out to be cracked solders for the discharge resistors. This is a 4 bay charger and tester with mah load test. The problem I am having is now bay 2 and 4 will not charge or discharge at all. It goes through all the steps as if it was working but my voltage never changes.
I have tested resistors and diodes and also tried to compare a good working bay with the non working bay but am at a loss.
Any help would be great

[stj]

Re: Opus BT-C3100 2.2 Problem

complete pcb pictures would help.
i didnt know the opus did a load test, i thought only the Litokala did that. - interesting to know.

[CapLeaker]

Re: Opus BT-C3100 2.2 Problem

each channel has it's own step down converter on the other side.
https://www.youtube.com/watch?v=JJxrDgEFxMU

[Jesse James]

Re: Opus BT-C3100 2.2 Problem

Thanks for the help guys. As far as I can tell the step down converters are ok. I am thinking the gate voltages are wrong on my AO3400A mosfets (not sure) but I think the voltage varies on the amount of charge determined so it confuses me. All I know is at any setting 200ma-1000ma three of the four bays are only charging 30-40ma on my volt meter and when selecting discharge I am getting 0.

[stj]

Re: Opus BT-C3100 2.2 Problem

fets are voltage not current driven,
maybe there is a bad resistor somewhere - or the computer is running at the wrong clockspeed - fucking the PWM up.

[Jesse James]

Re: Opus BT-C3100 2.2 Problem

Ok thanks. If I do find anything I will post it here. Who knows maybe someone else will run into this someday.

[Francesc V.]

Re: Opus BT-C3100 2.2 Problem

Hi:

This is the summary I've been facing on this charger so far.

Discharge problems:
- Replace the six 5.1R SMD resistors from the affected slot. They use to burn after long use.

Charging problems:
- Check step down converter might be the issue.

Wrong Voltage reading:
- I am having a slot that reads about the half of the voltage when inserting the battery. I am investigating it.......but so far with no success. If somebody might have an idea, I would appreciate it.

Regards,
Francesc.