BMS working even when I disable switcher

Status
Not open for further replies.
Hi,

I have a problem with my BMS and tried a lot of things trying to repair that but without succes for now.
Here is what i found out. When i discconect +5V_mES, switcher shoud be disabled and +5V should be 0V. But eventhough SHDN is on 0V, when i measure +5V line, there is 3-4V remaining. If i short +5V and GND, everything stops like it should be and remains in that stage. That happens also if i put 100ohm resistor from line +5V to GND. Now when you enable switcher, there is +5V, and when you disconnect +5V_mES, there is no more power in the board.
Also if I disconnect Q2 everything works. If i connect C and E of Q2 when its not soldered, problem returns.

I dont know where do i gat that 3-4V on 5V line when disabled so could could someone please help me?

 
You are making an elementary mistake - one I've also recently made

The SHDN pin shuts down the switcher, BUT it doesn't turn the output OFF, it only stops it switching - and even while it's not running you still get some residual output from it.
 
That makes sense. So how can I easily prevent that from happening, when I want to shut down regulator to turn off output as well?
 
You mean something like this?
Why dual gate FET is better than P FET in this case? ANd should i use some pull down resistor to "empty" output from LM16006?
 
Hopefully you'll try putting it the right way round Assuming the FET is happy working like that, and switches correctly, then fine, however it doesn't seem like it switches the right way?. If you're switching the output as shown?, then move C8 back to L1.

I used a dual device to get high side switching, as the switching is done from a lower supply rail, and I switched before the IC - as it's to minimise battery consumption. In your case it would correct the polarity as well (high ON low OFF)
 
Status
Not open for further replies.
Cookies are required to use this site. You must accept them to continue using the site. Learn more…