VBAT source switching circuit
This is VBAT switching circuit installed in one of my PCB for Timekeeping when VCC is not present.
The switching circuit is working as expected - when VCC is present, the current flows from D3 to VBAT and when VCC is not present, the current flows from J2 to Q6 to VBAT. After a month, we randomly checked the RTC coin cell battery (3V) and it was discharged to 1.6V and that's not normal because these batteries are supposed to work for years without changing.
Then I measured the current VBAT with VCC present = 3.117 and current is -2.251mA Without VCC, the VBAT is 2.8927V and current is 0.015mA (16uA)
In my observation, the negative current is flowing back to the RTC battery and hence degrading it. May be Q6 is not turned OFF properly.
I am not sure if that's what causing the issue. I tried simulating another circuit online. Can I use this circuit and replace the existing one ?
**Update - **
Here is the new circuit which is on the PCB and this circuit is causing battery drain.(ignore the top circuit)
2 answers
R39 keeps the FET gate permanently at 0 V. It seems you meant it to go high only when Vcc is present.
It would help if you showed the FET properly, with its body diode. The existing drawing doesn't give confidence that the FET is installed in the right orientation, or that it's the right polarity FET in the first place. With the scribbles on top of the symbol, I'm not going to guess what you think you have.
Olin already mentioned that Q6 is always on because R39 keeps the gate permanently low.
Power MOSFETs have a parasitic body diode. The way the MOSFET if drawn in your schematic, it can never fully cut off the battery from the VCC node, because the battery has a discharge path through the MOSFET body diode of Q6.
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