mosfetvoltage-regulator
Is it safe to use a MOSFET with gate connected to the low battery comparator output of this TPS61027 switching regulator in order to disconnect the load once batteries become low? The LBO is open-drain, and recommends a \$1 M \Omega\$ pullup to the 5V boost output line (I'm also wondering whether that resistor is a bit too big to effectively turn on the MOSFET when the open-drain is not active [no low battery condition]).
MOSFET is "upside down" as Mike said.
BUT
3 "fixes":
FET will be on when gate is LOW (~= 0 V).
FET will be off when gate is HIGH(~= 4.1V)
IF you can isolate the MOSFET 4.1V and gnd feeds from the rest of the PCB by cutting tracks then
Swap MOSFET 4.1V and ground
Reverse LED polarity
Take MOSFET
Bend leads UP
Invert MOSFET - solder MOSFET D to old S pad
Solder MOSFET S to old D pad.
Solder MOSFET G to G pad.
Or use wire tacked on leads to make it easier.
the maximum voltage output is around 3.3V and not at 5V. How would you explain that ?
The LM311 has an open collector output so the it does not "drive" the output high. For your circuit, the output high is given by:
$$V_{OH} = 5V(\dfrac{2.7k + 1k||2.7k}{2.97k + 2.7k + 1k||2.7k} + \dfrac{1k}{1k + 2.7k}\dfrac{2.97k}{2.97k + 2.7k + 1k||2.7k}) = 3.31V$$
how would you explain that the output is not exactly at 0V when the input is high?
Best Answer
It'll work with 1Mohm and you could use a 10k or 100k with out problems BUT consider this scenario: -
The battery gets low and switches off the load. The dumping of the load will inevitably cause the battery voltage to rise again because it no longer is under a load condition. This rise then re-enables the load FET and you are into a possible problematic situation. You really should be "latching" the battery low condition with a flip-flop BUT now we're into trying to work out how this situation remedies itself - what mechanism is acceptable for resetting the power?
The low-battery comparator does have hysterisis but it's only 10mV - it maybe enough but i wouldn't count on it. I'm sure the battery could rise (under load to no-load conditions more than 5% and this would be 25mV).
Another option is to have a resistor from the LB output to the LBI - this will increase hysterisis but it's a suck it and see approach. Having said that it'll probably work. You'll need more like a 10k resistor on the LBO and possible no smaller that 100k as feedback to LBI. Your normal LBI resistors will need to be about the 10k area to prevent too much hysterisis too.
EDIT following disclosure of circuit Q1 won't work how you imagine it to - if the source was connected to 0V and the drain to the negative end of the load I believe this would work. The 1Mohm pull-up would keep the load connected via the FET to 0V. As you now have it this won't work. If you are intent on a high-side switch (maybe because the load HAS to be always connected to ground) then there is a way but it involves two fets somewhat along these lines: -
Replace the BC547 with an n-channel FET and get rid of the 10kohm. Ignore the 5.5V - read it as 5V - it's getting late over here and i didn't have the perseverence to fully correct the diagram in paint!!! The MCU line is your LBO. The high-side FET is P-channel - choose one with very low Rds(on) and one that has a "logic-level" gate drive voltage. The n-channel is less stringent - it can't be a bioplar given the high impedance pull-up on the LBO pin BUT, given that you may use a 10kohm (previouse pre-edit comments) it could be a bipolar (just about). Stick with an n-channel logic gate driven device to be sure!