Purpose of MOSFET in this circuit











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I am failing to understand the purpose of the MOSFET in this circuit from Adafruit Feather board.
enter image description here



The FET is connected with source and drain reversed comparing to typical high-side pMOS switch, so this is not a power switch.



The combination of body diode with schottky should work like usual battery switch-over circuit, in which case what is FET doing there other than providing body diode?



My only guess is that it is there to avoid voltage drop on the diode when battery supplies power. If source is at drain voltage (via body diode) and gate is pulled to ground (via R12) then MOSFET should be fully open.






power mosfet







share|improve this question






















  • My guess would be the same as yours; I think it's to avoid voltage drop.
    – Hearth
    Nov 25 at 22:09















up vote
2
down vote

favorite












I am failing to understand the purpose of the MOSFET in this circuit from Adafruit Feather board.
enter image description here



The FET is connected with source and drain reversed comparing to typical high-side pMOS switch, so this is not a power switch.



The combination of body diode with schottky should work like usual battery switch-over circuit, in which case what is FET doing there other than providing body diode?



My only guess is that it is there to avoid voltage drop on the diode when battery supplies power. If source is at drain voltage (via body diode) and gate is pulled to ground (via R12) then MOSFET should be fully open.






power mosfet







share|improve this question






















  • My guess would be the same as yours; I think it's to avoid voltage drop.
    – Hearth
    Nov 25 at 22:09













up vote
2
down vote

favorite









up vote
2
down vote

favorite











I am failing to understand the purpose of the MOSFET in this circuit from Adafruit Feather board.
enter image description here



The FET is connected with source and drain reversed comparing to typical high-side pMOS switch, so this is not a power switch.



The combination of body diode with schottky should work like usual battery switch-over circuit, in which case what is FET doing there other than providing body diode?



My only guess is that it is there to avoid voltage drop on the diode when battery supplies power. If source is at drain voltage (via body diode) and gate is pulled to ground (via R12) then MOSFET should be fully open.






power mosfet







share|improve this question













I am failing to understand the purpose of the MOSFET in this circuit from Adafruit Feather board.
enter image description here



The FET is connected with source and drain reversed comparing to typical high-side pMOS switch, so this is not a power switch.



The combination of body diode with schottky should work like usual battery switch-over circuit, in which case what is FET doing there other than providing body diode?



My only guess is that it is there to avoid voltage drop on the diode when battery supplies power. If source is at drain voltage (via body diode) and gate is pulled to ground (via R12) then MOSFET should be fully open.






power mosfet




power mosfet






share|improve this question













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asked Nov 25 at 22:07









Maple

4,5812323




4,5812323












  • My guess would be the same as yours; I think it's to avoid voltage drop.
    – Hearth
    Nov 25 at 22:09


















  • My guess would be the same as yours; I think it's to avoid voltage drop.
    – Hearth
    Nov 25 at 22:09
















My guess would be the same as yours; I think it's to avoid voltage drop.
– Hearth
Nov 25 at 22:09




My guess would be the same as yours; I think it's to avoid voltage drop.
– Hearth
Nov 25 at 22:09










1 Answer
1






active

oldest

votes

















up vote
3
down vote



accepted










Your guess is correct. When VBUS is not present, the gate is pulled low, and the MOSFET shorts out the body diode, connecting VBAT directly to the LDO.



When VBUS is greater than VBAT, the MOSFET is cut off and the body diode is blocking, disconnecting VBAT from the circuit.






share|improve this answer





















  • Thanks. I wasn't sure, but now as I am looking at FET datasheet it comes down to about 14 mV drop at 200 mA, which is way below typical diode drops. Cool! I wonder why battery switch-over circuits with two diodes are dime-a-dozen while something as simple as this I newer saw before.
    – Maple
    Nov 25 at 22:46













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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
3
down vote



accepted










Your guess is correct. When VBUS is not present, the gate is pulled low, and the MOSFET shorts out the body diode, connecting VBAT directly to the LDO.



When VBUS is greater than VBAT, the MOSFET is cut off and the body diode is blocking, disconnecting VBAT from the circuit.






share|improve this answer





















  • Thanks. I wasn't sure, but now as I am looking at FET datasheet it comes down to about 14 mV drop at 200 mA, which is way below typical diode drops. Cool! I wonder why battery switch-over circuits with two diodes are dime-a-dozen while something as simple as this I newer saw before.
    – Maple
    Nov 25 at 22:46

















up vote
3
down vote



accepted










Your guess is correct. When VBUS is not present, the gate is pulled low, and the MOSFET shorts out the body diode, connecting VBAT directly to the LDO.



When VBUS is greater than VBAT, the MOSFET is cut off and the body diode is blocking, disconnecting VBAT from the circuit.






share|improve this answer





















  • Thanks. I wasn't sure, but now as I am looking at FET datasheet it comes down to about 14 mV drop at 200 mA, which is way below typical diode drops. Cool! I wonder why battery switch-over circuits with two diodes are dime-a-dozen while something as simple as this I newer saw before.
    – Maple
    Nov 25 at 22:46















up vote
3
down vote



accepted







up vote
3
down vote



accepted






Your guess is correct. When VBUS is not present, the gate is pulled low, and the MOSFET shorts out the body diode, connecting VBAT directly to the LDO.



When VBUS is greater than VBAT, the MOSFET is cut off and the body diode is blocking, disconnecting VBAT from the circuit.






share|improve this answer












Your guess is correct. When VBUS is not present, the gate is pulled low, and the MOSFET shorts out the body diode, connecting VBAT directly to the LDO.



When VBUS is greater than VBAT, the MOSFET is cut off and the body diode is blocking, disconnecting VBAT from the circuit.







share|improve this answer












share|improve this answer



share|improve this answer










answered Nov 25 at 22:26









Dave Tweed♦

116k9143255




116k9143255












  • Thanks. I wasn't sure, but now as I am looking at FET datasheet it comes down to about 14 mV drop at 200 mA, which is way below typical diode drops. Cool! I wonder why battery switch-over circuits with two diodes are dime-a-dozen while something as simple as this I newer saw before.
    – Maple
    Nov 25 at 22:46




















  • Thanks. I wasn't sure, but now as I am looking at FET datasheet it comes down to about 14 mV drop at 200 mA, which is way below typical diode drops. Cool! I wonder why battery switch-over circuits with two diodes are dime-a-dozen while something as simple as this I newer saw before.
    – Maple
    Nov 25 at 22:46


















Thanks. I wasn't sure, but now as I am looking at FET datasheet it comes down to about 14 mV drop at 200 mA, which is way below typical diode drops. Cool! I wonder why battery switch-over circuits with two diodes are dime-a-dozen while something as simple as this I newer saw before.
– Maple
Nov 25 at 22:46






Thanks. I wasn't sure, but now as I am looking at FET datasheet it comes down to about 14 mV drop at 200 mA, which is way below typical diode drops. Cool! I wonder why battery switch-over circuits with two diodes are dime-a-dozen while something as simple as this I newer saw before.
– Maple
Nov 25 at 22:46




















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