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.
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
asked Nov 25 at 22:07
Maple
4,5812323
add a comment |
up vote
2
down vote
favorite
I am failing to understand the purpose of the MOSFET in this circuit from Adafruit Feather board.
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
asked Nov 25 at 22:07
Maple
4,5812323
My guess would be the same as yours; I think it's to avoid voltage drop.
– Hearth
Nov 25 at 22:09
add a comment |
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.
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
asked Nov 25 at 22:07
Maple
4,5812323
I am failing to understand the purpose of the MOSFET in this circuit from Adafruit Feather board.
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
asked Nov 25 at 22:07
Maple
4,5812323
asked Nov 25 at 22:07
Maple
4,5812323
asked Nov 25 at 22:07
Maple
4,5812323
asked Nov 25 at 22:07
Maple
4,5812323
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
add a comment |
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
add a comment |
1 Answer
1
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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.
answered Nov 25 at 22:26
Dave Tweed♦
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
add a comment |
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.
answered Nov 25 at 22:26
Dave Tweed♦
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
add a comment |
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.
answered Nov 25 at 22:26
Dave Tweed♦
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
add a comment |
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.
answered Nov 25 at 22:26
Dave Tweed♦
116k9143255
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.
answered Nov 25 at 22:26
Dave Tweed♦
116k9143255
answered Nov 25 at 22:26
Dave Tweed♦
116k9143255
answered Nov 25 at 22:26
Dave Tweed♦
116k9143255
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
add a comment |
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
add a comment |
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My guess would be the same as yours; I think it's to avoid voltage drop.
– Hearth
Nov 25 at 22:09