Timeline for Adding nFET-driven reverse polarity protection to a switching regulator input
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
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| Feb 2, 2022 at 22:16 | comment | added | Kevin White | @BigOwls - I'm not sure what you are saying. As you have it drawn the switching regulator will get power even with reverse polarity. If you put the source pin feeding the load with the drain to the 24v input you will get the protection you want. | |
| Feb 2, 2022 at 22:04 | comment | added | user263983 | @bigowls The way you draw, MOSFET is always open and does not protect the circuit. The way the answered suggest MOSFET is always closed and resistance of opened transistor irrelevant. It used as diode for reverse polarity protection due to MOSFET nature. | |
| Feb 2, 2022 at 21:22 | comment | added | Louis | The internal diode is not added, it's an integral part of the channel N-P junction. The symbol arrow going to the middle is the representation of the body diode. | |
| Feb 2, 2022 at 21:13 | comment | added | Big Owls | @user263983 I'm not sure I understand your comment above starting with "To reduce...". The ground path is open during testing (to what degree I am not certain) | |
| Feb 2, 2022 at 21:10 | comment | added | Big Owls | @user263983 As far as my understanding goes, schottky diode loss: P=0.55V*1A = 0.55W. The FET loss: P=1*1*0.07= 0.07W. I am unsure if this is different for high frequency pulses to the gate vs constant ON. | |
| Feb 2, 2022 at 20:55 | comment | added | user263983 | @bigowls to reduce heat dissipation during regular working cycle meltable fuse in series and diode in reverse parallel to input. The answer recommend to use MOSFET as diode. It will not be open at any polarity. | |
| Feb 2, 2022 at 20:53 | comment | added | Big Owls | @KevinWhite I did want to touch on that in my original post but forgot to add that. This was my original intention but I goofed and got them flipped during the install, so I decided to publish the schematic of the applied version for conversation's sake. Are you saying you would expect the second regulator (onboard switching regulator) to energize? When I swap the leads to reverse polarity, it does not power on and seems to work correctly. Is this perhaps due to a limited amount of current flow in the wrong direction? | |
| Feb 2, 2022 at 20:50 | comment | added | user263983 | According to datasheet of some MOSFETs voltage drop is same as usual silicone diode. Why not use Schottky diode if voltage drop critical? | |
| Feb 2, 2022 at 20:47 | comment | added | Big Owls | @user263983 While I can afford the voltage drop of a diode, my device's current consumption during load conditions -- upwards of 1A @ 24V -- made it seem like the heat losses (read: losses into a small, enclosed space that is often placed outdoors in heat) could be better reduced via the FET. | |
| Feb 2, 2022 at 20:46 | comment | added | Kevin White | @user263983 - much less voltage drop and power loss | |
| Feb 2, 2022 at 20:40 | comment | added | user263983 | MOSFET as diode. Why just do not use diode? | |
| Feb 2, 2022 at 19:54 | history | answered | Kevin White | CC BY-SA 4.0 |