simplify reverse polarity protection w/ high side switch - Page 1
Forum section: electronics / projects

author: wergor
author_rank: Contributor
timestamp: October 17, 2021, 05:48:42 pm
content: I have come up with the following circuit implementing a reverse polarity protection and a high-side on/off switch:[ Specified attachment is not available ]M1 and M2 as well as D1 and D2 are of the same type, so I wonder, is it possible to simplify the circuit?OVP and OCP are not needed (Vin is monitored by a microcontroller and M2 is only switched on if Vin is in a range supported by the load. I_load is also monitored for overcurrent events)

author: DavidAlfa
author_rank: Super Contributor
timestamp: October 18, 2021, 02:32:50 pm
content: In fact, you need to add few things to make it stable and safe.There're no gate discharge resistors.Resistors values are too high, or at least in simulation had very slow switching times, several 10s of ms.Applying 12V at the gate 12V is ok, but it allows up to 20V, so I suggest using 16V for lowering Rds even more.I don't see how you could simplify the circuit. If you need reverse polarity protection, you need a second transistor, otherwise the current would pass through the body diode.

author: langwadt
author_rank: Super Contributor
timestamp: October 18, 2021, 03:12:56 pm
content: Quote from: DavidAlfa on October 18, 2021, 02:32:50 pmyou can skip R1,R2,D1 and connect the gates

author: Peabody
author_rank: Super Contributor
timestamp: October 18, 2021, 04:56:46 pm
content: I think the M1 circuit is ok. But I think M2 needs a gate pullup resistor to the source. Otherwise there's nothing the bring the gate back high when M3 turns off.I also think langwadt's suggestion is interesting. If you eliminate R1 and D1 and connect the gates, but add the pullup, then in normal polarity M1 would remain off until M3 turns on both M1 and M2. In reverse polarity, high input would flow through the body diode of M3, which brings both gates high, and both sources high through D2. DavidAlpha, could you simulate that?

author: langwadt
author_rank: Super Contributor
timestamp: October 18, 2021, 06:55:28 pm
content: Quote from: Peabody on October 18, 2021, 04:56:46 pmM1 and M2 is common source a single resistor is pull down for both gates

author: DavidAlfa
author_rank: Super Contributor
timestamp: October 18, 2021, 09:55:32 pm
content: Yes, but I made it like that because I expect some circuitry getting power from the first stage and checking everything is ok before enabling the second stage.Of course I simulate everything before posting it.

author: wergor
author_rank: Contributor
timestamp: October 19, 2021, 09:45:44 pm
content: Quote from: DavidAlfa on October 18, 2021, 02:32:50 pmgreat feedback, thanks!switching speed of the 1st stage is uncritical, the polarity and magnitude of the supply voltage are not expected to change during operation. M2 needs to be fast at switching off to be able to protect the load. with my original values simulations yield switching times in the order of hundreds of us and even faster with lower values. should be good enough for my application.Quote from: langwadt on October 18, 2021, 03:12:56 pmnice solution! thanksseems to be working fine, just like the 2 diode - 2 pullup version.Quote from: DavidAlfa on October 18, 2021, 09:55:32 pmI like the way you thinkthe output of M1 is indeed used to supply another part of the circuit, but I wanted to keep it simple. langwadt's solution also allows tapping off power after M1.

author: DavidAlfa
author_rank: Super Contributor
timestamp: October 20, 2021, 12:48:28 pm
content: Yep, you're right, for a small current, the body diode of M1 will be more than enoughBe careful with R2/R5, you're making a voltage divider, the gate voltage will drop to half if the supply is less than the zener diode threshold (ex. At 14V, the gate will get 7V instead 14V).That's why I used these resistor values, too.