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I would like to add overvoltage and reverse polarity protection to my circuit.

Searching on the internet, I saw that the solution that seems to be the most popular is this:

enter image description here

I have seen there is a lot of controversy about whether or not this is the best solution. I have seen that there are much more complex circuits or even chips dedicated to this specific protection. For example this: https://www.ti.com/tool/PMP10737

My application is very simple,

Vin: 24-28 VDC
Imax: 2 A

It is a microcontroller-based circuit and an analog front end part, nothing complex or particularly delicate.

It's the first time I've faced this problem and I'm a bit confused, what solution can you recommend?

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    \$\begingroup\$ That circuit doesn't provide any over-voltage protection at all. It only provides reverse-voltage protection. \$\endgroup\$
    – brhans
    Commented Feb 5, 2023 at 14:34

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As noted by others, the circuit you've implemented only provides reverse-polarity protection. I've seen this layout called an "ideal diode" MOSFET because the voltage drop is in the tens of mV, ten times lower than even a signal Schottky could provide. (Power diodes that could handle your load current, of course, are even more lossy.)

It works because of the body diode in the pMOS. Under proper polarity, the body diode conducts. The zener diode / resistor bridge then pulls down the gate voltage relative to the source, which biases the pMOS well into its operating region, where many power pMOS devices will have an on resistance of a couple dozen mΩ. Under reverse polarity, the body diode blocks, the pMOS gate voltage remains equal to the source, and no current can flow.

The solution to the overvoltage problem requires a second MOSFET. Because the pMOS in your circuit is "backwards", the body diode will always conduct if the drain-source voltage exceeds a minimal threshold. A second power MOSFET, connected "normally" and itself controlled by some zener diode reference voltage, can switch off the circuit when necessary.

schematic

simulate this circuit – Schematic created using CircuitLab

In this circuit, M1/D1/R1 form the reverse voltage protection circuit you already have. M2 is the load switching MOSFET which, under normal circumstances, is biased deep in its conducting region by the D3/R3 divider, with a source-gate voltage of 10 V.

M3 is a low-power MOSFET. Under normal circumstances, D2 is not conducting, R2 pulls M3's gate voltage up to its source voltage, and M3 is off. However, when the D2 cathode voltage exceeds its zener voltage (in this case chosen to be a little higher than 24 V), D2 starts to conduct, clamping the M3 gate voltage. Once M3's source-gate voltage drop is high enough, M3 conducts, D3 is bypassed, M2's source-gate voltage drops to zero, and the circuit shuts off.

Resistor values and zener voltages are for illustration. You'll need to match your resistors, zeners, and MOSFETs carefully to make sure your circuit doesn't turn off prematurely. A slight overvoltage combined with poorly-chosen components may result in a not-completely-off M2, which will significantly increase its on resistance, and thus its temperature, in a way you may not have expected.

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    \$\begingroup\$ Would adding a Schmitt trigger for the activation of M2 be the point at which a dedicated power protection IC is likely to make more sense? \$\endgroup\$ Commented Feb 5, 2023 at 17:06
  • \$\begingroup\$ Not sure. I'd have to think about how simple or complex the Schmitt trigger feedback would be. In theory it just needs to switch a different zener into circuit once overvoltage is detected. The hard part is probably going to be ensuring it powers on in the correct state every time. \$\endgroup\$
    – Matt S
    Commented Feb 5, 2023 at 19:36
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It all depends on the requirements of your application. The advantage of the shown circuit is that a reverse condition is not destructive (unlike blowing a fuse) and the nominal condition has very little forward voltage drop compared to other circuits like a single in-series diode.

By the way: The only thing that is protected against over-voltage in the shown circuit is the gate of the MOSFET. The load isn't protected against over-voltage at all.

If you have more requirements e.g. inrush current limitation or your load needs to work no matter how your battery is connected, the circuit will become more complex, of course.

Or you have a lot of capacitance in your load and - when disconnecting the battery - you want to prevent that capacitance from discharging into other loads that are connected in parallel, etc...

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  • \$\begingroup\$ Ok, very clear, but at this point I don't understand what a chip like LM74610-Q1 is for. I mean, what does this chip do more than the zener/resistor divider on the gate? Ok, very clear, but at this point I don't understand what a chip like LM74610-Q1 is for. I mean, what does this chip do more than the zener/resistor divider on the gate? \$\endgroup\$ Commented Feb 6, 2023 at 7:51
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    \$\begingroup\$ @FedericoMassimi The main advantage is that you can use an n-channel FET instead of a p-channel FET. A n-channel FET has significantly less power dissipation. This chip basically creates an ideal diode (very little forward voltage drop combined with reverse current protection). \$\endgroup\$
    – feynman
    Commented Feb 6, 2023 at 20:10

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