0

The schematic below uses an N-channel instead of a P-channel to block reverse-polarity voltage from occurring. This means the gate voltage (and Vgs) must be positive. But there are two options I've discovered so far to interpret how to calculate these values and I was wondering if you know which is correct (If any). If none of them is ok, then do let me know what is the right way to calculate this. In both cases we only look at if the polarity is right

System ratings: Vin=55V, Vz=35V, For use in a buck converter. NMOS is chosen over PMOS because they have a much lower Rds(on)

Credits to Tim for the idea

Schematic

Since the mosfet is pulled to VIN, 55V travel to gate, and for a small period, it allows GND to pass through the mosfet and reach the Zener's terminal. From here on, there are 2 options:

OPTION 1 (the only way I see this working):

If in the P channel version, Vz is subtracted from Vin and enters the gate, then here it must be something like -(-Vz), so Vg = Vin -(-Vz) = 55+35 = 80V (Again, not so sure...). This would make the Vgs positive since Vgs=Vg-Vs=80-55=35V (so the zener adds its Vz to the voltage) and the circuit would be turned on

OPTION 2 (the zener operates just like in the P-channel version)

Vz gets subtracted from Vin and sent to the gate, which in this case would mean Vg=Vin-Vz=55-35=20V. But then Vgs=Vg-Vs=20-55=-35, which would be false since the mosfet wouldn't get turned on (Vgs(th)>0) (interesting thing, Vgs sets the gate voltage, but if it's positive or negative I'm not sure)

So what do you think? Which option is right? (if any)

Has anyone actually tried a circuit like this?

Mito
  • 409
  • 9
  • Related: https://electronics.stackexchange.com/questions/675838/will-this-nmos-mosfet-reverse-polarity-protection-circuit-work – winny Jul 30 '23 at 16:35
  • @winny Yes, but that was the wrong version. The question received the answer that it wouldn't work and I moved on with the feedback to flip the MOSFET's position to a new one. – Mito Jul 30 '23 at 16:38
  • 1
    Yes, so related. Have you tried to simulate your new circuit? – winny Jul 30 '23 at 16:40
  • What does **Vz=35V** mean? What does **Since the mosfet is pulled to VIN**? These are both very confusing statements. – Andy aka Jul 30 '23 at 18:09
  • @Andyaka Vz is the zener breakdown voltage. *"Since the mosfet is pulled to VIN?"* highlights the gate to (what should be) Vin through resistor Rm23 (therefore pulled up when polarity is right) – Mito Jul 30 '23 at 18:18
  • A 35 volt zener diode won't protect virtually every MOSFET I've ever seen. It needs to be more like 15 volts. The gate gets nowhere near to the 55 volts due to the zener and the body diode of the MOSFET. Why don't you simulate like winny suggests @Mito – Andy aka Jul 30 '23 at 18:27
  • @winny I have just now using falstad with the following parameters: Vin=55V, Rm23=10kO, a 1kO dummy load and Vz=15 where exactly as you Andy thought, **the zener sets the MAX gate voltage (which *is* positive!) for the mosfet**. It conducted during normal polarity and stopped any current from flowing during reverse (well 700nA still went through but that's nothing!) I want to do a real world test myself tomorrow with a bench power supply (which unfortunately only has up to 16V, but still works). I'll be using Vin=14V, and any zener I can find at my local store with a Vz of <=5V – Mito Jul 30 '23 at 18:34
  • And yes, @Andyaka , most N channel MOSFETS have a Vgs of about 20V, so 15V should be safe enough (the ones with the smallest Rds(on) at least). For example (this one)[https://ro.mouser.com/ProductDetail/Infineon-Technologies/IPT007N06N?qs=jrjnRtnZuRt2bofTNjdFDQ%3D%3D] with an Rds(on) of only 750uOhms! – Mito Jul 30 '23 at 18:40
  • A comment about the other parts in the circuit - If you apply a reverse-polarity (-55 V) source to the input, at lease two of the diodes will blow. Vicor has a reverse-polarity device that does this on purpose with a much larger diode, the idea being that an upstream fuse will blow, protecting the downstream circuits. – AnalogKid Jul 30 '23 at 20:42
  • @AnalogKid Which diodes, through what current path? – Tim Williams Jul 30 '23 at 20:52
  • The drawing is unclear, but I'm assuming that because U15 is marked "VIN", it is a device being powered, not a source. That means the input is on the right. With a reversed connection, D24 is now forward-biased with 55 V at unknown current capability. Poof. D21 and D22 also are forward biased. They might both go poof at the same time, or one might fail before the other, fail in an open circuit mode, and protect the other like a blown fuse. If U15 is in fact the 55 V source, that changes things. Soooo . . . what is U15? – AnalogKid Jul 31 '23 at 00:31
  • @AnalogKid U15 is the 55V source. VIN is the (+) solder pad, and GND is the (-) solder pad for a battery/power supply to attatch to a buck converter. The circuit is made to serve as reverse-polarity protection – Mito Jul 31 '23 at 09:18
  • Does that mean that Rm22 represents the load on the circuit? – AnalogKid Jul 31 '23 at 12:44
  • @AnalogKid Yes, Rm22 is the representation of the load. I just placed it to know there's a load there. Its value doesn't matter – Mito Jul 31 '23 at 13:08

1 Answers1

2

Consider this connection:

Gate source zener connection zoom

Vgs is defined as Vz, when the zener is reverse-biased in breakdown.

You are subtracting from the wrong side, giving Rm23's voltage, not the gate. For further reading, please review Kirchoff's loop law.

Tim Williams
  • 22,874
  • 1
  • 20
  • 71
  • **Tested in real life, it's true.** Didn't fully understand the law, but the multimeter doesn't lie. I used the exact schematic above with an IRF840 mosfet, a zener with Vz=5V, Rm23 was 10kO, the other 2 simple diodes were classic 1N4007s, and a TVS with a working voltage of 30V. Vgs was constantly kept at the voltage of the zener. The load was one of those LEDs for Arduinos. Placing the voltage anywhere in the 5-10V range, Vgs was 4.9V. I could have gone a bit higher, but the LED would have caught fire – Mito Jul 31 '23 at 13:14