0
\$\begingroup\$

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?

\$\endgroup\$
14
  • \$\begingroup\$ Related: electronics.stackexchange.com/questions/675838/… \$\endgroup\$
    – winny
    Commented Jul 30, 2023 at 16:35
  • \$\begingroup\$ @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. \$\endgroup\$
    – Mito
    Commented Jul 30, 2023 at 16:38
  • 1
    \$\begingroup\$ Yes, so related. Have you tried to simulate your new circuit? \$\endgroup\$
    – winny
    Commented Jul 30, 2023 at 16:40
  • \$\begingroup\$ What does Vz=35V mean? What does Since the mosfet is pulled to VIN? These are both very confusing statements. \$\endgroup\$
    – Andy aka
    Commented Jul 30, 2023 at 18:09
  • \$\begingroup\$ @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) \$\endgroup\$
    – Mito
    Commented Jul 30, 2023 at 18:18

1 Answer 1

2
\$\begingroup\$

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.

\$\endgroup\$
1
  • \$\begingroup\$ 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 \$\endgroup\$
    – Mito
    Commented Jul 31, 2023 at 13:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.