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While learning how to make a buck converter, I realised that using P channel mosfets for reverse voltage protection is one of the best ways, but I think it could be better using an N channel using the circuit below ( I'm new to electronics, so it may not be correct and do let me know what, and if it's wrong). What do you think about it? Will it work to protect the buck converter from a reverse voltage? The reason I chose an NPN one instead of PNP is because of their Rds(on) resistance, in my case, the NPN has about 7X lower Rds(on) at 0.1mO compared to 0.7mO for the PNP, so lower power dissipation!

Buck converter ratings: Vin = 55V (MAX), Vout = 7V, Iout=31A. Vz=9V. Mosfet Vgs= +/-20V, Vds>=80V (yet to be picked) Schematic

**Circuit explained: ** When the polarity is RIGHT, the voltage goes through the Zener, subtracts 9V from 55, and feeds 46V into the linear voltage regulator (we're using the ZXTR2112FQ-7 with a fixed output of 12V), the LDO converts it to 12V and feeds it to the gate, thus opening the MOSFET's gate and allowing current to flow! Also the Vgs = Vg (gate is 12V) - Vs(source is connected to GND so 0V) = 12 - 0 = 12V, which is smaller than the 20V max of the Vgs. Perfect!

I'm unsure though if Rm8 at the other's diode end is even necessary? Something tells me it should be removed... I saw other schematics such as this one from Vince where he used a resistor to pull the diode to the line that's supposed to be VCC.

When the polarity is WRONG, the zener blocks any current from entering the LDO and causing any undervoltage since it's connected to GND (this only if Rm8 would not exist as that would plug VIN to GND... And voltage must be higher than the Vz = 9V to pass through it), and the gate is pulled to GND through Rm2, so it's off.

What do you guys think about this? Will it work?

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    \$\begingroup\$ Note the direction of the body diode. You might want to remedy that. You can use an N-FET as long as you don't care about having the same ground between your input supply and the output of the buck. \$\endgroup\$
    – John D
    Jul 29 at 17:45
  • \$\begingroup\$ @JohnD yes the diode should be reversed, but the circuit has multiple other design flaws too, such as not supplying the buck with any power if no load is connected, and closing immediately once the mosfet opened because the diode would stop conducting. But why is it bad to have all grounds tied together in a system? And also, how are we supposed to isolate the grounds? Say we connect a battery to the buck, the two already have a common ground... Or you mean when we have multiple components in the assembly to tie them up in one single point? \$\endgroup\$
    – Mito
    Jul 29 at 20:19

2 Answers 2

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Just use the P-ch circuit upside down. Everything is identical down to the symmetry of: swapping polarity (N <--> P). NPN transistors become PNP, diodes (PN) swap direction (NP), and supplies negate.

Reverse protection circuit plus TVS

This slightly elaborated circuit (including fast reversal protection and clamping) becomes like so when flipped,

Reverse protection circuit plus TVS, polarity swapped

which can be redrawn in the usual way, with grounds on the bottom and positive supply on top.

The main downside to an N-ch protector is, often sources and loads are common-ground outside of the converter in question, such as for many automotive, audio and computer peripheral applications. If this can be prevented by design, then an N-ch protector is indeed preferable, because N-ch perform about 2.5 times better than P-ch.

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  • \$\begingroup\$ Ok, but let me know if I got this correctly: the TVS diode only serves a transient voltage suppression purpose. When polarity is connected properly, the NPN is turned ON and current starts to flow. The zener and other simple diodes don't do anything in this case. If the polarity is connected wrongly, the NPN is pulled down to GND, and because voltage is 0V, D1 again doesn't conduct anything, and D4 and D2 again don't do anything. What am I missing here? What is the purpose and working of D1,D2,D4? \$\endgroup\$
    – Mito
    Jul 29 at 20:33
  • \$\begingroup\$ Also, what do you mean by "often sources and loads are common ground outside of the converter in question". Aren't all* electronics in a system supposed to have a common ground? And how would it be possible to isolate the source (Lipo battery/ power supply) from the converter when we connect the GND of the source to the solder pads of the converter in order to step it down? I think there's a concept I'm not getting. If you have a video/website that may explain what you mean I would appreciate it a lot! \$\endgroup\$
    – Mito
    Jul 29 at 20:42
  • \$\begingroup\$ The diodes are important in certain cases, see explanation: electronics.stackexchange.com/questions/675605/… | Wiring practices are out of scope here, but I suggest searching on the topic to learn more. \$\endgroup\$
    – Tim Williams
    Jul 29 at 22:00
  • \$\begingroup\$ Alright, I got it with the diodes, but there's a big problem with the N channel schematic (I think). The gate is simply pulled to VCC, or GND through a resistor, but if NO load is connected to the buck, then the voltage at G will be whatever Vin is, and most mosfets have a Vgs of +/- 20V. In our case, Vin(MAX) is 55V, which would instantly burn the N channel. How can this be avoided? In the P channel example, the Zener drops the Gate voltage to Vin-9V=46V, and Vgs would be 46-55=-9V, which is perfect for most gates to activate. But for the N channel I don't see what \$\endgroup\$
    – Mito
    Jul 30 at 9:06
  • \$\begingroup\$ Try "redraw[ing] in the usual way" as suggested. I think you will find more sense made that way. \$\endgroup\$
    – Tim Williams
    Jul 30 at 9:19
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What do you think about it?

If your load was removed then 55 volts won't feed the voltage regulator. You appear to be relying on 55 volts coming through the load on the right hand side of your diagram. Light-loads will cause this circuit to fail.

Also, as soon as you activate the MOSFET, the voltage regulator now receives an input close to GND and turns off thus, deactivating the MOSFET and you might get an oscillator.

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  • \$\begingroup\$ Also there's Rm2 pulling the gate to Vin. Linear regulators can't usually sink current, so the gate will likely be damaged by over voltage. \$\endgroup\$
    – John D
    Jul 29 at 18:04
  • \$\begingroup\$ Ok, so the diode should be reversed and Rm2 should be removed in order for the circuit to work as intended? @JohnD Also yeah, now I realise it. The circuit must have a load on the right hand side, else the regulator won't work at all. Wait, I think I got a solution. Instead of using the diode and then instantly the LDO, what if we used the (same schematic as Vince's)[bit.ly/43OBAxx] (the one before the last one, with a diode and resistor at the PNP diode's gate?), and attach the 12V regulator at the transistor's gate towards the mosfet, would that work? I'll post a new schematic soon \$\endgroup\$
    – Mito
    Jul 29 at 18:53
  • \$\begingroup\$ Here, what do you think about this updated schematic, or should I make a new post? \$\endgroup\$
    – Mito
    Jul 29 at 18:58
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    \$\begingroup\$ @Mito - Hi, (a) Re: "I posted a new schematic as you said as a temporary answer". You misunderstood what Andy meant about using a temporary answer. He was describing this technique. Note that the draft answer must not be submitted. It is used only to upload the image(s) & is then discarded. You are not allowed to post a non-answer as an answer, so it has been deleted. || (b) If you are unsure about an answer from person X then ask person X in comments on their answer. Do not ask person Y about an answer from person X. \$\endgroup\$
    – SamGibson ♦
    Jul 30 at 14:26
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    \$\begingroup\$ (cont'd) (c) It is too late to change the schematic in the question, as you have answers to the original question. If you want to ask about different schematics, then consider upvoting the useful answers here, áccepting your choice of the "best" answer, and then asking a new question about the new schematics (with a link back to this question for context, if relevant). Do not "move the goalposts" on a question by changing it, after receiving an answer. Thanks. \$\endgroup\$
    – SamGibson ♦
    Jul 30 at 14:26

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