24V-48V Reverse polarity protection P channel FET - Page 1
Forum section: electronics / beginners

author: PeLa
author_rank: Contributor
timestamp: August 24, 2020, 06:48:39 pm
content: Hi there,I cannot figure this one out. Can anyone point me in the right direction?I want to protect a load from reverse polarity using a P channel Mosfet. I am quite sure I know how to do that for 12V. (see image and spice simulation)Now I want to use a 24-48V input and still be gentle with the gate (Vgs +/-20V). I could probably use a voltage divider, divide my 24-48V by 3 and power the gate. Job done.But isn't there a way to use a Zener diode to always get e.g. 12V at the gate when the input voltage is positive and >=12V? My hope is that that the input protection would work across a wider voltage range, say 12-48V.No matter where I place the Zener, I can't seem to get it right (BS circuits not shown).Please don't tell me to use a rectifying diode. Even I know how to do thatCheers,Pete

author: Doctorandus_P
author_rank: Super Contributor
timestamp: August 24, 2020, 07:10:45 pm
content: This should work:The TVS is a "transient voltage surpressor" which is a sort of bidirectional zener diode.Ugs threshold voltage is often around 4V, so choose a TVS with a voltage above that, but below the 20V maximum for Ugs.

author: PeLa
author_rank: Contributor
timestamp: August 24, 2020, 07:17:29 pm
content: Thanks!It does indeed. Now I just have to figure out why. But I won't bother you with that.Thanks again!Pete

author: ArthurDent
author_rank: Super Contributor
timestamp: August 25, 2020, 01:19:02 am
content: Here's another way that should work just fine.https://youtu.be/7Tk5ghH_U2s

author: girishv
author_rank: Regular Contributor
timestamp: August 25, 2020, 01:33:42 am
content: Quote from: PeLa on August 24, 2020, 07:17:29 pmThe attached schematic is from a TI application note.

author: Jay_Diddy_B
author_rank: Super Contributor
timestamp: August 25, 2020, 03:36:37 am
content: Hi,The circuit proposed by girishv:This isONLYsuitable for application less than 12V. The MOSFETs maximum gate source voltage, Vgs, will be exceeded if this circuit is used at higher voltages. This will lead to MOSFET destruction.This is the problem that the OP was asking about.Jay_Diddy_B

author: Jay_Diddy_B
author_rank: Super Contributor
timestamp: August 25, 2020, 03:56:37 am
content: Hi,Try this circuit:I have attached the model.Jay_Diddy_Bp ch rp.asc(1.37 kB - downloaded 154 times.)

author: girishv
author_rank: Regular Contributor
timestamp: August 25, 2020, 07:06:04 am
content: Quote from: Jay_Diddy_B on August 25, 2020, 03:36:37 amI apologize for posting incorrect solution. There is a lot to learn. @Jay, Thank you very much for correcting.

author: PeLa
author_rank: Contributor
timestamp: August 25, 2020, 05:22:44 pm
content: Hi Jay,Thank you!I have played around with it a bit. What is D2 for?Cheers,Pete

author: PeLa
author_rank: Contributor
timestamp: August 25, 2020, 05:36:28 pm
content: Quote from: ArthurDent on August 25, 2020, 01:19:02 amThanks ArthourDent, I did not actually know that GreatScott Video. Also, It's explained well enough for me to understand

author: Jay_Diddy_B
author_rank: Super Contributor
timestamp: August 25, 2020, 10:12:51 pm
content: Quote from: PeLa on August 25, 2020, 05:22:44 pmD2 is to stop the B-E junction of the BJT from being reverse biased during normal operation.The BJT is to turn off the MOSFET quickly if the input polarity is reversed.Regards,Jay_Diddy_B

author: Zero999
author_rank: Super Contributor
timestamp: August 26, 2020, 09:51:41 am
content: How much current does the device use? At high currents, use an ideal diode controller, such as the LTC4357 and an N-channel MOSFET.

author: PeLa
author_rank: Contributor
timestamp: August 26, 2020, 11:10:53 am
content: Thank you, Jay!

author: PeLa
author_rank: Contributor
timestamp: August 26, 2020, 11:27:37 am
content: Zero999, it's a BLDC motor driver. Peak input current about 7A. So this would work for me.Until now I didn't even know of ideal diodes controllers! Thank you!The LTC4357 is not cheap and definitely nothing that I can just pull out of my drawer. But it is definitely something that I will be happy to have learned about once I start my next project.I am very positively surprised by the number of helpful replies! Thank you all!

author: PeLa
author_rank: Contributor
timestamp: August 26, 2020, 11:43:05 am
content: Quote from: Jay_Diddy_B on August 25, 2020, 10:12:51 pmhold on... how can there be a reverse bias between B-E when they are both connected to the same line, i.e. without D2?

author: Jay_Diddy_B
author_rank: Super Contributor
timestamp: August 26, 2020, 12:38:35 pm
content: Hi PeLa,You have replaced D2 with a short circuit. If you short the B-E junction the BJT can never turn on.Consider the circuit with D2 replaced with an open circuit.The B-E junction of the BJT will be reverse biased during normal operation.LTspice and almost all other SPICE simulators will not show the reverse breakdown of the B-E junction.Regards,Jay_Diddy_B

author: Peabody
author_rank: Super Contributor
timestamp: August 26, 2020, 02:08:57 pm
content: Quote from: PeLa on August 25, 2020, 05:36:28 pmThat's more than I can say. You're trying to protect against polarity reversal at the source (i.e. - the battery). But Great Scott's circuit is about reversing polarity at the load when the load is a battery. His mosfet appears to be in the opposite orientation to yours.I don't understand why for your purpose the simple zener circuit wouldn't work:

author: Jay_Diddy_B
author_rank: Super Contributor
timestamp: August 26, 2020, 02:31:29 pm
content: Quote from: Peabody on August 26, 2020, 02:08:57 pmThe issue with this circuit, with just the Zener, is that there is no discharge path for the MOSFET gate-source capacitance. The gate can remain charged. If you reverse the polarity with the gate charged, reverse current will flow until the MOSFET gate is discharged through the 100kresistor.The BJT circuit speeds up the gate discharge if the polarity is reversed.Jay_Diddy_B

author: Peabody
author_rank: Super Contributor
timestamp: August 26, 2020, 04:46:20 pm
content: Quote from: Jay_Diddy_B on August 26, 2020, 02:31:29 pmOk, I understand. But would the circuit below also work? It seems it would keep the source and gate at the same voltage, at least for a while. It would be slower to turn on normally, but might it not also solve the problem of the gate lagging behind. The capacitor would act like short-acting version of your transistor, which would be enough to get the gate charge to the right place. Well, I don't have any Spice stuff installed, but maybe someone could see how this performs. I don't know what the cap value would be. It seems it should be large as compared to the gate capacitance.I had also thought about putting the capacitor across the gate resistor, but the GS voltage limit would be temporarily exceeded every time you turn on power normally. The capacitor would act like a short, bypassing the resistor, and if your supply is 48V, then the GS voltage would be -36V for a short time. And the Zener current would spike. Temporarily bypassing the Zener with the capacitor doesn't have that problem since it makes the GS voltage zero for a short time.Edit: I guess what's confusing me about the transistor circuit is that it seems what you need to prevent in polarity reversal is the gate being more than a volt negative to the source, which would turn the mosfet on. If the gate is higher than the source, that doesn't matter - the mosfet is still off. But the transistor seems designed to discharge the gate down to the source voltage, which isn't needed - the Zener would do that anyway. If the gate is low enough to turn on the mosfet, the transistor collector will be at a lower voltage than the emitter, and I don't know what happens then. If polarity is reversed, all the points marked as ground will rise to 48V or whatever, and so will the source voltage. In fact, it may even go higher than that if there are big caps on the rail - they may act like a charge pump. So I think even if the transistor is needed, it maybe needs to be flipped so it charges the gate to the source voltage if its base is high.Either that, or I've gone completely Australian with everything upside down and turning the wrong direction and I'm hopelessly confused. It wouldn't be the first time.