Reverse Voltage Protection for Development Board - Page 1
Forum section: electronics / projects

author: Luis Estrada
author_rank: Contributor
timestamp: February 19, 2021, 12:45:13 am
content: Hello All,I am re-designing my development board for the PIC18F45K50 MCU and I have been meditating on appropriate voltage protection. I came up with the attached design schematic; my system is designed to handle 9 - 12V DC. Please let me know if this is an appropriate solution to reverse voltage and over voltage protection. Any feedback is welcome!

author: bdunham7
author_rank: Super Contributor
timestamp: February 19, 2021, 01:05:01 am
content: How will the overvoltage part work when the body diode still conducts? Just a simple 1N4001 rectifier would work just as well in this circuit and the L7805 would give you 35V + of overvoltage protection.

author: nctnico
author_rank: Super Contributor
timestamp: February 19, 2021, 01:48:50 am
content: I'd use a fat 12V TVS. Reverse polarity and overvoltage protection in one.

author: twospoons
author_rank: Frequent Contributor
timestamp: February 19, 2021, 01:58:01 am
content: Quote from: nctnico on February 19, 2021, 01:48:50 amAlmost - you'd want a PTC fuse in there too. Otherwise all the magic smoke might leak out.

author: Luis Estrada
author_rank: Contributor
timestamp: February 19, 2021, 03:01:47 am
content: Hmm I didn't think about that! I didn't consider diodes because of the pesky voltage drops, but I am assuming that the 7805 still will deliver ~5V DC regardless if the input voltage is off by 0.5 - 0.7 V, is this correct?

author: ajb
author_rank: Super Contributor
timestamp: February 19, 2021, 03:07:14 am
content: Quote from: Luis Estrada on February 19, 2021, 03:01:47 amA TVS goes in parallel with the input (after the fuse!), so has no impact on the voltage at the input to the regulator. HOWEVER, if you have the headroom, a series diode will actually reduce the power dissipation in the regulator because it reduces the difference between the input and output voltages. The total dissipation of the circuit remains the same, just some moves from the regulator to the series diode, which can be useful, although a a few hundred millivolts isn't likely to be terribly significant unless your design is marginal to begin with. But in any case, it's not really going to have a downside with a linear regulator unless you're already close to the minimum input voltage for your regulator, but if you're using a standard linear regulator you may be able to work around that by switching to a low-dropout regulator.

author: Luis Estrada
author_rank: Contributor
timestamp: February 19, 2021, 03:09:15 am
content: Yes! I was wondering about fuses, I didn't think you needed them in low DC voltage applications

author: Luis Estrada
author_rank: Contributor
timestamp: February 19, 2021, 03:11:53 am
content: Oh ok! I think I do have the headroom for that, I will prototype it first to make sure. The current design works fine, however, I want to make sure the magic smoke does not escape. Those PICs are expensive! Thanks

author: bdunham7
author_rank: Super Contributor
timestamp: February 19, 2021, 03:22:26 am
content: Quote from: Luis Estrada on February 19, 2021, 03:11:53 amThe current design works because of the L7805. You could literally eliminate every other component and it would likely still work fine, although I'd keep C1 and C2. The regulator will continue to work just fine down to below 7 volts input, so if you use a 1N4001 on the input, you'll be fine down to well below 8 volts. I don't know what your input is, but if it is a wall wart, you really don't need extra protection. The L7805 will provide that as well, at least for the outputs. If you use a TVS just to make it somehow bulletproof, use a higher voltage one, not 12 volts, as you don't want or need it to activate until the regulator is threatened.

author: ludzinc
author_rank: Unknown
timestamp: February 19, 2021, 03:52:48 am
content: Quote from: Luis Estrada on February 19, 2021, 12:45:13 amTo me it looks like your two PWR_FLAG nodes bypass the MOSFET altogether. You might want to have a look at that.

author: Gary350z
author_rank: Regular Contributor
timestamp: February 19, 2021, 06:11:03 pm
content: Here's some info:How to protect circuits from reversed voltage polarity - Afrotechmodshttps://youtu.be/IrB-FPcv1DcAutomotive MOSFETs Reverse Battery Protection - Section 3.3https://www.infineon.com/dgdl/Reverse-Batery-Protection-Rev2.pdf?fileId=db3a304412b407950112b41887722615Design Guide - PMOS MOSFET for Reverse Voltage Polarity Protectionhttps://components101.com/articles/design-guide-pmos-mosfet-for-reverse-voltage-polarity-protection

author: gnuarm
author_rank: Super Contributor
timestamp: February 19, 2021, 06:42:09 pm
content: Other than the FET being in backwards, the design looks good to me. Swap the drain and source pins and you are good to go. Of course, move the gate protection diode as well. I assume that's a Zener. Some people use a symbol with just one knee, but mostly the Zener symbol has two knees. I find this confusing, but searching a bit I see the one knee symbol used for an avalanche diode which is a higher voltage equivalent of the Zener, often used interchangeably. But it's also used for a step recovery diode...I see this sort of FET used in motor control designs to protect against reverse polarity.

author: AndersJ
author_rank: Frequent Contributor
timestamp: February 19, 2021, 07:09:30 pm
content: Why do you say the FET is connected backwards?Isn’t this how reverse polarity protection is done with a FET?

author: gnuarm
author_rank: Super Contributor
timestamp: February 19, 2021, 07:40:53 pm
content: Quote from: AndersJ on February 19, 2021, 07:09:30 pmThe body diode will conduct even when the FET is turned off. Look at your schematic. The diode is shown clearly. It is a real diode and will conduct in the direction shown.The FET conducts in both directions, but only when the gate has an appropriate voltage to the source. In fact, in your schematic the only way to turn on the FET is for the body diode to conduct to raise the voltage of the source. Swap the drain and source and this circuit is good.

author: SiliconWizard
author_rank: Super Contributor
timestamp: February 19, 2021, 08:03:59 pm
content: This is intended. Pretty typical setup for reverse polarity protection:https://components101.com/articles/design-guide-pmos-mosfet-for-reverse-voltage-polarity-protectionIt's meant to protect against reverse polarity. So in direct polarity, the body diode will indeed conduct, but as soon as the input voltage is high enough, the mosfet will conduct too and thus with no significant drop (as opposed to using a single series diode for protection.)Now in direct polarity, I can't see how this can protect from overvoltage, though, due to the body diode. I guess the OP is hoping the zener diode will do the trick - but unfortunately, when the voltage exceeds the zener voltage, this zener diode will conduct to ground through a 100k resistor. So the pull down effect will be negligible. So now it won't protect against over-voltage as far as I can see.

author: bdunham7
author_rank: Super Contributor
timestamp: February 19, 2021, 08:29:31 pm
content: Quote from: gnuarm on February 19, 2021, 07:40:53 pmI didn't mention all the problems with the original circuit and I didn't look at the detailed specs of the parts, but reversing it won't help at all.The way he has it, the MOSFET can't do diddly about overvoltage (nor does it need to because of the L7805) because of the body diode. However, even without that, the expectation that the zener will pull up the gate and turn off the MOSFET in case of overvoltage is misguided because the gate will always be VZlower than the source, and if the MOSFET requires more than that to turn on, it will never turn on at input voltages lower than VZ. Of course the circuit still 'works' because of the body diode. But there's no advantage to all of that over a 1N4001.If you flip the MOSFET, no aspect of the circuit will work. The MOSFET will be always on for any forward voltage over some threshold (no overvoltage protection) and the reversed body diode will now pass a reverse polarity input. Now your circuit is literally worse than no circuit at all (a straight wire).If you want reverse and overvoltage protection, I think you need two MOSFETs, no?

author: gnuarm
author_rank: Super Contributor
timestamp: February 19, 2021, 10:19:52 pm
content: Yes, the OP's drawing is correct. I was having a brain cramp getting reversed and direct mixed up. This is a P-FET and it is backwards from the usual way to control current flow. So whenreversevoltage is applied it will prevent all current flow, both the diode and the FET.Sorry for creating confusion.

author: Luis Estrada
author_rank: Contributor
timestamp: February 21, 2021, 11:30:49 pm
content: It is a P-Channel MOSFET, I followed a similar schematic

author: Luis Estrada
author_rank: Contributor
timestamp: February 22, 2021, 03:43:37 am
content: Hey All,I took all of your consideration into account and came up with a new protection circuit. This new circuit contains a PTC fuse, TVS diode (Bi-Directional) , and finally a 1N4001 rectifier diode. I don't know a lot about TVS Diodes or PTC fuses (School didn't teach us about this, shocker! I Know!) but I included the part identifiers so you can find the parts.

author: bdunham7
author_rank: Super Contributor
timestamp: February 22, 2021, 04:01:22 am
content: That looks robust enough, but unless you are wanting to use 24 volt inputs, I'd use a slightly lower voltage TVS, like a 1.5KE30CA to keep the maximum clamp voltage under the maximum input voltage of the LM7805. Notice that your parts count may have gone up, but the parts are dirt cheap.Edit: I spoke too soon. Your fuse is only rated for 9 volts, you want that to be higher. Try Bourns MF-SM050-2

author: james_s
author_rank: Super Contributor
timestamp: February 22, 2021, 04:50:33 am
content: Quote from: ludzinc on February 19, 2021, 03:52:48 amIIRC that isn't a node, power flag just flags a net as being power, it's only used for DRC. I don't even remember how it works exactly because I haven't bothered to use it in a long time.

author: gnuarm
author_rank: Super Contributor
timestamp: February 22, 2021, 05:26:50 am
content: I would use the 1.5KE24A as the clamping voltage is what the circuit can see in an over voltage situation. The other parts can allow too high a voltage to reach the regulator. I would also use a unidirectional TVS, not a bidirectional device.If you use a unidirectional TVS, it opens the fuse at a very low voltage (a forward diode drop) in the reverse direction and the diode is not needed. The data sheet shows the unidirectional diode limiting the voltage to 1.0V at 10A.

author: Terry Bites
author_rank: Super Contributor
timestamp: February 22, 2021, 04:35:30 pm
content: I am not familiar with that diode but it needs to be a Zener say 9 or 10V.Then it will work- you can even put a cap across the gate resistor for soft start.When the MOSFET is on there is a very low voltage drop drom source to drain and the body diode has too low a potential across it to get it into conduction.When the input voltage is reversed the MOSFET is off, the body diode is reverse biased. A tiny small leakage current will flow in it.Excellent app note from Infenonhttps://www.google.co.uk/url?sa=t&rct=j&q=&esrc=s&source=web&cd=&cad=rja&uact=8&ved=2ahUKEwji2f_k9f3uAhUkQEEAHd_VBeIQFjANegQIDxAD&url=https%3A%2F%2Fwww.infineon.com%2Fdgdl%2FReverse-Batery-Protection-Rev2.pdf%3FfileId%3Ddb3a304412b407950112b41887722615&usg=AOvVaw2eSUGgd99Sukx99QiUQxRfIgnore the naysayers!

author: Luis Estrada
author_rank: Contributor
timestamp: February 27, 2021, 08:47:12 pm
content: Thank you Terry! I did some simulations on falstad and it works great. Speaking of simulators, what simulator are you using?

author: gnuarm
author_rank: Super Contributor
timestamp: February 28, 2021, 02:21:19 am
content: Quote from: Terry Bites on February 22, 2021, 04:35:30 pmIt will certainly work, but it is overkill. Using a 12 V power source to drive a 5 V output linear regulator has much head room. So there is absolutely no reason to not use a simple diode. No need for three parts, one of them as big as the regulator being protected, when one small one will do.A simple diode is the perfect solution. An added advantage is you don't need to simulate anything. Amazing, no?I would add that putting a capacitor across the resistor will not produce a soft start, just the opposite making the resistor ineffective at limiting the current in the Zener which, btw, is not required. The input is 12 V and the Vgs max is 25 V on the part number shown here, so not likely to damage the gate. If you want a soft start, you would put the cap across the gate to source, in parallel with the diode.It is not a good idea to trust a simulator to simulate details it is not good at. For example, how does the FET get turned on in the first place? The transistor is off conducting a very tiny current and both the Zener and the 7805 conduct a tiny current at very small voltages. Which tiny current produces enough voltage to make the transistor conduct enough current to turn on??? The simulator only knows the values you use. It doesn't know how much resistance the load presents before the circuit powers up.One final note. The symbol on your schematic is for an N channel FET which won't work. The part number you give is a P channel device which will work. Hard to analyze a design when you use wrong symbols.So what is the down side of using a diode???

author: Luis Estrada
author_rank: Contributor
timestamp: February 28, 2021, 02:38:21 am
content: That's a KiCAD thing, I didn't even notice it lol! But yea, I see it is kind of an overkill, I just wanted to learn more about circuit protection as new EET grad.

author: gnuarm
author_rank: Super Contributor
timestamp: February 28, 2021, 04:26:21 am
content: Quote from: Luis Estrada on February 28, 2021, 02:38:21 amNothing wrong with that. Part of the protection problem is there are so many solutions, so many it is hard to know when to use what.A design I'm involved with is using a $4 chip and four FETs to select between two power sources. Yeah, a bit of overkill. The guy designing it never saw a Linear Technology part he didn't like. The parts are good, but not cheap. He had a $3 input protection chip and two FETs to protect against over and reverse voltage, when a diode would do pretty much the same job (everything being "protected" already had nearly as high a voltage rating). He is also using a $4 switching regulator when a $0.50 part would do the job. Why? He says he knows "the LT part will work" as if the other regulators don't. Whatever...

author: Luis Estrada
author_rank: Contributor
timestamp: February 28, 2021, 04:32:27 am
content: Yeah I understand, this project is not so critical. It's just a Development board for a PIC18F45K50 mcu, which works at max 5.1 V so I wanted to make sure it didn't get damaged. I did some calculations and the Diode will do fine.

author: Terry Bites
author_rank: Super Contributor
timestamp: February 28, 2021, 01:36:28 pm
content: I use 64 bit Microcap11. It would use 12 but it keeps crashing on win10 pro.https://www.spectrum-soft.com/download/download.shtm