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I need a simple circuit to protect my device from overvoltage (more than 13V). Typical working DC input voltage is 7-12V. Can I use this circuit:

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Will it work?

UPD. OK, it will not. I was too stupid. Yet another try:

enter image description here

(Sorry, I forget to change default names. My MOSFET is IRLML6402, transistors is BC817, zener 13V.)

When input < 13V, Q2 closed, Q5 open, MOSFET open. Else Q2 open, Q5 close, MOSFET close. OK?

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  • \$\begingroup\$ No, when the Zener fires it will maintain 13V source to gate, so the FET will stay on. \$\endgroup\$ – John D Jul 17 '14 at 15:05
  • \$\begingroup\$ All this circuit will do is prevent voltages of less than a half a volt or a volt from getting through. \$\endgroup\$ – Spehro Pefhany Jul 17 '14 at 15:08
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Your circuits (whether they work as you intend or not) seem to be based on the idea that when an over-voltage condition occurs, you will isolate the load with a MOSFET.

I think this is not a great approach. Consider, why did over-voltage occur anyway? You could say that it's because the impedance of the load is too high. Were the load impedance lower, then more current would flow, providing a way to sink whatever excess energy there is without the voltage becoming too high.

So, if you try to isolate the load with a MOSFET, that just effectively increases the load impedance. For a great many things that might have led to your over-voltage fault in the first place, they will simply respond by increasing voltage even more. This is problematic, because MOSFETs have relatively low maximum voltages. When a source-drain voltage is applied in excess of this, they go into breakdown and conduct (not unlike reverse breakdown of a diode). Consequently your load gets toasted anyway.

A different approach is a crowbar circuit. Instead of disconnecting the load, you short it out (usually with a TRIAC or SCR). If you put a short across the load, the voltage will be reduced (to 0V, for an ideal short). This provides very effective protection to the load from over-voltage.

Of course this also results in huge current drawn from the power supply. That blows the fuse, which isolates the power supply from the circuit, and does so much more effectively than a MOSFET can. Furthermore, the crowbar continues to conduct until there is no more energy which can result in a voltage across the load, which might take a little time, depending on just what caused the fault, and what reactive components are in the circuit.

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  • \$\begingroup\$ Not a fan of this answer because it doesn't actually answer his question. There are many reasons why you might have higher voltage than expected. I suspect the most common is using the incorrect wall-wart with a given device. I use a mosfet cutoff circuit with a zener for specifically this purpose in many of my designs. \$\endgroup\$ – akohlsmith Feb 5 '17 at 4:28
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You're close.

schematic

simulate this circuit – Schematic created using CircuitLab

I've chosen BSS84s only because they had low Vgsth which shows the effect more readily; you'll want to find more suitable devices for M1 and M3 as they carry all the current. Feel free to play with it in CircuitLab, I find it needs some tweaking for any given application.

M1 is for reverse voltage protection. It's intentionally installed "backward" (with the not-shown body diode cathode toward the right side of the circuit). When voltage of the correct polarity is applied, the mosfet is on and M1 is more or less invisible to the circuit. When reverse-polarity voltage is applied, M1 is off and its body diode is also blocking current flow. M1 must be selected to be able to withstand sufficient forward current as well as reverse voltage. For higher voltages you may run into trouble as many common mosfets tend to "spec out" at about 20V or so.

M3 is normally on via R2; R1 biases D1 and the voltage is applied to M4's gate. As the applied voltage rises we get closer and closer to M4's turn-on voltage and it starts to pull M3's gate back up toward the voltage at its drain, shutting it off. You'll have to select D1 such that M4 is fully-on at the desired overvoltage point, and M4 such that it's not starting to conduct too early.

The circuit isn't perfect; I've already mentioned that it's not suitable for voltages approaching 20V because it begins to get difficult to find mosfets which can withstand this voltage; I primarily use this for smaller, low-current battery operated applications where we're in the 3-10V range, where it has performed very well given how inexpensive it is.

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What you want to do first is Google "Over voltage protection IC", and then follow some links. http://www.onsemi.com/pub_link/Collateral/NCP346-D.PDF, for instance, provides a schematic and inner workings of a representative unit. If you don't want to spring a few bucks for the part, you can have a go at reverse engineering it.

"If I have seen farther than others, it is because I have stood on the shoulders of giants" - Isaac Newton

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  • \$\begingroup\$ This question has an education signification for me. Of course, today for almost everything there is a single IC that do it :) \$\endgroup\$ – user54579 Jul 17 '14 at 18:50
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No. If you apply 15VDC to the DC input yes the Zener will start to conduct, however this is a moot point since you have a large negative voltage from the gate to source of the PMOSFET. This will certainly drive it into saturation and DC IN will be connected directly to Protected DC with only Rds(on) between them.

A good way to picture it is if the zener is conducting with current flowing from cathode to anode, (which it will for DC inputs > 13V) replace it with a 13V battery with the negative terminal where the anode was and the positive terminal at the cathode and it will become clear that M1 will be driven to saturation.

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  • \$\begingroup\$ OK, I understand. But can I modify the circuit to get it work? I need no very expensive circuit with many components. The overvoltage is very very rare situeation for me, but just in case I need a simple protection. \$\endgroup\$ – user54579 Jul 17 '14 at 15:17
  • \$\begingroup\$ To offer a solution we need to know more details. What is the output driving? What are the load requirements? If it's little current and non-perfect regulation is needed then a Zener from GND to Protected DC might work. \$\endgroup\$ – ACD Jul 17 '14 at 15:20
  • \$\begingroup\$ Or a TVS/Zener to ground AFTER a PTC (Polyfuse) type device or even a conventional fuse. It all depends on your requirements. \$\endgroup\$ – John D Jul 17 '14 at 15:32
  • \$\begingroup\$ I add another attempt. Check please. \$\endgroup\$ – user54579 Jul 17 '14 at 16:20
  • \$\begingroup\$ Closer, but you will need some sort of hysteresis, otherwise as you approach the threshold you could get oscillation, or get the FET into the linear region. Just off the top of my head you could experiment with a resistor from the collector of Q5 to the base of Q2. \$\endgroup\$ – John D Jul 17 '14 at 17:17
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Have you considered using a Zener Shunt with a resettable fuse. If you need faster response, use a TVS instead. This will effectively sung the surge and isolate the supply from the current inrush.

A specific IC is unnecessary unless you need something very specific.

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