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I'm going to assume the obvious answer is no, but I figure I should check with the experts before I go off and waste a bunch of my time.

Say you have several MOSFETS in a row controlling the flow through different components like solenoids. If they all share a ground line, and the solenoids all share the same supply line, could too high of a voltage force the MOSFET open?

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ There's a proper schematic editor that's less confusing .. it looks like you have the MOSFETs connected directly across the supply, it's not clear where the solenoid you're controlling is? MOSFETs do have a maximum rated voltage which will destroy them, is that the question? \$\endgroup\$
    – pjc50
    Aug 6, 2014 at 14:19
  • \$\begingroup\$ i understand there is a voltage they can tolerate. I just want to make sure that anything below that wont cause it to leak at all \$\endgroup\$ Aug 6, 2014 at 14:36
  • \$\begingroup\$ @user3256725 I've edited to include my best interpretation of your ascii-art schematic. \$\endgroup\$
    – Phil Frost
    Aug 6, 2014 at 15:17

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could too high of a voltage force the MOSFET open?

If by "open" you mean "conducts current", yes. If the source-drain voltage is in excess of a maximum specified in the datasheet, the MOSFET will experience avalanche breakdown. If the current through the MOSFET is not somehow limited, this will result in failure of the device by overheating.

For an example, I've picked the very common 2N7000:

enter image description here

This particular MOSFET is not rated for avalanche operation, so you do so at your own peril. However, some power MOSFETs, especially those intended for motor drive applications, will specify a maximum energy the device can tolerate in avalanche breakdown conditions.

This is especially relevant in your situation, since you say that you are switching solenoids, which are very inductive. When you turn the MOSFET on, current builds in the solenoid until it is limited by resistance, and a strong magnetic field exists. When the MOSFET is switched off, the current must continue flowing somewhere, and the energy stored in the magnetic field must go somewhere else. Since you have not provided any other possible path for the current, when you switch one of your MOSFETs off, the inductive kick of the solenoid will drive the MOSFETs into avalanche breakdown, and quite likely damaging the MOSFETs, or any other components connected to the power supply.

The usual solution in this situation is a "flyback diode". I know this has been covered many times on this site, and someone will suggest a better reference, but in addition to searching for "flyback diode" generally, check out Questions About Inductors.

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If you apply a supply voltage higher than a MOSFETs breakdown voltage it will catastrophically fail. If you apply a voltage below this, then you need to ensure that your gate voltage is never floating or above the MOSFET threshold voltage unless you want it to turn on. If you want it off, the gate voltage needs to be below the threshold voltage referenced to the source voltage (in an NMOS). As the gate voltage gets closer to the threshold voltage, the more leakage current you'll get, but it should orders of magnitude below what your solenoids need unless you're getting close to the threshold voltage.

(the ends of the supply line is an open circuit to ensure that electricity only flows when the mosfet opens.)

This should be written "electricity only flows when the mosfet closes". Just like a switch the terminology is closed=short, open = open circuit.

connecting the source pin of each mosfet together, so when one MOSFET is open, current will flow directly through the source pins of all the others

This isn't a problem from a circuit standpoint. It's actually quite common. I just want to ensure you understand that no current will flow through the source pins (except for the MOSFET that is on). Instead the current will flow by the source pins using the wire you're going to connect them together with.

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