convert ground when closed triggers

I missed your last reply. Normally Open GND switches will almost always measure +12V on their outputs. More than a DMM voltage test is required to know what switching is involved. And the fact that that +12V cannot energise a grounded relay coil might prove my point. Instead try connecting the relay coil between +12V and the switch - that's how you'd use a relay on the Luggage Compartment switch (and how I reckon you should change your Hood switch wiring).

oldspark wrote: New to you? You nailed the (MOS)FET wiring!! The only additions - a 1M or 100k resistor from Gate to GND (assuming an N-ch FET) to ensure it turns off (it's too easy for the Gate to get its nA from floating outputs, so pull it down to GND). And add a pull-up resistor (10k? 100k - but significantly smaller than the Gds (G to GND) 1M etc resistor) if the trigger input is is foating etc (not normally "hi" = +12V). A series resistor between the source (switch) and Gate is often added but that's usually to protect the source (switch) from overcurrent in case of FET breakdown, but that's usually irrelevant for switches. (Sorry about "source" vs FET Source... maybe the latter should be referred to only as S?) However I'd prefer to swap the Hood Switch's #1 output to #2 so its GND is only connected when the Hood is open. Then add a normal car SPDT (changeover; 5-pin) relay to disconnects its 30-87a between GND and the HOOD AJAR #10 C2280F input. Otherwise the FET can drainlessly invert the standard OEM wiring/signal. There is no "resistor" to break any current. Only the relay, else as above for a MOSFET.

Thank you very much for the help. I'll buy the things needed, and report back how things go.

oldspark wrote: I missed your last reply. Normally Open GND switches will almost always measure +12V on their outputs. More than a DMM voltage test is required to know what switching is involved. And the fact that that +12V cannot energise a grounded relay coil might prove my point. Instead try connecting the relay coil between +12V and the switch - that's how you'd use a relay on the Luggage Compartment switch (and how I reckon you should change your Hood switch wiring).

The door triggers look like wired the same way as trunk and hood triggers, but I can use +12 from them to power door edge LED lights when doors opened. Not sure if they are differently regulated inside SJB.

Well, if they were wired like the trunk/boot switch, or if they were wired upside-down - ie, switch output to the SJB interchanged with switch pin #1 - then switch #2 powers extra loads/relays as normal. Otherwise ye olde transistor idea... (probably just one thru diodes).

Received MOSFET transistors today, but haven't received resistors yet. I tested with a 10k resistor and it works. I wonder if I can just use a 10k resistor or if I should wait until I receive 1M ones.

The MOSFET does not "need" any resistor. They are added to: - ensure MOSFET turn off when input (Gate) is "not ON". - limit damage to external "input" circuitry given the rare occurrence of a Drain-Gate breakdown thereby shorting +12V or whatever to the Gate hence shorting your input source to the MOSFET's supply (eg, +12V). - a resistor to pull up grounding inputs if required. And if more than one of the above (resistors) is used, consideration must be given to voltage-divider effects, ie the resistance of each inter-connected resistor should differ by some minimum ratio, eg one order of magnitude.

oldspark wrote: The MOSFET does not "need" any resistor. They are added to: - ensure MOSFET turn off when input (Gate) is "not ON". - limit damage to external "input" circuitry given the rare occurrence of a Drain-Gate breakdown thereby shorting +12V or whatever to the Gate hence shorting your input source to the MOSFET's supply (eg, +12V). - a resistor to pull up grounding inputs if required. And if more than one of the above (resistors) is used, consideration must be given to voltage-divider effects, ie the resistance of each inter-connected resistor should differ by some minimum ratio, eg one order of magnitude.

Thanks. I'll just use a 100k or 1M resistor for the gate. In this application, I think short is not a concern since drain is connected to a negative trigger wire similar to the the wire connected to gate.

I used a 100k resistor with a MOSFET for each of the two triggers. It's been several days, and there have been working well. Thanks you all for the suggestions, especially oldspark, who help me solved the problem.

Thanks appreciated. And thank you for the reply. That's one thing I like about MOSFETs - resistors are not critical. The main "optional" resistor is from G to GND to ensure the MOSFET stays off when the input (Gate) is not >+5V etc and that resistor merely needs to be less than 10 MOhm or maybe <1M - ie low enough to ensure any floating Gate voltage is grounded - and the currents involved (namely the Gate turn-on current (Ig?) with Vgs = ON) are typically well under 100nA. The only other resistor is "series" Gate resistor which is to protect the whatever that is connected to the Gate in case of a D-G short circuit. The 3rd resistor is an input issue - if the input to the Gate cannot pull the Gate voltage above Vgs-ON - eg, to ~+5V or higher - then a pull up resistor is required. The only complication or critical aspect is if the resistors from voltage dividers (which they will of more than one resistor is used). Then calcs will have to be done to confirm that Vg is high or low enough with the input ON and not ON.    However separating them by an order of magnitude or more (ie, at least 10 times bigger or 10 times smaller) is usually the simple solution since it means less than a ~10% voltage variance thru voltage divider activity. Simple "logic" does the rest... RG, so if the "stay OFF unless on" resistor from Gate to GND is 1M, then the series Gate "let's protect the switch/alarm/circuit input" resistor can be 100k or smaller. Sanity check... yes, the 100k current will swamp the 1M current, ie the 1M current will keep the MOSFET OFF until the 1M/100k = 10x higher input current turns it on. I'll avoid the pull-up resistor.    With BJT transistors (ie, not FETs) you have to derive the series-Base resistance range from transistor specs (ie, its gain which is usually specified as a range eg 80-500) and from the load size. FETS do not need input (Gate) current limiting whereas BJT (Bipolar Junction Transistors - commonly referred to as "transistors" even tho transistors include FETS & other types - needs to be of some minimum value (that limits current) but some minimum to ensure its resulting (Base) current times its gain (aka β = Beta) supplies enough current to the load. Then you add a Base to GND resistor and Base to input resistor that provide a Vb of about 0.7V blah blah blah.... Nah - transistors (BJTs) for simple amplifier designs, but (MOS)FETs for switching or digital applications like this. (The bonus is then the very high impedance (negligible current) of the FET's Gate etc.) PS - because vehicles are typically 12V or higher, same sized resistors may work. EG, if both our Gate-GND & input-Gate resistors were 1M and therefore a +12V "on" input (or 14.4V etc) means a Gate voltage of +6V (or +7.2V etc) which may be high enough to turn the MOSFET (fully) on. Certainly an L-type would - ie Logic-types needing input voltages under 5V for full turn on.