Electrical – Logic mosfet to switch 12v truck battery

You are looking for the Gate Threshold Voltage, marked Vgs(th) in the datasheet. There is a bit of a catch to watch out for here though.

For this particular MOSFET, if we look at the relevant bit of the datasheet:

We can see it's given as a minimum of 0.6V, and a maximum of 1.2V. This looks promising for your 1.8V logic output. However, the catch is that this is for a drain current of only 250uA, so it's not of much practical use, since we'll need more than that to pull the PMOS gate low with the 4.7kΩ resistor, we need to make sure the drain current is acceptable at 1.8V.
To get a better idea of what voltage we need at the gate to sink useful currents at the drain, we would usually need to take a look at the graphs for Id over Vgs and Id over Vds for different gate voltages is also useful:

Unfortunately both only goes as low as a gate voltage of 2.25V, so we don't really know how much current the drain will sink at 1.8V. Since at 2.25V Id is over 10A, it's a reasonably safe bet that at 1.8V we'll be fine sinking 3.3V / 4.7kΩ = 0.7mA, but we can't be certain, which is ideally what we need to be.

Solution options:

• Use a different N-ch MOSFET with a guaranteed drain current of over 0.7mA at 1.8V Vgs given in the datasheet.
• Use any small NPN transistor which has a Vbe of ~0.7V and will be turned on easily with 1.8V (with current limiting base resistor of course. For a gain of e.g. 100, assuming you want to pull as low as possible, calculate using (1.8V - 0.7V) / (0.7 / 100) = ~150kΩ, then halve this to compensate for gain reduction at saturation, e.g. 75kΩ or less. Another conservative rule of thumb is to assume a gain of 20-30)
• Do your own tests with a few IRLML2502s and confirm there is more than enough drain current at a Vgs of 1.8V.
• Increase R2, so less current is needed to pull the gate down.

Personally I'd go for using an NPN, since it's cheaper - you could get a suitable part for a couple of pence, compared to the 33 pence for the MOSFET.
For sticking with this part though, unless you need fast switching, or there is a lot of noise present, I'd probably go for simply increasing R2, to around 15kΩ, so you only have to sink 3.3V / 20kΩ = 220uA. This is less than the value given at a max of 1.2Vgs, so you can be certain it will be able to pull the gate down easily.

One other possible option which I use regularly is to use an IO in open drain mode - many microcontrollers have certain pins which are tolerant of higher voltages than their supply voltage, so if your micro has a 3.3V tolerant pin which can be used in open drain mode (you don't give the part number so I can't check this, although you may have done so already) then you can do away with Q1 completely and use this instead.

You are going to need to re-think how you drive the FB180SA10B FET. The VgsTH spec of 4V max is only going to assure you of a measely 250uA of drain current. One would have to guess that if you plan to use a monster FET looking like this:

...that you are planning to put some serious current through the part. For this you really need to be looking at Figure 2 from the data sheet to get an idea of the kind of gate drive voltage that you will really need to drive this FET:

For example if you wanted to operate the part at 20usec pulses of 100A and keep the package junction temperature at a safe level (assuming you heat sink the part properly) maybe you could live with 105 watts of device power dissipation. At 105 watts and 100A of drain current the device would be operating with a Vds of about 1.05 volts. Looking at the above graph you can see that this puts you at a VGS in the range of 6 to 6.5 volts. And you really would do well to increase the drive even more upwards.

There are no 5V logic gates that will provide the kind of drive needed here. If you have to use a 5V supply then you should be looking for a FET driver that has internal boost circuitry to produce a gate drive of 9 or 10 volts.

Also note that these devices have a HUGE gate capacitance. The part is speced with:

To reduce switching losses in the FET you want to swing the gate with fast rise times. Swinging this large capacitance fast requires a FET driver than can source a very respectable amount of current. Few standard logic gates will come close to the requirements for this.