In the modified schematic below I added D2 and D3 as freewheel diodes. Are these in the right places? Do I need both of them (or maybe just D2)?
One parallel with the FET is really redundant. The body diode is most likely good enough.
I'm pulling the gate to ground (so 11.7v from the source). Vthreshold is only -2.5 volts, so I think I'm OK on that.
Vth is irrelevant (another catch for the young players: it doesn't tell you much anything about when the channel is fully enhanced - you need to go look at the Vds/Id curves drawn at different Vgs). Of course, 11.7V is enough for any typical switching MOSFET, with a good marging.
With that 100k resistor, turn-off will be so slow that you will probably exceed the SOA.
The 100k (R2) is pulling the gate high, so keeping the MOSFET turned off. Q1 pulls the gate to ground to turn it on. How does R2 affect the turn-on time?
Please reread more carefully. But if you are 100% sure that you are always turning off at about Id = 1A and not 60A, this won't be a problem.
That's the main thing that worries/confuses me. The MOSFET is rated 80A drain current at 10v but Figure 3 of the datasheet seems to show it can handle it for only 1 millisecond!
Only? 80A at 10V is 800W of dissipation, in a miniscule die. You are using a tea kettle power to heat up a grain of sugar. Or similarly, you are using full nuclear reactor power to heat up one cup of tea!
Of course it can handle it for only 1 ms. But, sadly, these SOA curves can be totally unrealistic, so you need to apply derating for robust operation. So it's going to be a lot less than 1 ms.
Why would you want to operate the FET at Id=80A and Vds=10V? It would leave no voltage for the plane - you are powering a short circuit. Do you realize this condition only happens during switching?
So, switch faster!
1ms is a loooong time. Don't use the FET in linear mode for that long, it has no advantages. You get the EMI reduction / ringing mitigation robustness advantage of slow switching easier, you don't need to go that slow. Think about the range of about 1-10 us. Dissipate that 800W only for 1us, and the die will be OK!
But with that 100k resistor of yours, your switching may actually take 1ms, or even longer. That's why you should use a smaller resistor or a gate driver. You can approximate the switching time from the RC time constant caused by that resistor and the Total Gate Charge of the MOSFET.
The best N channel MOSFETs I can find have a Rds(on) only maybe 1/2 of what this part does - I'm not sure that's enough to solve the problem.
Any proper distributor should have several N MOSFETs with Rds(on) < 1.5 mOhm (when fully turned on) in stock. Parts below 1 mOhm start get a bit rare. You can parallel two.
Good lord. 60 amps?
Nothing special or difficult. No need to up the voltage, IMO, although it would allow some saving in wirings, thus weigh, but wiring is probably very short to begin with. Additionally, it's probably not doing full throttle all the time, so some efficiency loss on wiring is allowed.
I guess I'm confused - why is switching speed an issue? Switching will happen only when the current is low (< 1A). The 60A flows only when the airplane is in flight, not on the ground when it's turned on or off.
You didn't say this earlier, in fact, you explicitly asked how to switch 60A.
Now, with the specification significantly changed; if you can guarantee no switching at 60A, then it's ok to design it not to handle switching the full current. If you try to switch anyway, your switch will blow up as a short circuit and can not turn it off anymore, so you cannot use it as a safety shutdown.
If you really decide to design a system that cannot switch the full load current, then it's a decision which will make the design easier, and you can forget most of our SOA discussion. Even that 100k would be OK, although it's so horribly big that I'd still reduce it to at most about 4.7k.
And what does switching speed have to do with the SOA? I thought (maybe wrongly) the SOA had to do with how much power the MOSFET can pass without cooking itself. How does switching speed relate to the SOA?
While you switch, you are quickly dissipating huge amounts of energy, because the MOSFET is partially conducting, acting as a resistor, dropping a lot of voltage, producing losses P = U*I. Only when it's fully on, the U over the FET is small and losses defined by Rds(on). When it's fully off, the U over FET is the full voltage, there is no voltage left for the load, and the I must be 0. With many types of loads, the losses are biggest when the voltage is shared 50%/50% between the FET and load.