There are many losses associated with switching, but it sounds like you are most concerned about the additional thermal load introduced into the MOSFETs in the period transitioning between on and off. I thought it would be easy to find some application notes on this, but surprisingly it wasn't. The best I found was AN-6005 Synchronous buck MOSFET loss calculations with Excel model from Fairchild, the relevant parts of which I'll summarize here.
During the switching transition, the voltage and current in the MOSFET will look approximately like this:
![MOSFET switching current and voltage vs time](https://i.stack.imgur.com/8iIeQ.png)
The switching losses we are going to calculate are those in periods \$t2\$ and \$t3\$ due to the voltage and current in the MOSFET. The way to approach this is to calculate the energy of each transition, then convert this into an average power according to your switching frequency.
If you look at just \$t2\$, \$V\$ is nearly constant, and \$I\$ increases approximately linearly, forming a triangle. Thus, the power also increases linearly, and the total energy is the time integral of power. So the energy is just the area of that triangle:
$$ E_{t2} = t_2 \left( \frac{V_{in} I_{out}}{2} \right) $$
\$t3\$ also forms a triangle. In this case, the voltage is changing instead of the current, but still the power makes a triangle, and the calculation of energy is the same.
Since the calculation is the same for \$t2\$ and \$t3\$, then it's not really important how much time is spent in \$t2\$ vs \$t3\$; all that really matters is the total time spent switching. The energy losses from one switch are thus:
$$ E_{switch} = (t_2 + t_3) \left( \frac{V_{in} I_{out}}{2} \right) $$
And, your switching frequency is how many times per second you incur this energy loss, so multiplying the two together gets you the average power loss due to switching:
$$ P_{switch} = f (t_2 + t_3) \left( \frac{V_{in} I_{out}}{2} \right) $$
So, taking your calculation of the switching period being \$150ns\$, and the maximum current being \$330A\$, and the voltage \$12V\$, and the switching frequency \$30kHz\$, the power losses from switching are:
$$ 30kHz \cdot 150ns \left( \frac{12V \cdot 330A}{2} \right) = 8.91W $$
That's \$8.91W\$, ideally, shared between three transistors, so only about \$3W\$ each, which is pretty insignificant compared to your other losses.
This number can be checked for sanity with a simpler model: if you spent \$150ns\$ switching, and you do it \$30000\$ times per second, then you can calculate the fraction of the time you spend switching, and make the most pessimistic assumption of the full power of \$12V\cdot330A\$ being lost in the transistors:
$$ \require{cancel}
\frac{150 \cdot 10^{-9} \cancel{s}}{\cancel{switch}}
\frac{30 \cdot 10^3 \cancel{switches}}{\cancel{s}}
\cdot 12V
\cdot 330A = 17.82W $$
Of course, over the switching period, the average current and voltage is only half that of the maximum, so the switching losses are half this, which is what we just calculated.
However, I bet in practice, your switching times will be slower. A "\$2A\$ gate driver" isn't a constant current source as these calculations assume. The real picture is rather more complicated than this simple model. Additionally, the current will be limited by the resistance, and usually more significantly, the inductance of the transistor packages and the traces leading to them.
Let's just say the inductance of the gate driver, transistor package, and traces to it is \$1\mu H\$. If your gate drive voltage is \$12V\$, then \$di/dt\$ is limited to \$12V/1\mu H = (1.2\cdot 10^7)A/s\$. This may seem like a lot, but on the time scale of \$150ns\$, it's not. Keeping the inductance low will take some very careful layout.
So, I would say that these calculations show that your switching losses may be manageable, though you won't know for sure until you've made the layout and tested it. Even if you can't reach the ideal of a \$150ns\$ switching time, the losses are low enough relative to your other problems that you have some margin to do worse and still function.
Your bigger problem is probably getting the three MOSFETs to switch at the same time. Otherwise, one of them will get a disproportionate share of the total current, and thus heat, leading to premature failure.
Best Answer
This circuit is probably close to correct, but I can't tell without knowing what kind of load you are driving.
Most loads will cause the switching transistor to dissipate a pulse of energy during turn on. The transistor you selected can handle a pulse of 400mJ which makes it appropriate for use as a load switch. It has an on resistance of 0.3 ohms, which may be appropriate for a 1A load (although lower resistance wouldn't hurt).
Some things you should consider are...
1) How much capacitance does your load have? For example if you use this to turn on another power supply or some circuit card then most likely there will be capacitance on the front end of that setup.
a) During turn on M1 will dissipate 0.5 * C * (24V)^2 of energy. So 288uJ for every 1uF of load capacitance. Look at the safe operating area graph of the IRF9530, if the turn on time is several ms then you can dissipate up to 400mJ, which limits you to a load of 1.38mF max.
2) Your circuit doesn't control the rise rate at turn on. If your load has any capacitance whatsoever then its possible that you could exceed the 48A pulsed drain current rating.
You can fix this by adding a capacitor between the drain and gate of M1. The turn on rise rate then becomes...
dv/dt = ((24V - Vgs_th) / R2 -Vgs_th / R1) / C.
So for example C = 100nF Vgs_th=3.0V, R1=4.7K, R2=10K gives dv/dt = 14.6kv/s.
24V / (14.6kv/s) gives a rise time of 1.6ms (going from 0V to 24V).
3) The switch doesn't have any kind of short circuit protection. This may be OK if the 24V source is current limited. But if not, its worth considering.