Electronic – High power three phase ac motor controller

When using high-side N-channel MOSFETs, a MOSFET driver chip is generally used (e.g., an IR2130). These chips require a boost capacitor which allows the gates of the high-side MOSFETs to be driven with the required voltage. The downside of this is that the MOSFET driver has a restricted operating range (typically 10V-20V). The second down-side is that you can't drive the MOSFETS to 100% pulse-width; this is usually not a problem. Alternatively you could use P-channel high-side MOSFETs and N-channel for the low-side. With this configuration, you can PWM the low-side switches only and save some power. (You can do this with all N-channels as well, it's just trickier since you have to keep the boost capacitors charged). Go to Digikey and search for "mosfet driver", you'll want the 3-phase bridge drivers specified for external switches.

As for the MOSFETs themselves, I don't recommend you use them in parallel. This would increase your footprint as well as your gate charge. Instead, look for high-current mosfets on Digikey. My favorite packages for these are D-Pak, 8-PowerVDFN, and PowerPAK 1212-8.

Atmel has an app note for sensor-based and sensor-less motor control. Unfortunately AVRs don't have much processing power and your control algorithm will likely be restricted to fixed-point math because of this. Because of the processing power limitations, you may want to consider not using the Arduino environment and instead use straight C or assembly.

You may have some trouble isolating the I2C bus optically since it's a bi-directional bus. If you were to use a communication system with single direction lines, you may want to look at TI's line of isolator chips (e.g., the ISO7221).

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:

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.