For those who aren't sick of this, I'll post how I think Weber's design would work, and why the capacitors aren't optional as I first thought.
First consider only the U phase, during the first part of the cycle when the lower IGBT is on, and the upper one is off. I'll consider a PWM ratio of 67% (2/3), with 90 A @ 300 V being converted to 900 V at 30 A (voltages relative to DC-). The input inductors will be sized such that with 300 V across the inductor, the current ramp will be 30 A per PWM period.
Current (red arrow) flows from the pack through the inductor and lower transistor. The transistor is on for 2/3 of the PWM period, so the current will ramp up 20 A.
Now consider the second half of the PWM cycle, with the lower transistor off and the upper transistor on. The current through the inductor now flows through the upper transistor to the capacitor and the load. I've drawn the V phase in this state; the current is in blue. In the steady state, the output will already be at +900 V (relative to DC-). That means that the voltage across the inductor is now 600 V, and is of the opposite polarity to what it was in the first part of the PWM cycle. So now the current will ramp down twice as fast as it ramped up (so it will ramp down at 60 A per PWM period), but it will do so for only 1/3 of a PWM period, so that's 20 A. So the current will end up where it started, and the cycle repeats.
Of course, the very first cycle, the output voltage will be zero, but that means that the inductor still has 300 V across it of the same polarity as when the current was ramping up, so it continues to ramp up at about 30 A per PWM cycle. This will charge the output capacitor, which will reduce the voltage across the inductor in the second part of the next PWM cycle, until eventually the capacitor reaches over 300 V, at which point the current will start ramping down in the second half of the PWM cycle. It will eventually reach equilibrium with the output voltage at 900 V.
I'll temporarily consider the capacitor to be part of the load now. With equilibrium reached, and when the output current is 30 A (achieved by suitable adjustment of the load and output voltage, let's say), this is delivered as a "pulse" of 90 A for the last third of the PWM cycle, and no current (upper transistor is off) for the first two thirds of the cycle. So the average current during the last third of the cycle is 90 A, so it is actually ramping down from 100 A down to 80 A. During this time, the pack is supplying this current; it is in series with the inductor. During the first two thirds of the PWM cycle, the inductor current must therefore be ramping up from 80 A to 100 A.
In both parts of the PWM cycle, the average current from the 300 V pack is 90 A. Power and therefore energy are conserved, as of course they must. So we can see that both IGBTs are handling 90 A average (the output current). When the lower IGBT is off, it sees the output voltage (900 V). When the upper IGBT is off, it sees the output voltage (900 V) as well. So both IGBTs see full output voltage, and full input current.
So that's why the capacitors are needed; we can't have pulses of 3x current, then no current, into the load. So now we consider the capacitors to be part of the converter, not the load, and they supply the missing current when the upper IGBT is off, having absorbed the excess current in the second part of the PWM cycle. This is shown in green in the diagram (for the first part of the PWM cycle).
As mentioned earlier, in fact the output swing is only from 300 V to 900 V, further reducing the IGBT utilisation, compared to a traditional buck-only (but with regen) topology.
[ Edit: added paragraph about why the capacitors are part of the converter, not the load. ]
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