Electronic – Are there any caveats to running a boost converter at zero duty cycle

I'd go for a full H-bridge driving a transformer.

With a 24V supply you can drive nearly 48 Vp-p onto the primary and this reduces the secondary winding turns to a ratio that produces 600 Vp-p - then I'd use a fast bridge rectifier and smoothing to give a 300 V dc output.

Step up ratio would probably be about 14:1 and if you wanted to control the amplitude I'd consider feeding the H bridge from a decent efficiency Buck convertor. Maybe make the turns ratio 15:1 so that a 20V output from the buck would do the business - leave a little overhead for loads etc..

My main consideration is avoiding the secondary winding having too much self-capacitance and acting like a parallel tuned circuit. It should be quite a way off with this type of topology.

It's not going to be a small lump of ferrite and I'd use the best I can get my hands on such as 3F4 material from Ferroxcube - you should be able to get 300 watts from an E68 planar transformer. I'm getting 200 watts from a slightly smaller E58 - using two half cores rather than a half core and a plate.

Mine's running at about 600 kHz so I can use PCB tracks for the coils - a sandwich of PCBs does the trick for the primary (4 turns in total) - mine is 1:1 so it's the same PCBs on the output - yours will need thinking about as to whether you can get 60 turns at 1A from PCB. I think you should just about squeeze them in.

The main power losses in a boost converter can be summarized as follows:

1. Power switch switching losses (e.g. MOSFET, BJT. Hereafter I will refer to the Power switch as the MOSFET)
2. MOSFET conduction losses.
3. DIODE switching losses.
4. DIODE conduction losses.
5. Other conduction losses (e.g. inductor resistance)

The efficiency of a converter is given by: eff = Po/Pin = (Pin - Plosses)/Pin.

As the losses change the efficiency therefore changes.

One can not make a blanket statement as to why the efficiency reduces or losses increase as the duty cycle increases because then one would need to know all five loss parameters as a function of current, voltage and switching frequency.

However, a simplified explanation of this phenomena is that the MOSFET conduction losses are unequal to the diode conduction losses. As the duty cycle increases, the MOSFET will conduct for a longer period and the diode for a shorter period. This in turn alters the power losses in the circuit. If the DIODE happens to have higher conduction losses than the MOSFET for example, then as the duty cycle changes, causing the DIODE to conduct for a relatively longer period than the MOSFET, then the efficiency will decrease. This is a simplified explanation, but the main principle is that as you change the duty cycle, the operating conditions for each element in the circuit change. Since the losses for each device depend on it's specific operating point, then changing the duty cycle changes the losses.