How to get a 15V floating supply

[batee]

Yes, the energy the DC-DCs generate is cached in C1-C4 (470uF) continuously.

Instantaneously the Optos are rated for 2.5A. Even with 10 Ohm gate resistors, they don't heat up at all.

You should be able to calculate the energy required to switch your FETs once (E=0.5*Cgs*Vgs*Vgs) J, multiply that by the number of switches per second, and get the total energy the PS needs to supply. Then calculate the power needed: Energy (in Joules) = Power (in Watts) * Time (in seconds), so Power Required = Energy/time). To get power in Watts, set t=1sec, then the units will be Joules/Sec = Watts. Next calculate current required = Power/Voltage. In my case, I think I calculated power required for 20KHz switch rate at about 5mW. Remember that Vgs isn't 15V, it's the output of the Opto driver, which is about 1-2V less than the input.

Ignore the energy dissipated in the gate resistors (again, they don't get hot = they're only 1/6W). Or calculate the average current for the decaying exponential that results from the capacitor charging action and use it to calculate energy lost to the resistor. Opto driver circuitry current draw is available from the datasheet. Add this to the value you calculated above.

From measurements, I see about 123.3mADC draw on the +12VDC power supply input when the top FET is left on and the bottom FET is switched at a rate of 22KHz. The fact that it works at 400KHz backs up the calculation - even then power consumption should be at about 1W/bottom FET while switching, or about 0.5W / bottom FET averaged over time. I don't think I saw much change in draw between no switching events and full speed operation.

Also remember that in a full bridge operation, you're only providing power to switch the bottom FETs for 50% of the time, and the top FETs for 1/2*(switchrate) time, so the DC-DC converters have time to recharge C1-C4.

If I had it to do over, I'd combine the power supplies for the bottom FETs into one, and I'd use the 1W modules instead of the 2W DC-DC converter modules, as they're only about 1/2 the cost. I'd also eliminate the inductors at the output of the DC-DC. Poor manufacturing tolerances result in radically different voltage outputs at each of the Caps C1-C4. I'd also place a high value resistor across the outputs of the DC=DCs to bring their voltage outputs down to the rated level.

Bryan