Electronic – Connecting grounds when using an isolated DC/DC converter

When you have a large step-up voltage and a fair amount of power to transfer a full H bridge is the best way because your turns ratio is the least required of all the topologies. So thumbs up for that decision.

Also, when it comes to this sort of application, controlling the DC level and sticking with 50:50 squarewave control is not only simpler but more efficient. I've tried PWM but I've had resonance problems (leading to extreme losses and over-heating) with the high-turns secondary and had to despatch circuits to the graveyard. So, in my opinion thumbs up on the fixed duty cycle, variable DC method for control.

So, do you boost 12V to 48V and reduce turns ratio by 4:1 or go straight for the 12V control. Your turns ratio from a 12V supply is going to be based on a primary input of 24Vp-p yielding 700Vp-p with a turns ratio of about 30:1. If you used a little bit of capacitive resonance on your output winding to peak the voltage transfer, you might happily find that 25:1 will do for input voltages down as low as 10V.

My conclusion is that I'd stick with the full step-up from the 12V buck regulator because it's likely to be more efficient. Also, under no-load conditions you might find that you need to "buck down" to maybe 2 or 3V - that's maybe 20% of 12V - how would you get 20% of 48V from a boost - it would be switched-off and spluttering and you'd find that on very light loads you may not be able to control the voltage low enough to prevent the output dc rising significantly above 350Vdc.

From page 1 of the data sheet you see these two graphs (I have added max current figures in red just to emphasize things): -

So, when outputting 20 volts, the graph informs you that the maximum sustainable output current is 200 mA. It tells you this because the curve on the graph ends at 200 mA. Now, this is page 1 of the data sheet and if it could supply 300 mA (for instance) it would definitely have that curve going to 300 mA. Page 1 tries to promise as much as possible!

OK that's a power output of 20 volts x 0.2 amps = 4 watts.

Similar story for the 12 volt output, max current is about 330 mA and at 12 volts this means it can supply an output power of 3.96 watts.

What current are you going to get at 5 volts - you could say that maximum output power is 4 watts and this predicts an output current of only 0.8 amps.

You won't get 1.6 amps at 5 volts. Maybe try one of these: -

And yes, they are a little expensive so maybe try again on TI's site and narrow down your selection using their search engine.