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PSU design recommendations

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Both circuits provide isolation from the mains through the use of transformers. The first circuit (off line switcher) will use a relatively smaller transformer than the second (which will be very bulky for 400W). 'A' is more common because it is smaller, lighter, more efficient and cheaper. Circuit 'A' will feed more noise back into the mains than circuit 'B', which is why it has the filtering components on the input. You don't need PFC. I have seen integrated off line flyback SMPS controllers up to around 250W.

You can have a look at designs for computer ATX power supplies, which operate at your target power level and above, to see the topology they use and whether they include PFC or not. It seems that half-bridge (rather than flyback) w/o PFC is the way it's done, if these schematics are anything to go by https://www.google.com.au/search?q=...X&biw=1582&bih=781&sei=sZiLUZysGY-ViQehk4HgBA.
 
The PSU I want to build will have a PWM controller with variable output voltage (0-20V , 0-20A)
Both designs will not regulate at low voltage output. It is hard to get this type of error amplifier to work below 2.5 volts.

Design A was never designed as a 'bench power supply'.

Design B; I do not understand how VR1 sets the output voltage. There might be an error around VR1.
 
Both designs will not regulate at low voltage output. It is hard to get this type of error amplifier to work below 2.5 volts.
You're right about them not regulating down to 0V; the first could get down to around 3-4V (due to TL431 voltage of 2.5V, current >=1mA & ignoring series LED), and the second down to 3V due to the reference of 3V.

Design B; I do not understand how VR1 sets the output voltage. There might be an error around VR1.
VR1 is ok; adjustable range is 3-53V
 
I can't give you accurate efficiency values... an wild estimate for 20V/20A out might be:
A) major losses in mosfet, if peak current might be 4A then peak Pdiss is 4A^2 * 800m-ohms = 13W. Peak output diode loss might be 40W
B) peak PNP loss might be 0.25A * 50V = 12.5W for driver and 20A * 2V = 40W (if Vce-sat is 2V) for the transistor itself. Output diode same as for A. There's also loss in the input transformer + magnetising current.

The components will dissipate some fraction of the peak power due to each having <100% duty cycle, and because the current waveforms are not square. If 'B' is simpler for you, you can boost its efficiency by using a mosfet for the switching transistor and a synchronous rectifier (i.e. use a mosfet to replace the output diode). You can get controller ICs that have appropriate synchronous driver circuitry integrated.
 
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