R1=1,23^2/1000Ω=1.51mW
R2=12^2/10000Ω=14.4mW

There should be (12-1.23)^2 in R2 equation but shouldn't have much impact on the result(Power will get smaller anyway).

 

23?

 

1.23 is feedback voltage.I used 1.25 before because it was close enough.

 

I wanna replace R1 with 0.5W 5k6, much easier and cheaper than replacing the trimmer, there's multiple solder points that would be desoldered with the trimmer... (in fact 2, 2 too many)

11.5532142857V 5,600Ω 47,000Ω

 

Its up to you.

 

would you please see this thread: http://forum.allaboutcircuits.com/threads/down-now-up.104289/

 

what would happen to the psu as a 12V input if it has 150uH and not 100uH?

ISB123 are you sure 150uH is enough for 1 to 36V operation?
And the capacitor values? they recommend 680uF

EDIT moved the part about inductance to it's own thread
http://forum.allaboutcircuits.com/threads/inductance.104399/

 

Hi, back again from holiday, I want to build the 36V LM2576 buck now, I selected my resistor R1 at 750R, because I can get R2 VAR 22K, but now I see datasheet says R1 minimum 1K, what will happen if I use a 750R?
[http://appnote.avrportal.com/calculator/lm2575]