This small switcher is a fixed 12V\ 5W unit. with 77% efficiency 2575 can out put 5V 0.32A ?? correct ? (0.42A x 77% = 0.32A )

What is the input to the small switcher? If it is less than 40V then you can just supply the 2575 directly.

And the efficiency is power efficiency, so you can get 77% of the input power into the output drive, which would give you about 770 mA. Think of a DC-DC converter just like a transformer -- if you double the voltage at the output you can only get half the current, but if you cut the voltage in half then you can get twice the current. That's if it's 100% efficient. Whatever current you can get at 100% efficiency you multiply by the efficiency factor and that is what you can actually get. So you have a step-down factor of 12V/5V=2.4. So your 100% efficiency current output would be 420mA*2.4=1A. But with an efficiency of 77% you would only get 0.77A.

 

Hi,

There is no reason to stick with the LM2575 there are other buck regulators probably better.

But anyway, the max output current with an ideal buck circuit with 12v input and 0.41 amps input is less than 0.984 amps at 5v.
If you can get 90 percent efficiency then you can get about 0.886 amps out at 5v.
If you can get 80 percent efficiency then you can get about 0.787 amps out at 5v.
If you can only get 70 percent efficiency then you can only get about 0.689 amps out at 5v.

For a 'rough' estimate, assume you can get 1.00 amps out at 100 percent efficiency and with the percentage expressed as a fraction it is 1.00
Then at 90 percent efficiency the fraction is 0.90, so estimate 0.90 amps out.
Then at 80 percent efficiency the fraction is 0.80, so estimate 0.80 amps out.
Then at 70 percent efficiency the fraction is 0.70, so estimate 0.70 amps out.
So you see the available current out closely follows the percentage efficiency expressed as a fraction so it's quick and easy to estimate. The error when doing it this way is only about +1.6 percent so it's close enough just to get an idea what to expect.

 

Note that his figure of 0.41A came from poorly rounding 5W/12V, which is 0.4166.. A and so should have been rounded to either 0.417 A or 0.42 A. If trying to intentionally bias things conservatively, the 0.416 A would have been the better answer and 0.41 A a too-excessively-rounded alternative. Assuming that the 5W figure for the first switcher was power delivered to the load and not power consumed from the source, then a 100% efficient switcher taking it down to 5V would be able to deliver 1A.