Since no schematic was provided, how do you even know there is a capacitor on the output?

Bob

 

Since no schematic was provided, how do you even know there is a capacitor on the output?

The TS stated it was a computer power supply.
Difficult to get steady DC out of switching regulator without some output filter capacitance.

 

Below is the LTspice simulation of a simple buck converter that goes from 5V to 500mV.
Note that the average output current is about 133mA while the average diode current is about 97mA.

 

some formulas related #4 topology . . .
. . . it is still complicated - but possible to match required duty for .5V , 50A by last 3 formulas @ fig.2
- - - -
as it turns out "0"-use - it still goes to that "Diode - dilemma" as in #18 ←
or then not --

-- because of the erroneous definition of duty in #18

 

Hi,
I am just doing some reading on power electronics and I came across this question. They state the answer, which is 0.55. What I can't figure out is how they got the answer. The questions is:
"A dc-dc converter has an input voltage of Vg=5V, and it supplies a computer processor with V=0.5V at 50A (25 W). The converter is a buck converter, and it uses a diode having a forward voltage drop of 0.5V. All losses other than the diode conduction loss are negligible. Estimate the efficiency of this converter"
Efficiency is given by PowerOut/PowerIn. The Power out is 25W, information is given. I apologize if this seems vague but this is essentially all the information I have. I am assuming the Diode is acting like a parallel switch to the output and a MOSFET in series with the voltage source Vg. If someone can help me understand the process of getting the answer. I did a quick search on here but with no luck. Thank you in advance.