Hi everyone, I'm working with a Buck converter and I was wondering what should be the best duty cycle, I don't remember where I read that working lower than 50% is the best performance. At that time I didn't show to much interest about that because I was focused learning the general working, and the characteristic equations for the ripple current and inductance selection.

But now I'm interested on how to select the best duty cycle. Theoretically the output voltage depends by this equation. Vo = D*Vin. I did this chart to compare different input and different output voltages.



It can seen the more output voltage near the input voltage the Duty Cycle it's bigger. My common sense tells me that the bigger the Duty Cycle the bigger the on time by the switch, and the more the conduction looses.

Also at higher frequencies and a very low Duty Cycle we need a very high speed switch with a fast turn on and off.

So, in a real scenario when we need to manufacture a driver for a final product. Which is the better choice of the DC,
Lets say that I can chose or control the input voltage. And my output requirement is 55V I can decide.

See the above chart. What will be the best operation point. Duties above 50% or below?

Ty so much.

 

If you can choose you input voltage, then just choose it to be the output voltage.
Then you have 100% efficiency.

In a typical use of a buck converter you don't have a choice for the input voltage, so your question seems to be a little odd.

My common sense tells me that the bigger the Duty Cycle the bigger the on time by the switch, and the more the conduction looses.

Common sense doesn't always work in electronic circuits.
As the duty-cycle gets smaller, the input and switch current gets higher for a given output current, so the I²R losses due to the MOSFET and other resistances increase non-linearly.
For example, increasing the input voltage to reduce the duty-cycle by 1/2 will give twice the power dissipation for a given current output, due to those resistances.