The lab session was a success, and David left the lab feeling proud of what he had accomplished. He knew that mastering power electronics would take time and practice, but with his textbook, "Fundamentals of Power Electronics, 2nd Edition," and his own determination, he was well on his way to becoming an expert in the field.
where ΔVout is the output voltage ripple, Vout is the output voltage, Rload is the load resistance, ΔIL is the inductor current ripple, L is the inductance, and fsw is the switching frequency. --- Fundamentals Of Power Electronics 2nd Edition Solution
David knew that the switching frequency was 20 kHz, and he could calculate the inductor current ripple using the given values. He plugged in the numbers and began to solve for L. The lab session was a success, and David
ΔVout / Vout = (Rload * ΔIL) / (8 * L * fsw) David knew that the switching frequency was 20
As he worked through the problems, David encountered a challenging question: a buck converter with a input voltage of 12V, an output voltage of 5V, and a load resistance of 10 ohms. The question asked him to find the inductor value required to achieve a output voltage ripple of 1%. David wasn't sure where to start, but after re-reading the relevant section in the textbook, he remembered the formula for output voltage ripple: