Hi, I am going through Franco's "Design with Operation Amplifiers and Analog Integrated Circuits" (an undergraduate textbook) and I am on the chapter about regulators and my question is specifically about the buck kind of switching regualtors. As I understand the basic operation of a buck regulator to be when the switch is closed the magnetic field of the coil is charged. When the switch is open, then energry stored in the coil is used to power the load. However this requires the circuit to be closed as circuit with an open would have no current. But the switch was opened, so they way to get around this is to put a diode with the positive terminal to ground on the inductor terminal that connects to the switch. This way the diode will conduct with the switch is closed allowing current to flow from the coil to the load. So everything seemed pretty straight forward until I googled for a buck switcher datasheet to check some out. Almost all the ones I found had NFETs instead of diodes to conduct current when the switch is open. Now I am trying to figure out why. Well, more like I want to understand the tradeoff, diode vs NFET. I think it should be possible to make a diode with a lower on resistance than a FET so my only guess is that is has something to do with Vds. But this seems weird to me because to saturate the transistor Vds > Vgs. This makes it seem like Vds should be bigger than the forward drop of a diode which means the duty cycle of the switch will have to be longer to maintain the same output voltage, lowering efficency. Sorry for the long question but I wanted to show that I put some thought into this and unfortunately it seems like way too much. I should be doing my homework!