3 ms·
Unfortunately, this is not even close to correct. Resistive losses in wires are not the only cause of wasted energy, and in most battery-powered devices they ar
by AdamH12113 3y ago
Unfortunately, this is not even close to correct. Resistive losses in wires are not the only cause of wasted energy, and in most battery-powered devices they aren't very important at all. How is a superconductor supposed to help with e.g. switching losses in a buck converter, or reduce the amount of energy stored in a transistor's gate capacitance?
- jacquesm 3y ago> How is a superconductor supposed to help with e.g. switching losses in a buck converter That one I can answer for you: because it would reduce the time to go from 'on' to 'off' due to the much more rapid flow on account of reduced resistance. The currents in a buck converter can be very high because they are short, but that is also exactly where you'll find the losses: in the transition from 'off' to 'on' and vice versa. So this won't do anything for the MOSFET losses but it could seriously shrink inductor losses and those are the other major factor.
- AdamH12113 3y agoSure, the inductor conduction losses would go away (and that's a substantial improvement!), but I'm not sure what you're talking about with regard to reducing switching time. I don't think the MOSFET gate drive is limited by the wire that connects to it.
- jacquesm 3y agoIt isn't but the capacitor that is sitting on the far end of that circuit is limited by how fast you can pump current in or out. And if your current can go up considerably then the capacitor will charge that much faster and so your losses will be smaller because for the same 'on' duration of the MOSFET you can move more charge. Clearer now?
- AdamH12113 3y agoThe output capacitor in a buck converter is charged through the inductor, whose current is continuous. I think we must be talking about different things.