3 ms·
Switching losses are less than ohmic losses. A few chips cost less than a lot of copper. The crossover point depends on the cable lengths, cable size, load, a
by willis936 3y ago
Switching losses are less than ohmic losses. A few chips cost less than a lot of copper. The crossover point depends on the cable lengths, cable size, load, and voltage in question.
Let's do a quick example. 16 AWG carrying 12V. 16 AWG is 4 mOhm/ft. Let's say you have a 1200 W load (100 A, 0.12 ohm). Ohmic losses per foot of cable are 3%. So if you have a 15 foot cable it will use as much power as the load.
Edit: I think I responded to the wrong comment.
- amelius 3y ago100 amps through 16AWG? Man that's pushing it.
- Dylan16807 3y agoEspecially for a 15 foot cable. I'd expect a cable >10x the thickness with <1/10th the resistance.
- nine_k 3y agoAbsolutely. But with the voltage 10x as high, the current goes down 10x, too, and the power loss goes down, too, while the power stays the same. That's how things run on AC normally.