Electronic – Why are switchers more efficient than linear regulators


It is well-know that switching regulators are more efficient then linear regulators. I also know that linear regulator have to dissipate the difference between the input voltage and output voltage times the current as heat.

But why does this not apply to switching regulators with the same conditions: same input voltage and output voltage and current?

I know switchers can get hot; I have one on a board that gets so hot you can barely touch it, but then again it's only 2 1/2 millimeters on each side and looks like an ant compared to a through-hole 7805 with its heat sink.

Best Answer

Linear regulators work by effectively putting a controlled variable resistor between the source and load. All of the current for the load flows through this resistive element. And the voltage across it is equal to the difference between source voltage and load voltage. So the power dissipated is

\$P_{lin} = I_{load}\times{}(V_{src}-V_{load})\$.

Switching regulators work by changing the duty cycle of the current flow over a switching cycle, then averaging out the output using a filter. During part of the cycle a high current flows with a low voltage drop. During the other part of the cycle almost no current flows with a high voltage drop. Neither of these conditions dissipates much power as heat. Ideally the power lost becomes

\$P_{sw}= \mathrm{DC}(I_{on})(0\ \mathrm{V}) + (1-\mathrm{DC})(0\ \mathrm{A})(V_{off})\$,

which is, of course, 0 W. Typically much of the inefficiency in the real world is due to power lost during the very short switching inteval between the "on" and "off" parts of the cycle.