Electronic – Efficiency of a SMPS

What don't you understand?

Efficiency is simply a function of the ratio of power out to power in.

A Switch-mode power supply that draws 5W in and supplies 4W of output power has an efficiency of 80% (\$\frac{4}{5} = 0.8\ \$).

A linear regulator generally has very poor efficency. Take, for example, a linear regulator that is producing 5V from a 12V source, with 1A of draw:
\$P_{in} = 12V * 1A = 12W\$
\$P_{out} = 5V * 1A = 5W\$
\$\text{Efficiency} = \frac{5W}{12W} = 0.4166... = ~41.6\%\$

^{_{I'm ignoring the power consumption of the regulator itself here, since it will be small in proportion to the overall power.}}

Yikes! 40% efficiency! That's terrible. It also means the other ~60% of the power (7 watts!) will be dissipated as heat, so you'll need a big heatsink to get rid of the waste-heat.

Now, most switchers have efficiencies in the 70-90% range. Lets look at the above 12V-5V converter with a 80% efficient switcher.

\$P_{out} = 5V * 1A = 5W\$
\$P_{in} = P_{out} * \frac{1}{0.8} = 6.25W\$
\$I_{in} = \frac{6.25W}{12V} = ~0.52A\$

Well, since the efficiency has gone up, the current drawn on our input is now less then the current out the output! This is possible because, remember, energy is conserved (minus the efficiency losses), not current or voltage.

You can also see that the dissipation in the switch-mode converter is only 1.25W (\$P_{in} - P_{out} = \text{dissipation}\$).