Electronic – SMPS Efficiency Paradox

Here's a relatively simple solution, but not necessarily a cheap one (depending on your budget).

An oscilloscope is not likely to give you the accuracy you asked for because they typically use an 8-bit ADC, giving 0.2% in measurement uncertainty just from the sampling digitization.

Instead, consider simply using two benchtop multimeters, like Agilent 34401A's. I haven't looked at other models, but the 34401A can measure current at the accuracy you need (for example, 0.05% of reading + 0.005% of full-scale on the 100 mA range).

They can be triggered externally at 300 readings / second (for 5-1/2 digit resolution), so that gets you a sample window much shorter than your load switching cycle. If you hook the meters up to measure the input and output current, then trigger them simultaneously you'll be able to compare the results to determine the efficiency (assuming the input and output voltages are holding constant).

If the input and output voltages are also changing, you may need 4 multimeters to get all the information you need.

If you can synchronize your measurement to the moments when the load switches, you only need half as many multimeters, because you can first measure how the input current & voltage change in response to the load switch and then move the meters around and measure how the output responds to the switching event.

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}\$).