# Measuring switched mode power supply efficiency accurately for high currents

I hade implemented a switched mode power supply: * 1.2 kW * Input: 12 VDC, 100 A * Output 5 VDC, 240 A

I need to measure it's efficiency (power loss/total power) by accuracy of 1%. In other words: I need to know, if the accuracy is 98.1% or 98.2%.

I would prefer to use oscilloscopes and use them to integrate V(t)*I(t).

What is the most practical way: * Using shunt resistors? But, it seems, the lowest ones are 50 uOhm and then I need a preamplifier, which increases complexity. Or, if I increase resistance, then the measurement itself impacts the measurement. * Hall effect sensors (Allegro etc.). But, they do not seem to be geared for accurate measurements? * Something else.

Thanking in advance for any clues.

• Rent two expensive current probes for your oscilloscope. I am not sure if you can get to 1% accuracy, but it will be relatively easy to set up. Jul 24, 2016 at 7:11
• Can you just do DC measurements? Are you going to use an electronic load for your efficiency tests? Jul 24, 2016 at 7:28
• "... by accuracy of 1%. In other words: I need to know, if the accuracy is 98.1% or 98.2%" - Your second sentence implies a 0.1% accuracy, so which one do you need? 1% accuracy or 0.1%? Jul 24, 2016 at 11:21

You certainly won't measure to 1% with an oscilloscope, and you'll find staying within a 1% error budget will be fairly tough however you try to measure input and output power.

I don't normally suggest calorimetry, but have you considered that instead of trying to look at the small difference between two big numbers, you instead look at the small number?

Connect a power resistor and a thermometer to the heatsink, then thermally insulate the heatsink by wrapping it in bubble-wrap or similar. Switch on, and observe the temperature rise. Switch off before it's got too hot. Allow to cool, and then repeat with different small DC powers applied to the power resistor. Find the applied power that matches the rate of temperature rise in use, either by iteration or interpolation. Obviously the characteristics of the thermometer and heatsink do not need to be known, only the power measurement to the resistor.

On reflection, that's not going to work accurately. Not all losses are in the components on the heatsink, just most of them. So calorimetry will under-estimate total losses.

I misread your question. I did originally think you were going between 12v and 240v, 100A and 5A. The likelyhood of getting good scale transfer between the different ranges to better than 1% would be remote. However, re-reading, it's 5 and 12v, and 100 and 240A, so you can use the same ranges for input and output.

Splash out on a 4.5 digit meter. It's likely to have better linearity between half and full scale than a cheaper meter.

You have the opportunity to use a pair of nominally identical sense resistors for measuring input and output current. On one range, read voltages across both sense resistors. On another range, read input and output voltages. Compute the efficiency. Then swap the sense resistors, and do it again. Taking the average will eliminate any small residual difference between the sense resistors.

• @user117625 added a bit to the answer, might interest you Jul 24, 2016 at 9:40

Here's another incomplete answer but it may prompt some other ideas.

I would prefer to use oscilloscopes and use them to integrate V(t)*I(t). 