0
\$\begingroup\$

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.

\$\endgroup\$
3
  • \$\begingroup\$ 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. \$\endgroup\$ – mkeith Jul 24 '16 at 7:11
  • \$\begingroup\$ Can you just do DC measurements? Are you going to use an electronic load for your efficiency tests? \$\endgroup\$ – mkeith Jul 24 '16 at 7:28
  • 2
    \$\begingroup\$ "... 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%? \$\endgroup\$ – marcelm Jul 24 '16 at 11:21
1
\$\begingroup\$

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.

\$\endgroup\$
1
  • \$\begingroup\$ @user117625 added a bit to the answer, might interest you \$\endgroup\$ – Neil_UK Jul 24 '16 at 9:40
0
\$\begingroup\$

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).

enter image description here

Figure 1. Analog Devices AD633 four-quadrant multiplier.

When faced with a problem of making a true RMS measurement of a phase-controlled AC application I used an AD633 to square the current reading and feed it to an integrator. Looking at the specification of the AD633 I see that it quotes "Total Error within 2% of full scale" so it's not good enough for your application. That chip is quite old I have read in other threads here that there are better four-quadrant multipliers available now.

If my thinking is correct you need to measure the average power in and out. The analog multiplier technique will give you instantaneous VI. If you low-pass filter the output (a simple RC filter?) you will have the average power. Depending on your confidence of the linearity of your chosen multiplier you then only need a multimeter to read the average output power from the multiplier's low-pass filter.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.