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I've build a custom LED driver based on a DC/DC Buck converter with a current feedback loop. It works pretty well, the calculated efficiency shall be somewhere around 95%, but I'm struggling with measuring the real efficiency of the circuit. The basic driver schematic is below:

driver

There are some filters on the Vcc (capacitors, ferrite choke,..), the input current and voltage are quite stable, so the input power measurement is simple, multiply the current and voltage and you have the power. The part I'm struggling with is the output power.

The LED voltage and current have a triangular shape due to a switching supply design, what values shall I use in the efficiency calculation? Multiply of the effective values of the current and voltage?

I don't have a differential voltage and current oscilloscope probe and measuring the LED voltage with two ground referenced probes and subtracting the value is not very accurate.

The current measurement is also a big problem for me, the sense resistor is very small (18 mOhm), the 3 Amp current gives only 54 mV signal with a lot of noise and the current sense amplifier output is not directly usable as it has a significant error that cripples the measurement results (but is acceptable for the circuit function).

Is my approach (multiply effective values of current and voltage) correct for efficiency calculation in this case? And is there a better way how to measure these without paying several hundreds dollars for differential and current oscilloscope probes?

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    \$\begingroup\$ Why not to place a bigger shunt in series with Led to get about 100mV drop on it? Add the shunt power loss to output power during eff. calculation. \$\endgroup\$ Commented Nov 19, 2023 at 17:52
  • \$\begingroup\$ Ignore the triangle waveshape and measure average output current and voltage. Divide with input ditto. \$\endgroup\$
    – winny
    Commented Nov 19, 2023 at 18:16
  • \$\begingroup\$ The driver is changing frequency with load and voltage difference, additional drop of 100 mV would change the frequency by few dozens of kHz and affected the switching losses, changing the efficiency. \$\endgroup\$ Commented Nov 20, 2023 at 16:35

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