How to test driver efficiency

[outer limits]

What is the best and most accurate way to test driver efficiency? I only have standard multi meters and an oscilloscope

 

[brted]

You'd have to measure the voltage of the battery under load (V) and the current coming from the battery (I). Multiply those for Power In (volts x amps = watts). The tricky part is you also need to measure amperage between the driver and LED which means you have to disconnect something if the driver is already installed and the voltage drop across the LED. Multiply those for Power Out. Power Out divided by Power In is the efficiency (convert it to percentage).

You have to be careful because you are adding a lot of wires and connections that aren't usually there and if you are measuring current over about 1 amp, the readings will be sensitive to the resistance in the wires and connections. Since I only have 2 meters, I will measure V in and I in with a battery for just a couple of seconds and then reconfigure the connections and measure V out and I out (I will use 4 batteries at different voltages from 3.6V to 4.2V at rest so I get a good cross-section and can kind of fake efficiency as the battery runs down). Technically you should have 4 meters running at once. It would probably be even better to have a bench power supply so you don't have to worry about the battery being a variable.

In the threads of interest sticky there is a link to How to Use a DMM that is very good.

 

[outer limits]

Thanks for that but what if the driver does not give out good DC or some sort of sine wave. I have four meters. 2 of which are good for voltage and the other two tie up with each other for current. I also have a bench power supply which is great as I can look at different voltage. Is is the effect of the switched output that I dont know what to do with.

 

[brted]

That's tricky. A lot of drivers use PWM, so it is a rapidly pulsed DC signal. My cheap meter reads it as constant, so I don't have to worry about it. Can your oscilloscope read it? Maybe RMS would be an appropriate way to calculate the equivalent output? The peaks are full brightness, so you don't want to use that.

 

[MikeAusC]

You could just use a resistor-capacitor filter to convert the AC to DC for unambiguous measurement by the multimeter.

For current measurement, put a low value resistor in series as a current shunt e.g. 0.01 ohm and connect the filter to this.

Since LEDs are a resistive load, just averaging the DC will work.