LED bulb wattage query [closed]

I was trying to check the wattage of an LED bulb. How to know manufacturer's spec is correct?

closed as too broad by PlasmaHH, brhans, Bimpelrekkie, Harry Svensson, Leon HellerJun 27 '18 at 14:59

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• You do measurements – PlasmaHH Jun 27 '18 at 11:26

How to know manufacturer's spec is correct?

You confirm it by measurement.

Big Clive on YouTube often looks at LED lightbulbs and he uses a HOPI meter for measurements.

You could also use a cheaper equivalent like this one:

But it will probably be unable to accurately measure the 1 W of your bulb.

But what are you hoping to gain by knowing if a LED bulb meets its specification? It either emits enough light or it does not. If it is a bad design it will get warm and probably not last long. For me amount of light and heat output (and related to that: lifetime) are more important in a LED light bulb then a specification on a piece of paper.

You can use a multimeter to measure the forward voltage of the LED. Then measure the current being drawn by the LED.

You can calculate Wattage by using: W = A x V

It should generally be close to what the manufacturer specifications are if you are providing it a typical voltage.

• Assuming that these are ac measurements, what you have found is the apparent power with units of VA, not real power with units of watts. But, as you suggest, if the power factor is close to 1 then the apparent power should be only a little more than the real power. – Elliot Alderson Jun 27 '18 at 11:42
• @ElliotAlderson: depends on the multimeter that you have and the waveform that is driving the LEDs – PlasmaHH Jun 27 '18 at 11:53
• This answer applies to a LED (or some LEDs) running on a DC voltage. For a light bulb running on AC, things are more complicated. – Bimpelrekkie Jun 27 '18 at 11:57
• @PlasmaHH I'm not talking about rms vs. avg. vs peak voltage and current, or about sinusoids vs. other waveforms. If you measure current and voltage separately then multiply you get apparent power. To get real power you must multiply instantaneous voltage times instantaneous current. The difference is, essentially, the power factor of the load. A significant difference arises if the current and voltage are not in phase or if the shape of their waveforms is different. – Elliot Alderson Jun 27 '18 at 12:06
• @ElliotAlderson: some multimeters can measure power directly. For some waveforms its fine to measure truerms voltage and current to get a reasonable wattage. For some LED bulbs the voltage (and thus current) waveform is reasonably near enough to DC to ignore the waveform. – PlasmaHH Jun 27 '18 at 12:46