How can I convert lux to watts for CFL bulbs?

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SUMMARY

This discussion focuses on converting lux measurements to watts for CFL bulbs, specifically a 2700K bulb with an output of 1600 lumens. To convert lux to watts, one must first convert lux to lumens using the area of the light sensor, followed by applying the luminous efficacy of the bulb, which is approximately 60 lm/W for CFLs. Given the bulb's specifications, the direct conversion indicates that the power consumption is around 27 watts. Understanding the surface area of the detector rather than the bulb simplifies the conversion process.

PREREQUISITES
  • Understanding of light measurement units: lux and lumens
  • Knowledge of luminous efficacy and its application in lighting
  • Familiarity with the concept of power calculation using P = IV
  • Basic principles of light sensors and their area measurement
NEXT STEPS
  • Research the luminous efficacy of different types of light bulbs
  • Learn about the characteristics and specifications of CFL bulbs
  • Explore methods for measuring light intensity with sensors
  • Investigate the relationship between lux, lumens, and watts in various lighting scenarios
USEFUL FOR

Anyone involved in lighting design, electrical engineering, or energy efficiency, including lighting technicians, electrical engineers, and DIY enthusiasts seeking to understand light measurements and conversions.

cmkluza
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Hello all,

I've currently got a swirly CFL bulb that I lit up wirelessly and took some measurements of using a light sensor which returned the data to me in lux. How can I get from lux to watts? As I understand it, lux is a measurement of lumens per unit area, so do I have to somehow find the surface area of this bulb? That doesn't exactly seem simple. I've tried looking up specifications for this bulb specifically, but it doesn't include anything about surface area. The specifications available from the bulb are 2700K color and 1600 lumens. Not sure if that's entirely helpful.

I realize you can figure out power using P = IV, but I figure that doesn't help since I only know the voltage and current input to my slayer exciter, not the voltage and current that the bulb was picking up, and I can't imagine that the slayer exciter is 100% efficient.

Any ideas? Thanks for any help!
 
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cmkluza said:
Hello all,

I've currently got a swirly CFL bulb that I lit up wirelessly and took some measurements of using a light sensor which returned the data to me in lux. How can I get from lux to watts? As I understand it, lux is a measurement of lumens per unit area, so do I have to somehow find the surface area of this bulb? That doesn't exactly seem simple. I've tried looking up specifications for this bulb specifically, but it doesn't include anything about surface area. The specifications available from the bulb are 2700K color and 1600 lumens. Not sure if that's entirely helpful.

I realize you can figure out power using P = IV, but I figure that doesn't help since I only know the voltage and current input to my slayer exciter, not the voltage and current that the bulb was picking up, and I can't imagine that the slayer exciter is 100% efficient.

Any ideas? Thanks for any help!

You can convert lux to lumens by knowing the area of the detector, and convert that to Watts knowing the spectral distribution of the light (the luminous efficacy). Fluorescent bulbs and CFLs have a luminous efficacy of around 60 lm/W.

On the other hand, since you are given the excitance (1600 lm), you can convert that to W directly (27 W), or use that information to characterize your detector, etc. etc.
 
Andy Resnick said:
You can convert lux to lumens by knowing the area of the detector, and convert that to Watts knowing the spectral distribution of the light (the luminous efficacy). Fluorescent bulbs and CFLs have a luminous efficacy of around 60 lm/W.

On the other hand, since you are given the excitance (1600 lm), you can convert that to W directly (27 W), or use that information to characterize your detector, etc. etc.
Ah, so it meant the surface area of the perceiver, not the emitter? That makes this much more simple. Thanks for your help!
 

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