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I have been surfing the internet and my physics books for the relationship between the resistance of an LDR and light intensity (a formula), and so far I have found nothing.

If any of you have a formula and along with that a book I could cite, it would be helpful.

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    \$\begingroup\$ This is usually found in the specific component datasheet. \$\endgroup\$
    – Eugene Sh.
    Mar 7 at 19:49

5 Answers 5

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There are many reasons why LDR's compared to Photo Diodes (PD) are a poor choice for measuring LUX.

  • process tolerance is wide with poor 3 sigma variance of results.
  • spectral sensitivity is mismatched to human eye.

PD's are more sensitive to IR yet very repeatable in sensitivity. Thus for camera and light meters, the solution uses a PD with a log amplifier to expand the dynamic range from 1 to 100k LUX and adds an optical filter to results in duplicating the human spectral response centred around 550 nm green.

Sharp has pioneered this technology decades ago and the technology is now licensed to the Vishay brand name. Mouser carries these but not Digikey at one time.

enter image description here

The dynamic current range can be changed to a linear output voltage by the choice of the load resistor. These come in many shapes and sizes. This one is 5mm plastic THT 3 pin with a lens.

Similar parts were also developed by Infineon .

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Lux doesn't measure intensity, it measures illuminance. That means it measures how bright a light looks to the human eye.

The relationship you are asking for will depend strongly on the wavelength, or spectrum, of the light being measured. For IR or UV light, the relationship will be undefined, since the illuminance of IR and UV sources is 0 (their output is invisible to the human eye) but they might still cause a change in resistance of your LDR.

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  • \$\begingroup\$ I wanted to design a lux meter with an arduino controller using an LDR to then collect data for the light intensity as a function of resistance. Is there anyway to do so? \$\endgroup\$
    – Manessa
    Mar 7 at 20:07
  • \$\begingroup\$ An "I2C light sensor" (google it) module is much easier to integrate into your system that a bunch of components. Considering that a LDR is more expensive than these modules. \$\endgroup\$
    – D Duck
    Mar 8 at 9:28
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Typically the equation is

\$\frac {\log(R_A) - \log(R_B)}{ \log(E_A) - \log(E_B)} = \$ k

Where Ra and Rb are two resistances at illuminations Ea and Eb, and where the illumination spectrum is specified.

You can also simply extract an equation from the datasheet of a typical part such as this one. enter image description here

For which you can see that R = \$10^{1.905 - 0.5\log_{10}(E)}\$ (R in k\$\Omega\$)

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I wanted to design a lux meter with an arduino controller using an LDR to then collect data for the light intensity as a function of resistance. Is there anyway to do so?

Measuring lux requires you to measure the spectrum of a light source. There is no way to do that with a LDR, so no this is not possible.

If you are ok with measuring intensity, then you can do that, although an LDR is a poor choice. You can also measure intensity and then assume a spectrum (e.g. sunlight). This will compound the measurement errors from using an LDR with errors in the actual (unknown) spectrum, but if you only want a very rough estimate that might work. If accuracy is important, you should look into buying a spectrometer.

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According to a lengthy discussion in the German Wikipedia of user:Rotkaeppchen68 it is according to an old German physics book like this:

there is a relationship between resistance R and illumination level E like this: $$R \sim E^{\frac{\log R_1 - \log R_2}{\log E_1 - \log E_2}}$$

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    \$\begingroup\$ It would be best if you'd add a Wikipedia permanent link as source of your wisdom. \$\endgroup\$
    – U. Windl
    Mar 8 at 11:49
  • \$\begingroup\$ @U.Windl yes, u r right... i thought i already did it... but it seems like i forgot it... blush \$\endgroup\$
    – RRIDDICC
    Mar 9 at 14:37

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