Relation between light intensity, voltage and resistance

As the intensity of light decreases, resistance of LDR increases. So a higher resistance will pull higher voltage(as explained in the tutorial).

But when I expose the LDR to bright light, it gives a reading of 1.02 thousand. I understand it is the voltage reading coming out of the LDR captured through the A0 pin. But shouldn’t the voltage be lower in bright light conditions as reistance is lowest.

Also how do we read the output of A0. what is the unit when you say 1.02 thousand or 600. I know light intensity is measure is lux. How do we convert this values to lux?

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The 1.02 thousand value which you are getting is in terms of light intensity. That is it’s unit is lux. So, it is 1.02 thousand lux.

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As “light intensity” decreases, the LDR’s “resistance” increases, as does the potential “difference” across it. So, as distance increases, the potential “difference” increases. Since we want the potential “difference to” change in the same direction as the distance, the multimeter must go across the LDR, not the “resistor”

We connect the LDR between 3.3v pin and the analog input pin (A0), so that when light intensity increases, the resistance of LDR decreases so the voltage across the LDR decreases and as a result, the voltage on the analog input pin increases.

This means that as the light intensity increases, the voltage on the analog input pin also increases . The Bolt then converts that the voltage a 10 bit (10 places in binary number system) digital value that varies from 0-1024 (0 to 2 raised to 10).

This digital data is then sent to the cloud where it is plotted for visual representation.


We are taking output voltage of divider circuit in this configuration. That’s why the values are high for brighter and low for darker.

It is not in lux , the output of ADC is digital so it is binary number output displayed as bits (0 to 1024)

Your Analysis is absolutely correct. Except that the number which you see is the not the Voltage value but a proportional of that. Voltage across LDR decreases as you said but this also mean voltage input to the pin A0 increases.