Measuring and Calculating Lux Values
The resistance of a LDR depends on light intensity. At low light levels, the LDR has a high resistance. As the light intensity increases, the resistance decreases. It only means that the relationship between voltage and current changes with the We might be tempted to say that the intensity I of the light is. Linearity: Often a linear relationship exists between the actual value of the quantity w and the output signal of the sensor, e.g. the voltage U. In this case: 0.
Hence we have the coordinates of one point as well as the ascent of the line and for calculating any other point, we only need one coordinate.
A-level Physics (Advancing Physics)/Sensors/Worked Solutions
Meaning, if sensors' resistance RB is measured, it is possible to calculate from the equation of line, the intensity of light EB that falls on the sensor.
Finding EB from the equation of line: The resistance can not be measured directly with microcontroller. For this the photoresistor is in the voltage divider. The output voltage of this voltage divider is converted to a specific variable by the analogue-digital converter ADC.
To find the resistance, the output voltage U2 of the voltage divider must be calculated first, using the ADC value, also comparison voltage Uref of the converter must be taken into account: The formula is following: If circuit is used equipped with different components, respective variables need to be changed.
Photoresistor [Robotic & Microcontroller Educational Knowledgepage - Network of Excellence]
So somehow we have to compensate for the fact that two separate photosensitive elements—the human eye and the photosensor—are more or less sensitive to the countless different wavelengths of electromagnetic radiation concealed within the whitish light illuminating our environment. This seems hopeless, but the first thing to understand is that illuminance simply cannot be measured with the sort of precision we expect from thermometers or voltmeters or digital calipers.
For that we would need either 1 a photosensitive device with spectral response identical to that of the human eye or 2 separate narrowband photodetectors, each with known sensitivity, fine-tuned to numerous wavelengths within the visible spectrum.
Furthermore, such precision is by no means necessary or even possible. Who really cares whether their office is illuminated at lux or lux? Indeed, a high-precision lux measurement is almost a contradiction in terms, because illuminance is supposed to represent a human response to lighting conditions—and how often do humans agree on something so subjective?
Bring two people into a room and ask them to rate the ambient brightness on a scale of 0 to So calculating lux is an exercise in approximation, and in this case, an approximate value is just fine. The following image shows the spectral response of the OSRAM photodiode mentioned above superimposed on the luminosity function: Trying to compensate for such extreme discrepancy is simply not a good use of time. The Easier Way Instead, select a device with a human-vision-based spectral response and with output-to-lux conversion information provided in the datasheet.
Such devices are not hard to find; one example is the Fairchild phototransistor mentioned above. We should not need to adjust illuminance measurements based on the nature of the light source, because the process of calculating illuminance automatically accounts for variations in spectral composition. These are fairly consistent, but far from perfectly matched, so the light intensity detected by the phototransistor cannot be directly converted to lux.
Light Intensity and Voltage Blog
This graph shows how the photocurrent increases when the light intensity increases but the wavelength is held constant. The stopping potential is the same however, suggesting that the kinetic energy of the ejected electrons is the same and hence independent of light intensity.
This is the opposite of what would be expected classically.
- What is the relationship between light intensity and voltage?
- 2.2: Photoelectric Effect
- Relationship between INTENSITY of light vs. VOLTAGE (output) ?
If light was of the classical wave-like nature, we would expect a time lag which would increase as the intensity of the light decreased. The total energy would be spread across a wavefront striking the entire cathode and nothing could happen until sufficient energy were absorbed.
How an LDR (Light Dependent Resistor) Works - Kitronik
Also, why would the frequency of the light make a difference to the stopping potential? The amount of energy in a wave is carried by its amplitude. Here is a plot of the measured stopping potentials obtained for several light frequencies for two different metals.
The slope of this plot is the same in both cases and is equal to Planck's constant h.