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A device to measure the intensity of light should I do. Help Please? Answered

Actually thisis ahomework.Wouldyou help me.Ido not know how.What materialsshould I use.I wonder how can I dol. In fact, my homework to make an the sensor, example of rain sensor. That is something like I will a new sensor produced. I'm sorry I'm trying to write Google translate




The short-circuit-current of a solar cell should be roughly proportional to the intensity of light reaching the cell. 

A simple model to explain how this works is to assume each single photon hitting the cell,  contributes one electron to the short circuit current.  Light intensity is (photons/time/area)*(area of cell) ~ (electrons/time) = current

As an example:  a silicon solar cell, with an area of about 1cm^2, illuminated with dim indoor lighting, produces a short-circuit-current of about 20 microamperes of direct current (DC).  The picture is from:

20 microamperes (μA) is a  small amount of electric current, but easy enough for a typical, cheap, multimeter to measure.

In bright sunlight, this same setup would produce much more current.   I am guessing somewhere between 10 and 100 times as much current; i.e somewhere between 200 and 2000 μA, or 0.2 to 2.0 mA.

Anyway, the game here is that a solar cell can be used as a light sensor, where short-circuit-current is proportional to light intensity.
See also:


thanks jack .SUPER.
With this circuit so I can measure the intensity of light?
Based on the intensity of the sunlight to light the LCD screen as a percentage, Can I write. for example, the PIC circuit

Programmable Interface Controller talking to.Microcip Product. Can I print using the LCD to the temperature or arduino uno.


The analog inputs on a PIC(r) or Arduino(r) want a voltage signal , typically in the range from 0-5 volts.

The output from a shorted solar cell is a current signal, so you will need a circuit to convert current into voltage.  The usual way to do this is with an op-amp wired as a current-to-voltage converter (also called transconductance amplifier).  A diagram of such an amplifier circuit is attached.

For this circuit, Vout = I*R, and you choose R to give the desired gain.  For example R=100 KΩ gives 1 volt for every 10 μA of photo-current.  If you picked R=1K, then you would 1 mA = 1000 μA of photocurrent to get 1 volt of Vout.

The capacitor C, is placed there to limit the response to low-frequency signals.  I am guessing you just want to measure slowly changing light level.  So a good choice for C, would be that C which gives R*C= 1s (time).

The following Google image search will show you similar circuits:


small wood block with 2 drawing pins on it about 2 mm apart.

A wire to each drawing pin. Connect as diagram with an NPN transistor as a switch. The LED lights when rain bridges the drawing pins.

Did you not read the responses to your previous request?