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# Light sensor

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#### Vikky

##### New Member
Hi all,

I got a light sensor BPW34.

I read its datasheet but no where i could figure out how much amout of light will make it to change 1 lux.

Also i need the idea how to test it. I only got that in the dark room it will have no output and in the full sunlight it will give 100%. What about the intermediate range and how to test it?/

#### colin55

##### Well-Known Member
How to test the BPW34:

Connect it to a 9v battery with a 10k resistor in series with one lead.
Measure the voltage across the resistor as you subject it to different levels of light.

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#### Mikebits

##### Well-Known Member
How to test the BPW34:

Connect it to a 9v battery with a 10k resistor in series with one lead.
Measure the voltage across the resistor as you subject it to different levels of light.

It would be better to use a known voltage source as in a 9vds supply, who knows the condition of battery. Also how does one vary light source by 1 lux to calibrate the sensor, which was original question.

#### AllVol

##### New Member
Hi all,

I got a light sensor BPW34.

I read its datasheet but no where i could figure out how much amout of light will make it to change 1 lux.

Also i need the idea how to test it. I only got that in the dark room it will have no output and in the full sunlight it will give 100%. What about the intermediate range and how to test it?/

A lux is a lux is a lux, of course.

Wikipedia says:

"Lux is a derived unit based on lumen, and lumen is a derived unit based on candela.
One lux is equal to one lumen per square metre, where 4π lumens is the total luminous flux of a light source of one candela of luminous intensity:
1 lx = 1 lm·m-2 = 1 cd·sr·m–2. As with other SI units, SI prefixes can be used, for example a kilolux (klx) is 1,000 lux."

So now all you have to do is figure a good way of measurement.

Perhaps light falling on black and white patches of construction paper for a base of full reflected light and absence of reflected light, then voltage measurements as the distance from light source to test panel decreases.

According to Forrest Mims III, the "intensity or strength of light waves is inversely proportional to the square of the distance of the wave from its source. In other words, if the distance is 3, then the intensity is 1/9 the intensity when the distance is 1".

Have fun.

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