Light sensor always getting max value

I have build a simple voltage divider with a light resistor and a 5.6k resistor. I am reading the input on A0 on my arduino board but it is almost always at 1023, independent of light. If I cover it it might go down to 1020 but not more than that.

Do I have to use a greater resistor than 5.6k to achieve more difference in the sensor?

I think that the resistance is pending between 1k-10k.

The circuit is like this image: Where Vin is 5V from the arduino. R1 is the light sensor, R2 is 5.6k and Vout is connected to A0.

• Can you show a diagram of your exact circuit? – Peter Bloomfield Apr 22 '15 at 11:34
• A pointer to the datasheet for the photo-resistor would help, too. At the very least, we'd need to know its resistances when dark and bright. – JRobert Apr 22 '15 at 11:42
• 5.6k sounds good. Is A0 connected between LDR and Resistor? Can you check that LDR works using an Ohm meter? – user3060854 Apr 22 '15 at 12:18
• A circuit is essential. A description in words in the absence of a circuit is a logical minimum if you actually want an answer. LDR should go from eg A0 to Vcc and resistor from A0 to ground or vice versa - is that what you did? Where did you get the information? Datasheet for LDR? (There are zillions of types). ... – Russell McMahon Apr 22 '15 at 14:25
• Have you tested the voltages and resistances with a Multimeter? – Craig Apr 22 '15 at 15:45

What have you selected as the analog reference? And if 'external', what is connected to the Vref pin?

For the LDR resistances of 10K and 1K, you Vout should be ~ 1.8v at 10K, and ~ 4.2v at 1K, and giving A/D readings of somewhere around 370 counts and 860 counts, respectively, with a 5v analog reference.

If your analog reference selected to the 1.1v internal reference, either LDR resistance should give you the max reading, but an LDR = 20K could give just over a volt and pretty close to the reading you're getting.

Judging by a datasheet (that I found on the web) that has a resistance of:

• 100 k to 200 k in the light
• 1 M in the dark

Given that Vout = Vin * R2 / (R1 + R2) we can work out:

Light:

Vout = 5 * 200000 / (200000 + 5600) = 4.86 V

Dark:

Vout = 5 * 1000000 / (1000000 + 5600) = 4.97 V

Do I have to use a greater resistor than 5.6k to achieve more difference in the sensor?

Yes, I would measure the resistance in the light, and choose a similar resistor for the other side of the voltage divider. So if you measure 100 k on the sensor, another 100 k will at least divide the voltage in two nicely.