Photocell/Photoresistor Light Resistance

Question

I am new to electronics. I am looking at purchasing a photocell (photoresistor) for some of my projects. I have observed that there are different levels of light resistance for different photocells -- see "Light Resistance" Column below. What is the significance of the different Light Resistance levels? Same basic question for "Dark Resistance" column and "Sensitivity" column. I'm just trying to determine which model of photoresistor to purchase for my needs. Thanks for your input!

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ChuckMcM · Answer

"Dark Resistance" is the resistance your photocell will have when it is completely dark, "Light Resistance" describes how much the resistance will change based on the amount of light that lands on the photocell, and "Response Time" is how long it will take for the final reading to be reached. So lets start with things that are common to all of them, they are all rated at up to 150volts DC (You probably won't get close to that) they are also all rated for a maximum power dissipation of 100mW. Since power in a resistor is equal ti the current squared * the resistance, you want to be sure that when the device has its lowest resistance you are not putting too much current through it. Generally that isn't a problem either since with a reasonably low voltage (say 5V) you will be hard pressed to push more than a few microamps through the cell (5V/500000 ohms when dark = 10nA, or 50nW). Their resistance is measured at 25 degrees C so if you're going to be running hotter or colder than that and you want to measure small variations in light you may want to check for the change over temperature. They are all peaking at 540nm which is Greenish/Yellowish light (you'd think it was green). So lets say you wanted to use the first one on the list. If you hooked it up 5V -> photocell -> * -> 200K resistor -> Ground. And then measured the voltage between the * and ground. When it was dark, the Cell would have a resistance of 200K ohms, and the lower resistor would be 200K ohms, and in classic voltage divider fashion the voltage drop across the lower 200K resistor would be 2.5V (5V - 5V * R1/(R1+R2)) [1]. But if you had 10 lux of light shining on it, the Cell resistance would go up by 5K ohms, so the new voltage would be 5 - 5 * (205/405) or 2.47V. If you were reading that on a microprocessor with a ADC you would see the voltage go down slightly. More typically you might put the photocell divider to an op-amp, put 2.5V into the positive side, your divider into the positive side, and then build a feedback circuit (a resistor between the output and the voltage divider) that would adjust your gain. Then you could create a voltage from 0 to Vmax (what ever the OpAmp's power rail was). That is useful if you want to measure the light incident to the photocell. Lastly you might put the photocell input into a comparator and use that to change state when a certain level of light had been reached. It really depends on what your final application is. --Chuck

Photocell/Photoresistor Light Resistance

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