# How accurate is a photoresistor?

I'm doing a school project in which I need to measure the amount of light emitted by a specific object.

The first thing that came to my mind was to use a photo resistor hooked up to a voltage meter in resistance mode. After some research on the internet, it seems like photo resistors are not extremely accurate.

My question is exactly how accurate a photo resistor is against itself. I will be using the same one throughout the whole experiment, and would like to detect small changes with different light sources.

• What are the different light sources? How accurate do you need it to be? Sep 30, 2015 at 22:47
• Mainly different types of lightbulbs, but there could be others. As for accuracy, it just has to be accurate enough to prove that x is brighter than y. Sep 30, 2015 at 22:53
• "x is brighter than y" doesn't require accuracy. It only requires a monotonic response. That's much easier to achieve, and any photoreceiver ought to be able to do that as long as you don't dramatically over-drive it. Sep 30, 2015 at 23:09
• You only need accuracy if you want to measure the irradiance from the source with a specified error margin. Meaning if you want to be able to say things like "this source is producing 0.75 W/m^2, +/- 2%". Sep 30, 2015 at 23:11
• I think what he means is how sensitive* is a photoresistor
– Klik
Oct 1, 2015 at 2:00

Cadmium Sulphide (CDS) photo resistors were commonly used in film cameras for determining exposure, usually wired in Wheatstone Bridge configuration. These light meters were quite sensitive and repeatable. Here's an example circuit:-

Since you only want to determine which light source is brightest, absolute accuracy is not important. With a Wheatstone bridge circuit you can adjust the potentiometer to get a null reading (equal voltages on each side) with one light source, then switch to another source and note whether the reading swings positive or negative.

A photo resistor's value can change with temperature and aging, but this shouldn't be a problem so long as you do all comparative tests at the same time and don't let the light source heat up the sensor. At low light levels the response will be slow, so you may have to wait a few seconds for the reading to stabilize.

The biggest problem you may have is different spectral outputs of your light sources combined with the response of your sensor. Incandescent light bulbs emit a broad spectrum that is skewed towards Infrared, while LEDs and Fluorescents have sharp peaks at certain colors. CDS sensors are most sensitive to green light.

If you want to compare perceived brightness then you also have to consider the human eye's spectral response. CDS is a good match. Other sensor types can be quite different, leading to unexpected results.

• It's possible to get 'eye response' photodiodes e.g. BPW21 vishay.com/docs/81519/bpw21r.pdf - but a CdS photoresistor is cheaper. Oct 1, 2015 at 10:05

You should find out which kind of photo resistor you are using and consult the data sheet for the part. There, you should find some information on how accurate this device is. I'm not sure if there's a general rule on the accuracy. Just read the data sheet :-)

• I pulled the photo resistor out of a old night light so there isn't any data sheet to consult. Sep 30, 2015 at 22:54

A photo resistor is highly inaccurate and without optics will pick up ambient light, too. What you want to do is using a chopper that can shade the sensor for a zero-reading/adjustment. When you have 2 readings and can subtract the object reading with the zero reading. The chopper can be a simple magnetic relay that moves a piece of black cardboard mounted on it's anchor. Additionally place the sensor in a small tube (colored black) to avoid getting too much ambient light on it.

Forget the photo resistor, Your best bet will be a photodiode(PD). With a ~$5-10 PD and a DMM that has a 200uA current range, you can make a decent power meter. The best thing, is that the measurement is full of physics. Over most of the visible range a PD has a response that is close to 100% quantum efficiency(QE). (The exact response varies, but 70% to 90% QE is not uncommon.) 100% QE means that each photon that strikes the PD makes one electron-hole pair, measure the current, convert to electrons per second, (apply QE factor), and you've measured the number of photons per second. You can express this in watts/cm^2, if you know the wavelength, or just leave it at photons per second. (Trying to convert it to lumens is a pita.) If you need more sensitivity, comeback and we can talk about opamp circuits. Oh if the photodiode is too much$, You can use a cheap solar cell, but you'll have to guess at the QE. (And maybe measure the area.)