Some time ago I bought 50 photo resistors, 10 pieces of 5 different types each. Had no real need for that many but it’s very cheap if you buy them through AliExpress. It turned out to be quite difficult to find out which resistor does what. Tried looking for photoresistors, LDR, Light Dependent Resistor, photo cell but I could not find any explanation on the different types.
I did find the following information (the table is compiled from different sources):
| Model |
Light resistance
(10 Lux) KOhm |
Dark resistance
MOhm |
| GL5506 |
2-6 |
0.15 |
| GL5516 |
5-10 |
0.5 |
| GL5528 |
10-20 |
1 |
| GL5537 |
16-50 |
2 |
| GL5539 |
30-90 |
5 |
But how does that help in connecting these things? Well, I decided to connect these things, all at the same time and look at the readings.
So connecting, how do we do that? Well, we need to understand (somewhat) Ohm’s law. To measure the amount of light, we want to measure the resistance of the light sensor. Unfortunately the Arduino or the ESP8266 do not measure resistance. They can however measure the voltage on an analog port and this is where we need Ohm’s law. First, let’s look at a schematic:
We connect the photo resistor to +5V (or +3.3V) on one end and the other end on a analog port (the ESP8266 has one, the Arduino has many). We also connect that same end to a resistor (in my diagram a 10K resistor) and the other end of the resistor to ground. What does this do? Well, current will flow from +5V over the photo resistor, continue over the 10K resistor and than to ground. What we are going to measure is the voltage at the point between the photo resistor and the normal resistor. Ohm’s law states V=IR, voltage is current times resistance. The current flowing from +5V to ground over both resistors will be the same (the amount of current does not change midway a circuit). That’s the key to figuring out the current at our analog port and here is how.
We can calculate the total current that flows from +5V to ground by adding the resistors and plugging it in Ohm’s law. Let’s say the photo resistor currently has a resistance of 10KΩ. The total resistance of the circuit is 10KΩ+10KΩ = 20KΩ. V = 5V which means I = 5V / 20KΩ = 0.25 mA. So we now know that there is 0.25 mA running through the circuit. Now we can calculate V at our analog input. The voltage drop over the photo resistor is V = IR and we now know I and R. V = 0.25 mA * 10KΩ = 2.5 V drop so at our analog input we will measure 5V – 2.5V = 2.5V. With 2 equal resistances, it makes sense that in the middle you will have 2.5V. Let’s take the next step.
The resistance of the photo resistor goes down when there is more light, so let’s make it a little darker, say 30KΩ. Let’s recalculate:
Total R = 30KΩ + 10KΩ = 40KΩ
I = 5V / 40KΩ = 0.125 mAVoltage drop over photo resistor = 0.125 mA * 30KΩ = 3.75 V
We measure 5V – 3.75V = 1.25V at our analog input
So less light means less voltage at our analog input. Well this math is all fine, but what does this mean? Well, if we know the voltage at our analog input we can also do the reverse and calculate the resistance of the photo resistor, just do the reverse.
Let me quickly share the setup and the code I used to test the resistors. You can my setup at the top. It is an Arduino Nano (which has 5 analog ports) and the 5 sensors, the bottom is the 5506 and moving up all the way to the 5539 at the top. It might be a bit hard to see. All black wires are +3.3V, all the white are ground and the different colors are for each of the analog inputs. You can see that I followed the schematic 5 times and importantly, used five 10KΩ resistors. You cannot combine all of them on one. Please do not forget to connect 3.3V to ARef which gives a better ready is my experience.
The code is simple, initialize five ports and continously display the results of the ports. I have added a header row every 10 rows so I can copy and paste that:
Available on GitHub with more comment in the code. Direct download.
//
// The number of steps after which the header is repeated
//
#define STEPCOUNT 10
//
// The current step count
//
byte step = STEPCOUNT;
void setup() {
//
// Setup serial
//
Serial.begin(115200);
Serial.println("Ready");
//
// Set the reference for measuring the analog signals
// For some reason I got incorrect readings without connecting
// the 3.3v (on my Arduino Nano) to AREF and setting this
// reference
analogReference(EXTERNAL);
//
// Setup all analog inputs. We are using 5 which is the maximum
// for the Nano on which I tested.
//
pinMode(A0,INPUT);
pinMode(A1,INPUT);
pinMode(A2,INPUT);
pinMode(A3,INPUT);
pinMode(A4,INPUT);
}
void loop() {
//
// Increase our counter
//
step++;
//
// If we are over our STEPCOUNT then we repeat the header
//
if (step>STEPCOUNT) {
//
// The reference to the different photo resistors
// GL5506, GL5516, GL5528, GL5537 and GL 5539
//
Serial.println("* 5506 5516 5528 5537 5539");
step = 0;
}
//
// Print all analog values to serial
//
Serial.print(" ");
Serial.print(analogRead(A0));
Serial.print(" ");
Serial.print(analogRead(A1));
Serial.print(" ");
Serial.print(analogRead(A2));
Serial.print(" ");
Serial.print(analogRead(A3));
Serial.print(" ");
Serial.print(analogRead(A4));
Serial.println();
delay(250);
}
If we take the values it shows during daylight, inside the house:
* 5506 5516 5528 5537 5539
975 969 774 619 218
975 968 773 619 220
975 969 773 618 216
975 968 771 616 215
975 968 771 617 221
975 969 774 620 226
975 969 776 622 225
976 969 775 621 221
975 969 774 619 220
975 969 774 619 219
975 968 772 617 216
As you can see, the values are relatively stable. So let’s take one of the values of the 5528, 775. How do we calculate the resistance from that? The analog port has a 10 bit resolution, which means it can go from 0 to 1023. 1023 will be at 3.3V (in my case. If you use a Uno for example it will be 5V). The reading is 496 which means it is reading a voltage of 775/1023*3.3V = 2.5V. So we know the voltage drop was 3.3V-2.5V = 0.8V. So we know V. Now we need to know I and from that we can calculate R. We can calculate R because we know V and R for the other part of the setup, the 10KΩ. I = V/R = 2.5V / 10KΩ = 0.25 mA. Now we take that 0.25 mA and use it for calculating the resistance of the photo resistor.
R = V/I = 0.8V / 0.25 mA = 3,200 Ω. Now you can go to the GL5528 datasheet and see
that the amount of light is between 30 and 85 lux. Well, that’s quite a broad range and not really useful for calculating the amount of lux but the deviation of these sensors might vary piece to piece but do not vary that much per piece, so if your sample decides to be more towards the dark side, it will be there with higher or lower light settings as well. What this means is that the sensors are great for detecting deltas in light even though getting a precise lux reading is not really feasible. Ok, well than we need more readings:
light from a normal bulb:
* 5506 5516 5528 5537 5539
974 967 739 623 279
982 981 844 717 437
992 988 870 751 454
994 991 874 758 440
995 991 873 752 325
992 986 836 700 137
985 979 795 654 62
978 972 754 610 34
971 969 792 668 379
988 985 854 727 440
993 990 871 749 440
light from a mobile phone
* 5506 5516 5528 5537 5539
1006 1007 956 925 795
1006 1007 956 925 796
1006 1007 956 925 797
1006 1007 956 925 797
1006 1007 956 926 798
1006 1007 955 925 795
1006 1007 956 925 796
1005 1007 955 924 795
1006 1007 955 924 793
1005 1007 955 925 796
1005 1007 955 925 795
light at dusk
* 5506 5516 5528 5537 5539
834 834 385 208 20
840 838 394 216 22
840 840 402 222 23
840 840 405 225 24
839 839 404 225 24
839 838 404 224 23
840 838 405 225 24
841 839 405 226 24
841 839 405 226 24
844 840 406 227 24
845 840 406 227 24
light in the evening
* 5506 5516 5528 5537 5539
582 583 118 52 1
581 582 118 52 1
580 583 118 53 1
580 582 118 52 1
580 581 118 52 1
578 580 117 52 1
577 577 116 52 1
565 561 109 49 1
584 583 118 51 1
586 587 120 53 1
589 590 121 53 1
There are a few things to notice. First, I thought it was interesting to see that the values fluctuated with a normal bulb. This sample was not an incident, it was doing that the whole time. I think it might have to do with the fact that light bulbs fluctuate because of the way the alternating current alternates but I’m not sure. The second thing to notice is that the 5506 does not really vary that much with light, especially compared to the others. It’s not because it’s broken, but it’s because it’s resistance is small compared to the 10kΩ resistor which means it does not leave a lot of ‘bandwidth’ to show it’s capabilities.
Maybe I will do some more readings where I take a resistor to match each photo resistor and see what the results are than.
