# Photoresistor LDR in parallel

I need to sense light in 50 different points, but i can only arrange them in parallel because i need to use only 2 wires (this is limited by the application constrains).

I only need to know if any of the 50 points detects light, like 50 inputs with an OR gate. I would be using a micro controller with and ADC connected to a resistor and the 50 LDR in parallel in a voltage divider arrangement.

Could this work?

Resistance changes between 10Kohms to 100kohms between presence and absence of light.

• If you can make the math work then you can make the circuit work. – Ignacio Vazquez-Abrams Nov 7 '16 at 22:46
• Consider the resistance of 50 dark LDRs in parallel, and 49 dark, 1 lit. Can you reliably distinguish between those two resistances? – Brian Drummond Nov 7 '16 at 22:48
• 50 LDR in parallel at that range is 2K ohms, if one detects light, the total resistance becomes about 1.6K ohms. you wont see much of a difference, bit there will be one – ambitiose_sed_ineptum Nov 7 '16 at 23:24
• Unless you define the expected deviation of light on each source and the tolerance of the LDR, it is unlikely to give a good result. If you only want a digital result from each source rather than accumulate alll the light input from all sources, Use a \$2 optically correct and filtered light sensor to give a logic level out then combine as you see fit. – Sunnyskyguy EE75 Nov 8 '16 at 2:11

At first glance you might think that with 50 sensors dark and 100 kΩ each, you get 2 kΩ with all in parallel. With 49 sensors at 100 kΩ and one sensor at 10 kΩ, you get 1.7 kΩ. Going from 2.0 kΩ to 1.7 kΩ is a 15% change. You really should be able to detect that.

However, it's not that simple. CdS LDRs, like you're almost certainly using, are notorious for having wide tolerance. They won't all be nice and neatly 100 kΩ when dark. That value will vary considerably from part to part, with temperature, and with time. The lit resistance will vary too. In the end, I doubt you can find a overall parallel resistance threshold for reliably detecting at least one sensor lit.

If you were really forced to put all the sensors in parallel, then you probably need to calibrate to dark before each use.

Instead of just connecting all the sensors in parallel, I'd put a little active circuitry at each sensor. This could be as simple as one additional resistor and transistor. The transistor turns on when its cell senses light. It then pulls down on a common open collector line. This scheme makes use of the much higher signal range at each sensor, as apposed to all sensors averaged with one of them changing. The resulting digital signals are then ORed.

This does require three connections (power, ground, and output). You don't always get something just because you wrote a spec for it.

It really depends on what you want to accomplish, will you be able to detect a difference if the light changes? Yes. Will you be able to tell which one changed? No. And it will be hard to measure the ends of the range or the fully light and dark condtions

The values will be summed, its called integration:

$R_{total} = \sum\limits_{i=1}^{50} R_{sensor}[i]$

So lets consider if all of the sensors are around 100kΩ (fully on), your total will be 100k*50 or 5000kΩ. If you put your finger on one, then it will go to 10k and your total will be 5000kΩ-90kΩ or 4410kΩ. On the lower end its 10k*50 or 500kΩ.

Since you will probably be using an ADC, and use a voltage divider.

$V= I*R$

Lets see how much current the configuration will draw without a voltage divider. with a 5V stimulus:

$\frac{5V}{5000k} = 1uA \ and \ \frac{5V}{500k} = 10uA$

Now lets see if we put a 1000kΩ resistor in series with the total at 5V

$5V*\frac{500k \ to \ 5000k}{500k \ to \ 5000k + 1000k} = 1.66V \ to \ 4.166V$

That doesn't seem unreasonable, but realize at such a low current (uA's) its not going to take much to interfere with your measurement and create noise.

The voltage difference of one turning on if all others are off is

$5V*(\frac{500k \ to \ 50k}{500k \ to \ 50k + 1000k}-\frac{590k \ to \ 4191k}{590k \ to \ 4910k + 1000k}) = 37mV \ to \ 2.5mV$

The ADC will need to be sensitive to at least 2.5mV.