# Need help with a photocell

I'm trying to find a modern alternative for a discontinued photocell (a Clairex CL5M4 CdSe photoresistor.)

Its datasheet gives the following:

Illuminated resistance 1.5k @ 2ft-candles
Min dark resistance 400k @ 2ft-candles​
Max voltage 250V​

1: I believe 2ft-candles ≈ 21 lumen, but modern photocells are rated at 10 lumen.
How do I convert the given resistance to 10 lumen? Maybe (1.5k / 2.1 = .7k)?
But the light source will appear brighter when closer, so maybe (1.5k * 2.1 = 3.15)?

2: Modern photocells give Illuminated Resistance as a range, ie. "0.5 ~ 17kOhms".
How do I choose an appropriate photocell if I only have a single number as shown above?

3: How can I use a multimeter to measure the voltage applied to the Clairex photocell?

Any help would be greatly appreciated!

Just an update to this question: The responses I received in a different forum said to try a handful of different LDRs (light dependent resistors) as they aren't very precise and the circuit has two potentiometers to tweak the performance. One recommendation was a 5528.

Joined October 09, 2021      3
Saturday at 04:41 AM

Do you have schematic? any documents? it is quite difficult to understand without.

Joined February 12, 2018      696
Monday at 02:11 PM

These are just general questions which would apply to any LDR in any circuit. I wanted to learn how to choose one myself rather than asking "tell me which one I should buy"... teach a man to fish and all that.
​Because my questions are so generic, I felt that a schematic would just be unnecessary clutter:

1. I need a formula to convert 21 lumens to 10 lumens.

2. LDRs are sold in ranges but datasheets show a single number, and I asked how to buy a replacement one based on that.

3. I asked how to measure voltage through an LDR with a multimeter (this was answered on another forum)

Joined October 09, 2021      3
Saturday at 04:41 AM

Well,

1. For converting 21 lumens to 10 lumens in simple you need to devide it by 2.1. This is matahametical simple thing which anybody knows. Now, to replicate this in circuit you need a LDR and use two resistors to divide that voltage.

For example, you need a LDR that produces 21V (for example) at 21 lumens and 10V at 10 lumens. You need a voltage divider that could produce the 10V at 21v input. You could also use potentiometer to match this.

2. The single number is the nominal resistance. If you check there is always a range for the LDR. Do not look for the resistance then look for the lumen instead look for the lumen range and then get the resistance.

3. Well LDR is not a votlage source, you are having wrong parameters. A resistor works with current, LDR changes current flow (as it is a resistance) based on the light appered across it. You need a voltage source or current sourcce and need to get the voltage drop across it using a multimeter.

Joined February 12, 2018      696
Monday at 02:11 PM

Thank you for your help. (I had included links to the datasheet and schematic as you asked, but a moderator seems to have removed them.)

"This is matahametical simple thing which anybody knows"

Not necessarily. For example moving closer to a light source from 2 feet to 1 foot does not lessen the light's intensity, it increases it. So when going from 2 to 1 feet, dividing by two would seem to be obvious, multiplication is the correct method of determining the intensity. So while I knew that dividing by 2.1 is probably the correct solution to my first question, I needed verification from someone like you who knows better than I do.

The reason I want to measure the voltage applied to the LDR is because LDRs are rated at maximum voltage. This particular LDR is *rated* at 250 volts. But if I only look at 120 to 250 volt LDRs, there are very few choose from. My (unlikely) hope is that the voltage applied will be much less then 120 volts, so I can have more replacements to choose from.

Joined October 09, 2021      3
Saturday at 04:41 AM

Not necessaryly you need a 250V or 120V LDR. You can choose any. Always use a voltage divider to reduce the voltage. Make sure that the current is less. You are sensing the voltage which requires less current. You are not drawing current outof it. There is always a solution.

Joined February 12, 2018      696
Monday at 02:11 PM