I Need a Photocell

[Dingus]

Hi folks.
I have a relay that controls outside lighting.
It's a Maclean MCE83 Light sensor. Link to example
Unfortunately the sensor fails every winter (Probably ice) and I end up buying a replacement that includes the relay.
This time I decided to just get a photocell and not the relay also.
Here is my issue. I can't find the photocell online, and nor can I find Maclean that is supposed to be the maker.

Can anyone tell me where I can get a sensor or how to find the makers?
Or suggest an alternative photcell that would work?

 

[Diver300]

You could use an astronomical timer that works out whether the sun is up or not, and there does not need to be an outside element.

Here is an example:- https://www.ebay.co.uk/itm/306055578018

 

[Tony Stewart]

Ice? Did the relay toggle with the LDR blocked? Then it would be faulty relay contacts.
If not then it would be faulty sensor & relay driver.

 

[Dingus]

Diver ta, I’ll have a look.
Tony it’s definitely not the relay. I have 2 and I’ve tested both.
The photocell actually fell to bits

 

[augustinetez]

That sensor is just an LDR eg https://uk.rs-online.com/web/p/colour-light-sensor-ics/9146710 - you should be able to find it at any one of numerous component suppliers.

However, you will need to know the specs of the original ie light/dark resistance value as they come in various resistance ranges.

Then just a case of repairing one of the originals.

 

[Dingus]

Thanks for the info.
I just checked your RS link. It’s almost £11 when the one in my link is £5 for both.

 

[Dingus]

Forgot to say.
I’ve not been able to find any info on the original sensor.
Mostly due to not being able to contact Maclean

 

[augustinetez]

Thanks for the info.
I just checked your RS link. It’s almost £11 when the one in my link is £5 for both.

Doesn't surprise me, better off just buying the replacement for a fiver.

 

[Dingus]

The other point. I enjoy fixing electronic things. It’s not the few pennies. I’d rather fix than replace.
Probably like everyone on here.

 

[augustinetez]

OK, then the suggestion is spend the fiver first and measure the light & dark resistance values of the supplied sensor.

Once you have that, you can then start searching for an equivalent LDR - Ebay sellers have various ones at a much cheaper price than the RS link eg https://www.ebay.com.au/itm/284636904759 (I'm in Oz hence the AU site).

 

[Dingus]

Your a mind reader, but it’s more that my Mrs is nagging me to get them fixed sooner.

 

[camerart]

The other point. I enjoy fixing electronic things. It’s not the few pennies. I’d rather fix than replace.
Probably like everyone on here.

Hi D,
If I'm correct, then a potentiometer could be attached and turned until the light goes ON/OFF, then you would know the value of the LDR value.
C

 

[Dingus]

Thanks camerart but the cell was totally destroyed, so no measurement is possible.
Even looking for info online is fruitless.
Cheers for helping

 

[jaclement]

A Led can be used to sense light.

 

[Nigel Goodwin]

A Led can be used to sense light.

And is no use for replacing an LDR in an existing circuit.

 

[Nigel Goodwin]

Thanks camerart but the cell was totally destroyed, so no measurement is possible.
Even looking for info online is fruitless.
Cheers for helping

What happened to it?, is the housing not UV safe?, and rots away - leaving the LDR to get wet and rust to pieces?.

Regardless, you need to identify the specs of the LDR if you're going to make your own sensor - and that's not going to happen instantly - as suggested, buy a new one as before, and measure it before you install it (in total darkness, and in very bright light - make a note of the readings, for the future). That seems the fastest way to get it back working.

Or, order a selection of LDR's from China, such as these:

https://www.aliexpress.com/item/32833626809.html?spm=a2g0o.productlist.main.5.23e8tsBPtsBPU0&algo_pvid=94a7b0ac-bc8c-4154-a1c7-ae5a8ad8d6d3&algo_exp_id=94a7b0ac-bc8c-4154-a1c7-ae5a8ad8d6d3-2&pdp_ext_f=%7B%22order%22%3A%22605%22%2C%22eval%22%3A%221%22%7D&pdp_npi=4%40dis%21GBP%211.37%211.28%21%21%2112.17%2111.35%21%40211b813b17398002809124234e6e9d%2165040946169%21sea%21UK%211751606289%21X&curPageLogUid=zXX40i0iOp02&utparam-url=scene%3Asearch%7Cquery_from%3A

Five different types, and one of them (or more) will be suitable for your application. Either compare them to the reading from a new original sensor, or just try each type in turn, and see which is most suitable.

You then need to construct a safe casing for it, bearing in mind if you 3D print something (and glue a transparent window in it) make sure it's UV safe, by printing using ASA.

 

[Tony Stewart]

A CdS photoresistor has a wide range of values and a wide tolerance , and usually rated at 10 Lux. hence why they always use a pot. They are also sensitive to damage from full sunlight, condensation, shock, ESD, vibration, bent leads

When you design with photodiode, you the spec of 0.6mA/mW you choose a higher fixed resistor, for the comparator.
You do not need pots because of the consistent accuracy but may choose 10% hysteresis. A daylight sensor will have clear lens pointing at large plastic diffuser to capture a wider average light level. You might use a pot to design the threshold, but then PDs are known to be very consistent. I prefer using 1mm ^2 chip in a 5mm case.

But then I have made some just using an LED as a PD with a load resistor like 1 Meg and a comparator.
If you wanted to set the threshold after sunset or a nautical twilight 1 to 100 lux, you choose the ratio of diode voltage over current.

 

[Les Jones]

I find poto electric light switches similar to this one quite reliable.
https://www.screwfix.com/search?search=photocell+sensor
They seem to last about 10 years. (You can't change the light level at which the switch.)

Les

 

[Dingus]

Thanks Nigel.
The housing is perfect but the LDR is toast. I think it was the very low temperatures that did it. (Scotland)
H

 

[Dingus]

Les I know these sensors and they are good. However mine is in a decorative part and therefore these would be unsightly.
My setup is hidden and the sensor is also well hidden.
Thanks for the input.