Photo Resistor to Complete Circuit in Light

Question

I am trying to create a very simple circuit using a CDS photo resistor. The situation is simple, I need a circuit to be connected when its light, disconnected when its dark. The photocell will sit in complete darkness, when a camera flash is fired, I want the photocell to simply connect a circuit which will then trigger other flashes. The flash circuit doesn't carry any voltage, it just needs to be completed. Basically, I need the two photocell to simply connect two wires when its light and disconnect them when its dark.

I am using this CDS photo cell: https://www.creatroninc.com/product/7mm-light-dependent-resistor-photo-resistor/?search_query=light+sensor&results=60

This is my first experience with one of theses Photo Resistors LDR and I thought it would be really simply but I'm striking out this morning. Thank you!

Answer 1

As others have correctly said, a photo-diode would be the right part.

You can try something like this:

^{simulate this circuit – Schematic created using CircuitLab}

You need to choose a suitable opto-isolator for your trigger output requirements. That needs a comparator with a suitable output drive for it. As I don't know your trigger output requirements, I've left those.

The pot' adjusts the current at the trip point between 10 uA and 2 mA or thereabouts.

Answer 2

The cadmium sulfide photocell is too slow to trigger on a camera flash. But if it was fast enough, this is how you would do it: use the photocell as one leg of a Wheatstone bridge, then use a comparator to switch some variety of voltage-controlled switch (e.g. relay, FET, analog mux, etc) based on the photocell's resistance. Like this:

^{simulate this circuit – Schematic created using CircuitLab}

The values of R1 and R2 (specifically, R2/(R1+R2)) determine the sensitivity of the detector. You can put a potentiometer in there if you want to be able to set the light level at which it trips.