I have a light sensing voice chip module (http://www.invitebyvoice.com/light-sensor-voice-chip.html) which doesn't work quite as well as I'd like it to (the light sensor isn't sensitive enough for my use). I'm wondering if I can cut the light sensor out and replace it with a magnetic switch similar to this (http://www.amazon.com/gp/product/B0009SUF08). If so, would I need to solder, and what sort of wire should I get? I assume if I want the speaker to activate when the magnet is disconnected, I would need to wire for 'normal open'?

Hopefully this makes sense - I have no experience with circuits (only CS background), but can pick things up pretty quickly. Thanks!

 

That is a CdS (cadmium sulfide) photoresistor (also called LDR, light-dependent resistor) and it is not acting exactly as a switch. It exhibits a large change in resistance when it is exposed to light. So for instance its resistance might drop from 1MΩ in the dark to just 500Ω in full light. To replace it with a switch such as your magnetic switch, you'll want to also add a resistor in series. This will then resemble the original sensor.

You could snip the leads between the board and the CdS cell, or completely desolder the cell and its leads from the PCB if you are comfortable doing that.

 

The module makes a sound when daylight is detected,
you can cut the sensor off, and solder the wires to the sensor terminals, connect to the magnetic switch in Normally Closed, so when the magnet is present the module is quiet,(dark) and when the magnet is removed (light), the module makes a sound.

 

That is a CdS (cadmium sulfide) photoresistor (also called LDR, light-dependent resistor) and it is not acting exactly as a switch. It exhibits a large change in resistance when it is exposed to light. So for instance its resistance might drop from 1MΩ in the dark to just 500Ω in full light. To replace it with a switch such as your magnetic switch, you'll want to also add a resistor in series. This will then resemble the original sensor.

You could snip the leads between the board and the CdS cell, or completely desolder the cell and its leads from the PCB if you are comfortable doing that.

Thanks, that's great. Can you point me toward the kind of resistor I'd need?

Also, to clarify, if I desoldered the CdS, I could add the wiring to the magnet where that was (P0) and the resistor next to it (P1)?

Lastly, can you recommend a wire size to run to the magnet?

Thanks again

 

Any small guage wire like 22swg or 0.5mm will do.

 

If you want the module to activate when a door is opened, or some other situation where the magnet is removed from the switch, connect the module to the COM and NC switch terminals. For the other activation polarity, connect to the COM and NO terminals. In fact, you probably don't need to cut out the light sensor; tack the new wires onto the sensor leads and cover the sensor with a bit of black electrical tape. It is possible that the circuit depends on the minimum resistance of the CdS cell, or it might work fine with just a switch. To be safe, put a 470 or 1K resistor in series with the switch. The CdS sensor is too small to have a lower "on" resistance.

ak

 

Also, to clarify, if I desoldered the CdS, I could add the wiring to the magnet where that was (P0) and the resistor next to it (P1)?

No, the resistor needs to be between the switch and either of the spots currently occupied by the CdS. There will still be just 2 points of contact to the PCB. The resistor can be a standard 1/4W and these are widely available. Any value between 470Ω and 2.2KΩ would likely be fine. As AK notes, it might not be needed. But I'd use it for good measure, since we don't know the rest of the circuit.

 

Works perfect! Desoldered & used the 470Ω w/normal open so the sound plays when magnets are separated. Thanks to everyone who took the time to respond!