SteveinTijuana
Newbie level 2
how to use a photocell
Hello everyone. I have this project but I cannot figure out how to get the photocell to do what I need. Any suggestions will be greatly appreciated.
CHALLENGE: Integrate a battery-operated consumer (retail) smoke detector into a proprietary wireless security system, as economically as possible.
COMPONENTS: The first component is a 9V DC smoke detector WITH an ultra brilliant white escape light (LED) that turns on, along with the horn, during a smoke/fire alarm. The second component is a 3V DC battery-operated wireless transmitter, having a normally closed circuit (2.4V DC circulating), and featuring a 15” inch two-wire cable (usually connected to the leads of a magnetic contact). Finally, there is security console that receives an RF signal from the transmitter(s) and activates a security alarm.
METHOD: Utilize a single photocell soldered to the two wire leads of the wireless transmitter and, then, clear silicon the photocell (facing inwards) to the clear cover of the escape light of the smoke detector. Cost: $2.50.
PROBLEM: A photocell is a resistor that allows current to flow freely through it in the presence of light and restricts (blocks) current flow in the absence of light. Connected to my transmitter leads this would act as a Normally Open circuit in the absence of light--I need the photocell to do the opposite. In darkness, I need the 2.4 V DC passing through the wires of the wireless transmitter to loop (closed circuit) back to the transmitter. Then, when a smoke/fire alarm occurs, the escape light will light and the photocell will react by opening the circuit. The transmitter will sense the open and send a radio signal to the security console. The console will then react with a security alarm.
QUESTION: How do I reverse the action of the photocell? I have a suspicion I need to use a 5V DC SPDT (single-pole, double-throw) relay, but I don’t know what to wire to the coil and what to wire to the circuit (and it seems like I would need another power source to switch the relay). Any help would be greatly appreciated! Thank you for your consideration.
Hello everyone. I have this project but I cannot figure out how to get the photocell to do what I need. Any suggestions will be greatly appreciated.
CHALLENGE: Integrate a battery-operated consumer (retail) smoke detector into a proprietary wireless security system, as economically as possible.
COMPONENTS: The first component is a 9V DC smoke detector WITH an ultra brilliant white escape light (LED) that turns on, along with the horn, during a smoke/fire alarm. The second component is a 3V DC battery-operated wireless transmitter, having a normally closed circuit (2.4V DC circulating), and featuring a 15” inch two-wire cable (usually connected to the leads of a magnetic contact). Finally, there is security console that receives an RF signal from the transmitter(s) and activates a security alarm.
METHOD: Utilize a single photocell soldered to the two wire leads of the wireless transmitter and, then, clear silicon the photocell (facing inwards) to the clear cover of the escape light of the smoke detector. Cost: $2.50.
PROBLEM: A photocell is a resistor that allows current to flow freely through it in the presence of light and restricts (blocks) current flow in the absence of light. Connected to my transmitter leads this would act as a Normally Open circuit in the absence of light--I need the photocell to do the opposite. In darkness, I need the 2.4 V DC passing through the wires of the wireless transmitter to loop (closed circuit) back to the transmitter. Then, when a smoke/fire alarm occurs, the escape light will light and the photocell will react by opening the circuit. The transmitter will sense the open and send a radio signal to the security console. The console will then react with a security alarm.
QUESTION: How do I reverse the action of the photocell? I have a suspicion I need to use a 5V DC SPDT (single-pole, double-throw) relay, but I don’t know what to wire to the coil and what to wire to the circuit (and it seems like I would need another power source to switch the relay). Any help would be greatly appreciated! Thank you for your consideration.