NotatallTesla
New Member
I want to create a device that will detect when a certain LED lights up on my office phone (it's the French language line, actually); I want the device to then turn on a much brighter light that I will easily notice ….. This is what I’m hoping will work: I’ll place the photocell (shown here: https://www.futurlec.com/Pictures/PHOTOCELL2A.jpg) above the phone’s LED, in a surrounding “pocket” designed so that absolutely no ambient light could get to the photocell – only the LED’s light will get to it. When the LED's light hits the photocell, that will trigger the bright lamp to come on. But I don't believe the circuit will be too simple, as I will now explain. If you just set up a battery connected to the lamp by two wires, the lamp would be on all the time, of course. With the photocell added to the circuit, the lamp will never come on because the photocell always creates enough resistance that the electrical circuit cannot be completed. That is, the photocell's rsistance will never drop below about 5K or 10K Ohms, even when a light is shined on the photocell. So the photocell, in this case, would just be an “off switch.” From what I've heard of transistors, a transistor plugged into this circuit will somehow detect that the photocell is producing either the low 5K-Ohm resistance (i.e., when a flashlight is shined on the photocell) or high 100K-Ohm resistance (i.e., when photocell is kept in the dark). When the transistor detects a low resistance level in the photocell, it somehow “switches” the pathway of electricity in such a way that the resistance of the photocell is completely bypassed, allowing the electrical circuit to be completed. I imagine the circuit can be powered by a 9-volt battery.
Anyone care to offer an opinion on how exactly i can make this work? Help would be appreciated.
Anyone care to offer an opinion on how exactly i can make this work? Help would be appreciated.