This was originally used for my science project, where I had to test the time it takes for my parachutes to land. They fell at about the same speed so I had to come up with something to time them accurately, and I came up with this. It worked out very well and I like to share it with all of you.
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Thanks.
My understanding is that the stopwatch requries one impulse to start it and one impulse to stop it.
So the first circuit as the light shines on the photocell it generates current which is amplified by the transistor and delivered to the relay reed. This turn on the reed relay. When the light beam is cut it stops the current flow there by turning off the reed relay. Now what happens when then light beam comes back on after the object has gone through? This will cause more impulses to be sent to the stop watch so how does it work?...im a beginner so id really appreciate your help.
The photocell is a photo-variable-resistor. In other words, when light is shined upon it, the resistance in it decreases. When light is shined upon the photocell, it allows more current to flow between the positive side and ground (negative). This creates a "short" (in a way), preventing the transistor from receiving enough current to turn on. On the other hand, when less light is shined onto the photocell (laser beam cut off), the resistance of the photocell increases, stopping current from flowing to ground and forming a "short"; current now flows only to the base of the transistor and turning the transistor on. Once the transistor is turned "on", current flows from the positive, through the relay, through the transistor, and then to ground. The contacts in the relay closes as current flows through it, sending a impulse to the stopwatch, telling it to START.
When the light beam comes back on, it simply re-opens the contacts in the relay; it doesn't send another impulse.