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ArcticMyst Security by Avery

i really need help






Joined
Feb 1, 2007
Messages
810
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It is a very simple photodiode circuit to trigger a relay to switch, what do you find so hard to understand? if you say what you dont understand, then maybe me or someone else can help. If the relay is bothering you, then just consider it as a switch. The two vertical lines on the right hand side of the relay graphic are simply a switch, either open or closed to power another circuit of your design. The coil side of the relay simply requires a very low current signal , in this case supplied by the photodiode to make the actual switch. The diode bridging the gap on the relay is simply there to stop any reverse voltage occuring, when the relay is switched. (ie , a reactive load to the switching coil)

I am no electronics expert, but to me the inclusion of the transistor is to act as a simple amplifier to boost the signal current from the photo diode to enable the relay to switch, and the Potentiometer is there to adjust the sensitivity of the light (or absence of) to enable the switch to take place.


Jase
 
Joined
Jun 25, 2007
Messages
134
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ok, makes sence, but hoe do you attach the initial 6V?

there is one line in the diagram, but what about the positive and negative?

and what would you reccomend? a 6V dc? 4 AAAbatteries? cut of one thrid of a nine volt battery? i dont know :p
 
Joined
Feb 1, 2007
Messages
810
Points
0
Um. Connect the plus to the +6vdc at the top, the negative is connected at the sign for earth at the bottom (the arrow made up of lines).

Jase
 




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