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Registered Member #1225
Joined: Sat Jan 12 2008, 01:24AM
Location: Beaumont, Texas, USA
Posts: 2253
Hello there.
I recently bought a phototransistor/IR emitter LED pair, for a small project. Basically, i just need the signal to be interrupted as a spinning disk reached a certain point.
I will not post pictures rigght now, if you guys need one or do not know what is wrong without pictures, taking them is not problem.
Alright. Basically, i have two small circuit boards, with the LED's soldered onto them. To hold the two circuit boards away from each other, to give the LED's about 1/4 inch clearance from each other (to allow space for an obstruction to interrupter the signal) i used a screw with nuts to be able to adjust the clearance.
That is the schematic. I have the collector connected to V+ and the emitter is the output, with a 10kohm resistor connecting it to ground. Remember, this is two separate LED's with space in between them, to allow for an obstruction.
EDIT: I do have a 270 ohm resistor in series with the IR emitter. I forgot to add it on the schematic.
The problem is that no matter if there is an obstruction, the output of the phototransistor is always about 4.8 volts, which is about .3 volt drop from the source. Nothing heats at all. The output had not load other than a volt meter. When it works, the output will be on the base of a transistor for the switching.
When i had the emitter connected to ground, with the collector being the output with a 10kohm resistor to V+, it worked just fine. It is a very simple circuit, so i do not know how i could have possibly messed anything up... I just need a positive output to give a signal to the base of an NPN transistor.
Registered Member #902
Joined: Sun Jul 15 2007, 08:17PM
Location: North Texas
Posts: 1040
from what I can tell, the connection of the transistor via the 10k ohm resistor to ground essentially acts as a path for the V+ to go to your output, assuming your output also goes to ground...
then again, if that were the case it should be less than 4.8V on the output? try using a diode instead of a resistor to prevent V+ from flowing in the wrong direction... (e.g. in your schematic have it face to the left)
Registered Member #1225
Joined: Sat Jan 12 2008, 01:24AM
Location: Beaumont, Texas, USA
Posts: 2253
DaJJHman wrote ...
from what I can tell, the connection of the transistor via the 10k ohm resistor to ground essentially acts as a path for the V+ to go to your output, assuming your output also goes to ground...
then again, if that were the case it should be less than 4.8V on the output? try using a diode instead of a resistor to prevent V+ from flowing in the wrong direction... (e.g. in your schematic have it face to the left)
Gah, i am so stupid D':.
Lol, apparently i thought i used such high resistance the voltage drop would be MUCH more. Thanks, man.
Update: Killed it. Apparently the base of a darlington pulls more than 50ma at 5 volts.
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IR photodetector and emitter pair.
