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Topic: Powering IR emitter (Read 164 times) previous topic - next topic

Levister2121

I recently purchased IR emitters and receivers. The receivers when connected to 5v work as I expected and I am able to read data from them on an analog pin using the serial monitor. The problem I am having is with the emitters. I can't seem to turn it on. I don't have specific information about the emitter I am using as far as the voltage requirements but from some research I did it looks like they typically require less voltage and are more sensitive than the receivers. I tried using the 3.3v pin with a couple of resistors which brought the voltage down to just over 1v and nothing so I tried straight 3.3v and also straight 5v. Still can't seem to get it to give off any IR light at least not that my receiver is seeing. My understanding of electronics is somewhat basic and I am not sure how to troubleshoot this. Any help is greatly appreciated.

raschemmel


herbschwarz

Hello Levister2121,
The IR receivers you have are modules, not just photo transistors
or diodes, right? If so, they have circuitry that rejects any light
that does not conform to what they are supposed to receive. Most are
designed to respond to a 38 KHz (or so) signal.

Herb

Levister2121

#3
Nov 03, 2019, 03:10 am Last Edit: Nov 03, 2019, 03:48 am by Levister2121 Reason: Reread response
Hey Herb thanks for the reply. They are not modules. I want to use quite of few of these pairs in a project and it wasn't cost effective to use the premade modules. I don't know all the technical terminology but I think they would be described as just diodes. They look like an LED and do not come connected to any other components. The receivers are not what I am having trouble with though it the emitters that I cannot seem to get working properly.

gilshultz

They are a LED that emit light you cannot see! The ones I use require about 50mA  or more for proper operation. If you are using the 3.3 Volt you will need a resistor in the 50 Ohm range.  To determine if it is on look at it with your cell phone camera:-)
Good Luck & Have Fun!
Gil
This response is to help you get started in solving your problem, not solve it for you.
Good Luck & Have Fun!
Gil

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