Hello, I am a beginner and trying to understand the board I am using. I have created a circuit with a nano and a laser sensor board to detect a laser beam and it's working but I'm not sure why they set it up with a resistor from VCC to signal.
From my understanding, if this were a pull-up switch design, the sensor would connect/disconnect ground but it's not doing that. When no laser is present, I'm measuring 4.5v from VCC to GND and 6mv from OUT to GND. When a laser is detected, the out to gnd jumps to 4.5v. I expected the laser to pull OUT to low but it's pushing it HIGH instead.
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The sensor seems to be connecting vcc to out when activated. Why is the 10k resistor (R2 in my schematic) there in this case with ground constantly connected?
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Does anyone know where I can get just the sensor itself? I haven't been able to find the bare sensor without the board and the numbers printed on the board haven't helped me.
Thanks in advance.
There is no datasheet I could find but I've created a schematic that I've attached. If anyone knows a simple circuit drawing app, I would appreciate letting me know.
Laser Sensor Board Schematic.pdf (278 KB)
Most likely the sensors OUT is an open collector --> will pull to ground, but not to VCC, so you need to "add" Vcc via resistor. But you can do it using the interal pullup of the nano, too 
Thank you for the response and please bear with me while I stumble through this. If the sensor will pull to ground when activated, then wouldn't OUT be 4.5v while not activated?
This is why I'm confused. I thought it was pull-up but then OUT would be 4.5v while inactive and 0v while active. Since it is the opposite, it must be pulled to ground while inactive and ground is disconnected when active?
But then if that's the case, why is the OUT to GND voltage 6mv? I can't wrap my head around this and don't know what to research.
dirtyam:
Thank you for the response and please bear with me while I stumble through this. If the sensor will pull to ground when activated, then wouldn't OUT be 4.5v while not activated?
This is why I'm confused. I thought it was pull-up but then OUT would be 4.5v while inactive and 0v while active. Since it is the opposite, it must be pulled to ground while inactive and ground is disconnected when active?
It's a matter of sensor design which way it works.
I think I've got this. If this is an open collector, then the sensor is like a npn bjt. By default, the "transistor" is active and pulling OUT to ground. When the laser hits it, the transistor switches off and then the 5v vcc load through the resistor I'm asking about provides 5v to the GPIO.
Thank you both.
Does anyone know where I can get the individual sensors or similar ones without the boards?
dirtyam:
Does anyone know where I can get the individual sensors or similar ones without the boards?
Try the usual suspects: Digikey, Mouser, RS Components, Element14, Taobao, Aliexpress, E-bay, Amazon, ..