Help me understand this photogate (Robotics)

Tonyr1084

Joined Sep 24, 2015
7,853
Do not feed 24V in at point 67 (in the left yellow zone). Let 24V come from the sensor board, which is always powered (in the right yellow zone). The 1.2KΩ resistor will keep the current low for the HL4 LED. When the beam is blocked pin 15 should see a high of 24V. When it is unblocked the photo transistor (EE-SV3-B) should ground pin 15 minus whatever the forward voltages of the LED and transistor are. If your schematic is correct then I don't see why it wouldn't work.
1601322753542.png
 

MisterBill2

Joined Jan 23, 2018
18,178
What I see now is that the input terminal 15, is a current sinking input. So the sensor needs to be a current sourcing sensor. That is quite possible if the emitter connection of the photosensor can be separated from the cathode of the LED and terminal X40-22 A resistor would be connected between X42-1 and X42-2 and the connection between X40-20and X42-1 would be opened. The emitter would be connected to terminal X40-20.
The input would be close to 24 volts except when the beam was broken, at which time it would be close to zero. So there would need to be a logic inversion added, but hopefully that is possible.
 

Thread Starter

mcardoso

Joined May 19, 2020
226
What I see now is that the input terminal 15, is a current sinking input. So the sensor needs to be a current sourcing sensor. That is quite possible if the emitter connection of the photosensor can be separated from the cathode of the LED and terminal X40-22 A resistor would be connected between X42-1 and X42-2 and the connection between X40-20and X42-1 would be opened. The emitter would be connected to terminal X40-20.
The input would be close to 24 volts except when the beam was broken, at which time it would be close to zero. So there would need to be a logic inversion added, but hopefully that is possible.
Bingo,

That's exactly what I just tried 10 minutes ago. All it took was a bit of quick soldering. Unfortunately, I did not account for one thing... The current limiting resistor. With that there - before the emitter - the voltage at the power supply pin (X42-2) is only the forward voltage of the LED, 1.2V. The circuit did switch on and off properly, however it only gave 1.2V High and 0.02V Low. This is annoying!

I wish I could easily change the PCB at the base of the robot to put the resistor "below" the emitter, but that is going to be a whole lot more work than replacing the sensor with an "active" kind where you give it 24V and it handles driving the LED and amplifying the output with a transistor.

Plenty of options in this category exist, but I have yet to find one that matches the physical dimensions of the one I have. Still working on it.
 

MisterBill2

Joined Jan 23, 2018
18,178
One more option, if there is room someplace to put another transistor, is to add an amplifier stage . Then 2 or 5 mA from the opto-switch could easily control 20 or 30 mA into the other module.
 

Thread Starter

mcardoso

Joined May 19, 2020
226
All,

Thanks for the wonderful discussion and help thinking through this. I ended up buying some Panasonic PNP active photo sensors to replace the original ones. I will need to machine a small adapter, but that will be easy (I have a home machine shop). The sensor is powered directly by the 24V line and has amplification built in to drive the output. They were $11 each (I need 3) and an additional $20 for Molex connectors and pins. I am replacing the 800 ohm resistors with 1A fuses to allow the full 24V to be provided to the sensors located throughout the robot.
 
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