Hello,
When I apply voltage to the emitter of a photo-interuppter I took out of a copy machine, the signal wire goes to ground/low. There is voltage between the signal and the positive equal to ground and the positive voltage. When I interrupt the path between the emitter and the detector the signal wire goes off, meaning neither high nor low. When path is interupted There are no continuity or voltage between signal and either of the positive and negative lead.
I have tried it with half dozen more I had gathered from the large copy machine and they all the same.
The ones I have seen on YouTube all conviniently show that when the path is interrupted their LED light up that is connected to signal output wire.

I can not picture in my mind as to where the signal wire is connected.
3 wires are coming out of the gadget.

 

One wire is common the others are power & signal.

 

One wire is common the others are power & signal.

That would also be my guess. Most photo interrupter modules are NPN devices with an open collector output. The NPN emitter and light emitter share a common. The light emitter anode will need a resistor to limit current and the NPN open collector will need a resistor to develop a voltage across at the output.

This is an example of a 1mm APERTURE OPTO-ELECTRONIC SINGLE CHANNEL SLOTTED INTERRUPTER SWITCHES WITH TRANSISTOR SENSORS.

Ron

 

That would also be my guess. Most photo interrupter modules are NPN devices with an open collector output. The NPN emitter and light emitter share a common. The light emitter anode will need a resistor to limit current and the NPN open collector will need a resistor to develop a voltage across at the output.

This is an example of a 1mm APERTURE OPTO-ELECTRONIC SINGLE CHANNEL SLOTTED INTERRUPTER SWITCHES WITH TRANSISTOR SENSORS.

Ron

=============================================
I don't know what all that means in terms of question that I am asking in my post. Just want to know where the signal wire is connected to, to behave that way.
Anyway, I took this thing apart after it popped in my head to just take the darn thing apart. With that, I couldn't help but to laugh as to what had held me back from doing it at once and immediately days ago.

It didn't take much to see what is what and what is connected to what, it is in plain site. But yet it doesn't explain as to why the signal wire go negative. I do not know if you can see in the picture for yourself, you just have to take my word for it that the signal wire is not connected to anything except the one leg of the detector.
By the way both the emmiter and the detector look exactly alike when I looked at them with my magnifying glass.

Both the emmiter and the detector have just two legs. They both look exactly like an LED.
Here is what is connected to what:
The left leg of the detector is connected to the right leg of the emitter. The right leg of the emitter is positive.
The right leg of the detector is not connected to anything, it is the signal wire.
The left leg of the emitter is negative ground.

So how do I know which leg of the emitter is positive and which one negative? That is a real honest to goodness question.
The answer to that one is simple straight forward. I put my multimeter on diode setting to get readings. Only when I hooked up the positive to one wire and the negative to another wire is when I got a little over 1 volt. So that was a dead give away. Not only it showed which one of the two was the emitter but also which one was positive and which negative.

And that is how I hooked up the power to the emitter in the first place and as which lead to what leg.
In pictures:
One leg of the detector is connected to the negative of the emitter. It's other leg is connected to nothing, it is the signal wire.
The other leg of the detector is connected to the negative of the emitter. And that is it.

QUESTION: WHY DOES THE SIGNAL WIRE GO NEGATIVE WHEN THE LIGHT IS EMITTED? ONE MORE TIME, WHY?

 

Connect a 1K resistor from the battery + to the signal wire. Now use a voltmeter to read the voltage on the signal wire both with and without the interrupter blocked. Then you will understand.

Bob

 

I don't know what all that means in terms of question that I am asking in my post. Just want to know where the signal wire is connected to, to behave that way.

Your first post did not identify what each wire was? Now you have identified the wires. I gave you a link to a typical slotted optical coupler as an example. Less knowing the pinout that was about all I could come up with. Knowing the pinout helps. I agree with BobTPH as to how to configure it. You are sure about the polarity in your last image?

Ron

 

This is one way it could be wired. (see post #11) However, I don't know the specifics of yours. When the beam is NOT interrupted the transistor is ON and "Sig" (Signal) is HIGH (at 5V). When the beam IS interrupted then "Sig" goes open. Not ground, not some specific value - OPEN. It's like having a wire that is connected to nothing.

[ THIS POST HAS BEEN EDITED FOR CORRECTNESS. SEE POST #11 FOR CORRECTED DRAWING ]

 

This is one way it could be wired. However, I don't know the specifics of yours. When the beam is NOT interrupted the transistor is ON and "Sig" (Signal) is HIGH (at 5V). When the beam IS interrupted then "Sig" goes open. Not ground, not some specific value - OPEN. It's like having a wire that is connected to nothing.
View attachment 258398

Tony, as drawn the LED cathode goes to Vcc via a resistor and the LED anode is going to ground. Don't we want the LED cathode low and anode high? As drawn I do not see the LED source as ever being On.

If I were to guess it would look like this:


NPN Open Collector output.

Ron

 

Don't we want the LED cathode low and anode high?

Yes. Absolutely. [edit] I failed to pay attention to the question. See post #11

 

I added a dwg ro my post as to what my guesstimate is.

Ron

 

When I apply voltage to the emitter of a photo-interuppter I took out of a copy machine, the signal wire goes to ground/low.

Sorry I didn't pay closer attention to the details in your question. Below is the result of what you describe. When the beam is blocked the signal goes OPEN. When the beam is Uninterrupted signal is held to GND. Depending on the voltage on your board and the LED, the resistor will potentially be 500Ω. That will limit current to the LED at 6mA. The way that figure is arrived upon is estimating the forward voltage of the LED to be close to 2 volts. I used 2 volts specifically to achieve a low current. 5V - 2Vf = 3V. So 3V ÷ 0.006A = 500Ω. Of course this is numbers I chose and may likely not represent the true voltage and current of your device. The data sheet will give you all the information you need. Don't assume my numbers are god's absolute truth. They're probably not.

 

Sorry I didn't pay closer attention to the details in your question. Below is the result of what you describe. When the beam is blocked the signal goes OPEN. When the beam is Uninterrupted signal is held to GND. Depending on the voltage on your board and the LED, the resistor will potentially be 500Ω. That will limit current to the LED at 6mA. The way that figure is arrived upon is estimating the forward voltage of the LED to be close to 2 volts. I used 2 volts specifically to achieve a low current. 5V - 2Vf = 3V. So 3V ÷ 0.006A = 500Ω. Of course this is numbers I chose and may likely not represent the true voltage and current of your device. The data sheet will give you all the information you need. Don't assume my numbers are god's absolute truth. They're probably not.
View attachment 258405

============================================
Now, what is the answer to my question?
I am still waiting for someone to answer my question instead repeating what I have already said repeatedly.
One of the fellows up there tellling me to hook up 1k resistor to signal wire, then I will understand. That is what I call pure unadulterated expert explanation.
Perhaps time for Hitchcock mystery hour.

 

There are a few things we don't know.

1) We don't know what is inside that device.
2) We don't know how to wire it for it to be functional.
3) We don't know what are the voltage and current requirements.

However, we can make some wild guesses and experiment, assuming that the components inside have not already been destroyed. Assuming that the emitter is an LED and the detector is a phototransistor then the first thing to do is to use two 470Ω resistors.

If a voltage source was applied directly to the emitter then it possibly could be blown already.
Connect a 470Ω resistor between your 5V supply and the emitter wire.

If we assume that the detector is a phototransistor, connect another 470Ω resistor from the signal output to the +5V supply.

 

=============================================
QUESTION: WHY DOES THE SIGNAL WIRE GO NEGATIVE WHEN THE LIGHT IS EMITTED? ONE MORE TIME, WHY?

It is difficult to decipher what you have written in such a long post.
The bottom line is: WE DON'T KNOW.
Because we are not there looking over your shoulder to see what you are doing.

We don't know what power source you are using.
We don't know what you are using to measure voltage.
We cannot see how your device is connected and how you are taking a measurement that would produce a negative voltage.

 

Your first post doesn't ask a question. The question you ask in post #4 (where do the wires go) appears answered by Tony1084 quite comprehensively in post #11.

QUESTION: WHY DOES THE SIGNAL WIRE GO NEGATIVE WHEN THE LIGHT IS EMITTED? ONE MORE TIME, WHY?

The signal goes negative because the phototransistor shorts the wire to the negative line when the emitter is powered on.

 

You may have heard of the fellow who ordered 10 LEDs and when they arrived he decided to test one of them.
He wired it to his power supply and it blew instantly.
Thinking that it must have been defective he proceeded to test the remaining nine LEDs.
They all behaved exactly as the first.
He concluded that he must have received a batch of bad LEDs and demanded a refund from the seller!
We don't know if he succeeded in getting back his money.

 

The signal goes negative because the phototransistor shorts the wire to the negative line when the emitter is powered on.

If the test meter shows a negative voltage then the TS is doing something unexpected that we cannot see.

 

If the test meter shows a negative voltage then the TS is doing something unexpected that we cannot see.

From his pictures, by 'negative' it seems the TS is referring to negative in the context of a battery, that is the opposite of positive. Not the standard nomenclature when voltages might go negative from a base reference but common enough with beginners.

So I read his use of negative as meaning the line of meter common reference.

 

I gave you a link of what is typical. Now with a part number or manufacturer things can get better. Some like yours have three wires normally indicating the ground sides are tied, they would be the LED cathode side and transistor emitter side are tied The LED emitter side will need a resistor to VCC for current limiting and the transistor collector side a resistor to Vcc. If you know how to use and read an ohmmeter you can measure across Ground and the LED Cathode side and LED Anode side as well as the transistor emitter and collector. Less a part number that is as good as it gets. A Google of "slotted opto couplers" will bring up a dozen of them. You can also look at where it was removed from. Note if one lead was ground and if resistors were on the remaining two leads. Your descriptions of what you are doing lack clarity making it hard to guess what you are doing.

Ron

 

Now, what is the answer to my question?

I'll be frank! Then I'm bowing out. The answer your question - - - I don't know. I would need all pertinent information about the device and exactly how you're testing it. From what I've given you - I can take one of those that I have and wire it up as shown in the diagram in post #11. And that's the best I can do. Beyond that - is also beyond my skills.

Have a good 2022.

Unwatched.