Chaining Optical Endstops

[Lilithet]

So I looted a bunch of optical endstops from a printer, and wanted to use them as limiter switches, but to do this I need to connect two of them to one output pin, as seen below. Unfortunately I can't figure out what I am doing incorrectly. When I measure signal to VCC I get a full 5V but when I chain them as seen in the diagram, I only get 2.5V. I don't understand why this is happening, nor how to fix this issue.

Any help would be appreciated.

 

[KlausST]

Hi,

You must not chain the LEDs. They need to be supplied with 0V and 5V.
Thus - in any case - you need to connect
* 0V to either GND and
* 5V to either VCC.

now it depends on whether you need
* wired AND or
* wired OR
for the both SIG signals.

I don´t know, it depends on your application.. thus you have to know.

Klaus

 

[Lilithet]

Thanks for the fast reply,

I still don't understand though... I am chaining the LED to the light sensor, and not an LED to an LED...

Also I am trying to get a logical 'OR' signal as a result.

Thanks

 

[Lilithet]

Power is connected to the leads with Vcc.
High is the default, I need Low if one or both are interrupted.

Why is it not possible to chain them as I have. Why does the voltage drop to half on the signal?

 

[FvM]

The schematic in post #1 makes no sense. Where's the power supply connected?

Also I am trying to get a logical 'OR' signal as a result.

What's logical OR for you? Do you mean interrupting either light barrier should give a high output level? In this case, the photo transistors must be series connected but without the LED supply. It's not possible with the given three terminal wiring.

- - - Updated - - -

High is the default, I need Low if one or both are interrupted.

That's not how these devices work. The output is default low and goes high when interrupted. Why not use external diodes or logic gates to achieve the intended OR operation.

Power is connected to the leads with Vcc.

Power has + and - leads. + should be connected to VCC, - to GND, otherwise the LEDS are not correctly supplied.

 

[Lilithet]

That's not how these devices work. The output is default low and goes high when interrupted. Why not use external diodes or logic gates to achieve the intended OR operation.

My bad, you are correct, LOW is the default... I was testing with an LED which was ON when not blocked, and got confused.

The schematic has both Vcc and Gnd noted... figured that would be enough... But yes, + to Vcc, - to Gnd

How would I use diodes to achieve this?

Thanks.

 

[FvM]

Connect like below

 

[Lilithet]

Thanks I will try this. How much greater should R3 be in this case? is there a rule regarding this?

 

[FvM]

Depends on the expected output high level. I'm sure, you can calculate it.

 

[Lilithet]

I'm sure, you can calculate it.

I would, but as I am more of a beginner than even a beginner, I have no idea what formula to use, since I honestly don't know how this will even work.

 

[KlausST]

Hi,

two resistors in series.

upper leg = input (5V?)
lower leg = GND
center tap = output.

Only you know what output voltage you need ( something between V_IH_min and V_I_max of the receiving device).

The formula you need is just the most basic formula in electronics: Ohm´s law.

Klaus

 

[d123]

Hi,

It's a voltage divider.

R1 (10k or 1k)
R2 (10k or 100k)

Calculation: R2/(R1 + R2). e.g. 10k/(10k + 10k) = 0.5. Or, 100k/101k = 0.99

If you put say 10k/10k, you'll only get 2.5V out (of a 5V supply).

If you put 1k/100k, you'll theoretically get 4.95V out of a 5V supply at the resistive divider junction.

Then, subtract the diode forward voltage drop. Let's assume it's around 0.7V. So, 4.95V - 0.7V = 4.25V out. Theoretically, it will be a little different, especially depending on type of diode used, also, unlikely the BJT is ideal so it will leak a tiny bit when off ("off" state is not always truly "off" with transistors), I have no idea how much but maybe a few mV or so will be lost there, so let's make up a final value of e.g. 4.2V out of the resistive junction after the diode.

Hope I haven't missed anything pertinent to your calculation.

 

[Lilithet]

Ahhhhh... ok, I do know about the existence of a voltage divider, but I didn't recognize it with the diode in the middle. Thanks, that helped a bunch.