How should we wire the photoswitchs?

[Itamar]

Hi,
we just got our 2011 kit, and in that kit were three Photoswitch sensors. None of us has ever used this kind of sensors so I'd really appreciate if you could explain how it works, how we should wire it and where to.

Thanks.

[Al Skierkiewicz]

Itamar,
There is a section on the First Website under Competition Docs that explains the sensors. I believe it is under the KOP section. I just tried to log on but the site looks a little busy.

[Jon236]

Here is what I had downloaded from the FIRST website. Hope it helps. They are photoswitches and will give you a digital response. Follow the wiring diagrams carefully!

[Jon Stratis]

http://usfirst.org/roboticsprograms/....aspx?id=18530

This page contains, among other things, links to the datasheets for a lot of the KOP items. I believe the sensors your referring to are the "Rockwell Automation Line Sensors":

http://www.ab.com/sensors/photoelect...eral/42ef.html

The links on the right side of the page to the catalog and installation PDF's are probably the most help... although I don't know the specific part we got in the KoP, and the checklist doesn't have a part number! Without those, I can't give too much useful information... although your questions are exactly those I know my students will be asking me!

[Bot190]

The Line sensors given to us in the KOP are model: 42EF-D1MNAK-a

[mutantblond]

ok, so we've looked at the wiring diagrams, and I have to say, it's not helping us very much. Under the 10.8 - 30V DC Sensors, there are 6 different configurations. Does the 'Load' mean that we have to use resistors in the circuit? I'm gonna display my ignorance here - how do we wire them to the robot? Is the digital sidecar involved here?

[Alan Anderson]

I don't have the datasheet handy, so I'm going from memory.

DO is the "Dark Output". It is active (i.e. connected to ground) when the sensor is looking at "dark", and inactive (i.e. allowed to float) when the sensor sees "light".

LO is the "Light Output". It is active when the sensor is looking at "light", and inactive when the sensor sees "dark".

Connect ground to ground, power to +12 volts, and either LO or DO to a digital input pin on the Digital Sidecar. The input pin will read a logic 1 when the signal is inactive, and a logic 0 when it is active.

[kamocat]

We need to put a resistor in series, don't we? The 9403 can withstand a 30v signal, but use at this voltage will degrade the life of the module.
The standard voltage it specs is -0.5v to 5.25v.

If possible, I'd prefer to use an inline resistor, instead of a voltage divider. At 250 micro-amps, it would take a 34k resistor to drop the voltage from 12v to about 5v.

[Alan Anderson]

Originally Posted by kamocat

We need to put a resistor in series, don't we?

It looks like the output of the sensors is just like a Banner sensor's NPN output. It supplies no power. It is effectively an on/off switch that either connects the output to ground or leaves it open. The Digital Sidecar's inputs will happily take that input without modification; it's basically what they were designed for.

[kamocat]

Oh, okay. That's why they gave us NPN then. Cool!

[bakketti]

Originally Posted by Alan Anderson

DO is the "Dark Output". It is active (i.e. connected to ground) when the sensor is looking at "dark", and inactive (i.e. allowed to float) when the sensor sees "light".

LO is the "Light Output". It is active when the sensor is looking at "light", and inactive when the sensor sees "dark".

Just as a sanity check... Looking at the Wiring Diagrams for the NPN ouputs: Dark Output is the Black line and Light Output is the White line. Is that a correct statement?

[Tem1514 Mentor]

Lets be very careful with those 'line sensors' and the electrical hookup.

Some facts that we do know. The sensor will work just fine on 12 volts DC which generally means at the output (ie: LO,DO ) MAY have a voltage range from 0VDC to 12VDC. The Digital sidecar does have a 'general purpose' I/O lines which opearte in the 0VDC to 5VDC range!!!!!

That means if the sensor sources a 12VDC level you can kiss your digital sidecar goodbye as the blue smoke comes out. NO, a series resistor will only limit the current, not the voltage and either or both will kill a digital sidecar.

To be on the safe side I would use the analog sidecar to input the sensors.

I have not looked at the sensors tech sheets yet but will sure err on the side of caution.

Of course one way to find out is to look at the 'sample code' and see where the information for the sensors is read.

[Alan Anderson]

Originally Posted by Tem1514 Mentor

The sensor will work just fine on 12 volts DC which generally means at the output (ie: LO,DO ) MAY have a voltage range from 0VDC to 12VDC. The Digital sidecar does have a 'general purpose' I/O lines which opearte in the 0VDC to 5VDC range!!!!!

The sensor outputs do not source anything. They will never have a voltage greater than what they are connected to. If you connect LO or DO to the Digital Sidecar, it will either have 5 volts (when inactive) or zero volts (when active) on the output.

I have not looked at the sensors tech sheets yet but will sure err on the side of caution.

If you haven't looked at the documentation, you shouldn't be giving advice about how to use things.

Of course one way to find out is to look at the 'sample code' and see where the information for the sensors is read.

It reads it from the Digital Sidecar DIO pins.

[Al Skierkiewicz]

The light sensor outputs will provide a closure to power common or will remain open circuit. The digital side car digital inputs have pullup resistors to +5 volts. So when the sensor is open, the output will be pulled up to 5 volts through the DSC and when the sensor is closed, the output will be pulled down zero volts through the NPN transistor in the sensor. No other components are needed. The 12 volt input is required to power the emitter and other internal electronics.

[caffel]

The original question is quite direct, but to rephrase it:

What should be connected to:
1. White
2. Black
3. Blue
4. Brown.

We have read all of the Allen Bradley docs.

They seem to show that once the two bleeder resistors are in place that the unit has two leads left.
Since one is supply at 11-30 volts DC and the other must be common, there are no leads left to serve as the output.
Where and how does the output signal get connected to the CRIO I/O?