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Unread 12-01-2011, 12:54
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Bryscus Bryscus is offline
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Re: How should we wire the photoswitchs?

Quote:
Originally Posted by Tem1514 Mentor View Post
Bench test hookup as per instructions;
Brown wire to + 12 volts (VCC) from the power distrabution board.
Blue wire to 0 volts (Ground) from PD board.
***Two 10K (10,000) ohm resistors (load), have one end connected to VCC.
***The other end of one resistor goes to the White wire, call this TP1 and then the last free end of the other resistor goes to the Black wire, call this TP2
Tem1514 Mentor,

I have starred the issues above. The Rockwell sensors use what's called an open-collector output.

http://en.wikipedia.org/wiki/Open_collector

I won't go into the gory details. However, the way it works is that it has only two output states - high impedance (or disconnected) and low. What this means is that a pull-up resistor to any (reasonable) voltage may be used to pull up the signal when outputting a "high" (which is the output's high impedance state). In the case of a low, the output will drive the signal low. This allows devices running at different voltages to "talk" to one another.

What others have been trying to say is that this pull-up resistor is included for each input of the digital side car. Thus, one could (conceivably) use any open-collector output source to connect to the DSG regardless of the core voltage the device uses to operate. An extra external pull-up resistor (or "Load" in the datasheet) is not required.

I hope I have shed some light on the situation that has been causing some distension between FIRST participants. Thanks.

- Bryce

P.S. For a BJT (transistor type), the output is called open-collector. It's MOSFET (transistor type) counterpart is called an open-drain output.

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