View Full Version : Wiring the 0/16/16

09-26-2008, 11:27 AM

I am new to electronic wiring and have a quick question about wiring the 0/16/16 interface. Here is what I have:

9 switches
9 LEDs (6v)
1 6v power supply

All that I need to do is connect the switches to the inputs and the LEDs to the outputs.

How do I wire this? Can I use the 1 6v power supply for everything or do I need another power supply for the switches?

I know this should be really simple, so extra thanks to anyone who will take the time!

09-26-2008, 12:19 PM
Hi, Welcome to the TRC!

Couple questions:

Do your LEDs already have pull-up resistors built-in? If not, you'll need them.
Your 6v power supply - how many milliamps (or amps) is it able to deliver? In order to answer your question, the current that can be delivered is every bit as important as the voltage.

Switches themselves don't generally "take power" (unless they have lighting built-in to them). Switches are like cutting a piece of wire. Either the electrons flow through the wire (switch) to the device being powered - or they don't. So we need to know how much current we have to source for your LEDs.

Feel free to link to the specific LEDs and power-supply you're using. :)

Again - welcome to the forum!

09-26-2008, 01:56 PM

Thanks for the warm welcome. My LEDs do NOT have pull-up resistors. But I did buy an assortment of resistors. What flavor should I use? My power supply is 600mA and exactly 5.94V (tested with a multimeter).

I am actually using lighted pushbuttons, (not that it makes a difference, because the LED and the switch are separate, so it is essentially the same as have buttons and LEDs). I am using the Happ small round pushbutton


with the optional LED replacement bulb


it is the 91-1157-00 one.

To be clear, I do not want to use the switch to turn the LEDs on directly. I want to take the switch input into the computer, do some processing with Max/MSP and then light one or more LEDs. In the manual it shows the switch connected to a 9V battery, so I assumed the switch needed power to pass through to the input. So, my question was, can I grab that power from the same source as the light or should I get a separate power source for the switches.

Thanks so much for your help.

09-26-2008, 04:23 PM
The inputs on the 0/16/16 are the numbered contacts on the block. On either side of the row of 8 inputs are two ground contacts (labeled "G"). One lead on your switch goes to the numbered contact, the other goes to ground. When the switch closes the input value goes from 0 to 1. No power source is required.

Best way to test it is to use the Phidget monitor program for the device. Just use a piece of wire and short one of the input contacts to ground. The indicator for that input should change (it's a checkbox I think).

09-26-2008, 04:26 PM
Oops, sorry. Ignore what I just said. That's how digital inputs work on the 8/8/8. The 0/16/16 is different.

09-29-2008, 04:26 PM
I had a long detailed post not too long ago that had details for calculating pull-up resistors. We need a FAQ engine. Hey Alex: We need a FAQ engine. :P

Aha. There it is: http://forums.trossenrobotics.com/showpost.php?p=13746&postcount=21

We don't have the specs on your LEDs so we'll just have some fun with "safe" numbers instead:

((6vcc-2vf) /18) * 1000 ~ 222.22 = 220 Ohm 10% resistors. Required size is 0.08 Watts, so you're more than safe with 1/4 watt resistors- or even 1/8th watt if you want to keep the size even smaller.

You're fine running the inputs from the same power - they'll sink up to 30vDC with 4v at the bottom end [according to the specs here: http://www.trossenrobotics.com/store/p/3201-InterfaceKit-0-16-16.aspx] and they'll output up to 30vDC (from the same specs).

So, yes, you have your 9 LEDs drawing about 0.08 * 9 watts = 3/4 watts = 0.75watt / 6v = 125mA. The board itself can sink up to 2A, so it's safe. There's no notation in the manual for how much the board sucks-up for when you apply a voltage to the input, but we'll assume it fine since that's a basic design. Given that your power supply is call it a half an amp, you're more than double the requirement which is good design theory.

The long lead on your LED (The "Anode", aka "a", aka "+", aka "positive") goes to the limiting resistor, the output of the resistor goes into the + side of the connector on the board's output.
The short lead on your LED (The Cathode, aka "k", aka "-", aka "negative") goes to the minus side or ground side of the connector.

On the switch side, the - side of the battery goes into the - side of the input. The + side goes through the switch to the plus side of the connector.

Does that help?

09-29-2008, 06:12 PM
Yes, that helps, thanks. I'll put together a circuit diagram and post it up here before I wire anything, if anyone would be so kind as to check it for me.

Thanks again!

09-30-2008, 10:38 AM
I have no idea if this affects the answer (news flash: I'm a software, not hardware guy;)), but there is a new version of the 0/16/16 that has been released:

The new version of the 1012 PhidgetInterfaceKit 0/16/16 (http://www.phidgets.com/products.php?product_id=1012) has built-in filtering on the digital input, to eliminate false triggering from electrical noise. The digital input is first RC filtered by a 15K/100nF node, which rejects noise of higher frequency than 1Khz. This filter generally eliminates the need to shield the digital input from inductive and capacitive coupling likely to occur in wiring harnesses.

Full Press Release:

I don't think we have them in yet, but they'll be here soon.

09-30-2008, 11:24 AM
Thanks, Alex. I don't think it should make any difference in the case of a switch on the input. Might lessen debounce, but that's really a different problem than what they're shooting at.

09-30-2008, 11:52 AM
OK. So I think I have the circuitry figured out. Could someone please take a look at this for me?

http://www.flickr.com/photos/[email protected]/2901620207/ (For easy reading, I only show 4 of the 9 switches and LEDs here)

From post #6 on this thread you said "The long lead on your LED...goes to the limiting resistor, the output of the resistor goes into the + side of the connector on the board's output.
The short lead on your LED...goes to the minus side or ground side of the connector."

Did you mean the short lead goes to the connector or the power supply? Since the 0/16/16 board itself isn't powered, it would have to go the power supply, right?

Thanks again!!!!

09-30-2008, 12:27 PM
Right now, it appears you have the anode of the LED going to the positive side of the transformer, and into the output of the board. As wired, it appears that it'll be a race to see which releases its magic smoke first - the board or the LEDs. :eek:

Let me download the manual for the 0/16/16 and take a peek. I wish I had some of this phidgets stuff - makes it hard to answer questions without it. ;)

09-30-2008, 01:01 PM
Yeah, I see now that I might fry my board with this configuration. It's a little confusing because I've never used a board that didn't have it's own power supply. Thanks so much for looking into this for me.

09-30-2008, 01:13 PM
I think all I'm really doing is flipping your LEDs around... Hand drawn sketch to follow as soon as my camera finishes...

The LEDs would have fried, board would have been fine, btw. You are correct, the outputs are effectively a "digital relay" - just a switch to ground.

09-30-2008, 01:21 PM
Thank you so so much. Let me know if you ever need any non-electrical wiring help!

09-30-2008, 01:22 PM
I think this addresses the design, if you can read my very hurried chicken scratching...

[Click twice on the image for full zoom, once to medium, then on the medium to get the original image...]

P.S. No problem! Just hang around the community and "pay it forward". It all comes back around, right? :)