Turning LEDs on/off

[rsisk]

I have an LED wired to the digital side car power and ground pins. I want to be able to turn it on/off programatically, but it always stays on. I am using Labview to open a digital output sand setting it to false, but the LED is always staying on.

I have the jumper on the DSC's port.

What am I ding wrong? Should the LED be wired to the ground and signal pins?

[connor.worley]

Wiring to the signal pin sounds like it would work.

[Ether]

Originally Posted by rsisk

I have an LED wired to the digital side car power and ground pins. I want to be able to turn it on/off programatically, but it always stays on. I am using Labview to open a digital output sand setting it to false, but the LED is always staying on.

I have the jumper on the DSC's port.

What am I ding wrong? Should the LED be wired to the ground and signal pins?

The power stays on. Only the signal turns on/off. Whether or not the LED will illuminate when you connect it between signal and ground depends on how much current the LED requires to illuminate, and how much current the signal can source. What's the part number of the LED?

[theawesome1730]

I think you are right when you say that you should wire to signal and ground, just don't mix up positive and negative on the LED, they make a nice pop sometimes, you can also run LEDs on spike relays if they are 12 volt

Edit: Also the signal may not carry enough current to power LEDs in which case you should use a spike

[rsisk]

LEDs from SuperBrightLED

Qty | Product / Options | Price/ea |
================================================== ==========
12 | RL5-G7532: 5mm Green LED | $ 0.54 |
RL5-G7532: Green LED
----------------------------------------------------------

Resistors from www.MOUSER.com

299-82-RC
299-82-RC
82ohms

Assuming the flat side of the LED is the negative side.

Going to try the ground/signal setup now.... thanks

[Ether]

Originally Posted by rsisk

Resistors from www.MOUSER.com

Why did you select 82 ohm?

[flameout]

Originally Posted by theawesome1730

don't mix up positive and negative on the LED, they make a nice pop sometimes

For anyone who reads this thread in the future:

Reversing polarity on a LED will not damage the LED (or the Digital Sidecar, for that matter) -- it will simply not night. LEDs are diodes. Feeding them too much power, however, can make them pop.

[theawesome1730]

Originally Posted by flameout

For anyone who reads this thread in the future:

Reversing polarity on a LED will not damage the LED

Only reason I said that is because I actually did make one pop when I wired it backwards.

And like flameout said it won't hurt the sidecar or the LED at that voltage. An LED is a diode so it only lets current through in one direction and also happens to light up.

[Ether]

Originally Posted by flameout

Reversing polarity on a LED will not damage the LED (or the Digital Sidecar, for that matter) -- it will simply not night.

Caution: the above is true only if driving the LED with a source whose open-circuit voltage is less than the reverse voltage which would damage the LED.

A high voltage high impedance source might work fine in the forward direction but damage the LED if polarity is reversed.

[rsisk]

Originally Posted by Ether

Why did you select 82 ohm?

I used the calculator linked below to determine the resistor. Forward current was 20mA, source voltage was 5v and the forward voltage was taken from the LED specs.

http://led.linear1.org/1led.wiz

When I switched to the signal tonight, there was no light when I set the DIO to true or false. When the LED is on power and ground, it comes on

[Ether]

Originally Posted by rsisk

When I switched to the signal tonight, there was no light when I set the DIO to true or false. When the LED is on power and ground, it comes on

Sounds to me like the output impedance of the DIO signal line is too high to provide enough current to light the LED (as mentioned in my earlier post).

Disconnect your LED circuit and put a voltmeter (a cheap $10 one will do) between signal and ground at the DIO output. Flip the DIO true/false and confirm that you are getting 5 volts and zero volts.

Now connect your LED circuit (leaving the voltmeter connected between signal and ground). My guess is that you will no longer see 5 volts. The load from the LED acting on the impedance of the signal source will drop the voltage below that necessary to provide the current to light the LED.

[rsisk]

Originally Posted by Ether

Sounds to me like the output impedance of the DIO signal line is too high to provide enough current to light the LED (as mentioned in my earlier post).

Disconnect your LED circuit and put a voltmeter (a cheap $10 one will do) between signal and ground at the DIO output. Flip the DIO true/false and confirm that you are getting 5 volts and zero volts.

Now connect your LED circuit (leaving the voltmeter connected between signal and ground). My guess is that you will no longer see 5 volts. The impedance of the signal source will drop the voltage below that necessary to provide the current to light the LED.

I'll check as soon as the show is over (the robot is part of the play Something's Afoot). If that is the case, what is the solution?

If I use a spike to control it, I assume I will need a different resistor to handle a 12v input? Or I run 5v throught the spike?

[Ether]

Originally Posted by rsisk

I'll check as soon as the show is over (the robot is part of the play Something's Afoot). If that is the case, what is the solution?

Instead of using the signal as a power source to drive the LED, you could use the signal to control a transistor which switches the DIO power line.

I'll draw a sketch and post it here in a few minutes.

[rsisk]

Originally Posted by Ether

Instead of using the signal as a power source to drive the LED, you could use the signal to control a transistor which switches the DIO power line.

I'll draw a sketch and post it here in a few minutes.

Very cool thanks Ether.

[Ether]

Originally Posted by Ether

I'll draw a sketch and post it here in a few minutes.

Something like this.

edit: corrected resistor value