LED Reactive Lighting Circuit

I'm a little rusty, but from what I recall as long as you have sufficient forward voltage to make the thing work (i.e. the required 'activation voltage' drop across the diode) its the current flow through it that determines the brightness, the more current the brighter it'll glow, till you reach the point that it melts!

Connect it in series with a high ohm low wattage resistor for "standby" mode, and replace this with a low ohm high wattage resistor for the "on" mode.

As for running it on a lower voltage, I don't think it will hurt it.

IPG

Thanks, Zaphod!

That is what I was thinking as well. I'm presuming the switch has the current-limiting resistor for the LED internally. My research indicated that a typical red LED has a forward voltage of about 2.1V @ 20mA. Assuming I use the 4V standby value and current through a diode squares(?) with voltage, that gives me 400mA @ 4V. That gives me a voltage across my "standby" resistor of 8V and a current of 400mA. Using P=IV, my resistor will see 0.4A * 8V = 3.2 W. Up-size to 5-6 watt for service. Does that sound right? Intuitively, I was expecting a much smaller value.

Thank you!

Those numbers are much more in line with the values I was expecting! My understanding of the LED / internal resistor relationship was imperfect. At least this gives me a starting point to prototype from. Thanks again for your help...I'll let you know what my results are.

Hi CSM,

Please do, unless of course you enbd up blowing yourself up!

It was fun using those rusty braincells again!

IPG

You are right.

Don't think of LED's as Voltage driven, think of as Current driven

R=(V_cc-V_LED)/I_LED

Most of the LED's I have used were run under recommended current, 12-16mA instead of the recommended 20mA

So, therefore, you can have E-stop running 18-20mA and Other state being 10-12mA

Then using your "Alarm Relay" drop a resistor in parallel to the bias resistor in Alarm conditions, then, you could have that relay pulsing at 1Hz so the E-stop flashes.

<What are the consequences of driving a LED at lower voltages?>

Lower current, and therefore lower brightness. To sustain the brightness the current must be increased, by reducing the value of the resistor in series with the LED.

Thank you all for your feedback.

I stopped by RadioShack on the way home last night and picked up some 2.2 kOhm, 1/4 W resistors and wired the circuit up this morning. (Can you believe that RadioShack doesn't carry a test breadboard anymore?!) Circuit seems to work as designed, but I'll let it run all day to make sure the magic smoke is securely contained.

Zaphod, looks like your maths worked out. Thanks for dusting off the rusty grey matter, I appreciate the guidance. You're right, this was kind of a fun departure from the everyday! Thanks again!

Hi CSM,

Glad it worked out. I guess school wasn't a complete waste of time after all