12v LED on 120v Power

[JayInJersey]

So...I need a quick and easy way to light up a 12v LED for a 110v circuit (basically an LED light that lights when I have the power switch ON...boring toggles though not LED switches.


Can I use a Doorbell 120v to 12v transformer or do I need to get a resistor and diode in there?

 

[thadass]

A 120v to 12v transformer will work, but you will want a resistor as well. Check out http://led.linear1.org/1led.wiz to find out what resistance you will need (ballpark).

 

[opqdan]

The LED will require DC and a doorbell transformer would just step down the 110v AC to 12v AC. Even with a DC source, you will need a current limiting resistor based on the voltage and max current for the LED. You could build a rectifier to convert the AV to DC, but it requires a few more parts.

Why not just pick up a 120v panel lamp? They should only cost a few dollars.

EDIT: "require" above isn't quite right. You *can* run an LED off of AC, but if it comes down to needing that, there are better methods to get this done (the panel lamp for example).

 

[JayInJersey]

Was trying to do it this week and can't find any place local that can get it in time

 

[thadass]

Oh, I wasn't aware doorbell transformers might be AC->AC. Yeah, that'd be a problem.

I've got so many wall worts around I always seem to be able to find one of those for projects like this (same with resistors I probably bought at radio shack when I was 11).

 

[JayInJersey]

All my warts are more than 12v sadly and the one I did have...let's just say there was a minor benchtop accident when I was testing because some numb nuts connected the led to the input and the 110v to the output.

 

[ajdelange]

The D in LED stands for Diode so you don't need one of those. If the LED consists of a single diode (as opposed to a back to back pair) and you connect another diode with the wrong polarity then you will prevent any current flow. If the LED is a single diode it will only conduct on one half cycle. If you want it to conduct on both halves install a diode bridge in the AC line. This converts the AC to a sequence of single polarity DC pulses. You don't need to do this if the 'lamp' consists of a pair of LEDs or has its own built in bridge. Check with your ohm meter in 'diode test' mode.

A transformer is not necessary unless you want the voltage in the LED circuit to be low. Otherwise determine (from the specs) the current draw of the led. If it is i then you want a total resistance of R = E/i kΩ in the circuit. So, for example, if i = 20 ma and E = 120V R = 120/20 = 6 kΩ. If the LED is a 12V 20 ma LED then it already has a resistor of about 12/20 = 600Ω in the package and you only need 5.4 kΩ in the external cirucuit. Clearly, you can probably ignore this internal resistor and just use 6 kΩ. The final thing you need to do is check the power dissipation in the resistor. It is i*i*R. In this example, 20 ma is 0.02 amps and that squared is 0.0004 amps^2. Multiplying by 6 kΩ gives 6 E3*0.4 E-3 = 2.4 watts. Be sure to get a fat resistor.

If you use a transformer (a doorbell transformer from Radio Shack is the most likely choice) just use 28 V in the calculations instead of 120. A smaller resistor (both ohms and dissipation rating) will be required.

 

[thadass]

Hahaha oh yikes, did you get a pretty arc light show out of it at least? Pushing 1000V+ through small DC components has a _potential_ to be entertaining at least

 

[JayInJersey]

Yeah...hence my lovely trip to "The Source" [old Radio Shack slogan back in the day before they only had bad TVs and Cellphones] and grab a new project box...one without a pretty burn mark on it.


That's where I saw the transformer and thought "maybe that will work"