Help to reduce 19 VDC to 12 VDC to power LED

Need some help guys. Trying to power a 12v LED bit I have 19v source. I believe that I need to attach a resistor to the negative terminal but I am not sure which one or if there is a better way. Another option is a 24v LED, then will 19v power the light. Any help great. Thanks Santo

The 12V LED will already have a resistor built in.
See this article for how they calculated it.
http://www.electronics-tutorials.ws/diode/diode_8.html

From the data sheet, or by applying 12V to the LED, determine what current that the LED is running at.
Add an extra series resistor, doesn't matter which lead you connect it to.
It's value is.
1) R=(19V-12V)/ I (current at 12V).
Power rating of resistor.
w> 49/R (where R is the result from 1) above)
If you use a 24V LED at nineteen volts, expect a drop in brightness, but it may be acceptable, you may be able to work out what the intensity drop will be if you have sufficient data for the LED, otherwise do it by observation,
cheers,
Richard

Hi there. I think using a resistor will waste electricity and add on to your electric bill (or if you're using a battery as source, will quickly deplete it). Why not try a single channel step-down converter (LDO or buck) IC or chip? It is quite cheap and will save you the energy dissipated by the resistors. Just connect the 19V to the input of the converter and the 12V output to the LEDs. No worries in polarity.

Hope this helps.
Justin

Can you post a picture and details of the LED unit ? (Power and/or current rating.)
Some 12V units have three LEDs in series plus a resistor.

Richard's additional resistor solution is simple and easy for a low-powered LED, but Justin's step-down convertor would be more suitable for a high-poweed device.