whats limiting LED voltage?

[MtLuke]

I'm currently building a Triple Cree setup much like the "nightrider" setup.
I'm using a 3021 Buck Puck for a driver run @1amp, but I don't know what will limit the actual voltage to the LEDs. I'm running a 14.4v pack and am concerned about over volting the LEDs and burning them out.
The Cree run 3.7v @ 1 amp X 3 = 11.1v. I added 3 volts for the buck pucks slight loss of efficiency. Does this math work?
This is my first build so if this is a dumb question forgive me.

Eric S

 

[evan9162]

The buckpuck will supply the proper current/voltage for your LEDs. Since it is a buck converter, designed specifically to drive LEDs (i.e, it's a constant current source) then you shouldn't have anything to worry about. It looks like you've done your homework to ensure that your power source (14.4V battery) will work properly with the buckpuck's electrical requirements.

So all you need to do is connect the LEDs to the output of the buckpuck, and it will properly regulate the current to the LEDs.

 

[FrontRanger]

I'm currently building a Triple Cree setup much like the "nightrider" setup.
I'm using a 3021 Buck Puck for a driver run @1amp, but I don't know what will limit the actual voltage to the LEDs. I'm running a 14.4v pack and am concerned about over volting the LEDs and burning them out.
The Cree run 3.7v @ 1 amp X 3 = 11.1v. I added 3 volts for the buck pucks slight loss of efficiency. Does this math work?
This is my first build so if this is a dumb question forgive me.

Eric S

Hi Eric,

As evan9162 already said, you should be fine. To get more insight into why that is true, consider the following:

An LED has a specific relationship between current (I) and voltage (V). So if you drive a particular current into the LED, only one value of voltage can appear across its terminals. Conversely, if apply a particular voltage across the terminals, only one value of current can appear. (The details of the I-V relationship vary with temperature and manufacturing process, but not enough to change the conclusion here.)

Now, your LED is a semiconductor diode, and you're operating it in the forward biased region with a reasonable current. "Forward biased" means that current is going into the anode (+ terminal) and leaving the cathode (- terminal), and "reasonable" in this case means that it's within the range specified by the manufacturer. In this operating region, the diode current is a very strong function (exponential) of the applied voltage. Conversely, the diode voltage is a very weak function (logarithmic) of the applied current. Even a large variation in current will produce only a small variation in voltage. So when you drive a well-defined (regulated) current into the diode, you get a very well-defined voltage appearing across it.

So to answer the question in title of your post, it's the nature of the diode's I-V relationship combined with the regulation of the current in your driver that limits the LED voltage.