What's the maximum distance using a remote Driver?

posted in: Constant Voltage Controls Drivers Products Constant Current

Was this answer helpful?

Let us know!

It is important to spell out what you mean when using "Driver".  
The industry has gotten into a bad habit of calling constant voltage power supplies, "drivers" - .therefore it is important to verify 

LEDs are current driven devices, so there are huge advantages to using remote constant current drivers.  One of these is minimal voltage drop in comparison to driving LEDs using a remote Constant Voltage device.

Voltage Drop = (Amps) x (Wire run distance x 2) x (AWG Resistance/ft) 
Voltage drop is the product of Total Resistance and amperage (or drive current).  With a constant current driver, the drive current is stable, so easily calculated.

As an example, powering Seven 350ma (0.35A) LED fixtures wired in series with a total distance of 500 feet of 18AWG wire (rounded up to .007 ohms/foot)
therefore using the formula above
VDrop = 0.35  x  (500 x 2) x (.007) = 2.5V
Power/Watts lost in that cable = 0.35A x 2.5v = 0.9W

14AWG (rounded up to .003 ohms) wire cuts that voltage drop and cable loss by more than half, so you can see that distance and cable size are the primary variables of voltage drop and will not impact a constant current driver as long as the driver operates within specifications - has capacity for this additional load.

Expanding the example above further, notice the Forward Voltage (Fv) of each of those Seven fixtures is not included since it had no bearing on voltage drop when using a CC driver.  

However, this info is needed to size the driver.  The fixture should be marked with tested Fv at rated drive current.  In this example, all Seven fixture's Fv are ADDED together and totals 25.5 VDC.

If using Model RPK-120-40D, its rated to drive 10 - 40 VDC of load at 350ma.  

The 25.5VDC load is less than maximum of the driver (40V) and greater than the minimum load (10V).  The driver also has the room for the additional 2.5 volts of voltage drop.  25.5v + 2.5v = 28VDC total load.

If the total load of all LED fixtures added to 38 VDC instead, then voltage drop could not exceed 2 volts and wire size and distance would need to be adjusted accordingly or the -60 driver would be required for this installation.