OT: LED lighting configuration... - test-drb@ccmp.vtda.org

29 Apr 2016

...
  On Apr 29, 2016, at 4:10 PM, Bill Sudbrink
<wh.sudbrink at verizon.net> wrote:

 Paul Koning wrote:
   On Apr
29, 2016, at 3:32 PM, Bill Sudbrink <wh.sudbrink at verizon.net>  wrote:

 ... The "bulbs" are labeled:

 15F18120-45 15 watt 36vdc constant current

 I'd like to put four in a fixture and I'm trying to
 understand what kind of driver I need and how to wire
 it.  I was thinking of using a Mean Well LPF-60D-36
 like this:

 http://www.ebay.com/itm/Mean-Well-LPF-60D-36-AC-DC-POWER-SUPPLY-  Dimmable-LED
  -DRIVER-36V-60W-CLASS2- 
/161857068172?hash=item25af6edc8c:g:9hQAAOSwA4dWHVn5

 and wiring the "bulbs" in parallel to it.  But after
 realizing that I'm not completely sure what a "constant
 current" power supply does and doing a little "googling"
 I don't know if that's the right approach.  
 A constant current supply is one that delivers a constant
 current to a varying load (within limits) just as a constant
 voltage supply delivers a constant voltage to a varying load.  
 Ok, I figured that much.  The problem/question is why there
 are no Amp ratings on anything?  Assuming the DC equation:

 Watts = Amps X Volts

 I want a constant current supply that "pushes" 0.41 Amps.
 A little more googling reveals that the above supply is
 rated "1.67A output".  This seems to support the W=AV
 theory.  So, do I want a PS labeled "15 watt 36vdc",
 regardless of how many bulbs I want to drive?  You say
 "within limits".  What specification do I look for to
 understand the limits? 
15 watt 36 V is an odd spec for a device that needs constant current.  What it seems to
translate to is 400 mA device current, 36 volt nominal operating voltage.  That's
perhaps 10-12 LEDs in series, since each has a forward voltage around 3 volts, perhaps a
bit more.

If you have a supply rated for constant current operation, 36 volt or so, settable
current, you could use that, crank the current setting down to 400 mA.  If it's a
fixed supply (36 volts 60 watts, i.e., 1.66 A) then that would not work.

	paul