It should output full power without anything connected to the dimming wires.
All you need to set it up is a 10v source and a 10k potentiometer, It shouldnt cost too much..
Pot: Can be of any value ,as total value is of no importance in a "voltage divider " .
But better to choose a pot of low value like say 10K ,because if you choose a large resistance ,
current flowing through will decrease and at low currents ,"RFI /EMI / digital noise pick-up" can be an issue.
May I suggest an expensive (~$10 )
multiturn ,linear ,high precision,low tolerance one ...
10K is fine.
Low limiter resistance: It's not a 'must' ,but better to be there .Set a low limit and not absolute zero.
(a 5K6 ,or a 6K8 or even a 10K ..)
-Turning pot CW ,more and more volts are applied to
DIM+ .
Output current of led driver is increasing .
When pot wiper is at end (at +10 VDC ) then Io=max .
(it's like DIM+ is 'open ' )
-Turning pot CCW ,wiper moves towards DIM-/
common ground ( COM ) .
Output current of led driver is decreasing .
DIM+ AND DIM- tend towards 'shorting'...
When wiper reaches end then Io= 0 (if a low limiter resistance is not set ,of course )
Shorting DIMs is a bad idea,thus the low limiter res.
Caution: The 10VDC source must not exceed 11 volts at any case .
The source also must be able to 'sink ' about 500
μA (
microAmperes )
per driver dimmed.
Cheers.
