Wattage usage - Dimmer function

Hi,

I just built my first cob setup, 5x citizen 1212 4000K on a MW 240-C1400B.

The driver spreadsheet says it operates from 700mA up to 1400mA. Great, so I thought, if completly dimmed (100k potentiometer), the lights will pull around 125W from the wall as they will be operating at 700mA, 5 units x36V (cobs voltage) x700mA equals 125W, if not dimmed then it will operate at 1400mA, thus pulling 250W from the wall, 5 units x36V x1400mA equals 250W.
However when taking some measurements I realised they are actually pulling only 11W from the wall when completly dimmed, they still turn on. When not dimmed they pull 220W, instead of the theoretical 250W.

Any clues about this?

Thanks.

Your driver will dim 90% not 100% so that is the 11w. When not dimmed the 100k plus 10k resistor reduces the total wattage below the 250w. To get the 250w you will need to use a switch to open the circuit allowing max watts or remove the potentiometer and resistor.

The 700ma to 1400ma range is for the 240-C1400A with internal dimming pot, your "B" version (240-C1400B) has 10% to 100% dimming.

MW drivers need 100k for 100% output. Most pots actuality have less resistance than the rated spec so the the driver won't go to 100%. I just measured 4 pots from ebay, 88k, 95k, 97k and 94k. People here are adding a 10k resistor in series to get at least 100K.

Hey guys, thanks for the help.

I didn't add a 10k resistor after the pot, if not dimmed is it equivalent to open circuit, or the 10k is the internal resistance of the driver ?
I don't need more dimming than I actually have, but would like to understand what's going on.

I still don't get how 90% dimming pulls from the wall only 11W. The remaining 10% of 250W is 25W.

So you are saying I might need more that 100k to achieve the max rated current from the driver?