Rahz
Well-Known Member
It will work, you just won't get any dimming until you've turned the knob half way. 100K ohm / # of drivers, so 50K ohm for 2 drivers to get full range, or maybe 55-60K ohm to insure you're hitting 100% brightness when open.
MeanWell LED Drivers: 3 in 1 Dimming Function.
- Thread starter stardustsailor
- Start date
littlejacob
Well-Known Member
Bonjour
No
Need a 50k pot I guess
CU
No
Need a 50k pot I guess
CU
Getgrowingson
Well-Known Member
Use 50k divide 100 by the number of drivers
dionysus4
Well-Known Member
so in case of using the 2ok pot with 6 drivers does this mean that the lowest dim value will not be off?
like if i use a 50k for 3 drivers? what dif does that make?
stardustsailor
Well-Known Member
6 pairs of Dim wires connected in parallel ..
20K ( linear ) pot ...
100/6 =16,666 k
That means the max output will not be at the far end of the pot ,
but a tad before that ...
At the other end of the pot (min value ) you will get 0K resistance ,
but that does not mean actual "LEDs off " ( Although the LEDs/COBs will go completely dark ) ..
It's just the drivers on protection mode .
The drivers should not go lower than 10% of their max output ...
Thus ..16,666K / 10 = 1,66 K ...
A resistor set of 1,66 K shoul be connected in series with the "start"( min Io ) or the " wiper " .
to ensure that the minimum value of the pot is 1,66 K and not 0 K .
( the max in that case is 21,66 K _ ideally .. )
Thus the dimming range of the drivers is : from 10% (min Io = 0.1* Io max ) to 100% = Io max
( from 100mA to 1 A ,
from 140mA to 1400 mA
from 200mA to 2A
and so on ... )
20K ( linear ) pot ...
100/6 =16,666 k
That means the max output will not be at the far end of the pot ,
but a tad before that ...
At the other end of the pot (min value ) you will get 0K resistance ,
but that does not mean actual "LEDs off " ( Although the LEDs/COBs will go completely dark ) ..
It's just the drivers on protection mode .
The drivers should not go lower than 10% of their max output ...
Thus ..16,666K / 10 = 1,66 K ...
A resistor set of 1,66 K shoul be connected in series with the "start"( min Io ) or the " wiper " .
to ensure that the minimum value of the pot is 1,66 K and not 0 K .
( the max in that case is 21,66 K _ ideally .. )
Thus the dimming range of the drivers is : from 10% (min Io = 0.1* Io max ) to 100% = Io max
( from 100mA to 1 A ,
from 140mA to 1400 mA
from 200mA to 2A
and so on ... )
Last edited:
dionysus4
Well-Known Member
you da man!
stardustsailor
Well-Known Member
TP1 to TP6 : Dim + #1 to Dim+ #6 wires
TP7 to TP12 : Dim - #1 to Dim- #6 wires .
Far counter clock wise = min setting ( pot = 1,66 K ) =output 10% of max .
Far clock wise =( over range ) max setting (pot = 21,66 K )
Actual max output, may be ~ 110-115 % of the rated max output )
Fror example ,HLP drivers rated at 1400mA max ,the actual max output is about 1650-1660 mA ..
dionysus4
Well-Known Member
sorry i dont really understand all that
i will now just connect 1 ELG 150c to a pot
so i need a 100k pot and resistor 10k?
dionysus4
Well-Known Member
ok did a bunch more reading and i think i got it for resistance dimming
BUT i found some interesting looking drivers in china that can run a cxb3590 up to 1700mA
the catch is they only have 0-10v dimming
can i connect a 10v power suply in parallel to dim and also run fans?
BUT i found some interesting looking drivers in china that can run a cxb3590 up to 1700mA
the catch is they only have 0-10v dimming
can i connect a 10v power suply in parallel to dim and also run fans?
robincnn
Well-Known Member
You can use 10V power supply. I have seen some on ebay. Fans should run fine at 10V
You can also use 12v power supply and step it down to 10V using a LM7810 voltage regulator.
To achieve dimming you can do a 2 resistor to create variable 0-10V
See implementation 2. Their V aux is just like 12V
They use 1kohm resistor and 5k ohm pot to get variable voltage across dim + and dim -
You can actually use 12V and no need for 7810. Just with correct resistors you can get 0-10V
Calculate like this
Current = voltage/ resis
=12 V / (1K+5K) = 0.002A
Then 5k × 0.002a = 10V
Now reduce pot to 1k ohm by turning it down
12 V / (1K+1K) = 0.006A
Then 1k × 0.006a = 6V
See same happened in implementation 2
It says 1k ohm 60%
So you can do dimming but it is not linear.
If you have external 12V I would put 12V + on that 1k resistor and - on the dim -.
Last edited:
EfficientWatt
Well-Known Member
dionysus4
Well-Known Member
You can use 10V power supply. I have seen some on ebay. Fans should run fine at 10V
You can also use 12v power supply and step it down to 10V using a LM7810 voltage regulator.
To achieve dimming you can do a 2 resistor to create variable 0-10V
View attachment 3587782
See implementation 2. Their V aux is just like 12V
They use 1kohm resistor and 5k ohm pot to get variable voltage across dim + and dim -
You can actually use 12V and no need for 7810. Just with correct resistors you can get 0-10V
Calculate like this
Current = voltage/ resis
=12 V / (1K+5K) = 0.002A
Then 5k × 0.002a = 10V
Now reduce pot to 1k ohm by turning it down
12 V / (1K+1K) = 0.006A
Then 1k × 0.006a = 6V
See same happened in implementation 2
It says 1k ohm 60%
So you can do dimming but it is not linear.
If you have external 12V I would put 12V + on that 1k resistor and - on the dim -.
thanks so much for your reply, if only i understood it lol
here is what im thinking
robincnn
Well-Known Member
This looks good too. Should give a linear dimming with a linear pot.
Looks like this way you can use any value of potentiometer. Like 25k , 33k, 50k ohm or 100k ohm pot
dionysus4
Well-Known Member
cool tnx
really any value? huh....?
but i will be dimming the voltage with resistance from the pot not dimming the power supply
boostedhonda
Well-Known Member
i have the hlg 185h c700b, will a 100k pot be fine for a single driver, can the driver be ran without a pot?
Fastslappy
Well-Known Member
if u leave the leads open (capped off ) then the drivers runs @ full cap & that can be approx 106+ 0r - %
with a dimmer it will only go to 100%
and you can add a switch to the circuit & get the best of both worlds
PyspherE
Well-Known Member
Hey RIU!
Im putting together a light system for my 4x8 tent. Ive decided on 28x CXB3590 36v 3500k powered by 7x Meanwell HLG-185H-1400b and mounted on 7x HeatsinkUSA 5.886" profile x 44". Im pretty sure I have the dimming down but just want to confirm im doing things right.
With 7x drivers on a single pot i need a resistance of:
100kohms/7drivers=14.29kohms
My thinking is to use a 5k resistor in series with a 10k pot for a total resistance of 15Kohm.
This will leave a small portion of the dial unusable but its miniscule and should allow me to dim down to about 34%
Can anyone confirm if im on the right track?
Be Safe! Much Love!
PyspherE
Im putting together a light system for my 4x8 tent. Ive decided on 28x CXB3590 36v 3500k powered by 7x Meanwell HLG-185H-1400b and mounted on 7x HeatsinkUSA 5.886" profile x 44". Im pretty sure I have the dimming down but just want to confirm im doing things right.
With 7x drivers on a single pot i need a resistance of:
100kohms/7drivers=14.29kohms
My thinking is to use a 5k resistor in series with a 10k pot for a total resistance of 15Kohm.
This will leave a small portion of the dial unusable but its miniscule and should allow me to dim down to about 34%
Can anyone confirm if im on the right track?
Be Safe! Much Love!
PyspherE
JavaCo
Well-Known Member
Math looks correct to me, You really don't want to dim much lower than 33% anyways, efficiency goes to hell
PyspherE
Well-Known Member
nevergoodenuf
Well-Known Member
Don't forget to add a kill switch to the dimmer. This will give about an 8+% boost in output. Get a watt meter so you can confirm the output. Also the cheaper the pot, the less accurate the pot. I use the watt meter to put 50 watt incruments on the dial. They will not be evenly space.
boostedhonda
Well-Known Member
u can wire in a resistor with the pot instead. will do the same thing but u will have full range for dimming
BobCajun
Well-Known Member
What makes you think shorting the dim leads on a driver with 3 in 1 dimming is bad for the driver? In data sheets it says less than 1v OR SHORT results in minimal brightness. Nowhere in anything have I ever seen something like "do not short the dim leads, will cause driver damage". I had an Arduino wired to a couple drivers and it was shorting and leaving open the dim leads at many times per second. Thing should be toast by now, by your reasoning.