yodagrower
New Member
And if I don't want to go below let's say 40% current should I simply add a 40k resistor?
MeanWell LED Drivers: 3 in 1 Dimming Function.
- Thread starter stardustsailor
- Start date
Timothypaul26
Well-Known Member
Yes, that should work. Can check it with a volt meter too to verify, you should get ~1v per 10k.
swiftkillpapa
Well-Known Member
Ok so I'm late to the game as usual. I read sds tutorial and as usual I'm lost as fuck. Every time I try to learn something new because I finally got the last thing down, I realize how ignorant I am in this area of work.
So I want to run 4 hlg185hc700b powering 168 cree xpgs. According to the chart I need 25ΚΩ 20mW 2.5K. Which means I need a 25ΚΩ pot and a 2.5K resistor wired in series with the dim-?
So I want to run 4 hlg185hc700b powering 168 cree xpgs. According to the chart I need 25ΚΩ 20mW 2.5K. Which means I need a 25ΚΩ pot and a 2.5K resistor wired in series with the dim-?
VegasWinner
Well-Known Member
yep.
swiftkillpapa
Well-Known Member
O well that was easy I guess. Thank god ppl like @stardustsailor are here to help out noobs
Last edited:
boostedhonda
Well-Known Member
Keep It Simple Stoner
swiftkillpapa
Well-Known Member
Ok so then I have another question and don't know where to ask it but I figure y'all would know.
If I wanted to put in a in off switch for each driver would I add the switch to the a.c. Input cord of each driver?
If I wanted to put in a in off switch for each driver would I add the switch to the a.c. Input cord of each driver?
stardustsailor
Well-Known Member
swiftkillpapa
Well-Known Member
I don't get it? They're just smaller drivers?
JavaCo
Well-Known Member
It would make it so you couldn't dim lower then 20%.
swiftkillpapa
Well-Known Member
So I was at my local fry's electronics looking for resistors to fit my to my 2 drivers at 50k ohm pot. It says 5k resistors. Are these it?
VegasWinner
Well-Known Member
Exactly they use transistors resistors and a power source all on a small board. Instead of building them I just buy them for $4. They are made to drive motors so they have fine tuning
JavaCo
Well-Known Member
Those are 5.1 ohms you need 5k or 5000 ohm resistors for making it so you cant dim lower then 10% with two drivers connected to the one Pot.
swiftkillpapa
Well-Known Member
Got it. Thank man
Hi to all, sorry for my englisj. I' m in a trouble with my potentiometer on a HLG-240H-C1400B.
I have read all the 3d and i took lots of useful informations but I'm still confused..
Because I need to drive 5 COBs( CLU 048 1212) wire in series at 1.4A, i need of 36.7V for each cob(183.5V) i need ALL the courrent the driver can do, so an open circuit.
In another 3d a RIU user sugegsted me to put a swith to cutt off the dimmer to have an opern circuit and some extra VOLTS (182v instead of 179v whit potentiometer at 100%)
I decided to buy a potentiometer with integrate switch.
Here the link of potentiometer with datasheet
http://www.digikey.com/products/en?keywords=P170S-FC20BR10K
here the link for resistor
http://www.digikey.com/products/en?keywords=od103je
Are those component right for my lamp?
i put also a little scheme postproducing on my own a scheme from datasheet trying explain better my issues.
Hoping in a your answer, best regards.
I have read all the 3d and i took lots of useful informations but I'm still confused..
Because I need to drive 5 COBs( CLU 048 1212) wire in series at 1.4A, i need of 36.7V for each cob(183.5V) i need ALL the courrent the driver can do, so an open circuit.
In another 3d a RIU user sugegsted me to put a swith to cutt off the dimmer to have an opern circuit and some extra VOLTS (182v instead of 179v whit potentiometer at 100%)
I decided to buy a potentiometer with integrate switch.
Here the link of potentiometer with datasheet
http://www.digikey.com/products/en?keywords=P170S-FC20BR10K
here the link for resistor
http://www.digikey.com/products/en?keywords=od103je
Are those component right for my lamp?
i put also a little scheme postproducing on my own a scheme from datasheet trying explain better my issues.
Hoping in a your answer, best regards.
Attachments
JavaCo
Well-Known Member
Those should work fine. You just need to take one of the DIM leads and run it to the switch first, doesnt matter what Dim + or minus. Then out of the other side of the switch to one of the pot leads.
now i take trying to understand your clerar instuction..sure it easy but not so much for me
resistor i buy seems wrong, cheking the colours code...isn't it?
https://www.digikey.com/en/resources/conversion-calculators/conversion-calculator-resistor-color-code-4-band