Quick potentiometer question

[Airwalker16]

If I have one B version driver on the same power cord as some other Chinese drivers that aren't dimmable I still use a 100 k pot right? And if I use 2 that are on the same cord it's 50K, even if there's other Chinese drivers too?
is it basically just relevant to how many of the dimmable drivers are on the single power cord? Also if they happened to be different currents or maybe even different wattage drivers, does the same still apply? 50K for 2 33k for 3?
But to keep everything simple, the easiest thing to do is regardless of whether the drivers are on the same cord or not, just use a 100K pot and dim each driver individually right? Haha

 

[1212ham]

it has nothing to do with the power cord

 

[GBAUTO]

I think the formula Meanwell uses is R=100kohm/n, where n is the number of driver dimming lead pairs connected together. 1 driver=100k, 2 drivers=50k, 3 drivers=33.3k...

 

[Airwalker16]

I think the formula Meanwell uses is R=100kohm/n, where n is the number of driver dimming lead pairs connected together. 1 driver=100k, 2 drivers=50k, 3 drivers=33.3k...

So do varying models matter? Like a 120 & a 320 on one dimming lead. Hell even different currents too. Say 120-1400 & 320-2100? Is that a factor?

 

[trojanvirus]

u know potentiometers cost like $2 a pop... why not buy a couple to alleviate problems?

 

[GBAUTO]

So do varying models matter? Like a 120 & a 320 on one dimming lead. Hell even different currents too. Say 120-1400 & 320-2100? Is that a factor?

I would think that as long as you're using the same manufacturer, then you should be able to daisy-chain as many as you want.