Question on resistors for pot on B series drivers.

[Airwalker16]

Like the title says, I have a question regarding the resistors I've used in previous builds on the white wire before soldering to the pot. My question is that I don't have any on hand right now, so could I just solder the wires directly to the pot and leave out the resistors or do I need to get a 10K resistor or any combination of resistors to get close to 10K to use the dimmer correctly?

 

[NGA]

You don't need resistor, they just help you get the max out of driver, some pots don't reach 100% ,

 

[whytewidow]

I thought people used 100k pot on external 7 pin drivers? That's what I'm using. And I know if it's not atleast 100k you dont get full power when you crank it open. And I have mine soldered straight to the pot tabs.

 

[NGA]

Adding a resistor before the pot allows you to get max ,far as I can remember been months since I've done some builds

 

[Airwalker16]

I thought people used 100k pot on external 7 pin drivers? That's what I'm using. And I know if it's not atleast 100k you dont get full power when you crank it open. And I have mine soldered straight to the pot tabs.

It's a 100K pot on an HLG-240H-C2100B but I don't have resistors handy so I went ahead and just soldered it. Hope it doesn't fuck me over.

 

[CoB_nUt]

It'll work fine. As stated, the rssistor just helps get the full potential out of the pot if it's not the full 100k ohm. If your pot isn't, and no resistor, you won't max the driver out.

 

[CobKits]

It's a 100K pot on an HLG-240H-C2100B but I don't have resistors handy so I went ahead and just soldered it. Hope it doesn't fuck me over.

breaking the circuit manually or with a switch to create infinite resistance is the same thing really

 

[Airwalker16]

breaking the circuit manually or with a switch to create infinite resistance is the same thing really

What do you mean?

 

[CannaBruh]

What do you mean?

An open circuit is similar as infinite resistance. If you want full output break the dim circuit or put it on a switch so you can flip it in/out dim/full output.

One might look into running with a 20% 120k pot sans 10k resistor.

 

[Airwalker16]

An open circuit is similar as infinite resistance. If you want full output break the dim circuit or put it on a switch so you can flip it in/out dim/full output.

One might look into running with a 20% 120k pot sans 10k resistor.

So to get the full potential power of a B Version driver, you just cap off the blue and white wires separately?
And for a potentiometer I'd just wire it up but use a switch on either of the wires and use it to break the connection to the potentiometer?

 

[CannaBruh]

So to get the full potential power of a B Version driver, you just cap off the blue and white wires separately?
And for a potentiometer I'd just wire it up but use a switch on either of the wires and use it to break the connection to the potentiometer?

Say you use a pot that has only 90k resistance max. You won't get the maximum output out of your driver without your 10k resistor in series for the 100k sum.
You can get around this with the switch, if you want the full output (why buy dimming drivers..)

By using a switch in series with your pot you can effectively switch out the pot (open circuit) making the resistance infinity.

I'm not sure on these drivers if you can run the dim circuit open with nothing or capped off as you put it or if the driver wants to see some current here and 100k is ideal, someone else might have a better idea.

 

[Airwalker16]

Say you use a pot that has only 90k resistance max. You won't get the maximum output out of your driver without your 10k resistor in series for the 100k sum.
You can get around this with the switch, if you want the full output (why buy dimming drivers..)

By using a switch in series with your pot you can effectively switch out the pot (open circuit) making the resistance infinity.

I'm not sure on these drivers if you can run the dim circuit open with nothing or capped off as you put it or if the driver wants to see some current here and 100k is ideal, someone else might have a better idea.

So basically what I said...right?

I bet @CobKits has the answer to your question though.

 

[CobKits]

youre both right

capped leads are fine if you dont need to dim

101kohms, 110 kohm, infinite ohms, all the same. were just trying to get it above 100

airwalker, what canna is trying to say is pots have a tolerance.
a 100k 10% can be anywhere in the 90-110K range. if you get one thats 100-110K, adding resistance will do nothing. if you get one that is in the 90-100K range, adding 10K resistance (or opening circuit to infinite resistance) will give you a little bump in output to the max the driver is capable of

so for example the formula for dimming drivers is 100k divided by number of drivers

so 100k for 1 driver
50 k for two drivers dimming together
33.3k for 3

well i dont stock 33.3k pots and if someone wants to dim 3 drivers i jsut tell em to use a 50. why? because it can operate in the required range of 0-33K. doesnt matter if it goes over 33K.

the only difference is 33K-50K on the knob will look the same (full power), but you only get 2/3 of the travel of the pot for adjustment

like lets pretend the pot is like a clock, 7 oclock is zero ohms, 5 oclock is max resistance

on a 33K pot,
7:00 = 0 ohms (10% output)
12:00 = 16.5 ohms (~50% output)
5:00 = 33 ohms (100% ohms *if youre lucky and your pot is actually 33.3 or slightly over *)


on a 50K pot,
7:00 = 0 ohms (10% output)
10:00ish = 16.5k ohms (~50% output)
12:00 = 25k ohms (~75% output)
2:00ish to 5:00 = 33-50k ohms (full output)

so you have a little area on the pot which does nothing, but you are guaranteed to hit your 33.3k resistance to get full output

we just use 20,50,100k ohm because they are cheap and common. as canna mentioned 110-120k pots would be ideal if they were more common

5:00 = 33 ohms (100% ohms *if youre lucky and your pot is actually 33.3 or slightly over *)

 

[Airwalker16]

on a 50K pot,
7:00 = 0 ohms (10% output)
10:00ish = 16.5 ohms (~50% output)
12:00 = 25 ohms (~75% output)
2:00ish to 5:00 = 33 ohms (full output)

Did you mean 2:00-5:00 = 33-50Kohms?

 

[CobKits]

yes, corrected. but its the same to the driver.

 

[Airwalker16]

yes, corrected. but its the same to the driver.

So you're saying by using a 50K on 3 drivers, you're at Max output around 2:00?

 

[CannaBruh]

So you're saying by using a 50K on 3 drivers, you're at Max output around 2:00?

On an ideal pot but these have wide tolerances. It could be 10-20% more in position to the left or right.

 

[Isawthelight]

Adding a resistor before the pot allows you to get max ,far as I can remember been months since I've done some builds

Adding a 10 k resistor in series with the dimmer pot also limits the dim signal to 1 volt at minimum dimmer pot setting. Dimmer signal can't dim to zero but this is not an issue for users of dimmers with the 1-10v dimmer design.

 

[CannaBruh]

Adding a 10 k resistor in series with the dimmer pot also limits the dim signal to 1 volt at minimum dimmer pot setting. Dimmer signal can't dim to zero but this is not an issue for users of dimmers with the 1-10v dimmer design.

This is hard to follow even after 1 cup of coffee

I think you're saying with the 10k r that the driver will never DIM to 'no output' ??

Will it dim to 'no output' when DIM terminals are shorted?

 

[whytewidow]

This is hard to follow even after 1 cup of coffee

I think you're saying with the 10k r that the driver will never DIM to 'no output' ??

Will it dim to 'no output' when DIM terminals are shorted?

If you short out the dimming leads on straight to each other. I believe meanwell only dims down to 10% my hlg320 I use 100k pot. It reads 103k when turned wide open and runs 325.4w @24.4 All the way turned down my light is at 11.85w still. If I hook both dimmer wires together direct short. My light runs 3.08w