Dimming LED ...
- Thread starter Therrion
- Start date
WillieP
Well-Known Member
I'm guessing you are using a 50K ohm pot, not a 50 ohm pot??
If that is the case, because one driver uses a 100 k ohm pot, so you take the 100 K ohm divided by the number of drivers (2) and get a 50 K ohm pot.
The pot is then wired in parallel, not series.
Hope this helps
Cheers,
WillieP
If that is the case, because one driver uses a 100 k ohm pot, so you take the 100 K ohm divided by the number of drivers (2) and get a 50 K ohm pot.
The pot is then wired in parallel, not series.
Hope this helps
Cheers,
WillieP
Therrion
Well-Known Member
Yeah , I misspoke, it is wired in parallel and is also a 50k pot.
WillieP
Well-Known Member
Try wiring each individual driver to the 50 k pot one at a time.
You won't get full range dimming, but it will prove that both drivers are able to be dimmed.
Just a suggestion.
WillieP
ChiefRunningPhist
Well-Known Member
Maybe this possible explanation will help...
You're right about reducing the POT size to 50kΩ, just have to wire in parallel.
V = IR
^^^-same-vvv
V = AΩ
If there's a constant current being supplied through the dim leads (say 100 nA), the greater the R in its path, the greater the resulting V measured across that section of the cct. The greater the V that's needed to pass that small constant current through the dimming leads, then the greater the brightness. When no current passes, the V would be the greatest possible across that section of the cct, or 10V.
POT turned all the way up to 100%...
V = (0.0001A) × (100,000Ω)
V = 10V, 100% brightness
Turning down POT to 50%...
V = (0.0001A) × (50,000Ω)
V = 5V, 50% brightness
Turning down POT to 10%...
V = (0.0001A) × (10,000Ω)
V = 1V, 10% brightness
When you wire in parallel the supply currents are additive but the supply V maintains. If the POT isn't resized, the resulting increase in V noticed across the POT from passing the increased amount of current will reduce the effective range of the POT. When current doubles but the R doesn't change, the V doubles (if current doubled) which means you only have dimming range for half a turn, or half a revolution.
Assuming 2 drivers dim leads = 0.0002A not 0.0001A...
POT turned up all the way to 100%..
V = (0.0002A) × (100,000Ω)
V = 20V, 100% brightness
Turning down POT to 50%...
V = (0.0002A) × (50,000Ω)
V = 10V, 100% brightness
Turning down POT to 10%...
V = (0.0002A) × (10,000Ω)
V = 2V, 20% brightness
After 50kΩ resistance was reached on the 100kΩ POT, the resulting dimmer lead V would be maxed, so any further turning or rotating of the POT will be redundant or useless. If the POT size were adjusted to the difference in current, the resulting V will be maintained..
POT turned up all the way to 100%...
V = (0.0002A) × (50,000Ω)
V = 10V, 100% brightness
Turning down POT to 50%...
V = (0.0002A) × (25,000Ω)
V = 5V, 50% brightness
Turning down POT to 10%...
V = (0.0002A) × (5,000Ω)
V = 1V, 10% brightness
^^^By using a smaller POT the resulting V is maintained to a max of 10V allowing full dimming range on the POT..
POT's actual Ω are typically a couple % under their rating (ie a 100kΩ rated POT is typically only ~95kΩ - 100kΩ). So using a slightly over rated POT (ie a 110kΩ instead of a 100kΩ) will ensure you're able to reach max brightness, whereas with an exact rated POT you could be missing out on some of the top end...
You're right about reducing the POT size to 50kΩ, just have to wire in parallel.
V = IR
^^^-same-vvv
V = AΩ
If there's a constant current being supplied through the dim leads (say 100 nA), the greater the R in its path, the greater the resulting V measured across that section of the cct. The greater the V that's needed to pass that small constant current through the dimming leads, then the greater the brightness. When no current passes, the V would be the greatest possible across that section of the cct, or 10V.
POT turned all the way up to 100%...
V = (0.0001A) × (100,000Ω)
V = 10V, 100% brightness
Turning down POT to 50%...
V = (0.0001A) × (50,000Ω)
V = 5V, 50% brightness
Turning down POT to 10%...
V = (0.0001A) × (10,000Ω)
V = 1V, 10% brightness
When you wire in parallel the supply currents are additive but the supply V maintains. If the POT isn't resized, the resulting increase in V noticed across the POT from passing the increased amount of current will reduce the effective range of the POT. When current doubles but the R doesn't change, the V doubles (if current doubled) which means you only have dimming range for half a turn, or half a revolution.
Assuming 2 drivers dim leads = 0.0002A not 0.0001A...
POT turned up all the way to 100%..
V = (0.0002A) × (100,000Ω)
V = 20V, 100% brightness
Turning down POT to 50%...
V = (0.0002A) × (50,000Ω)
V = 10V, 100% brightness
Turning down POT to 10%...
V = (0.0002A) × (10,000Ω)
V = 2V, 20% brightness
After 50kΩ resistance was reached on the 100kΩ POT, the resulting dimmer lead V would be maxed, so any further turning or rotating of the POT will be redundant or useless. If the POT size were adjusted to the difference in current, the resulting V will be maintained..
POT turned up all the way to 100%...
V = (0.0002A) × (50,000Ω)
V = 10V, 100% brightness
Turning down POT to 50%...
V = (0.0002A) × (25,000Ω)
V = 5V, 50% brightness
Turning down POT to 10%...
V = (0.0002A) × (5,000Ω)
V = 1V, 10% brightness
^^^By using a smaller POT the resulting V is maintained to a max of 10V allowing full dimming range on the POT..
POT's actual Ω are typically a couple % under their rating (ie a 100kΩ rated POT is typically only ~95kΩ - 100kΩ). So using a slightly over rated POT (ie a 110kΩ instead of a 100kΩ) will ensure you're able to reach max brightness, whereas with an exact rated POT you could be missing out on some of the top end...
Last edited:
ChiefRunningPhist
Well-Known Member
Quick gif of parallel and series...
.......Really?
This is the winning answer I think. Haha.