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Help with potentiometer in series with led

BlinkingLeds

Feb 23, 2013
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The problem is as follows:
I have a 220v led strip which draws 200ma and I want to put an potenitometer in series with it so i can regulate the light output. How many watts should the potentiometer be rated?
Thanks.
 

GreenGiant

Feb 9, 2012
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does this wire directly to the mains?
I would recommend getting a dimmer switch rated for things of that nature.


if you are dead set on a pot though you would want to get a pot and a set resistor (1-1.2K) so you dont go above the rated current.
 

BlinkingLeds

Feb 23, 2013
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does this wire directly to the mains?
I would recommend getting a dimmer switch rated for things of that nature.


if you are dead set on a pot though you would want to get a pot and a set resistor (1-1.2K) so you dont go above the rated current.

Dimmers don't work with leds unfortunately.
If i put a resistor i will drop it's light output and i don't want that i want to put a pot to regulate from full brightness to zero and the leds have their own driving circuit.
 

GreenGiant

Feb 9, 2012
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Dimmers don't work with leds unfortunately.

Yes and no, if they run off the mains then you should be able to use a dimmer even with the internal driver, yes there are some older dimmers that will not adequately work with LED's but there are many new ones that will.

If i put a resistor i will drop it's light output and i don't want that i want to put a pot to regulate from full brightness to zero and the leds have their own driving circuit.

If there already is a current limiter then you dont need that resistor, I was just covering all the bases, you should be able to get away with using a fairly low wattage one, since when you drop the resistance the voltage will be so low that there is almost no current, and when you increase it the resistance will be high enough that there is very little current passing through it. You are probably going to want a fairly high resistance value (something like 50-100k)

Now I give you all this advice assuming that you have done your research and know about the internal driver circuit and how to safely change things.
 

BlinkingLeds

Feb 23, 2013
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Yes and no, if they run off the mains then you should be able to use a dimmer even with the internal driver, yes there are some older dimmers that will not adequately work with LED's but there are many new ones that will.



If there already is a current limiter then you dont need that resistor, I was just covering all the bases, you should be able to get away with using a fairly low wattage one, since when you drop the resistance the voltage will be so low that there is almost no current, and when you increase it the resistance will be high enough that there is very little current passing through it. You are probably going to want a fairly high resistance value (something like 50-100k)

Now I give you all this advice assuming that you have done your research and know about the internal driver circuit and how to safely change things.

they do run of the mains but they can't run with a regular triac dimmer nor does any other led can even the regular bayonet/e27 believe me i tried. To work with a dimmer you need a special "Dimmable Led"

So if i want to be able to adjust it from 0v to 220v ,is 1w or 2w potentiometer enough?
 

BobK

Jan 5, 2010
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LEDs don't work that way. You cannot dim them by varying the voltage from 0 to their operating voltage. The way to dim LEDs is by PWM or a variable current source. Not to mention that we have no idea of what kind of driver is being employed in your device.

Bob
 

BlinkingLeds

Feb 23, 2013
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Ok let's assume it's not an led lamp but an regular incandescent lamp that draws 0.2A how many watts should the potentiometer be to be able to adjust the volts from 0 to 220v?
 
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(*steve*)

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Jan 21, 2010
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Ok let's assume it's not an led lamp but an regular incandescent lamp that draws 0.2A how many watts should the potentiometer be to be able to adjust the volts from 0 to 220v?

The wattage of a pot essentially defines the maximum current that can flow through it.

And that current ir defined by the power rating and the total resistance of the pot.

The relationship is I = Sqrt(R/P) (from P = I^2.R)

In your case we use the formula from which it is derived. P = I^2.R.

If you want a 1k pot, then the power rating is 0.5*0.5*1000 = 250 watts.

(And that is yet another reason why we don't use pots in applications like this)

Yes, I know the rating of the pot is *way* in excess of the total load. What you need to remember is that the whole winding can dissipate 250W, but the last 10% can only dissipate 25W, and the last 1% only 2.5W -- and it is near the end of the pot's range that you start to have interesting issues,
 
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