# Potentiometer question?

#### TheFauxFox

Joined Aug 20, 2013
8
Hey everybody! Recently I purchased some LED lights off of eBay and BOTH of the controllers didn't work! Then I purchased a standalone controller/remote and STILL nothing. I looked inside all of them and sure enough, flux all over the board, and the leads were in the wrong place, sadly the damage was already done to the components and they were all unusable (got my money back though) so now I have LED strips that WORK, but don't have anything to control them. And my idea was to build a basic wired controller where they all share the same + wire, but each light color has a pot. wired up to control individual colors... I don't need strobing patterns, just a solid light.

IDK what pot. to get or how to calculate it but I'd love to learn how

My power supply is (pretty weak but it works) 12v at 100ma and the lights are (of course) also 12v... How can I find what pot. to use? I know that as of now there are 120Ω resistance in this circuit, but I'm slightly confused my potentiometers being marked with K at the end... Is that just thousands of ohms?

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#### R!f@@

Joined Apr 2, 2009
9,861
POT's won't work. U need PWM drive or a constant current source

#### TheFauxFox

Joined Aug 20, 2013
8
POT's won't work. U need PWM drive or a constant current source
Could you explain why? I would assume it would as it would just limit the current going through? Would it not?

And I've SEEN it work in two places, I just don't know what pot. to get for myself

and the source is a constant 12v from a power supply, sorry I diagrammed it as a battery...

#### inwo

Joined Nov 7, 2013
2,419
Yes, k is for thousand.

If you draw 100ma now then 120Ω is close enough to guess at a value.

Each leg if sharing equal is 360Ω. (33ma)

A good place to start then. Try 250-500Ω 1+ watt.

Or if you have a 1Kpot or fixed resistors, set it to minimum brightness and measure Ω value.

#### TheFauxFox

Joined Aug 20, 2013
8
Yes, k is for thousand.

If you draw 100ma now then 120Ω is close enough to guess at a value.

Each leg if sharing equal is 360Ω. (33ma)

A good place to start then. Try 250-500Ω 1+ watt.

Or if you have a 1Kpot or fixed resistors, set it to minimum brightness and measure Ω value.
That's what I needed! Thank you! My next q was gonna be about wattage spec but you answered that. I assumed it would be heavier duty as it does draw 1.2w......

Are these potentiometers gauged on their minimum or maximum resistance capacities? Lemme reword that... when I get a 5k pot. does it act as a 5k Ω resistor on it's minimum or maximum setting? minimum, right? And can I find the max as well?

Thanks again for the great answers and help!

Wow 250ohm pots are hard to find...I keep finding 250k

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#### samuel.whiskers

Joined Mar 17, 2014
95
Maximum. The resistance range for a 5k pot will be from near zero to 5kOhm.

#### TheFauxFox

Joined Aug 20, 2013
8
Maximum. The resistance range for a 5k pot will be from near zero to 5kOhm.
oh ok! That makes sense when I think about.

Thanks everybody!

#### wayneh

Joined Sep 9, 2010
17,201
Note the zero ohms at one end of the dial. That'll cause your LED to go poof, and it WILL happen. So, add a fixed resistor in series so that the LED is at max current when the pot is turned hard to one end.

#### samuel.whiskers

Joined Mar 17, 2014
95
The pot may go poof too down the zero end.....

#### R!f@@

Joined Apr 2, 2009
9,861
A resistor is better than a POT.
Sorry about what I said, what I meant to say was standard POT won't do, you would need a round 2 to 5 Watt pots. But be careful about how u set it, without a sereis limiting resistor you are bound to burn the POT at 12V. Leds might survive if they are rated for 12V.

But you cannot adjust brightness to a useful level with just a POT, as far as I have seen

#### wayneh

Joined Sep 9, 2010
17,201
I tend to agree. You might make something work to an acceptable level with just a pot and some resistors to tweak its behavior, but it will be costly due to the power rating of the pots, and it will never give you the nice smooth control over the full 0-100% brightness range.

I haven't priced things out, but you might find a genuine LED controller for less than the price of the pot, and it will be far more efficient and useful. Your experience with failed controllers is unusual. Plenty of folks have better luck.

#### MrCarlos

Joined Jan 2, 2010
400
Hi all

Let me put a little bit, my hands on this issue.
As far as I understand, the originator of this topic wants to control the light intensity of the LED's via a potentiometer. He also mentioned that he would like to know how to calculate the value of the potentiometer that he intends to use.

Well, it would be more appropriate to use a rheostat instead of a potentiometer.
The potentiometers are for low current applications while Rheostats are for more high current applications.

And yes, of course, a potentiometer might work; but, probably, the lifetime of this device would be very short.

Now:
The brightness of these devices (LED’s) is very close between the minimum and maximum light. It's not like the filament bulbs.

I would like to have on hand a chart where you could see the plot of luminous intensity (Lumens) versus applied current (milliamps) to LED's.
Certainly you will discover what I mentioned in the previous paragraph.

Anyway.

It is very likely that you already have well known what I'm going to mention, but should have a look again.
We can say the following statement: When the LED is crossed by an X current, we shall have a voltage drop Y on its terminals.
The X and the Y -parameters- are provided by the manufacturer of LED in their data sheets.
These specifications should be taken into account when we work with LED's to avoid damage.
(These data have not been provided by the originator of this topic).

Let me assume that the LED's that TheFauxFox uses have the same electrical parameters as those who have ISIS Proteus Simulator.
X = If = 10 mA.
Y = Vf = 2.2 Volts.
When a current = 10 mA. Is flowing through the LED, there will be a voltage drop on its terminals = 2.2 Volts.

What is our next step?
To investigate: How much current must apply for just the LED lights ?
We will assume that is 3 mA. since we do not have this information at hand.
We also need to know how much voltage we will to polarize our circuit.
TheFauxFox tells us that has a power supply of 12V at 100 mA.

OK, we already have almost all the data to calculate the value of the current limiting resistor and the value of the potentiometer. or probably a rheostat.
just is missing the voltage drop across the terminals of the LED when it is biased with a 3 mA. current.
I will suppose, to save time and words, like 0.6 Volts.
But you can find the value for this parameter in the LED data sheets.

We will handle the LED to change its brightness by a current from 3 mA. Up to 10mA. applying a bias voltage to the circuit, of 12 Volts. . . Right?

Keep in mind that our circuit is: 12 Volts power supply, a current limiting resistor in series with an LED.

Now the formulas:
(Vcc - Vf) / If) = Rx (Current limiting resistor value).
12 - 2.2 = 9.8 Voltage drop across the terminals of the current limiting resistor.
9.8 / 0,010 = 980 Ohms. Value of the current limiting resistor.
So when we polarize our circuit with 12 volts. which consists of an LED in series with a resistance of 980 Ohms, the LED light at full brightness

To calculate the value of the current limiting resistor when the LED is biased with a current of 3 mA. We could use the same formula above.
(Vcc - Vf) / If = Rx.
12 - 0.6 = 11.4 Voltage drop across the terminals of the current limiting resistor.
11.4 / 0,003 = 3,800 Ohms. Value of the current limiting resistor.
So when we polarize our circuit with 12 volts. which consists of an LED in series with a resistance of 3,800 Ohms, the LED light at minimum brightness.

Now The Potentiometer, or perhaps the rheostat
So far we know that:
At low light intensity, with a current limiting resistor of 3,800 Ohms.
At high light intensity, with a current limiting resistor of 980 Ohms.

When the potentiometer is at its minimum value, near zero ohms, Should prevail current limiting resistor with a 980 Ohms value.

When the potentiometer is at its maximum value should not exceed 2,820 Ohms.
3.800 - 980 = 2,820 Ohms.
Because the LED could be off. . . Could You Please explain to me, Why ?

You decide whether to use a potentiometer or rheostat.
The power of the device, which will adjust the LED current is:
E x I = W
We will choose the worst case:
11.4 x 0.003 = 0.0342 Watts.
9.8 x 0.010 = 0.0980 Watts. This is it.
It is prudent to increase a safety margin for the device, which the LED current is adjusted, so that no burning it.

This is calculated by multiplying by 4, the value obtained in the worst case.
0.0980 x 4 = 0.392 Watts. Commercial value could be 1 Watt or ½ Watts.

All these calculations are based on assumptions.
It is necessary to know the electrical parameters of the LED TheFauxFox are using.
Otherwise we will be working for the Test-Error method.

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#### R!f@@

Joined Apr 2, 2009
9,861
@ Carlos

Do you consider the fact that OP has LED strips which differ from standard LED(S).

If I am right on what OP has, means he cannot do with a rheostat much.

Still a rheostat is far too expensive than this which is far better to control the LED strips at even higher voltages
Use this and u do not need a series resistor actually. As it can control the current been supplied.
And if I am correct about the RGB strips OP has. It will have resistors soldered on to the flex cable and it will Run fine with 12V. And they can be cut at standard lengths which are marked on it.

by the way Carlos. Ur post is informative. I admire you took the time to explain things that a n00b needs in lighting single LED's.

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#### MrCarlos

Joined Jan 2, 2010
400
Hello R!f@@

Yes of course.
The OP mentioned having LED strips. and probably not get any progress with the potentiometer or rheostat.

But the intent of my response is to answer the OP this phrase:
what pot. to get or how to calculate it but I'd love to learn how.
Coupled with trying to describe how to get to that calculation.

Sure, in some cases, it is better and faster to buy something already done, to try to do it ourselves.
Especially if the one is trying to do has little knowledge of electronics.

Joined Oct 4, 2013
472
ive done this exact thing on a roll of 200 LEDs, with 12v. I used a 1.5k resistor on the 0 ohm end to prevent burnout and used a 0-5k 25 turn pot on each leg Of the RBG LEDs and was able to mix about any color I wanted,,

#### TheFauxFox

Joined Aug 20, 2013
8
Yea... I was thinking about wiring in a small resistor, then a pot... does anyone know where I can find a pot higher than 1/2 watt? I cannot seem to find any...