Potentiometers to control each RGB LED strip color

[blksith0]

I have this LED strip with these specs "Input voltage: 12V DC; 60w/300leds" and it's a common anode. Hooked it up to a molex connection (12v) from my PSU and wanted a simple way to control each brightness completely, however there's a few problems.



1) It seems no matter how large of a potentiometer I used (20Ω, 5kΩ, 10kΩ), it was never enough to completely dim the LED. There was always a little bit of light left. How can I avoid that? I want the potentiometer to range from completely dark to full brightness as smooth as possible.
2) If only a single color is turned on, then that particular potentiometer gets really hot.
3) The red one just stopped getting to full brightness, but it looks like the connections are good. Did the pot die?

How can I fix this? Should the extra terminal on the pots be used?

 

[Colin]

You need to connect the other end of the pot to something and add 3 transistors and a few other things.
Then it will work good.

 

[ronsimpson]

3) The red one just stopped getting to full brightness, but it looks like the connections are good. Did the pot die?

Maybe the pot is damaged or maybe the LEDs are damaged. You can trade pot-red and pot-green to see what is bad.

If you connect the unused lead on the pot to (+) you will be able to get to no light. (the pot will be across the power)

How big are your pots? This is a place for a "power" potentiometer. You have 20 watts of green current so a "audio" pot (built for the input of a power amplifier) will not do.

 

[JonSea]

The LED strip is drawing a lot of power - 60 watts if you have the full 300-LED length. A standard pot isn't rated for anywhere near this power!

You need to use PWM (pulse width modulation) to control the brightness. Look on eBay for "LED strip controller" and you'll find many options. You need to find one that will handle the current your strip draws; there should be a spec for amps per foot to determine this, or just check the current drawn from your power supply without the pots installed.

One option is shown below. This is the simplest, and similar to what you're trying to do. A pot controls a PWM circuit to drive the LEDs. You'd need 3 of these, one for each color.

Various "electronic" controllers are available too. These have many effects, like flashing and varying the colors in different patterns. The downside of these controllers is if you just want to vary the color, all the options can make it complicated to get what you want.

 

[blksith0]

Thanks. I guess pots weren't the best option. In the future, how can I predict that 60W will be too much for a pot to handle?

 

[djsfantasi]

You calculate the watts required by the load. Call that X.

You get the pots rated power (or calculate it) from the datasheet. Call that Y.

If X is greater than Y, the pot can't handle it. In reality, It should be that Y > 2*X.

 

[JonSea]

The spec sheet for the pot will give a power rating.

Here's a typical pot like I'm guessing you used. It's rated at half a watt.

This wirewould beast is rated at 25 watts....and it costs 50 bucks! It was a reasonable guess when you said you tried a bunch of values that they weren't rated quite this high.

 

[audioguru]

A huge expensive high power pot was used as a light dimmer 100 years ago. Today we use Pulse-Width-Modulation controlling an inexpensive high power Mosfet and a cheap little low power pot.

 

[MrAl]

Hi,

The power in a resistor or pot Rp used to drop the voltage (and current) in another fixed resistance RL is:
P=(Rp*Vcc^2)/(RL+Rp)^2

where Vcc is the power supply voltage, RL is the load resistance, and Rp is the pot resistance, P is the power in watts.

 

[MrAl]

Hello again,

I did not get a chance to finish yesterday, so here is something to think about...

If instead of using a pot, we use a string of resistors all with the same value resistance and all with the same power rating, and use a single pole multi throw switch to select only part of the string so that we can reduce the current in the fixed resistor.

For an example, the fixed resistor is 10 ohms, and the fixed Vcc source voltage is 10 volts.
We want a pot value of 100 ohms so that we can reduce the current to roughly 1/10 of the full load.
The full load is 1 amp and because of the 10 ohm resistor the power is 10 watts.
If we use 10 resistors in series and a 10 position switch, what is the resistance of each resistor and what is the power rating requirement of each resistor, and note that they all must be identical.

First, the resistance is easy:
Take the 100 ohm and divide by 10 and we get 10 ohms for each resistor.
But what about the power for each resistor?
Remember they must all be the same value and have the same power rating.
Also, when just one resistor is selected only that resistor is in series with the 10 ohm load resistor, and when all are selected that means all 10 are in series with the load resistor of 10 ohms (the fixed load resistor).

 

[JonSea]

....But what about the power for each resistor?
Remember they must all be the same value and have the same power rating.
Also, when just one resistor is selected only that resistor is in series with the 10 ohm load resistor, and when all are selected that means all 10 are in series with the load resistor of 10 ohms (the fixed load resistor)....

Ahh...no.

If the load is 10 ohms, and the voltage is 10 volts:

I = V/r = 10/10 = 1 amp.

If a 10 ohm resistor is inserted in the line:

I = V/r = 10/20 = 0.5 amp

The power dissipated by R1 = I^2 R = 0.5*0.5*10 = 2.5w

If another 10 ohm resistor is inserted in series (20 ohms total plus the 10 ohm load):

I = 10/30 = 0.333 amp

The power dissipated by R2 = I^2 R = 0.333*0.333*10 = 1.11w

If a third 10 ohm resistor is inserted in the circuit (40 ohms total)

I = 10/40 = 0.25 amp

The power dissipated by R3 = I^2 R = 0.25*0.25*10 =0.62w

It's clear to see that all the resistors in this circuit do not dissipate the same power and therefore don't need to be the same wattage.


At any rate, this idea could work, but with less flexibility and at greater cost then the PWM controller I showed above.

 

[MrAl]

Ahh...no.

If the load is 10 ohms, and the voltage is 10 volts:

I = V/r = 10/10 = 1 amp.

If a 10 ohm resistor is inserted in the line:

I = V/r = 10/20 = 0.5 amp

The power dissipated by R1 = I^2 R = 0.5*0.5*10 = 2.5w

If another 10 ohm resistor is inserted in series (20 ohms total plus the 10 ohm load):

I = 10/30 = 0.333 amp

The power dissipated by R2 = I^2 R = 0.333*0.333*10 = 1.11w

If a third 10 ohm resistor is inserted in the circuit (40 ohms total)

I = 10/40 = 0.25 amp

The power dissipated by R3 = I^2 R = 0.25*0.25*10 =0.62w

It's clear to see that all the resistors in this circuit do not dissipate the same power and therefore don't need to be the same wattage.


At any rate, this idea could work, but with less flexibility and at greater cost then the PWM controller I showed above.

Hello there,

That's very good, but the power requirement for ALL resistors to be the same is not a question, it is a specification. In other words, i am specifying that all the resistors must be the same wattage because we dont have a choice about that when it comes to the next step.

The next step is to make a pot from those values, but the pot will have 10 detents. That means on the first click we get 10 ohms, second click 20 ohms and so on, and the resistive element inside is the same as a regular continuous motion pot. The only difference is we only get values that are fixed, first 0 ohms, then 10 ohms, theh 20 ohms, up to 100 ohms max.

The next idea after that is to calculate the total power rating that pot would have to have in order to be used in the manner we have been working with, which is in series with a 10 ohm fixed resistor that has 1 amp max through it and dissipates 10 watts. As we turn the pot shaft, the power dissipation in the 10 ohm fixed resistor (RL) will go down.
The real question now though is what would the pot have to be rated for, in watts, given the above discussion?

Since the power dissipated starts at 2.5 watts for the first 10 ohm section, then goes down, it seems apparent that the max for this setup is 2.5 watts because when we click it farther the total power in any one resistor is always less. But we cant make that 2.5 watt resistor any less or it will burn up when we decide to only click it once (to 10 ohms).

So what is the required rating of the ten-detent pot, given the max power for any single 10 ohm resistor section is 2.5 watts and we have 10 resistors in series?

See what happens?

 

[JonSea]

For the particular case at hand, there is no reason to require all the resistors be the same wattage, as you can calculate the wattage needed for each step.

Yes, you can make a crude pot this way, but it's not a very practical solution unless you have a bunch of power resistors and a high-current rotory switch around.

Unless the original poster wants to go this route, I think this is confusing the issue.

 

[JonSea]

There's a bigger issue though....the assumption that N equal steps in voltage provide N steps in brightness. This is not true for LED strips.

The typical RGB LED strip set for 12 volt operation has groups of 3 LEDs in series with a resistor for each color; 3 red LEDs and a resistor selected to provide the right current, 3 green and a different value resistor, and 3 blue with the appropriate resistor. The resistors are different for each color because the forward voltage of each color is different.

The forward voltage for the red LEDs is about 2.2 volts each, or 6.6 volts for the series atring of 3. The resistor is sized to drop 5.4 volts at the desired current. The forward voltage of the blue LEDs will be 3.3 volts or even higher, making the voltage drop forvthe strip nearly 10 volts.

So why is this important? If a pseudo-variable resistor is set to provide 10 equal voltage steps from the 12 volt source, as soon as the voltage is reduced below the foward voltage of the series string of 3 LEDs, they won't illuminate! FAIL.

A PWM controller doesn't vary the voltage to the LEDs, it varies the percentage of time the LEDS are on. PWM is the only effective way to control brightness on LED strips.

 

[MrAl]

For the particular case at hand, there is no reason to require all the resistors be the same wattage, as you can calculate the wattage needed for each step.

Yes, you can make a crude pot this way, but it's not a very practical solution unless you have a bunch of power resistors and a high-current rotory switch around.

Unless the original poster wants to go this route, I think this is confusing the issue.

Hello again,

I thought by now you would see the 'reason' but you dont seem to be getting the point. Just because you dont see a reason for something does not mean there is no reason

Take another look at what i had posted before, then try to relate that to the topic of this thread. You'll see the line of reasoning.

The resistors MUST be the same, because that is the way i specified them. If a company says they want all the resistors the same, you make them all the same. You might suggest that they dont have to do that, but once you've said it you dont have to imply that the resistors MUST be all different, because they dont want them all different they want them the same. One reason not related to this is that they might want to stock only one part number. One 10 ohm resistor at whatever wattage is required. But in any case, for my purpose and the purpose of this thread, the resistors must be all the same. Same value, same wattage.

If you followed my lead instead of shooting off on a tangent, you would have noticed that when we follow those specifications we end up with a 'pot' that has 10 indents, and this roughly approximates a regular pot with no indents, so what we find out about the power rating we can later apply to a regular continuous pot which was part of the subject of this thread. What is interesting is the pot rating seems to have to be higher than expected for a job like this one.

See next, we figure out the wattage rating of the 'pot' which would be the required rating of a pot *IF* we were to follow this rough procedure. Since a pot is made of a linear resistive element, the element is like that of 10 resistors in series, which is in the style of a finite element model of the resistance of a linear pot with 10 elements. We do this just to get the idea how the wattage of one section of the resistive element (one 10 ohm resistor in our case) affects the rating of the whole pot. We do this because it's harder to see the required rating of the pot when we look at the whole thing, but looking at one small section reveals that the wattage must be higher than expected.

For example, around the web some have suggested using a pot that is rated the same as the resistor. This would mean that we would go out and buy a 100 ohm pot rated for 10 watts (because the fixed resistor is rated for 10 watts assuming good cooling conditions). However, it appears that the 10 watt rating would not be enough, which is part of the point of this analysis. Using a finite element model of the pot helps to see the reason behind a higher rated pot.

To save time, i'll take it to the next step so as to illustrate the way this works...

The next step is to note that when we buy a pot, the linear element has a certain width and length. When we turn the pot shaft, we make the resistance smaller and the length shorter but the width stays the same. So what we do is we make the resistance less but we also make the surface area of the resistive element smaller because we are only using part of the element now. This smaller area has to be able to dissipate enough power to keep the pot from overheating.
This is approximated with 10 resistors, each 10 ohms. Since we found out that using just one of those resistors requires a power rating of 2.5 watts, that means all the resistors must be rated for 2.5 watts because the resistive element inside common pots does not vary in wattage along the length. Since we have 10 resistors, this adds up to 25 watts. Thus for that one small detent section to be able to handle the 2.5 watts, the pot must be rated for 25 watts. If we buy a 10 watt pot, that one small section will only be able to handle 1 watt so when we turn it to 10 percent of the travel it burns up that section of the pot element and renders the pot useless.

So you see how this works now? We dont really intend to build a pot with 10 resistors, we just want a model we can work with that reveals the reason for the higher pot rating requirement.

The next step after that would be to improve the finite element model by lowering the value of the first resistor down from 10 ohms to just 1 ohm. We then construct the 'pot' from 100 resistors each 1 ohm , then calculate the power requirement of some of the resistors and find the one with the max power. We then know that we need 100 times the power rating of that one max powered resistor.

Continuing with this idea, we let the value of the incremental resistance drop to zero as the number of resistors goes to infinity, and what we end up with is a more normal continuous pot and a calculated power rating.
But it's interesting enough just to look at the 10 ohm increment and the 1 ohm increment and see what becomes of the power rating of a pot that was made with those resistances in series. It comes out higher than most people seem to expect. When i look around the web i see lots of lower estimates which cant be correct, and the finite element model helps to explain this.

Try it and see what you get, it's very interesting to do it this way.

Of course it is usually more practical to use a PWM or even a transistor in the linear mode. Either way we loose power though so the efficiency isnt that great. I have several high power rheostats around but only use them for loads for testing different things.

 

[JonSea]

I'll concede the point - for what you have in your mind, the resistors must all be exactly the same and colored pink. Or purple.

Sadly, it's not a solution to the original poster's problem and a rather expensive waste of time.

 

[blksith0]

The math behind it all is very interesting. I'll probably just buy something like this 3-way PWM device.

Though I'd have more fun making it myself. I need to research PWM circuits.

 

[JonSea]

Perfect! How much does that cost?