Hey everyone, fairly new to the circuit side of things. I teach lighting design at a university and have a soldering project in mind for my students. They will make a circuit to control an RGB led. I have everything sorted out except for the three potentiometers that will each control one color of the led. The problem is, I have no idea how to choose the right resistance of potentiometer. I will have two coin batteries providing 6v. Maybe this is another question for you all because I increase from just 3v because these are the specs I have for the LED's:

• Forward Voltage:Red:1.9-2.1v Blue/Green:3.0-3.2v(Low Voltage for DIY PCB Circuit);Current:20mA

I was worried that the forward voltage of the B/G is equal to the voltage in so that would cause an issue (also that I don't know what resistor to put in at that point). So if you have help on that I would welcome it.

Back to my original question, how do I figure out what resistance on a pot I need? I see them at 10k and 20k which I think is pretty common but I don't know if it will do the job I need it to. Any help would be fantastic. Thank you all!

 

LEDs of course are non lienar devices and want to be driven by current
rather than voltage.

One of the better ways to do LED driving is via a PWM. In your case since
you are not coding a UP a led driver module that takes a V from a pot and
that is handled in module to convert to PWM settings to control the duty cycle
which controls brightness.

Another way is a simple controlled current source.

Here you would use your pot in series with the R. The R would set max current,
and the pot would reduce the current. Its not a linear approach, amybe a log
taper pot might work for that problem. Note the load is the LED.

Instead of a LM317 I would use this because its low dropout -

https://www.onsemi.com/pub/Collateral/LM2931-D.PDF



Regards, Dana.

 

Hey everyone, fairly new to the circuit side of things. I teach lighting design at a university and have a soldering project in mind for my students. They will make a circuit to control an RGB led. I have everything sorted out except for the three potentiometers that will each control one color of the led. The problem is, I have no idea how to choose the right resistance of potentiometer. I will have two coin batteries providing 6v. Maybe this is another question for you all because I increase from just 3v because these are the specs I have for the LED's:

• Forward Voltage:Red:1.9-2.1v Blue/Green:3.0-3.2v(Low Voltage for DIY PCB Circuit);Current:20mA

I was worried that the forward voltage of the B/G is equal to the voltage in so that would cause an issue (also that I don't know what resistor to put in at that point). So if you have help on that I would welcome it.

Back to my original question, how do I figure out what resistance on a pot I need? I see them at 10k and 20k which I think is pretty common but I don't know if it will do the job I need it to. Any help would be fantastic. Thank you all!

Ohm's Law is your friend. You need a resistance that cannot allow damage to the LEDs. The Red will be the most at risk, so let's work on that one.

For the red LED, you need the resistor to drop 4V when it's running at 20mA. V = I•R so 4V = 0.04A • R and R = 100Ω. For safety (of the LED) you might choose a 120Ω resistor. On the other hand, the battery itself will have internal resistance and you might find (from measurement) that the resistor could be even less than 100Ω because the battery voltage drops (to say 4-5V) when it's loaded by an LED. The Blue and Green LEDs could also get by with a little bit lower resistance because they have slightly higher forward voltages, meaning the resistor needs to drop less voltage at 20mA.

OK, so you need roughly 100Ω external resistor no matter what setting of the pot, otherwise you risk blowing up the LED. This is your max brightness setting, as if you short across the pot (because that's what happens at one end of the pot). To dim the light, you're going to want to get the current well under 1mA. It'll still glow dimly at 0.1mA. If you want to hit that low level, you need V = I•R so 4V = 0.0001A • R and R = 40K Ω

You can find pots at 10K and 50K, maybe 20K as well. Not sure about other values.

The trouble is that a very small turn at the bright end will have an enormous effect while a wide change at the low end will have very little effect. You can improve this a bit by going towards the low end, say a 5K. But that will limit how low in brightness you can go.

This is one reason why LED dimmers use pulse width modulation (PWM), because it allows smooth linear dimming from 0 to 100%. (Another reason is that using a pot requires all the power to the LED goes thru the pot. That's inefficient and would require very expensive pots for high wattage.) A linear controller like that kindly provided by @danadak solves the range issue but is still inefficient. That's not much of a concern for your project but it matters in others.

 

The trouble is that a very small turn at the bright end will have an enormous effect while a wide change at the low end will have very little effect. You can improve this a bit by going towards the low end, say a 5K. But that will limit how low in brightness you can go.

You can mitigate that to some extent by using log pots which have non-linear track resistance.

 

You can mitigate that to some extent by using log pots which have non-linear track resistance.

Yeah I didn't want to dig into that but if you're looking to buy a pot for this purpose, I suppose that would make sense. I can't say I've ever tried to dim an LED that way. Maybe I'll give it a try.

 

Thank you all so much. Extremely helpful.

 

How about a 1000 ohm pot in series with the LED and then a resistor from one end to the slider that will limit the maximum current to whatever the data sheet states is the maximum. That will allow adjustment without the chance of doing damage, as long as your calculations are close to right.

 

As a reasonable simple step up from just a pot and resistor, I would use an NPN transistor as a simple current-source to drive the LED, which will give a linear change in current with resistor wiper position, and also allows the current to go down to zero.
LTspice simulation below:
Q2 is a diode-connected transistor that generates an offset voltage to cancel the Vbe of Q1.
This allows LED current to start as soon as the pot is adjusted from minimum to give a full 0% to 100% adjustment range.
The transistor can be just about any small NPN bipolar junction transistor (BJT).
The value of R2 may need to be changed to get the desired maximum LED current, depending upon the battery voltage you have.

Edit: Circuit modified to eliminate change in current for different color LED voltage drops.

 

You can mitigate that to some extent by using log pots which have non-linear track resistance.

It took a while to sink in but your post triggered a memory. I once simulated in Excel what you could do to dim an LED with a pot. I'd completely forgotten I already had the exact answer for any combination of resistors.

This example was with a 300Ω resistor in series with the pot, a 9V supply and a typical 20mA LED with Vf = 3V.