# Design review - RGB LED color mixer using PWM

I'm trying to make an RGB color mixer. LED strip runs on 0.5-2A. It has internal resistors. LED Strip needs 5v, has one pin for that, one for red, one for green, and one for blue. Each LED module that you can cut off was about an inch an a half, actually a little less, and there's 5.5' of the strip, I think there's around 40 LEDs in the strip. I found this video online, but this is the exact LED strip I got. https://www.youtube.com/watch?v=sN5_Hjq_9Jg.

simulate this circuit – Schematic created using CircuitLab

Original schematic.

I will be using a total of 6 volts. i saw a video where a guy made a PWM circuit, https://www.youtube.com/watch?v=x4RmIzUd2lk, and it looked something like this.

simulate this circuit

555-based PWM control.

simulate this circuit

Tidied-up schematic.

I meant for the threshold to connect to the trigger, but I couldn't get it to. I'm confused on how to pick my resistors, my pot, and my capacitor, because in a different video, it said something like 1.44/(R1+R2+R2)*C1=Frequency. I assume frequency is how fast the LED flickers? In this schematic, there's 2 capacitors, which do I use in the equation? Does higher frequency mean faster flickering? How fast would be too fast for the naked eye? How do I control which I way I turn the pot to make the lights brighter. And how do I know how much current is coming out of the the output? How can I control that so it doesn't burn out my LED?

• Are you sure you are using $3 \Omega$ resistors? – copper.hat Feb 9 at 3:24
• @jsotola I never said 1V? – EmeraldVolts Feb 9 at 7:16
• @copper.hat No actually. I realized my mistake (I used the max current for the strip in ohm's law, when I should've used the minimum) while making the schematic, but I was going somewhere and forgot. I actually need a 9.8 ohm resistor, as with the 500 ohm pot, it would result in max of 9.61 resistance (using a parallel resistance calculator), and I need a 9.6 ohms resistor. – EmeraldVolts Feb 9 at 7:19
• Your schematic says 1 V on the power supply. What is the power rating of the pots? – Transistor Feb 9 at 7:53
• When you move the wiper up to the top of the pot you will short out the 12 ohm resistor and apply 6 V to the LED. It will be lovely and bright for a very short time. – Transistor Apr 28 at 22:45

As shown your pots are in PARALLEL with your fixed resistors so will REDUCE the resistance to below 3 Ohms.

500 Ohm pots are too large - you will have to set the wiper very near one end and have minimal control.

Each LED colour has a different voltage drop UNLESS they already have internal resistors. Red LEDs about 2V.
Green and blue closer to 3V.

The AA supply using Alkalines will be 4.9V when new, will rapidly fall to about 4V and then over time fall to about 3.3V.
At 9.6 ohms and even 0.5A you will drop V=IR = 0.2 x 9.6 = 1.92V.
At even 0.5A if you drop 1.9V you will have too little voltage except when the battery is very new.

Especially LED specs.

• Yea, reducing the resistance was what I was going for since I couldn't find any cheap low resistance pots. – EmeraldVolts Feb 9 at 7:10
• I saw that the strip has those chip resistors. or something. – EmeraldVolts Feb 9 at 7:10
• I know. When I was making the schematic, I remembered that something that made me realize I was supposed to use the minimum current for ohm's law. I was going somewhere, so I forgot to change it. – EmeraldVolts Feb 9 at 7:12
• Sorry, I bought the strips from walmart. And what exactly is voltage drop? – EmeraldVolts Feb 9 at 7:15

If the strip draws 2 A with all colours at maximum then you need about 700 mA per colour. That means that you need about $$\ \frac 1 {0.7} = 1.4\ \Omega \$$ per extra volt of your supply.

Now let's look at your 500 kΩ pots. These are typically rated 1/8 W being dissipated along the whole resistance track. We can calculate the maximum current from $$\ P = I^2R \$$. Rearranging we get $$\ I_{max} = \sqrt {\frac P R} = \sqrt {\frac 1 8 \frac 1 {500k}} = \frac 1 {4,000,000} \ \text A \$$. It should be clear why this isn't going to work. (Don't think that you can run higher current if you turn the pot to a lower value.)

At 500 Ω the result is $$\ I_{max} = \sqrt {\frac 1 8 \frac 1 {500}} = \frac 1 {4,000} = 0.25 \ \text {mA} \$$. That's out too then.

The right way to do it is to use PWM control as does the controller that came with the LEDs.

Figure 1. PWM signal transitioning from high pulse width (75%) to low (25%) and back again. Note amplitude remains constant. This will result in an apparent brightness of 75% and 25% of rated current. Source: LEDnique.

PWM, pulse width modulation, is very efficient as the LEDs are either fully on or fully off. If the PWM frequency is high enough you won't notice the flicker.

simulate this circuit – Schematic created using CircuitLab

Figure 2. A linear dimmer for one channel. R3 is an approximate value for all the resistors on one channel of the strip.

You could try something like Figure 2 but it has a few problems. You will lose a couple of volts across Q2 so you'll never achieve full brightness with a 6 V supply.

As explained already, the right way to do this is by PWM and there are thousands of examples on the web. Just make sure you chose a circuit for common cathode LEDs, if that's what you've really got.

• wow, the confusion is real. But ok, I'll look at PWM. – EmeraldVolts May 2 at 20:08
• Wait, I said a parallel resistor with the potentiometer. – EmeraldVolts May 2 at 20:09
• But I still don't really understand why a pot wouldn't work. It would change the resistance, the voltage is the same, so the current should change. – EmeraldVolts May 2 at 20:20
• The individual LED chips on the strip have 6 pins, 3 on each side, but the strip itself has 4 pins. – EmeraldVolts May 2 at 20:23
• I found this video online, youtube.com/watch?v=YmPziPfaByw – EmeraldVolts May 2 at 20:36