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I've got a 3-9W RGB LED (https://www.adafruit.com/product/2524).

It's connected to a breadboard with a power supply that tests as giving 4.9V and 2A on a short circuit.

I've got 5W resistors at 2ohms for the green and blue leads (the datasheet says they have VF=3.6) and a 3.3ohms for the red lead (VF=2.5V).

The datasheet says the max continuous forward current is 700mA, so I'm trying to get as close to that as possible for each color. According to https://www.digikey.com/en/resources/conversion-calculators/conversion-calculator-led-series-resistor these resistor values should be drawing ~700mA per die on the LED.

But if I hook up the circuit so the common anode lead goes to positive power, then the individual color leads are connected to their resistors, then the resistors go to ground) and put my multimeter in series with the anode, it shows it's only drawing 0.5A. I was expecting a lot more. I want to get this LED really bright. I mean, it's pretty bright as is, but I'm thinking it could be a lot brighter if it were drawing more current.

I'm trying to understand why it's not drawing the full 2A available... In looking at the datasheet again, I'm guessing it has to do with the forward current curves? For example, the red curve shows that at 2.5V it's only going to draw 200mA. Does that mean that if I take my supply voltage of 4.9V and subtract the VF of 2.5V (for red) that I should look on the curve at 2.4V to see how much forward current it will use? If so, does that mean I need to up my power supply from 5V to something higher to get it to draw more current?

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  • \$\begingroup\$ If you want to understand, do you know Ohm's Law? if so show what you think is the equivalent circuit using the schematic editor, It is 3W max ( with a heatsink) and not 2A \$\endgroup\$ – Sunnyskyguy EE75 Mar 25 at 20:17
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    \$\begingroup\$ "It's connected to a breadboard with a power supply that tests as giving 4.9 V and 2 A on a short circuit." That's not how you check the output current. If you don't have the power supply specifications then load it gradually until the voltage starts to drop. That is the rated current. What is the PSU voltage when all the LEDs are on? \$\endgroup\$ – Transistor Mar 25 at 20:18
  • \$\begingroup\$ @Transistor the breadboard power supply is one of those little prototype boards that give 5V or 3.3V depending on jumper setting. I've got it set to 5V. Looking at the specs for it (which I hadn't thought to do), it looks like it would have maximum output of 700mA. So is my first step here to get that out of the way and get something like a bench power supply or a PSU wall wart with a set rating? \$\endgroup\$ – mlibby Mar 25 at 20:31
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It sounds like you have an inadequate PSU.

"It's connected to a breadboard with a power supply that tests as giving 4.9 V and 2 A on a short circuit."

That's not how we check the output current. That tells you the current delivered when the output is 0 V. You want to know the maximum current delivered at 5 V.

If you don't have the power supply specifications then load it gradually (with more and more load - lower resistance) until the voltage starts to drop. Then measure to current and you will have the maximum that the PSU can deliver. Note that this value may droop due to thermal limiting as the PSU heats up so it might be wise to assume 80% of that figure as the rated output.

Looking at the specs for it (which I hadn't thought to do), it looks like it would have maximum output of 700 mA.

Most USB phone chargers will supply 2 A at 5 V. Get an old USB cable, cut the 'B' plug off and use the red (+) and black (-) wires.


I've got 5 W resistors at 2 ohms for the green and blue leads (the datasheet says they have Vf = 3.6 V) and a 3.3 ohms for the red lead (Vf = 2.5 V).

schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1. Schematics are so much better than words. This is what you've got. The voltmeters show the way to measure what's going on.

If your reading of the datasheets is correct then you should measure 2.5 V across R1 which would indicate that you have \$ \frac {V}{R} = \frac {2.5}{3.3} = 750 \ \text {mA} \$ through the red and \$ \frac {1.4}{2} = 700 \ \text {mA} \$ through the green and the blue. So my calculations agree with yours.

You may be losing some voltage elsewhere in your circuit due to high resistance wiring or a bad connection so check the voltage as shown with VM3 at the anode and at the bottom end of the resistor.

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  • \$\begingroup\$ Thanks a bunch for this... I've gone straight to a USB device charger that says it has 5.2V/1.8A for output and using the leads from a USB cable as you describe. Now when I hook up the circuit, the LED looks a bit brighter and the multimeter says it's using 0.87A. So I'm still a bit stumped as to why it's not drawing more. Is this the point where I have to model the circuit better and understand the rest of the voltages and currents involved? \$\endgroup\$ – mlibby Mar 25 at 20:59
  • \$\begingroup\$ See the update. \$\endgroup\$ – Transistor Mar 25 at 21:30
  • \$\begingroup\$ Thank you so much, I really appreciate it. With what you've described mathematically and shown in the schematic, it really helps reinforce what I thought was going on in the circuit. Thanks as well for the additional trouble-shooting information, that will really help. \$\endgroup\$ – mlibby Mar 25 at 21:42

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