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I have a RGB LED for photo-detection in my design with each color controlled by a MOSFET to manage light intensity of each color. The green and blue LEDs work great, but my red LED appears to slowly fade out over time. I measured my forward voltage (Vf) to be 2.6V on the red LED at the beginning, but it drops to ~2.56V over the next few minutes (in spec).

Here's my schematic. The left is my tricolor LED and MOSFET set-up and the right is the PWM expander, controlled by I2C:

enter image description here

PCB:

enter image description here

Here are the specs for my tricolor LED (I circled the correct one):

enter image description here

When I measure the light intensity coming from the LED over time, I can see the red intensity dropping while green and blue stay steady. Can a solder joint affect the LED? I wouldn't have thought twice about this, but I have an extra RGB LED so I removed one of my old (dead) ones and connected the LED with wires instead of soldering to my board. I have the LED on 'red' and the intensity isn't going down.

This has happened to 4 LEDs now...

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    \$\begingroup\$ I would guess it's temperature. What frequency and duty cycle do you use? I would use an inductor to make current more or less constant instead blinking with high frequency. It's possible that you experience some phenomenon related to edges of square wave... \$\endgroup\$
    – user76844
    Commented Nov 14, 2017 at 17:53
  • \$\begingroup\$ Can you measure V(Rs) for each at 100% to measure current? How much is heat dissipated by design? Vf of Red may be getting too hot to touch and dropping Vf by Schockley Effect and losing efficacy by heat. \$\endgroup\$
    – D.A.S.
    Commented Nov 14, 2017 at 18:02
  • \$\begingroup\$ @GregoryKornblum I have my PWM frequency set at 1600Hz, that was the frequency that the Adafruit PWM library came with (powering the PWM by arduino). Anything I can read up upon with the edges of the square wave affecting performance? I'm curious. I don't quite understand how hooking it up with wires instead of solder contact isn't killing the LED either..hmm \$\endgroup\$
    – Jordan Lewallen
    Commented Nov 14, 2017 at 18:03
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    \$\begingroup\$ How hot was it to touch? and where is the before/ after data and why do I have to ask twice? You have a thermal issue and we need better data to help you understand, to fix it.. Photos, temp, V,I \$\endgroup\$
    – D.A.S.
    Commented Nov 14, 2017 at 19:27
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    \$\begingroup\$ Serious design issues with this application.... I can fix if you have specs. \$\endgroup\$
    – D.A.S.
    Commented Nov 14, 2017 at 20:28

2 Answers 2

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Just because the forward voltage is in spec it doesn't automatically mean the forward current is too. 2.6V across the LED leaves 2.4V across the resistor/MOSFET which could mean 96mA through the LED and hence 250mW dissipated, which is well above the absolute maximum of 150mW.

I think you're killing your red LED.

The forward voltage in the datasheet simply gives the value you would expect to see when passing a given current through it, I presume this is the 50mA it gives in the column header but it will say somewhere.

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    \$\begingroup\$ We're seeing a current of 66.2 mA into the red LED, so this would mean ~ 100mW dissipated, which should be fine i think \$\endgroup\$
    – Jordan Lewallen
    Commented Nov 14, 2017 at 18:20
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    \$\begingroup\$ The ammeter shunt resistance may introduce enough further resistance to reduce the current to an apparently safe value. Instead, just measure the voltage across the 24.9 Ω resistor and calculate the current from that. \$\endgroup\$
    – Transistor
    Commented Nov 14, 2017 at 18:40
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    \$\begingroup\$ @JordanLewallen No it isn't, refer to the comment in my answer. 2.6V at 78mA gives just over 200mW, way over the maximum. You have to stay within ALL maximum specifications. \$\endgroup\$
    – Finbarr
    Commented Nov 14, 2017 at 20:12
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    \$\begingroup\$ You would be wise to have good specs , margins , ambient deratings and stability specs. Right now I do not see any and you are exceeding the Pd for the chip. I suggest cumulative current cannot exceed 50mA average for this chip. Epoxy is a good thermal insulator. These are also very sensitive to solder durations > 5 seconds ( fail or marginalized) and to ESD >-5V You may be far better off with SMD RGB with lens due to better heatsink. Right now your heatsink is the skinny Cathode lead \$\endgroup\$
    – D.A.S.
    Commented Nov 14, 2017 at 20:15
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    \$\begingroup\$ You would even be better off with excellent 5mm RBD LEDS focused on your target. I can send you RGB 10~16Cd 30 deg@ 20mA with some mods to your aperture housing. This Bivar part is n.g. I have the best 5mm parts in the industry. Far more than this 45 deg part @ less current. and very accurate. then I would add a Panasonic Photo 5mm 5V detector that measures CIE corrected intensity to calibrate your results. \$\endgroup\$
    – D.A.S.
    Commented Nov 14, 2017 at 20:23
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Your problem appears to be the variation in the RDS(on) for the device you are using. The datasheet for the BSS84 shows that the normalized RDS(on) will increase as the FET gets warmer.

enter image description here

The RDS(on) for the device could be quite high (10 Ohms) compared to your series resistor.
Since in your comments you say the voltage across the 24.9 Ohm is 1.946 V ....added to the Vf of the LED this leaves about 494 mV across the FET (depending how accurate your 5 V supply is), so it's not an ideal device (RDS(on) is about 6.3 Ohms) for your application.

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