I'm designing a circuit to power a high-brightness UV LED. I'd like to safely drive it as close as possible to maximum irradiance, and I'm having trouble figuring out whether all the components are properly chosen to avoid damaging/stressing anything (especially the LED, which is relatively expensive).

A Raspberry PI will be used to provide the on/off logic. The PI is powered separately, but shares a common ground with the LED power system.

Ideally I'd like the LED to be driven directly by a N-FET, where the gate is hooked up to a raspberry pi GPIO pin:


simulate this circuit – Schematic created using CircuitLab

The first question I have is: am I reading the UV-LED spec (here) correctly?

  • Forward Voltage: 3.45V
  • Maximum Current: 1400mA
  • Electrical Power (Max): 4830mW

The above three make sense, because 3.45V * 1400mA = 4830 mW. Then I see some peculiar items:

  • LED Output Power: 1500mW (min), 1700mW (Typical)
  • Test Current for Typical Power: 1400mA

How can the "Typical" LED output power be 1700mW when the test current for "typical" power is 1400mA?

Essentially, I'm not sure whether my resistor value should be calculated based on the 1400mA figure (5V-3.45V / 1400 mA = 1.11 Ohms) or whether I should be deriving a current from the 1700mW figure (1700mW = 3.45V X I; I = 492mA; R = 3.1 Ohms) and then using a larger resistor.

The second question I have is regarding the NFET. The NFET I have is rated for 1.7A and 30V (datasheet here). But the package is scarily tiny (SOT-23) and my gut feeling says I should ask before proceeding. I will be running this at 1.4A which is pretty close to the 1.7 rating. Will it work, and do I actually need to get a tiny heatsink for this NFET? Or is it better to get a beefier NFET?

Other than that, I'm making sure to get a hefty power resistor rated at 1-2 Watts, all wires will be 22 gauge, and all PCB traces will be nice and fat. Any other power-related gotchas I should be aware of? Power source details also listed below.

Thanks so much!

Datasheet for the NFET (1.7A / 30V): https://www.fairchildsemi.com/datasheets/ND/NDS355AN.pdf

Datasheet for the UV LED: https://www.thorlabs.com/drawings/ca01256cc12b40f8-35473EF6-5056-0103-7951A15FE4DCB58B/M405D2-SpecSheet.pdf

2A+ Power Board and battery: https://www.adafruit.com/product/2465 https://www.adafruit.com/products/353

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    \$\begingroup\$ For that kind of current a constant current LED driver power supply is preferred. \$\endgroup\$
    – Majenko
    Sep 2, 2016 at 17:34
  • \$\begingroup\$ Thanks! Something like this? mouser.com/ProductDetail/Mean-Well/LDD-1500L/… \$\endgroup\$ Sep 2, 2016 at 17:38
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    \$\begingroup\$ Ideal. Even has an on/off input to it so you can lose the MOSFET as well. \$\endgroup\$
    – Majenko
    Sep 2, 2016 at 17:42

2 Answers 2


You probably want to consider the recommendation in the datasheet: "We recommend using Thorlabs’ DC2200 or LEDD1B LED current drivers"

If you still want to use voltage mode control, I would ask: do you really need to run it at full power?

VF is a typical number than can fluctuate with temperature and vary part to part. So if your application permits it, I would run at lower currents (maybe 700mA).

The LED output power number quoted is for light output power. Since the test current used to generate this output is stated as 1400mA you can estimate the efficiency of your LED and plan for the generated heat accordingly.

Your NFET is close to max rated spec. I would double the current rating if possible. If you don't then I would definitely use a heat sink with some thermal paste.

One last thing that occurs to me is to add a resistor in series with the gate of the FET. Since you will be running high currents (relative to the drive capability of the GPIO), you don't want any glitches in your LED or power supply to shove a ton of current into the GPIO. A 10K resistor should do.

  • \$\begingroup\$ Thanks so much, the 10K resistor is a great suggestion and I'll add it to the diagram. Light-output power makes perfect sense, thanks for clearing that up! The Thorlabs DC2200 is about two thousand dollars so... yeah. We don't need to run the LED at full power, but the light goes through 50 micron pinhole, so most is absorbed. The remaining coherent light is then captured by a CMOS camera. The more intensity, the lower the camera's exposure time can be (yielding better images). My thinking is that we're paying for that current capacity rating after all, and may as well get the most out of it. \$\endgroup\$ Sep 2, 2016 at 18:05
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    \$\begingroup\$ "so most is absorbed". Have you considered to place some kind of UV lens between the LED and the pinhole? In this way you could increase the effective power passing through the hole without driving the LED full throttle. \$\endgroup\$ Sep 2, 2016 at 18:55
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    \$\begingroup\$ I mean concentrators like these or these. \$\endgroup\$ Sep 2, 2016 at 18:58
  • \$\begingroup\$ That's a great observation, Lorenzo. In our case, the research direction is to pursue a very low-cost imaging method using lens-less holography - just a CMOS camera and LED. Eventually we'll move away from this expensive LED towards a cheap SMT led with a similar current. Adding a lens would be cheating :) \$\endgroup\$ Sep 3, 2016 at 1:16

3.45 volts is given as the typical forward voltage but the data sheet doesn't explicitly say that this forward voltage is measured at such and such a forward current. The table that gives this number (page 1 of the DS) kind of implies it might be at 1 amp (given that the peak wavelength measurement uses that current) so where are you with deciding what value of resistor to current limit the device?

Not close unfortunately - your circuit shows a supply of 5.1 volts in series with a 1.1 ohm resistor. If the forward voltage is exactly 3.45 volts then the current will be 1.5 amps and bang your LED is likely to die after a short while.

The 3.45 volt could be at a test current of 1 A and therefore a resistor of 1.65 ohms is more appropriate. Better still, use a constant current source as recommended - you can make your own quite easily - they are not rocket science.

How can the "Typical" LED output power be 1700mW when the test current for "typical" power is 1400mA?

If the current (NOT power) is 1.4 A then the power into the LED is 1.4 A X 3.45 V = 4.83 watts. With 1.5 watts output light power, that's an efficiency of 31% and pretty good.

  • \$\begingroup\$ Oh sorry, I meant to distinguish the two 5V sources by calling one 5V1 and the other 5V2! I will change the names, they don't correspond to fractional voltages -- but you do bring up a good point, the actual power supply could be outputting a little more. Yes, I noticed the 1000A figure too and that was also a bit concerning. Maybe I should just play it safe and put a resistor to drive it around that current... \$\endgroup\$ Sep 2, 2016 at 18:11
  • \$\begingroup\$ No, use a constant current circuit is my advice. It's not like they are hard to implement so there's no big deal. \$\endgroup\$
    – Andy aka
    Sep 2, 2016 at 19:04

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