Electrical Engineering Stack Exchange is a question and answer site for electronics and electrical engineering professionals, students, and enthusiasts. It's 100% free.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I just bought some UV LEDs

I bought them for testing banknotes and identity papers. I read the documentation and saw some warnings about UV light in which the vendor advises not to look directly at the light and not to expose skin to the light.

Can I use these LEDs safely ?

share|improve this question
Even modern white LEDs can constitute an eye hazard and more so at the blue end of the spectrum. We had some 50 mA LEDs tested by Nichia and they formally advised that these were considered hazardous at the blue end of the spectrum. | In practice - and this is a comment only and NOT professional advice, I'd expect that your LEDs would be safe if used sensibly with precautions taken not to allow them top be direct viewed. |If UV damage does occur it will generally lead to short term but potentially very painful eye irritation and not to permanent damage. [Ask me how I know ... :-(.] – Russell McMahon Aug 1 '14 at 18:00
When in doubt, use protection. If you don't need to see it, cover it. If you do, protect you vision. Even reflected emissions can cause vision damage. – Enemy Of the State Machine Aug 2 '14 at 6:00
up vote 18 down vote accepted

The limits of "safe" emitted light are very complicated. You can read about the basics here.

A rule of thumb I have heard is if the emitted power is over 5mW, protection should be used. Since the LEDs you linked are capable of 10mW, yes they can be harmful. Do not use them until you understand how they can be harmful and how to prevent it.

UV is particularly dangerous because we can't see it so our blink reflex won't help. To safely work with these LEDs you should get a pair of glasses that block the possible wavelengths the LED can emit, 390 to 405nm ±2.5nm. Examples here.

This paragraph answers an edited out question in the OP wondering why he had a UV pen with no warnings. As for your UV pen light question, it is likely the power was low enough that it was not harmful. Less than 0.39mW (roughly) is considered eye safe so under that no warning would've been required.

share|improve this answer
Ok thank for your answer, but i don't think that those leds can emit at 120mW, it's the power dissipation, the emitting power it's in the Electrical Characteristics table, and the minimum is 5mW. Am i right ? – Superdrac Aug 1 '14 at 16:17
You are right, edited. that is electrical power dissipation. They still can emit >5mW which is a rule of thumb anyways so it's better to be safe than sorry. – ACD Aug 1 '14 at 16:20
Yes ok, thank for your answer! – Superdrac Aug 1 '14 at 16:22
The provided reference refers only to LASER sourcers classification, not to non-coherent light exposure. It does not provide irradiance data which can be extrapolated to the question asked. – jose.angel.jimenez Aug 6 '14 at 8:56
The indicated glasses are not suitable for UV non-coherent light protection. The glasses refered to are designed specifically for LASER protection. – jose.angel.jimenez Aug 6 '14 at 8:59

The emitted light power of those LEDs is UVA typically 10mW with a 15° 50% power angle. There's no maximum specified, but probably we could say more than 15-20mW would be very unlikely.

That's within the range where caution is called for. Here is a very detailed safety report from a large University. I'll reproduce a chart below, but note that it is based on exposure with 'breaks' to allow cellular repair to take place. Also, I would expect it's conservative.

enter image description here

If you get a bit of distance from then end of the LED, the power density drops with the square of the distance. If you allow an irradiance of 0.1mW/cm^2 (30 second exposure limit according to the chart), then you need to be at least distance \$d\$ from the LED:

$$d = \left(\sqrt\frac {P_{mW}}{Ir_{MAX}}\right) \frac{1}{\tan(7.5°)}$$

So if you allow 20mW at the LED you'd need a distance of more than about 1m to be safe for a ~30s exposure (and assuming the power is fairly well distributed over the center 'spot').

share|improve this answer
Ok thank ! I will buy the glasses, and don't expose my skin ! – Superdrac Aug 2 '14 at 2:10
@Superdrac and especially don't expose your eyes or leave this accessible to children. – Technophile Aug 2 '14 at 9:22
Also, to take into account, 400nm non-coherent light is in the limit/border of UVA and visible light spectrum. That means, it is the least "dangerous" UV radiation. – jose.angel.jimenez Aug 6 '14 at 9:02

When in doubt or dealing with health and safety matters, I strongly advice to avoid at all means opinions, rules of thumbs or any other common sense wisdom. Do not accept any advice which is not backup by an authoritative reference. Good practice in science and engineering mandates going in first place to the source of contrasted and accepted knowledge.

For this case, I refer you to the excellent ICNIRP (International Commission on Non-Ionizing Radiation Protection) paper on the matter:

Guidelines on limits of exposure to incoherent UV ratiation between 180 and 400nm

The paper may be a little bit daunting for those lacking a little bit of background on optics and power calculation, however, it is very clear and well written.

Here follows a summary of my interpretation of these guidelines, for your specific case:

  • A LED source of incoherent light at ~390nm is at the limit of what we call visible light. It is barely entering the UVA region of the spectrum, the least dangerous UV radiation. In fact, at around 400nm is what we (humans) call and perceive as "purple" color.
  • The ICNIRP guidelines indicate a maximum recommended exposure of 10KJ/m2.

    • The data values shown at table 1 of the paper are used for calculating the (weighted) aggregated exposures of a UV source composed of several "wavelengths". By doing you, you may incorrectly reach a maximum of 680KJ/m2 @ 390nm.
    • Correcting the previously arrived value, on page 174, it is clearly stated that the exposure limit for the eyes should not exceed 10KJ/m2 in case of unweighted UV exposure in the region 315-400nm.


  • As indicated in the datasheet, the UV3TZ-390-15 LED will typically output 10mW @ 20mA with a 50% power angle of 15 degrees.
  • If the LED source was Lambertian (120 degrees of 50% power angle), the peak irradiance (at 0 degrees, front view) would be the total emitted power divided by pi (3.1415...)
  • However, accounting for the 15 degrees 50% power angle of the actual LED, we will apply an additional conversion factor, arriving to the peak irradiance:

$$ I_p = \frac{P_{tot}}{\pi} \frac{\sin^2\theta_1}{\sin^2\theta_2} = \frac{10\text{mW}}{\pi} \frac{\sin^2 (60^{\circ})}{\sin^2 (7.5^{\circ})} \approx 140 \frac{\text{mW}}{sr} $$

Lets's calculate two cases:

  1. The operator stares at the LED at 0 degrees (frontal view) at 1m of distance:

    • At 1m, 1sr is 1m2, so that Ip = 140mW/m2 @ 1m.
    • To reach the ICNIRP exposure, the operator will have to stare at the LED (at 1m) during,

$$ T_{\text{max}} = \frac{10\text{KJ}/\text{m}^2}{140\text{mW}/\text{m}^2} \approx 20\;\text{hours} $$

  1. The operator stares at the LED at 0 degrees (frontal view) at 10cm of distance:

    • At 10cm, 1sr is 0.01m2, so tha Ip = 14W/m2 @ 10cm.
    • To reach the ICNIRP exposure, the operator will have to stare at the LED (at 10cm) during,

$$ T_{\text{max}} = \frac{10\text{KJ}/\text{m}^2}{14\text{W}/\text{m}^2} \approx 12\;\text{minutes} $$

share|improve this answer
Wow, what can i say except, thank you for your demonstration! – Superdrac Aug 6 '14 at 15:55

protected by Nick Alexeev Oct 15 '15 at 23:38

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site.

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.