UV-C dose required to kill microorganisms

Below is a list of microorganisms and the UV-C dose at 254 nm (displayed in J/m²) required to achieve a 90% (1 log reduction).

UV dose (J/m²) = UV irradiance (W/m²) x exposure time (seconds)

If you know the UV irradiance value then you can use our online UV dose calculator to determine the UV dose applied, which varies with time.

In order to calculate further log reductions, you could – as a guide – multiply the D⁹⁰ value to further reduce contamination as follows:

• D⁹⁰ multiplied by 2 could provide you with a 99% (2 log) reduction
• D⁹⁰ multiplied by 3 could provide you with a 99.9% (3 log) reduction
• D⁹⁰ multiplied by 4 could provide you with a 99.99% (4 log) reduction
• D⁹⁰ multiplied by 5 could provide you with a 99.999% (5 log) reduction
• D⁹⁰ multiplied by 6 could provide you with a 99.9999% (6 log) reduction

However, it is recommended that trials and evaluation are conducted to determine the required UV dose levels required to achieve different log reductions of your target microorganisms.

Coronavirus (SARS-CoV-2)

Since the start of the SARS CoV-2 pandemic, Dr. Anthony Griffiths, Associate Professor of Microbiology at Boston University School of Medicine and his team have been working on developing tools to support scientific advancement in this field.

During their research, they have treated inoculated material with different doses of UV-C radiation and assessed the inactivation capacity under various conditions. The team applied a dose of 50 J/m², resulting in a reduction of the SARS-CoV-2 virus of 99%. Based on the data, it was determined that a dose of 220 J/m² will result in a reduction of 99.9999%.

“Our test results show that above a specific dose of UV-C radiation, viruses were completely inactivated; in a matter of seconds we could no longer detect any virus,” said Dr. Anthony Griffiths.

Bacteria

Viruses