Written by phil lamass
  • Share
  • Tweet
  • Share
  • Pin
  • Email
It's a challenge, but you can make a simple DIY ultraviolet (UV) light source. (Led blu image by Davide D. from

You can make a simple DIY ultraviolet (UV) light source consisting of a number of UV LEDs in a matrix, a power supply and a switch. Each matrix column is made up of a string of LEDs wired in series and connected to the power supply via a series-limiting resistor. The number, power supply voltage and current requirements are determined by the number of LEDs in a matrix column and the number of columns.

Skill level:

Things you need

  • UV LEDs
  • Current limiting resistor
  • Power supply
  • Single-pole single-throw (SPST) switch

Show MoreHide


  1. 1

    From the specification sheet for the LED you plan to use, find the minimum voltage required to turn on the LED (forward voltage) and the normal operating current (forward current) of the LED.

  2. 2

    Calculate the total voltage required to drive the LED matrix by multiplying the forward voltage of a single LED by the number of LEDs in a matrix column. Next, add 20 per cent to the result by multiplying by 1.25. This will be the minimum power supply voltage. For example, if you are building a 3-by-3 matrix using an LED with a forward voltage of 3.3 volts, then the minimum required power supply voltage is 3.3 volts x 3 x 1.25, or approximately 12 volts.

  3. 3

    Calculate the total current load of the LED matrix by multiplying the current through one column (equal to the forward current of a single LED) by the number of columns. For example, if the forward current of an LED on a 3-by-3 matrix is 25 milliamps, the total current load is 75 milliamps. Multiply this value by 1.25 to arrive at the minimum power supply current requirement, which in this example is approximately 95 milliamps.

  4. 4

    Calculate the series-limiting resistor value and the power dissipation. To calculate the resistance value, subtract the total voltage by the LED matrix from the supply voltage and divide the result by the forward current. Using a power supply voltage of 12V, the value of the limiting resistor is (12 volts - 9.9 volts)/.025A = 84 ohms.

    To calculate the required power dissipation multiply the voltage across the resistor by the current through the resistor: (12 volts - 9.9 volts) x .025A = .0525W or 52.5 milliwatts.

    The closest standard resistor is an 82 ohm quarter watt (250 milliamp) resistor. However, the conservative choice would be the next highest standard value, which is 91 ohms. The lower current resulting from the higher resistance will cause the LED to be slightly dimmer.

  1. 1

    Connect the positive terminal of the power supply to one terminal of the switch.

  2. 2

    Connect the other terminal of the switch to one lead of the series-limiting resistor.

  3. 3

    Connect the other lead of the limiting resistor to the positive lead (usually the longest of the two leads) of the first LED in the column.

  4. 4

    Connect the negative lead of the first LED to the positive lead of the next LED in the column. Repeat this step for the remaining LEDs in the column.

  5. 5

    Connect the negative lead of the last LED in the column to the negative terminal of the power supply.

  6. 6

    Repeat Steps 1 through 4 for the remaining columns in the LED matrix.

Don't Miss

  • All types
  • Articles
  • Slideshows
  • Videos
  • Most relevant
  • Most popular
  • Most recent

No articles available

No slideshows available

No videos available

By using the site, you consent to the use of cookies. For more information, please see our Cookie policy.