Rams DT Aaron Donald Named 2018 NFL Defensive Player of the Year

Aaron Donald won the award for the second year in a row.

Los Angeles Rams defensive tackle Aaron Donald was named the 2018 NFL Defensive Player of the Year on Saturday night.

Donald won the honor in back-to-back years and led the league this season with 20.5 sacks, 25 tackles for loss and 41 quarterback hits. His 20.5 sacks set a new franchise record for the Rams.

The five-time Pro Bowler becomes the third defensive player to win the award in back-to-back seasons, joining Hall of Fame linebacker Lawrence Taylor (1981, 1982) and Houston Texans defensive end J.J. Watt (2014, 2015).

.@RamsNFL DT @AaronDonald97 is the 2018 Defensive Player of the Year! #NFLHonors pic.twitter.com/JMVQRJmpxP

— NFL (@NFL) February 2, 2019

Donald and the Rams are preparing to face off against the New England Patriots on Sunday in Super Bowl LIII. The Rams clinched the NFC West this year for a second consecutive season and finished the regular season 13–3.

Kickoff for Super Bowl LIII in Atlanta is set for 6:30 p.m. ET on CBS.