Los Angeles Rams defensive tackle Aaron Donald was named the Associated Press NFL Defensive Player of the Year for the second straight season.
Donald was dominant yet again in 2018, leading the league with 20.5 sacks and 25 tackles for loss. He also forced four fumbles and recovered two.
The star interior lineman received 45 of a possible 50 votes for the award. Chicago Bears edge-rusher Khalil Mack received the other five, according to NFL.com's David Ely.
A first-round pick in 2014, Donald has been named to the Pro Bowl in each of his five seasons and was a First-Team All-Pro selection in his last four.
Donald solidifies himself as the NFL's top defensive player with the award. His presence as an interior rusher is a major reason why the Rams will be playing for the Super Bowl on Sunday.
