Aaron Donald named AP Defensive Player of the Year, wins Deacon Jones Award

ATLANTA — What has been expected for months is finally official — defensive tackle Aaron Donald is the 2018 AP Defensive Player of the Year.

Donald recorded a season that could only be described as dominant in 2018, leading the league with 20.5 sacks, 25 tackles for loss, and 41 quarterback hits. His 20.5 sacks set a new single-season record for an interior lineman and a new single-season franchise record. The defensive tackle also set a new league record with 183.5 sacks yards on his quarterback takedowns. Plus, only five players had more forced fumbles than Donald’s four.

Donald is now the eighth player to win multiple AP Defensive Player of the Year awards, and only the third to win it in back-to-back seasons. Hall of Fame linebacker Lawrence Taylor accomplished the feat in 1981 and 1982, and Texans defensive end J.J. Watt did it in 2014 and 2015.

Look back at Aaron Donald's 20.5 sacks of the 2018 regular season.

While Donald did not attend the NFL Honors ceremony in Atlanta as he’s getting ready to play in Super Bowl LIII, he did say this week that winning the award would mean a good deal to him.

“Well, anytime you accomplish something you’re going to be happy, you’re going to be blessed,” Donald said. “So putting the body of work in and seeing it all pya off, it’s the type of stuff you dream about. But I’m just happy and blessed to be at the Super Bowl right now. But definitely a special year thus far.”

Additionally, as the league’s leader in sacks, Donald has won the Deacon Jones Award. Created in 2013, the Deacon Jones Award was made to permanently honor the legacy of one of the greatest players in league history. The Hall of Famer played 11 seasons as a Ram, becoming one of the famed “Fearsome Foursome” along with Rosey Grier, Lamar Lundy, and Merlin Olsen.

Advertising