The Los Angeles Rams have signed defensive lineman A'Shawn Robinson to a two-year deal worth $17 million, according to Tom Pelissero and Ian Rapoport of NFL Network.
Robinson spent the first four years of his career with the Detroit Lions and is coming off of a 2019 campaign in which he played 13 games and registered 40 tackles, 1.5 sacks, a forced fumble and a couple of fumble recoveries.
The 24-year-old, who played his collegiate football at the University of Alabama, was originally selected by the Lions in the second round (46th pick overall) of the 2016 NFL Draft.
He appeared in every contest during his rookie year, finishing with 30 tackles, a pair of sacks and seven passes defended. Robinson then showed marked improvement in 2017, totaling 53 tackles, a half of a sack, an interception, a forced fumble, six passes defended and a defensive touchdown.
After playing two straight full 16-game campaigns to begin his career, Robinson was limited to 13 games in 2018, recording 49 tackles, a sack, a forced fumble and a fumble recovery.
Robinson will join a Rams defensive line that already includes Aaron Donald, who many regard as the best defensive player in all of football.
Los Angeles was one of the more disappointing teams in the league this past season, as it went just 9-7 and missed the playoffs one year after winning 13 games and making it all the way to the Super Bowl.
It doesn't help the Rams' case that they play in a loaded NFC West division that also includes the defending conference champion San Francisco 49ers and the Seattle Seahawks.
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