Former Alabama defensive tackle A'Shawn Robinson has signed a new NFL contract with a new team. Robinson will be joining the Los Angeles Rams. The Rams and Robinson came to terms on a two-year contract worth $17m according to NFL Network's Tom Pelissero.

Robinson was drafted by the Detroit Lions in the second round of the 2016 NFL Draft. He has spent the entirety of this NFL career thus-far with the Lions. In four seasons Robinson has accumulated 172 tackles, 16 tackles for a loss, five sacks, three fumble recoveries and one interception returned for a touchdown.

As a member of the Crimson Tide, Robinson won a National Championship, two SEC Championships and was named a consensus All-American in 2015. He was also named to the SEC-All Freshman team in 2013. He was a five-star recruit and a USA Today High School All-American out of Arlington Heights High School in Forth Worth, Texas.

Robinson will join the Rams and play alongside vaunted defensive lineman Aaron Donald. The Rams lost in the 2018 Super Bowl and did not make the playoffs in the 2019 season.

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