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Former Alabama defensive lineman A'Shawn Robinson became the first Crimson Tide player to switch teams in NFL free agency when reportedly signed a two-year deal with the Los Angeles Rams on Wednesday night. 

According to Tom Pelissero of the NFL Network, the deal is worth $17 million.

After helping Alabama win the 2015 national championship, Robinson was a second-round selection by the Detroit Lions.  

Last season he tallied 40 tackles, 1.5 sacks and three pass breakups over 13 games.

Robinson's breakout game was the 2014 SEC Championship, when he racked up nine tackles, including three for a loss, against Missouri.

His final year, though, Robinson was on the of the clear leaders of the defense, which was known for making hits during practice even when the drills may have been considered no-contact. 

"It’s crazy to me how many times Jarran [Reed], or A’Shawn or Jonathan Allen tackle a guy in practice," said Kirby Smart, when he was still Alabama's defensive coordinator. "They want to go live. They want to go hot people. I’m like, ‘Whoa, slow down.’ You almost have to hold them back. You worry about them injuring a scout-team player or injuring themselves, but they’re not worrying about that. They’re thinking about ‘We need to go play good, so we need to go practice good.’ 

"When they set that tempo they turned around to the linebackers and they jumped on them, they jumped on the DBs when they’re not out thudding guys. So the demand is there and when you’ve got that as a coach you just step back and let them go."

Robinson was a consensus All-American and named a finalist for the Outland Trophy in 2015. He was credited with tallying 46 tackles, 7.5 for loss, and 3.5 sacks in 15 games while helping lead the No. 3 defense in the nation. 

Coming out of Alabama, Sports Illustrated had Robinson No. 16 on its list of the top 50 NFL prospects, calling him "One of the draft’s most polarizing players, who, if put in the right NFL scheme, has the potential to do just about everything."