Former Alabama star OJ Howard signs deal with Super Bowl contender

The Buffalo Bills have signed tight end OJ Howard to a one-year deal, per NFL reporter Ian Rapoport. The deal is reportedly worth up to $5 million when including incentives, with $3.5 million guaranteed.

A former Alabama tight end, Howard was selected 19th overall in the 2017 NFL Draft by the Tampa Bay Buccaneers. He played each of his first five NFL seasons with the team, including winning a Super Bowl in 2021. His rookie contract expired this offseason.

Howard started nine games this past season to finish with 14 catches, 135 yards and a touchdown. He posted at least 400 yards receiving in each of his first three years in the league, but hasn’t reached 200 in the two seasons since. Tampa Bay brought in Rob Gronkowski to play with Tom Brady beginning in 2020, which bumped Howard down on the depth chart.

The tight end should see more opportunity to shine with the Bills, where he’ll himself competing with Dawson Knox for playing time. Knox is coming off a career season in 2021 after recording 49 catches for 587 yards and nine touchdowns.

Buffalo finished the year with an 11-6 record to win the AFC East division and make the playoffs. However, they were knocked out by the Kansas City Chiefs for the second season in a row during an incredible 42-36 overtime thriller. Bills quarterback Josh Allen finished seventh in the NFL with 36 passing touchdowns in addition to 4,407 yards. With the addition of Howard, he now has one extra weapon to look to as well.

Howard wasn’t the only big free agency signing from Buffalo, however. The Bills defense also made a big splash, landing former Denver Broncos linebacker Von Miller to a six-year deal worth $120 million. The soon-to-be 33-year-old show he still has some gas left in the tank last season, finishing with 50 tackles (19 for loss) and 9.5 sacks.

With Howard and Miller now in the fold, could next year be the year Buffalo finally advances to a Super Bowl?