The Buffalo Bills have been connected to a few different pass-catching tight ends in the rumor mill early this offseason.
They’ve finally landed one.
The team announced on Wednesday that OJ Howard has signed with the Bills.
ESPN’s Adam Schefter adds that Howard’s deal is a one-year contract, worth $3.5 million deal. With incentives, it could reach up to $5M.
Howard, 27, was a first-round pick of the Tampa Bay Buccaneers in 2017. He spent the first five years of his career with the Bucs, putting up 119 catches, 1,737 yards and 15 touchdown–but he never really fully broke out.
In the first three years of his career, Howard started decent. Over the past two, things have dipped.
That has to do with the position depth in Tampa over that period.
There were less snaps to go around for tight ends once the Bucs signed Rob Gronkowski, a player Buffalo was also connected to this offseason. In the end, Gronkowski’s teammate joins the Bills instead.
In those first three seasons for Howard, he put up no less than 26 catches a season. The last two saw him grab 11 and 14, respectively.
His most productive year was 2018, when he had 34 catches for 565 yards and five touchdowns in only 10 games.
Howard has sporadically been in and out of the lineup due to injury in his career, but he did appear in all 17 games in 2021. Of those, nine were starts.
While Howard joins the Bills who do also have Dawson Knox, he should still see more opportunities than he had in Tampa.
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