JACKSONVILLE, Fla. – The rebuild is in full effect for the Jacksonville Jaguars.
ESPN’s Adam Schefter reported on Sunday that the Jaguars have agreed to trade Calais Campbell to the Baltimore Ravens for a 2020 fifth-round draft pick when the NFL’s league year begins on March 18.
Campbell was the Jaguars Walter Payton man of the year award recipient for the Jaguars. He also started all 16 games last season for the Jaguars and had 56 combo tackles. According to the NFL Network’s Ian Rapoport, the Ravens and Campbell were working toward a new contract worth roughly $27 million.
The #Ravens and new DL Calais Campbell are finalizing a 2-year new deal for $27M, sources say. It gives him $20 guaranteed after the trade from the #Jaguars. It’s not completely done yet, but sounds like it will be.— Ian Rapoport (@RapSheet) March 15, 2020
The trade continues Jacksonville’s offseason housecleaning and no doubt pushes the franchise toward rebuilding the roster from last year’s 5-11 team.
The Jaguars agreed to trade cornerback A.J. Bouye to the Broncos for a fourth-round pick earlier this month. They also declined defensive tackle Marcell Dareus’ option as part of their offseason housecleaning. Jacksonville used the franchise tag on defensive end Yannick Ngakoue last week, although Ngakoue has made it clear that he doesn’t want to be in Jacksonville any longer.
The non-exclusive franchise tag means it’s possible for Ngakoue to negotiate with other teams, with the Jaguars having the ability to match any offer. It’s also possible he could be traded.
Jaguars and Ravens have agreed to a trade in principle that would send Pro Bowl DE Calais Campbell to Baltimore for a 2020 fifth-round pick, per sources. Campbell will try to finalize an extension with Baltimore.— Adam Schefter (@AdamSchefter) March 15, 2020