Jaguars DL Calais Campbell wins AFC Defensive Player of the Week

Jacksonville Jaguars 12th-year defensive lineman Calais Campbell has been named AFC Defensive Player of the Week for his performance in Jacksonville’s Week 3 victory over the Tennessee Titans.

In the team’s Thursday Night Football victory, Campbell made seven solo tackles, including 3.0 sacks, and an additional tackle for loss. This marked the second career game for the Jaguars with at least 3.0 sacks, as he also owns the franchise record after registering 4.0 sacks against the Houston Texans in Week 1 of the 2017 season.

Campbell is now the only player in franchise history to post at least 3.0 sacks on multiple occasions and his six multi-sack games are tied for fifth in the NFL since he signed with Jacksonville in 2017.

Since joining the Jaguars prior to the 2017 season, Campbell’s 28.0 sacks rank third in the NFL during that span. His 173 games played and 155 games started are both the most among active defensive linemen in the NFL.

In Week 3, Campbell also became the only player 33 years or older to tally at least 3.0 sacks in a game over the last three seasons (2017-19). Since turning 30, Campbell has totaled 153 tackles and 28.0 sacks in 35 games.

Overall, this marks Campbell’s fourth time earning Player of the Week honors — AFC Defensive Player of the Week in Week 3 of the 2019 season, AFC Defensive Player of the Week in Week 1 of the 2017 season, NFC Defensive Player of the Week in Week 2 of the 2012 season and NFC Special Teams Player of the Week in Week 2 of the 2009 season.

Campbell is the first Jaguars player to earn AFC Player of the Week honors since punt returner Jaydon Mickens earned AFC Special Teams Player of the Week honors in Week 14 of the 2017 season, and he’s the first Jaguars defender to earn a weekly award since cornerback A.J. Bouye was named AFC Defensive Player of the Week in Week 10 of the 2017 season.