Joe Schobert named AFC Defensive Player of the Week (11.27.19)
For Immediate Release
Nov. 27, 2019
BEREA, Ohio – Cleveland Browns LB Joe Schobert has earned AFC Defensive Player of the Week for games played Nov. 21-25 (Week 12), the National Football League announced Wednesday.
During a 41-24 victory against Miami, Schobert recorded five tackles, two interceptions and four passes defensed. He became the sixth linebacker in NFL history to have at least two interceptions in consecutive games and the first to accomplish the feat since Cato June in 2005. Schobert led all AFC players with two interceptions and tied for the lead among AFC players with four passes defensed.
A fourth-round pick in 2016 and a 2017 Pro Bowl honoree, Schobert is leading the team this season in tackles (97), interceptions (four) and passes defensed (nine). He ranks tied for fourth in the NFL in interceptions and sixth in tackles.
Schobert is the first Browns player to win AFC Defensive Player of the Week since LB D’Qwell Jackson in Week 12 of 2012. Schobert joins Jamie Gillan (AFC Special Teams Player of the Week in Week 2) and Nick Chubb (AFC Offensive Player of the Week in Week 4) as Browns weekly award winners this year. This marks the first time the Browns have won AFC Offensive, Defensive and Special Teams Player of the Week honors in the same season since 2010 when RB Peyton Hillis (Offensive) and LB David Bowens (Defensive) and P Reggie Hodges (Special Teams) earned the awards.