CLEVELAND, Ohio (WOIO) - Cleveland Browns Linebacker Joe Schobert has been named the AFC Defensive Player of the Week.
The award comes after the Browns’ 41-24 victory against the Miami Dolphins.
The linebacker disrupted the Dolphins’ passing game all day.
Schobert had five tackles, two interceptions, and batted down four passes.
He became the sixth linebacker in NFL history to have at least two interceptions in consecutive games.
Schobert is leading the team this season in tackles (97), interceptions (four) and passes defended (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.