Browns’ Joe Schobert named AFC Defensive Player of the Week

CLEVELAND, OHIO - NOVEMBER 24: Middle linebacker Joe Schobert #53 of the Cleveland Browns celebrates with teammates after his second interception of the game during the second half against the Miami Dolphins at FirstEnergy Stadium on November 24, 2019 in Cleveland, Ohio. The Browns defeated the Dolphins 41-24. (Photo by Jason Miller/Getty Images)

BEREA, Ohio (WJW) –– Browns linebacker Joe Schobert has been named AFC Defensive Player of the Week.

The NFL said it’s for games played Nov. 21-25.

The Browns said Schobert had five tackles, two interceptions and four passes defensed during the Browns’ 41-24 victory against Miami.

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, according to a press release from the Browns.

The team said Schobert, a fourth-round pick in 2016 and a 2017 Pro Bowl honoree, is leading the Browns this season in tackles (97), interceptions (four), and passes defensed (nine).

Schobert is reportedly the first Browns player to win AFC Defensive Player of the Week since LB D’Qwell Jackson in Week 12 of 2012.

**More stories on Schobert, here**

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