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A.J. Klein named AFC Defensive Player of the Week

This marks the fourth time this season a Buffalo player has won a Player of the Week award.
Credit: AP
Buffalo Bills' Brandin Bryant, right, and A.J. Klein, left, tackle New England Patriots' Cam Newton (1) during the first half of an NFL football game Sunday, Nov. 1, 2020, in Orchard Park, N.Y.

BUFFALO, N.Y. — On Thursday the Buffalo Bills announced that linebacker A.J. Klein has been named AFC Defensive Player of the Week after his performance in the win over the Los Angeles Chargers. 

Klein led the team on Sunday with a career-high 14 tackles. He also tallied 1.5 sacks, three tackles for a loss, and a pass breakup. 

The eighth-year linebacker out of Iowa State is in his first season with the Bills. This is the fourth Player of the Week honor a Buffalo player has received this season, and the second DPOW after defensive end Jerry Hughes won earned the award in Week 7 against the New York Jets. 

"To be honest I know I’ve been playing good football, but I know there’s more out there for me," Klein said. "Even on Sunday I know left some missed players out on the field. 

"I felt like I have always played at a high level, and I prepare at a high level every single week regardless if that’s reflected in the stat line or not. My mentality and my preparation hasn’t changed over my entire career. 

"But it is nice to string these games together because I wasn’t to be able to contribute to the team to help this team win. That’s the most important to me."

He is the first Bills linebacker to earn the honor since Matt Milano back in 2018. Klein is also the third player from the Bills to be named AFC player of the week this season, joining with Josh Allen and Jerry Hughes.