The NFL has announced that Cleveland Browns linebacker D’Qwell Jackson has been named AFC Defensive Player of the Week for games played on November 22-26.
Jackson tallied a team-high nine tackles and had a hand in two turnovers as he forced one fumble and recovered another during the team’s 20-14 victory against the Pittsburgh Steelers at Cleveland Browns Stadium last Sunday. Jackson led the Browns’ defensive unit which limited the Steelers to 242 total net yards, including just 49 rushing, which marked Cleveland’s lowest rushing total allowed since 2003. In total, the Browns defense forced eight turnovers, which were the most by an NFL team since 2001 and the most by the Browns since 1989.
This is the first AFC Defensive Player of the Week award forJackson. He is the first Brown to win AFC Defensive Player of the Week honors since linebacker David Bowens in Week 7 of the 2010 season. Jacksonwas named AFC Defensive Player of the Month in September of 2011.
“I’m not about personal accolades,” said Jackson of the award. “It’s great, don’t get me wrong. I put a lot of pride into what I do and I’m passionate about what I do. I love what I do and I love playing here in the city of Cleveland. It’s all about everyone in this locker room and the coaches. It’s a little bit better when you win games and as you guys know, there have been some rocky years at times. We had a great week last week hopefully we can carry it over to Oakland, get another win and start a streak here.”