Woodson Bags AFC Defensive Player of the Week Honors

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Oakland Raiders safety Charles Woodson has earned AFC Defensive Player of the Week honors, the National Football League will announce Wednesday.

Woodson, now in his 17th NFL season and 10th with the Silver and Black, totaled nine tackles (six solo), including three tackles for loss, one sack and one pass defensed in the Raiders' 24-20 victory over Kansas City on Thursday night.

Woodson made history in front of a national television audience, recording a third-quarter sack of QB Alex Smith to become the first NFL player since the sack became an official statistic in 1982 to rack up 50 career interceptions and 20 sacks. The safety was part of three tackles for loss on the game, stopping RB Jamaal Charles for a 9-yard loss in the first quarter and getting to Charles behind the line of scrimmage twice more for 1-yard losses in the second period.

The 6-foot-1, 210-pound veteran also returned his first punt since 2009, becoming the second-oldest player in NFL history return a punt at 38 years and 44 days old. Former Raider Tim Brown holds the distinction of being the oldest player to return a punt at 38 years and 94 days old.

Woodson increased his team-leading tackle total to 116 (77) on the season, marking his second-straight campaign with more than 100 tackles. He has also recorded two interceptions, seven passes defensed and one fumble recovery on the year. The AFC Defensive Player of the Week honor is the first for a Raider since Woodson took home the honor following a Week 5 win over San Diego last season.

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