Woodson named AFC Defensive Player of Week

Oakland Raiders safety Charles Woodson has been named AFC Defensive Player of the Week, the National Football League announced 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’ first win, defeating the Kansas City Chiefs 24-20 on Thursday night.

The 38-year-old defensive player also made history in front of the national audience when he sacked quarterback Alex Smith in the third quarter, becoming the first player in NFL history to record 50 interceptions and 20 sacks.

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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