FS Charles Woodson helped the Raiders defeat the Chargers with a fumble return for a touchdown and an interception. Photo by Tony Gonzales.
Oakland Raiders safety Charles Woodson has earned the AFC Defensive Player of the Week award, the National Football League will announce Wednesday.
Woodson, a 6-foot-1, 210-pound veteran in his 16th NFL season, posted a fumble return for a touchdown and added an interception in the Raiders' 27-17 win over San Diego Sunday night.
Woodson recovered a third-quarter fumble and returned it 25 yards for his 13th career defensive touchdown, which is tied for first in NFL history with Darren Sharper and former Raider Rod Woodson. He sealed the win with his first interception of the season, picking off a Philip Rivers pass with less than a minute remaining in regulation. Woodson was also credited with eight tackles (six solo).
Woodson, who returned to Oakland in 2013 after seven seasons in Green Bay, currently ranks third on the team with 39 tackles (28) and has added one sack.
The AFC Defensive Player of the Week honor is the first for a Raider since Lamarr Houston earned recognition in a Week 7 win over Jacksonville last season, and the first for an Oakland defensive back since Michael Huff in 2010.
