Charles Woodson named AFC Defensive Player of the Week

Following Charles Woodson's defensive effort in a 27-17 win over the Chargers Sunday night, he has been named AFC Defensive Player of the Week.

The former Defensive Player of the Year earned the honor by recovering a fumble for a touchdown as well as a game clinching interception and four solo tackles.

His 25-yard fumble return for a touchdown tied him with Darren Sharper and Rod Woodson for the most defensive touchdowns (13) in NFL history. And with the Chargers down two scores, desperately hoping to make something happen, he ended the game by intercepting the ball. It was the third interception of the day for the Raiders after the team had come into the game with no interceptions on the season through four games.

After seven seasons in Green Bay, Woodson returned to Oakland where he spent the first eight years of his career. The Packers moved on from the 16-year NFL veteran and his signing with the Raiders was thought by some to be simply to please the fans who wanted to see a former Raider Legend return to Silver and Black. Woodson's addition has become much more than that. He ranks third on the team with 39 tackles (28 solo) with a sack.

The last Raider to be named AFC Defensive Player of the Week was Lamarr Houston in a Week 7 win over Jacksonville last season.

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