Programming note: Tony Sparano and Derek Carr are scheduled to address the media at 1:40 p.m. this afternoon. Watch the live stream right here.
Raiders safety Charles Woodson was named the AFC Defensive Player of the Week, the NFL announced on Wednesday morning. Woodson was excellent in the team’s first win, a 24-20 victory over the Kansas City Chiefs.
He had eight tackles, including a sack, against Kansas City. The sack made him the first player to record 50 interceptions and 20 sacks, a moment that was not lost on a 38-year old in the midst of a football renaissance.
“I’ve been playing a long time, but I think what it also says is that I’ve been a productive player,” Woodson said after the game. “I’ve been able to do a lot of things. I’ve been put in position to make plays for the teams that I’ve been on and they’ve put me in a lot of different positions and a lot of different scenarios where I could come up with plays, and I’ve come up with a lot of them.
"This team is no different. The coaching staff, even though I’m 38 years old, they’ve trusted me to go out there and do those things. I’m just happy to be a part of it. This win was huge for this team and for this organization. It was huge for us.”
Woodson is playing excellent football this season. He has 82 tackles, two interceptions, a fumble recovery and a sack in 12 games. He’s already played 807 snaps this year despite advanced playing age.
This is the team’s first defensive player of the week honor since Woodson won the award following a Week 5 victory over the San Diego Chargers last season.