Charles Woodson is the NFL Defensive Player of the Year
Green Bay Packers cornerback Charles Woodson is having a career season, and is a big reason – if not the No. 1 reason – that the Packers’ defense jumped to the top overall ranking this week after the team’s 34-12 trouncing of the Detroit Lions on Thanksgiving.
According to the Journal Sentinel’s Tom Silverstein, the last time a Packers defense ranked No. 1 was Week 3 of the 2001 season, when Ed Donatell was defensive coordinator. The team finished No. 12 overall, mostly as a result of giving up 524 yards in a 34-25 victory over the New York Giants in the final week of the regular season.
As we noted earlier this week, Woodson is having the best season of his career at age 33. He’s on pace for 78 tackles, three sacks and 10 interceptions. All of those would be career highs. Woodson has already registered a career high with four forced fumbles. He’s tied his career high with two touchdowns. And Woodson is a guy who’s made five Pro Bowls and been named All Pro three times.
Barring injury or a disastrous end to the season, Woodson definitely deserves to be the NFL Defensive Player of the Year. Hands down.
The last time a member of the Packers won the honor was 1998, when all-pro defensive end Reggie White registered 16 sacks and led the team to an NFC wild card appearance. White’s defense finished the season ranked No. 4 that year. This year’s Packers unit is on pace to finish at that level or better under Woodson’s leadership.
Now in his 12th season, Woodson is tied for second in the league in interceptions with seven. He’s been the NFC’s Defensive Player of the Week, was named NFC Defensive Player of the Month in September and was the only Packer to make Pro Football Weekly’s midseason All-Pro team.
NFL Defensive Player of the Year would be a fitting way to cap what has so far been a spectacular season.
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