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Derek Wolfe is the Broncos' 3rd AFC Defensive Player of the Week this year

Derek Wolfe, run-stopping monster.

Chris Humphreys-USA TODAY Sports

This Denver Broncos defense boasts a lot of playmakers, especially against the pass. But it's for his tremendous play against the run that Broncos defensive end Derek Wolfe has earned the AFC's Week 8 Defensive Player of the Week award.

Via the Denver Broncos in a release -

Wolfe led the Broncos with a career-high seven tackles, including one for a loss, against the Packers on Sunday as Denver stayed perfect on the season improving to 7-0 for just the second time in team history.

His seven tackles, which represent the most by a Broncos defensive lineman this season, helped limit the Packers to 140 net yards—224 yards below their season average entering the game. Green Bay's passing game produced just 50 net passing yards and the Packers' running backs combined for just 49 yards on 18 carries (2.7 avg.).

It is the 3rd such honor for this Broncos this season, as Aqib Talib won the award in Week 1 and T.J. Ward took it home in Week 4. It is Wolfe's first such honor of his career.

Read more: Game Balls for the Broncos offense following Packers win

PFF graded Wolfe with a team-high +4.9, bolstered by a +5.2 grade against the run. It's rare that the NFL recognizes someone for their less-splashy play in run support, so kudos to the NFL for doing so this week, and kudos to Wolfe on a tremendous effort against the Green Bay Packers.