ENGLEWOOD, Colo.— After a dominant performance in the Broncos' 29-10 win against the Green Bay Packers, defensive end Derek Wolfe was named AFC Defensive Player of the Week.
The award represents the first conference award of his career and the third AFC Defensive Player of the Week honor for the Broncos this season (CB Aqib Talib – Wk. 1; S T.J. Ward – Wk. 4).
Wolfe led the Broncos with a career-high seven tackles, including one for a loss, against the Packers in the 'Battle of the Unbeaten' 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 is 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.).
This marks the 30th time a Broncos player has taken home the award since it was initiated by the NFL in 1984. Wolfe is the first Broncos defensive lineman to receive the honor since defensive end Elvis Dumervil in 2007 (Wk. 14).
