Las Vegas Raiders defensive end Yannick Ngakoue was named the AFC Defensive Player of the Week for Week 7, marking the first time he has earned the honor in his six-year career. He is the first former University of Maryland player to win the award since linebacker D’Qwell Jackson in Week 13 of the 2014 season.
Ngakoue registered four tackles (two solo), two passes defended, and two third-down sacks in the Raiders’ 33-22 win over the Philadelphia Eagles at Allegiant Stadium. His final sack of the day helped seal the victory as he took down Jalen Hurts with just over a minute left to play.
Ngakoue’s multi-sack performance was his second of the season and the 12th of his career. He also notched two sacks in the Raiders’ Week 5 loss to the Chicago Bears.
Ngakoue’s two sacks on Sunday tied for the second-most of any AFC player and third in the NFL. He additionally accumulated two tackles for loss and two quarterback hits in Week 7, bringing his season totals to three and 10, respectively.
Ngakoue now has four sacks in seven games this year, which is good for second on the team behind Maxx Crosby’s five. Ngakoue is well on pace to reach the double-digit mark in sacks, which he has accomplished only once in his career (12 in 2017).
Along with consistently bringing pressure off the edge, Ngakoue fared well in coverage, setting a new career high with two passes defensed. He was the only player to record at least two sacks and two passes defensed in Week 7.
With Crosby previously earning AFC Defensive Player of the Week honors in Week 1, the Raiders joined the Buffalo Bills and Arizona Cardinals as the only teams this season to have multiple players win the award.
The Raiders now are on a bye before returning to action in Week 9 against a depleted New York Giants team that is middle of the pack with 14 sacks allowed this season. It represents another favorable matchup for Ngakoue, who will look to continue his recent tear on quarterbacks.
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