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Raiders defensive end Yannick Ngakoue wins AFC Defensive Player of the Week

This is the third PoW honor for Las Vegas this season

Philadelphia Eagles v Las Vegas Raiders
Yannick Ngakoue
Photo by Ethan Miller/Getty Images

The Las Vegas Raiders are winning — and their accomplishments are being noticed.

For the third time through seven weeks of the NFL season, a Raider has been named the Player of the Week. This time, it’s defensive end Yannick Ngakoue, who was tabbed as the AFC Defensive Player of the Week for his performance in Week’s 7’s 33-22 victory over the Philadelphia Eagles.

Ngakoue had two sacks, including one that clinched the game in the fourth quarter, against the Eagles. He was active all game and it was the best game of the season for the Raiders’ free-agent prize —- and he’s been good this season. Ngakoue has four sacks and has formed a viscous pass-rush tandem with fellow defensive end Maxx Crosby, an NFL Defensive Player of the Year candidate.

Crosby and kicker Daniel Carlson have also won Player of the Week awards. The Raiders won five Player of the Week awards in 2016.

Also, Las Vegas quarterback Derek Carr has won the Offensive Player of the Month award for September. As long as the winning continues for the 5-2 Raiders, expect the awards to keep piling up.