Yannick Ngakoue named Week 7 AFC Defensive Player of the Week
New Las Vegas Raider, DE Yannick Ngakoue, has been named Week 7’s AFC Defensive Player of the Week. Ngakoue made himself known time and time again as the Raiders took on the Philadelphia Eagles at home on Sunday.
Ngakoue had 4 tackles, 2 sacks, and 2 deflected passes during the Raiders win.
Ngakoue propelled the Raiders to their fifth win of the season in Week 7, second straight under interim head coach Rich Bissacia, when his second sack of the game came with just over a minute left in the fourth quarter.
Ngakoue now has 4 sacks and 32 total pressures this season. This is the first time Ngakoue has won Defensive Player of the Week.
Ngakoue’s partner in crime, DE Maxx Crosby, took home the award Week 1. The Raiders are one of only three teams to have multiple players win the award this season.
