Raiders DE Yannick Ngakoue wins AFC Defensive Player of the Week
Sunday against the Eagles, Yannick Ngakoue had his best game as a Raider. His numbers jump off the stat sheet with two sacks, two tackles for loss, two QB hits, and two pass breakups. Quite a complete game for the veteran pass rusher. And today that effort has earned him AFC Defensive Player of the Week honors.
Ngakoue joins teammate Maxx Crosby who took home the award for his effort in the team’s season opening win over the Ravens.
It’s befitting considering after the win over the Eagles, Ngakoue spoke of how he and Crosby are always competing with each other.
“We always compete with each other,” Ngakoue said of he and Crosby. “From our get-offs to how clean we eat, all the way to how much extra work we put in after practice. If you watch the practice, we’re the last two guys to leave. Always notice him looking at me and trying to figure out how he can compete and I’m always looking at him to figure out how I can compete. That’s what I love about him. He’s a guy that busts his ass every day, I see a lot of myself in him, and that’s why he’s having tremendous success.”
It was just last week, in the Raiders’ win over the Broncos in Denver, that Crosby got to quarterback Teddy Bridgewater five times for three official sacks. Certainly sending the message to Ngakoue that he was up. Yan answered the call, putting up his second multi-sack game of the season and moving to second on the team in sacks (4.0) behind Crosby (5.0).
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