Star defensive end wins 6th AFC Player of the Week award in his 6th season
The Las Vegas Raiders would have not pulled off their stunning Week 2 26-23 upset win at the Baltimore Ravens on Sunday without the brilliant play of star defensive end Maxx Crosby.
And the NFL noticed.
Wednesday morning, the league announced Crosby is the Week 2 AFC Defensive Player of the Week for his performance in the game.
And he’s just getting started
Congratulations to @CrosbyMaxx on being named the AFC Defensive Player of the Week! #RaiderNation
— Las Vegas Raiders (@Raiders) September 18, 2024
Crosby had two sacks of reigning NFL MVP Lamar Jackson, including one in the fourth quarter that was a major part of the Raiders’ comeback from a 10-point deficit in the fourth quarter. He had five tackles and three for tackles for loss.
Crosby has had a tackle for loss in nine straight games, which is the longest current streak in the NFL. He has 92 career tackles for loss. He is one of seven players since 1999 to have his that mark with his first six season and he has a whole lot of season to go.
This is the sixth time Crosby has been named a player of the week, which is tied for the third most in team history. He has done it five times on defense (the most in team history) and once on special teams.
Crosby has been amazing thus far this season and he has to be considered a contender to win the Defensive Player of the Year award, which could be his first since joining the Raiders as a fourth-round pick in 2019.