Monday night was a good night for defensive end Maxx Crosby and the Raiders.
Crosby had a sack, four tackles for loss, and four quarterback pressures as the Las Vegas defense paved the way for a 17-13 victory over the Packers. The NFL announced on Wednesday that Crosby has been named the AFC’s defensive player of the week in recognition of that performance.
It’s the fourth time that the conference’s weekly honors have gone to Crosby, who has the second-most tackles for loss in the league behind T.J. Watt since entering the league in 2019.
After the game, Crosby said that he has to be “even better” in Week Six as he pursues his goal of being the best defensive player in the league.
