Connect with us

NFL

Raiders DE Maxx Crosby Named AFC Defensive Player of the Week

Published

on

Steelers Raiders DE Maxx Crosby

Las Vegas Raiders defensive end Maxx Crosby was named the AFC Defensive Player of the Week for his dominant performance against the Baltimore Ravens.

Crosby was a menace Monday night recording two sacks, two tackles-for-loss and four hits on quarterback Lamar Jackson.

A fourth round selection of the Raiders back in 2019 out of Eastern Michigan, Crosby totaled a combined 17 sacks and 30 tackles-for-loss over his first two seasons.

Crosby and the Raiders now travel to Pittsburgh, where they will look to spoil the Steelers’ home opener.

Crosby will primarily lineup against right tackle Chukwuma Okorafor, providing a significant test to determine just how much progress he has made from a season ago.