Myles Garrett wins AFC defensive player of the week for Week 3

Browns DE Myles Garrett wins AFC defensive player of the week for Week 3 for his outstanding performance against the Chicago Bears

As if there was any doubt…

Myles Garrett and his monstrous performance against the Chicago Bears in Week 3 earned the Browns defensive end the AFC defensive player of the week honors. The NFL revealed the honor on Wednesday.

Garrett sacked Bears rookie QB Justin Fields 4.5 times, setting a Browns team record for an individual game in the process. No. 95 finished with seven total tackles. It was part of the Browns’ incredible defensive effort that sacked Fields nine times and allowed just one net passing yard to Chicago.

It’s the second time Garrett has captured the weekly award. He also won it in Week 4 of the 2020 season against the Dallas Cowboys.