BEREA, Ohio (WJW) – Defense end Myles Garrett has been named the AFC Defensive Player of the Week.
The NFL announced Wednesday the Browns star won the award.
Garrett set a Browns single-game record with 4.5 sacks.
The team handed a big loss to the Bears on Sunday 26-6.
Overall, the Browns’ defense finished with 9 sacks and held the Bears to 1-for-11 on third downs.
Garrett previously won the award in Week 4 last season.