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.