Myles Garrett named AFC Defensive Player of the Week (11.26.25)
Nov. 26, 2025
Myles Garrett named AFC Defensive Player of the Week
BEREA, Ohio — Cleveland Browns DE Myles Garrett has earned AFC Defensive Player of the Week for games played Nov. 20-24 (Week 12), the National Football League announced Wednesday.
Garrett anchored a defense that helped the Browns to a 24-10 victory at Las Vegas. Garrett recorded five tackles, four tackles for a loss, three sacks and two forced fumbles. He led all NFL players in Week 12 in sacks and forced fumbles. During the contest, he set the Browns franchise record with 18 sacks. The Browns defense recorded 10 sacks, the most of any NFL team this season and the second-most by a Browns team in a game. Garrett leads the NFL this season in sacks (18) and tackles for a loss (26).
This is the fifth career weekly league award for Garrett and first since Week 12 in 2024. He joins K Andre Szmyt (AFC Special Teams Player of the Week in Week 3) and S Grant Delpit (AFC Special Teams Player of the Week in Week 7) as the third Browns player to be named AFC Player of the Week this season.
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