Dolphins’ Jordyn Brooks named AFC Defensive Player of the Week

Bills Dolphins Football Miami Dolphins linebacker Jordyn Brooks (20) gestures as he runs onto the field during player introductions before an NFL football game against the Buffalo Bills, Sunday, Nov. 9, 2025, in Miami Gardens, Fla. (AP Photo/Doug Murray) (Copyright 2025 The Associated Press. All rights reserved.) (Doug Murray/AP)

MIAMI GARDENS, Fla. — The NFL announced Wednesday that Miami Dolphins linebacker Jordyn Brooks has been named AFC Defensive Player of the Week for his Week 11 performance in Sunday’s 16–13 overtime win over the Washington Commanders in Madrid, Spain.

It is the second AFC Defensive Player of the Week honor of Brooks’ career – also earning the award in Week 8 in Miami’s dominating victory over the Atlanta Falcons.

Brooks led the Dolphins with 20 tackles (10 solo) in the first-ever NFL regular season game played in Spain.

His 20 tackles, which led the league in Week 11, are the most in a single game in the NFL this season and the most by a player in an international game in league history, according to the team.

Brooks is the first NFL player to record 20 tackles in a game since 2023 when linebacker Roquan Smith had 21 for the Baltimore Ravens against the Cleveland Browns.

He is also the first Dolphins player to record 20-plus tackles in a game since Hall of Famer Zach Thomas had 21 against the Buffalo Bills in 2006.

The Dolphins, who acquired Brooks as an unrestricted free agent in 2024, named him a first-time team captain before the start of the 2025 season.

The team returns to the field at Hard Rock Stadium Nov. 30 against the New Orleans Saints at 1 p.m. ET.

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Brandon Liguori

Brandon Liguori

Brandon Liguori is a Floor Director and Web Contributor for WPLG Local 10 News.