Miami Dolphins linebacker Bradley Chubb has been named AFC Defensive Player of the Week for his performance in Sunday's game against the Buccaneers, the NFL announced Wednesday.
It is the second Defensive Player of the Week accolade of his career and the fourth for Miami this season as linebacker Jordyn Brooks earned the award in Week 8 (at Atlanta) and Week 11 (vs. Washington), while cornerback Rasul Douglas did so in Week 14 (at N.Y. Jets). The Dolphins are the only team in the NFL with four Defensive Player of the Week recognitions this year. Additionally, the Dolphins have been recognized on offense, as running back De'Von Achane was named AFC Offensive Player of the Month for November.
Chubb tallied three tackles (two solo), 2.0 sacks and one forced fumble in Miami's Week 17 victory over Tampa Bay. His 2.0 sacks were tied for the most in the AFC this week. His forced fumble came on a strip-sack of Buccaneers QB Baker Mayfield late in the fourth quarter, which was recovered by the Dolphins. He was the only player in the league to record 2.0 sacks and a forced fumble in Week 17.
His performance marked the 10th multi-sack game of his career and the first of this season. It also tallied Chubb's 15th-career forced fumble while marking his 13th-career strip-sack and his second of the season.
Chubb, a team captain, has started all 16 games of the 2025 season. He leads the Dolphins with 8.5 sacks and has recorded 46 tackles (23 solo), eight tackles for loss, two forced fumbles and one fumble recovery this season.
