Steelers' Jalen Ramsey Earns NFL Honor for Dominant Performance vs. Buccaneers

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Steelers' Jalen Ramsey Earns NFL Honor for Dominant Performance vs. Buccaneers originally appeared on SportsNet Pittsburgh. Add SportsNet Pittsburgh as a Preferred Source by clicking here.

Jalen Ramsey made his presence felt in the Pittsburgh Steelers' overtime win over the Tampa Bay Buccaneers, and the veteran defensive back is receiving recognition for his efforts.

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The NFL named Ramsey AFC Defensive Player of the Week after he delivered a standout performance in Pittsburgh's victory. The Steelers announced the honor Wednesday after Ramsey played a key role in shutting down Tampa Bay's offense.

Ramsey recorded 10 tackles, including nine solo stops, four tackles for loss and two sacks. He also played through a wrist injury that limited him in practice leading up to the game.

The veteran's stat line put him in a category of his own. According to the Steelers' announcement, Ramsey became the first player since 2023 to record at least 10 tackles, four tackles for loss and two sacks in a game. He is also the first defensive back to accomplish the feat since 2000.

His second sack proved to be the difference in overtime. With the Buccaneers facing fourth down, Ramsey brought down quarterback Jalon Daniels for an 11-yard loss to end Tampa Bay's possession and secure the Steelers' win.

The award is the third AFC Defensive Player of the Week honor of Ramsey's career. He previously won it with the Jacksonville Jaguars in Week 16 of the 2016 season and the Miami Dolphins in Week 11 of the 2023 season.

Ramsey is also the second Steelers defender to earn the honor this season, joining T.J. Watt. The recognition adds to a strong start for Pittsburgh's defense, which will look to carry that momentum into its upcoming games.

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