CLEVELAND, Ohio (WOIO) – Cleveland Browns linebacker Devin Bush was named the AFC Defensive Player of the Week for games played Jan. 3–4.
Bush recorded a game-high 14 tackles and returned an interception 97 yards for a touchdown in the Browns’ 20-18 victory over the Bengals to end the season.
Bush’s return was the second-longest interception return in the NFL this season and is tied for the fourth-longest in franchise history.
The Week 18 performance capped a standout season for Bush. He led the NFL with two interception-return touchdowns and finished the year with a career-high 124 tackles and three interceptions.
This marks the second time Bush has earned a weekly league honor; he previously won the award in Week 6 of the 2019 season.
Bush becomes the fourth Browns player to be named an AFC Player of the Week during the 2025 season — joining kicker Andre Szmyt (Special Teams, Week 3), safety Grant Delpit (Special Teams, Week 7) and defensive end Myles Garrett (Defensive, Week 12).
This was the final year of Bush’s contract, he is now a pending free agent.
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