LB Devin Bush has earned AFC Defensive Player of the Week for games played Jan.3-4 (Week 18), the National Football League announced on Wednesday.
In the final minute of the first quarter as the Bengals were deep in the red zone, QB Joe Burrow threw a pass intended for WR Tee Higgins. However, Bush jumped the route and secured the ball for the interception, then took off running down the field. With an open field in front of him, Bush ran the length of the field into the end zone for a 97-yard pick six.
His interception return for a touchdown was the first of two defensive touchdowns in the Browns' 20-18 win over the Bengals in Week 18. It was the second-longest interception return in the NFL this season and tied for the fourth-longest in Browns history. Bush led the NFL this season with two interception-return touchdowns and finished with a career-high in tackles (124) and interceptions (three). Bush also recorded 14 tackles in Week 18.
This is the second career weekly league award for Bush, who also won in Week 6 in 2019. He joins K Andre Szmyt (Special Teams in Week 3), S Grant Delpit (Special Teams Week 7) and DE Myles Garrett (Defensive in Week 12) as the fourth Browns player to be named AFC Player of the Week this season.
