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Steelers rookie ILB Devin Bush wins AFC Defensive Player of the Week award for Week 6 performance

The Pittsburgh Steelers rookie was given a high honor after his dominant performance on Sunday Night Football.

NFL: Pittsburgh Steelers at Los Angeles Chargers Jake Roth-USA TODAY Sports

The Pittsburgh Steelers beat the Los Angeles Chargers in Week 6 by the score of 24-17, but the score didn’t really indicate just how badly the Steelers beat the Chargers on their home turf. One of the reasons for their win? None other than rookie inside linebacker Devin Bush.

Bush had 7 tackles, an interception and a fumble recovery which he returned 9 yards for the touchdown.

Talk about a day.

To top off Bush’s tremendous week, he was recently named the AFC Defensive Player of the Week.

Prior to the 2019 NFL Draft it became clear if the Steelers were going to get one of the ‘Devins’ they would need to make a trade to do so. Pittsburgh showed aggressiveness moving up 10 spots to take Bush 10th overall. I think I speak for all of Steelers Nation when I say the trade was worth it.

Bush has had his rookie moments, but he is slowly starting to come into his own as a playmaker. Fans are starting to see Bush flying around the field from sideline to sideline like they did when he was in Michigan. Less thinking and more reacting is the name of the game for Bush, and he is just cracking the surface of his potential.

On top of this weekly honor, Bush leads the defense with 32 total tackles, and of the team’s 15 takeaways he has been a part of 6 of those. Let’s just say I can this being one of many awards like this for Bush in his time in the league.

Stay tuned to BTSC for the latest news and notes surrounding the black-and-gold throughout the bye week as they prepare for the Miami Dolphins in Week 8.