Thanks to a huge performance against the Tampa Bay Buccaneers, Pittsburgh Steelers defensive back Jalen Ramsey was named the AFC Defensive Player of the Week for Week Six. Ramsey made the first big play and he made the last one for the Steelers to seal a thrilling overtime win, to help Pittsburgh improve to 4-2 on the season.
Ramsey stopped the Bucs opening drive with a sack and he called ballgame with his second sack of quarterback Jalen Daniels in overtime on fourth down.
It was clear that defensive coordinator Patrick Graham saw something in what Tampa Bay was doing and knew he could use the veteran savvy of Ramsey to take advantage of the rookie quarterback. For the game, Ramsey had 10 total tackles, two sacks, four tackles for loss and one quarterback hit.
AFC DEFENSIVE PLAYER OF THE WEEK: JALEN RAMSEY
— Pittsburgh Steelers (@steelers) October 21, 2026
📝: https://t.co/547nRSLwPtpic.twitter.com/ljdilp80pr
For up-to-date Steelers coverage, including any roster news and injury updates, follow us on X @TheSteelersWire and give our Facebook page a like.
This article originally appeared on Steelers Wire: Steelers DB Jalen Ramsey named AFC Defensive Player of the Week