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LB Foyesade Oluokun Wins AFC Defensive Player of the Week

This is Oluokun's fourth time winning the award and second with the Jaguars.

Published: Wednesday, 7 October 2026 at 4:00 PM
LB Foyesade Oluokun Wins AFC Defensive Player of the Week
Key Things to Know
  • Foyesade Oluokun was named AFC Defensive Player of the Week for his Week 4 performance against Cincinnati.
  • Oluokun recorded seven tackles, a forced fumble, a quarterback hit, and an interception against Joe Burrow.
  • He became the fourth active NFL player to achieve at least 10 sacks, 10 interceptions, and 10 forced fumbles.
The Story in 60 Seconds

Jacksonville Jaguars linebacker Foyesade Oluokun has been named the AFC Defensive Player of the Week following his performance in a Week 4 victory against the Cincinnati Bengals. Oluokun recorded seven tackles, a forced fumble, a quarterback hit, and intercepted Bengals quarterback Joe Burrow. The performance earned him his fourth career Player of the Week honor and second with Jacksonville.

With his interception, Oluokun reached 10 career interceptions, making him just the fourth active NFL player to record at least 10 sacks, 10 interceptions, and 10 forced fumbles. Alongside the award announcement, the Jaguars made multiple roster moves, including placing defensive end B.J. Green on injured reserve and signing linebacker Jack Kiser to the active roster.

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