JACKSONVILLE – LB Foyesade Oluokun has been named AFC Defensive Player of the Week after his performance in the Jaguars' win at Cincinnati in Week 4, the NFL announced today.
Oluokun totaled a team-high seven tackles, one forced fumble, one quarterback hit and one interception of Bengals QB Joe Burrow in the victory. The interception marked his 10th career interception. He became the fourth active NFL player with 10+ sacks, 10+ interceptions, and 10+ forced fumbles, joining S Harrison Smith, LB Fred Warner, and LB T.J. Watt. Oluokun's two interceptions in 2026 are tied for the third most in the NFL this season.
Oluokun is helping lead a Jaguars defense that ranks tied first in interceptions (six), second in points allowed (53) and run defense (74.3), and tied second in takeaways (eight) in 2026.
This marks his fourth Defensive Player of the Week honor of his career and second as a Jaguar. He won the AFC award in Week 1 of 2025 and won the NFC award twice as a member of the Atlanta Falcons (Week 9 of 2020 and Week 16 of 2021).
