Jacksonville Jaguars linebacker Foye Oluokun has been named the AFC Defensive Player of the Week for his performance Oct. 4 against the Cincinnati Bengals.
Oluokun totaled a team-high seven tackles, one forced fumble, one quarterback hit and one interception of Bengals QB Joe Burrow in the Jaguars' 22-17 victory. The interception marked his 10th career interception.
He is one of four active players with 10 or more sacks, 10 or more interceptions and 10 and forced fumbles, along with Vikings safety Harrison Smith, 49ers linebacker Fred Warner, and Steelers linebacker T.J. Watt.
Oluokun's two interceptions in 2026 are tied for the third most in the NFL this season. His first came in Week 1 against the Cleveland Browns.
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For the season, Oluokun has 24 tackles, 16 solo, along with the two interceptions and forced fumble.
As a whole, the Jaguars' defense is tied for first in the NFL interceptions (six), second in points allowed (53) and run defense (74.3), and tied for second in takeaways (eight) through four games this season.
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).
The Jaguars next play Oct. 11 in London against the Philadelphia Eagles.
This article originally appeared on Florida Times-Union: Jaguars linebacker Foye Oluokun named AFC defensive player of the week
