Jacksonville Jaguars linebacker Foyesade Oluokun has been named the AFC Defensive Player of the Week following the Jaguars 22-17 victory over the Cincinnati Bengals.
This is the fourth time he has been named AFC Defensive Player of the Week and his second time with the Jaguars. He is the fifth player in franchise history to earn the honor multiple times, and joins Ryan Fitzpatrick, a Harvard alumni who won eight times, as the only former Ivy League players to win a Player of the Week award four or more times.
. @foyelicious has been named AFC Defensive Player of the Week after his performance in the Jaguars’ win at Cincinnati in Week 4. 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… pic.twitter.com/QbHwvhYwnW
— JaguarsPR (@JaguarsPR) October 7, 2026
In the game against the Bengals, Oluokun racked up a team-high seven tackles, one forced fumble, one quarterback hit, and one interception of Bengals QB Joe Burrow in the victory. The interception was his 10th career interception, making him one of just seven linebackers drafted since 2018 to have at least 10 career interceptions. His two interceptions in 2026 are also tied for the third most in the NFL this season.
The Jaguars have built a big part of their defensive identity on takeaways, and Oluokun’s performance was another example of why he remains at the center of that identity. With the Jaguars continuing to emphasize takeaways and situational football, their veteran linebacker delivered both when they needed them most.
This article originally appeared on Jaguars Wire: Jaguars' Foyesade Oluokun named AFC Defensive Player of the Week
