JACKSONVILLE, Fla. — Jaguars linebacker Foye Oluokun has been a tackling machine for years, but this week, he’s earning national recognition for his standout play.

For the first time in his NFL career, Oluokun has been named the AFC Defensive Player of the Week.

The honor comes after a dominant performance in Jacksonville’s 26-10 win over the Carolina Panthers on Sunday.

Oluokun filled the stat sheet, racking up 10 tackles, grabbing an interception, and forcing a fumble as the Jaguars' defense smothered Carolina from start to finish.,

While Oluokun has quietly been one of the league’s most consistent linebackers, leading the Jaguars in tackles each of the last two seasons and surpassing 100 tackles in five straight years, this marks the first time he’s received weekly honors from the

See Full Page