Jacksonville Jaguars linebacker Devin Lloyd was named the AFC Defensive Player of the Week for his performance against the San Francisco 49ers.
Lloyd recorded eight total tackles, two interceptions and a tackle for a loss in the win.
He led the charge as the Jaguars picked up a west coast victory ahead of their Week 5 Monday Night Football game against the Kansas City Chiefs.
Lloyd is off to a strong start in his fourth season with the Jaguars. He’s recorded 23 total tackles, with three interceptions, three pass deflections and a fumble recovery this season.
The former first-round pick is in the final year of his rookie deal and is playing the best football of his career.
“We always practice taking the ball away, punching the ball out and catching the ball,” Lloyd said after the game about the takeaways. “Everybody always has the ball on their mind and I think whenever you’re thinking about it all the time, naturally, you will be ready for those plays and those opportunities them whenever they present themselves.”
Lloyd joins fellow linebacker Foye Oluokun as Jaguars who’ve won AFC Player of the Week honors this season. Oluokun won Defensive Player of the Week in Week 1.
Jacksonville’s defense leads the league with 13 takeaways this season. They’ve already surpassed the total number of takeaways from last season through four games.
The Jaguars will face the Kansas City Chiefs on Monday Night Football on Oct. 6 at 8:15 p.m. from EverBank Stadium.
This article originally appeared on Florida Times-Union: Jaguars’ Devin Lloyd named AFC Defensive Player of the Week ahead of Monday Night Football
Reporting by Juston Lewis, Jacksonville Florida Times-Union / Florida Times-Union
USA TODAY Network via Reuters Connect
