Jaguars' Josh Allen Named Week 18 AFC Defensive Player of the Week

The fourth-year pass-rusher was dominant against the Tennessee Titans, helping the Jaguars clinch the AFC South for the first time since 2017.

The Jacksonville Jaguars have picked up a number of accolades during their stunning late-season playoff push, and this week is no different.

After the Jaguars knocked off the Tennessee Titans 20-16 in Week 18 to punch their ticket to the playoffs, outside linebacker Josh Allen has been awarded the prestige of AFC Defensive Player of the Week.

Allen recorded six tackles, one tackle for loss, one forced fumble and recorded the game-winning 37-yard fumble-return for a touchdown with 2:51 remaining in the Jaguars’ 20-16 win over Tennessee. Jacksonville completed the “worst-to-first” turnaround and clinched the AFC South division title for the first time since 2017.

The win means Allen, who the Jaguars drafted in the first-round of the 2019 NFL Draft, is not only on his first winning NFL team, but he is set to make his playoff debut on Sunday vs. the Los Angeles Chargers in the Wild Card.

"I mean, when you lose so long, man, it teaches you a little something about opportunities. I think we capitalize on every opportunity that we had, man. I'm blessed," Allen said on Saturday. 

"I'm thankful to even be in the position that we are here today. I love my teammates. We are brothers, keepers no matter what situation arises. We just play until the clock hits 00:00, and we're going to find a way to be victorious."

This is the second-career Defensive Player of the Week award for Allen, having previously won the award in Week 9 of the 2021 season. He is the second Jaguar to earn Defensive Player of the Week this season, joining safety Rayshawn Jenkins in Week 15.

“He’s really picked it up. It’s good to see. It’s good to see him play his best football this time of the year," Jaguars head coach Doug Pederson said on Tuesday. 

"Great players do that, and we’ve just got to keep him coming that way.”