Jacksonville Jaguars

Jaguars Telvin Smith gets AFC Defensive honors.

(Photo by Zach Bolinger/Icon Sportswire)


The recognition of the Jaguars success continued Wednesday, as linebacker Telvin Smith was named AFC Defensive Player of the Week for his role in their shocking 30-9 victory over Pittsburgh in Week 5.

Smith’s 28-yard interception returned for a touchdown changed the momentum of the game and was the first pick-six of two for Jacksonville’s defense.

The Jaguars finished the game with a franchise record-setting five interceptions against Steelers quarterback Ben Roethlisberger.

Smith is also setting records of his own. It marks the third time the Florida State alum has received the player of the week honor, which is the most by a defender in franchise history.

Smith was named AFC Defensive Player of the Week for his performance against the Cleveland Browns in November 2014 as a rookie, and for a 2015 performance against the Bills.

Jaguars head coach Doug Marrone told reporters on Wednesday that Smith has further developed his skill set this season.

“He’s done a nice job,” Marrone said. “I think that he’s improved in the run game, I really do. Not that he was poor in it, but I think he has done a very good job in the run game. I see him getting better and better each week becoming more physical, becoming more instinctive and being able to make plays.”

The four-year veteran recorded a game-high 10 tackles against Pittsburgh and has forced three turnovers over the last four games. He’s produced 40 tackles, two interceptions and a fumble recovery in the previous five.

The Jaguars, as a defensive unit have become a force early in the year. Their record 10 sack performance week one in Houston, their demolition of the Ravens in London, and now, holding the Steelers to just nine points last week has the league buzzing.

Now, can they sustain it starting by winning for the first time at home, this week vs. the L.A. Rams.

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