Former Packers S Darnell Savage will sign 3-year deal with Jaguars

Former Packers S Darnell Savage, a 2019 first-round pick, will sign a 3-year deal with the Jaguars.

Darnell Savage, a first-round pick of the Green Bay Packers in 2019, is leaving Green Bay and will sign a three-year deal with the Jacksonville Jaguars, according to Ian Rapoport of NFL Network.

Savage was the 21st overall pick and the first defensive back selected in the 2019 draft. The Packers traded two fourth-round selections to move up from No. 30 and get him.

Savage went on to play five up and down seasons with the Packers. A member of the All-Rookie team in 2019, Savage broke out as a difference maker to end the 2020 season and looked like one of the NFL’s best young safeties entering 2021. But his impact never ascended and eventually plummeted under defensive coordinator Joe Barry.

Between 2021 and 2023, Savage intercepted only three passes and recorded 15 passes defensed. In 2020 alone, he had four picks and 12 passes defensed. Missed tackles and missed assignments became common, and Savage was eventually benched during the second half of the 2022 season.

The Packers will begin rebuilding the safety room. Savage will attempt to reignite his career in Jacksonville.